User:EGM4313.f14.Team1.Linehan/Report 6

=Report 6=

Problem Statement
Continuation of R5.2


 * 1) For n = 4, 6, repeat the steps in R5.2 Part 1.
 * 2) Redo the solution for n=2,4,6, but using the projection of log(1+t) on the larger interval [0,2] as done in R5.1. Compare the results to those in the previous questions (i.e., projection in interval [0,1]).

Part 1
From R5.2:


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$$ f(x)=log(1+x) $$
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Following the same steps as the preceding problems, we know that it is necessary to develop the vector, d. We will start by applying the same method of integration by parts.


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$$ \int u dv=uv-\int v du $$
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For the first integral, we will use u= log(1+x) and dv=1
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$$ \langle x^0,\log(1+x) \rangle = \langle 1,\log(1+x) \rangle = \int_0^1 \log(1+x) dx = x\log(x)-\int_0^1 \frac{x}{1+x} dx $$
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In order to solve the integral, we need to make a "U-substitution". We will use u=(1+x) and du=dx:
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$$ x\log(x)-\int_0^1 \frac{x}{1+x} dx= x\log(x)-\int_0^1 \frac{u-1}{u} dx = x\log(x)-\int_0^1 1-\frac{1}{u} dx=x\log(x)-u+\log(u) $$
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Substituting back in and simplifying:
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$$ x\log(x)-u+\log(u) = x\log(x)-(1+x)+\log(1+x)= (1+x)(\log(x)-1)|_0^1= 2\log(2) = 0.38629436111 $$
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http://integrals.wolfram.com/index.jsp?expr=log%281%2Bx%29&random=false

We apply the same integration process for the d1 except that it quickly gets very complicated:
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$$ \langle x,\log(1+x) \rangle=\int_0^1 x^1 \log(1+x) dx= \frac{x^2\log(1+x)}{2}-\frac{1}{2}\int\frac{x^2}{1+x}dx $$
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Instead of performing the unnecessary algebra to solve the remaining integrals


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$$ \langle x,\log(1+x) \rangle=\int_0^1 x^1 \log(1+x) dx= \frac{1}{4}(2(x^2-1)\log(1 + x)-x(x-2)) = \frac{1}{4} = 0.25 $$
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http://integrals.wolfram.com/index.jsp?expr=x*log%281%2Bx%29&random=false


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$$ \langle x^2,\log(1+x) \rangle=\int_0^1 x^2 \log(1+x) dx= \frac{1}{18}(x(-6 + (3 - 2x)x) + 6(1 + x^3)\log(1 + x))= \frac{2}{3}log(2)-\frac{5}{18} = 0.184320342595519 $$
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http://integrals.wolfram.com/index.jsp?expr=x%5E2*log%281%2Bx%29&random=false


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$$ \langle x^3,\log(1+x) \rangle=\int_0^1 x^3 \log(1+x) dx= \frac{x}{4} - \frac{x^2}{8} + \frac{x^3}{12} -\frac{x^4}{16} - \frac{\log(1 + x)}{4} + \frac{x^4\log(1 + x)}{4}=\frac{7}{48} = 0.145833333 $$
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http://integrals.wolfram.com/index.jsp?expr=x%5E3*log%281%2Bx%29&random=false


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$$ \langle x^4,\log(1+x) \rangle=\int_0^1 x^4 \log(1+x) dx= \frac{-x}{5} + \frac{x^2}{10} - \frac{x^3}{15} + \frac{x^4}{20} - \frac{x^5}{25} + \frac{\log(1 + x)}{5} + \frac{1}{5}x^5\log(1 + x) = 0.120592205 $$
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http://integrals.wolfram.com/index.jsp?expr=x%5E4*log%281%2Bx%29&random=false


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$$ \langle x^5,\log(1+x) \rangle=\int_0^1 x^5 \log(1+x) dx= \frac{x}{6} - \frac{x^2}{12} + \frac{x^3}{18} - \frac{x^4}{24} + \frac{x^5}{30} - \frac{x^6}{36} - \frac{1}{6}\log(1 + x) + \frac{1}{6}x^6\log(1 + x) = 0.102775 $$
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http://integrals.wolfram.com/index.jsp?expr=x%5E5*log%281%2Bx%29&random=false


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$$ \langle x^6,\log(1+x) \rangle=\int_0^1 x^6 \log(1+x) dx= \frac{-x}{7} + \frac{x^2}{14} - \frac{x^3}{21} +\frac{x^4}{28} - \frac{x^5}{35} + \frac{x^6}{42} - \frac{x^7}{49} + \frac{1}{7}\log(1 + x) + \frac{1}{7}x^7\log(1 + x) = 0.0895386 $$
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http://integrals.wolfram.com/index.jsp?expr=x%5E6*log%281%2Bx%29&random=false

n=4
The basis is


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$$ \{ 1, x, x^2,x^3, x^4 \} $$
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so the function f(x) can be approximated as


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$$ f(x) \approx c_0b_0+c_1b_1+c_2b_2+c_3b_3+c_4b_4=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4 $$
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Now, determine the known computable right hand side (using the values determined from the quad function in Matlab):
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$$ \mathbf d = \begin{bmatrix} \langle x^0,\log(1+x) \rangle \\ \langle x^1,\log(1+x)\rangle\\ \langle x^2,\log(1+x) \rangle \\ \langle x^3,\log(1+x)\rangle \\ \langle x^4,\log(1+x)\rangle \end{bmatrix} = \begin{bmatrix} 0.3863\\ 0.2500\\   0.1843\\    0.1458\\    0.1206 \end{bmatrix} $$
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Using the Hilbert matrix of order 5 calculated prior, we can plug into the equation (1) and solve for c:


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$$ \boldsymbol \Gamma \mathbf c = \mathbf d $$
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$$ \mathbf c = \begin{bmatrix} -0.0179\\1.3181\\1.8119\\2.2059\\-1.011\end{bmatrix} $$
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Therefore, the function can be approximated as:
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$$ \log(1+x) \approx -0.0179+1.3181x-1.8119x^2+2.2059x^3-1.011x^4 $$
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Modal Equation 1
Modal applied forces as given in sec.53f p. 53-48:
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$$ r(t) = \begin{bmatrix} 2\\0 \end{bmatrix} \log(t+1) $$
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Modal equation 1 is given as the L2-ODE with initial conditions:


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$$ y_1'' + \frac12 \, y_1' + 4 \, y_1 = 2log(t+1) $$
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$$ y_1(0) = 1, \ y'_1(0) = 0 $$
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Solution for modal coordinate $$ y_1(t) $$ is the summation of

the homogeneous and particular solutions


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$$ y_1(t) = y_{h1}(t) + y_{P1}(t) $$
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To find homogeneous solution we assume trial solution


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$$ y_{h1}(t) = e^{\lambda_{1} t} $$
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in which we retrieve the characteristic equation


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$$ \lambda_1^2 + \frac12 \,\lambda_1 + 4 = 0 $$
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which generates the following complex-conjugate roots


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$$ \lambda_{11} = -\frac14 + i \frac{3 \sqrt7}{4} $$
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$$ \lambda_{12} = -\frac14 - i \frac{3 \sqrt7}{4} $$
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to which the homogeneous solution then takes the form


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$$ y_{h1}=e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)] $$
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Now, to find the particular solution $$ y_{P1} $$

The particular solution needs to be subbed in to modal equation 1 and set equal to the new approximation solved for in R4.4


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$$ y_{P1}'' + \frac{1}{2}y_{P1}' + 4y_{P1} = 2log(1+t) $$
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Sub in the logarithmic approximation from R4.4
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$$ y_{P1}'' + \frac{1}{2}y_{P1}' + 4y_{P1} = 2[0.00014+0.99543t-0.46407t^2+0.21641t^3-0.05486t^4] $$
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We know that solving for a 2nd order differential equation requires a 2nd order homogeneous solution in the terms of
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$$ y_p(t)= A + Bt + Ct^2 + Dt^3 +Et^4 $$
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And, differentiating twice,
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$$ y'_p(t)= B + 2Ct + 3Dt^2 +4Et^3 $$
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$$ y''_p(t) = 2C + 6Dt + 12Et^2 $$
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Plugging into the original differential equation,
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$$ (2C + 6Dt + 12Et^2) + \frac{1}{2} (B + 2Ct + 3Dt^2 +4Et^3) + 4(A + Bt + Ct^2 + Dt^3 +Et^4) = 2[0.00014+0.99543t-0.46407t^2+0.21641t^3-0.05486t^4] $$
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Equating the coefficients on both sides,


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$$ 1: 2C + 0.5B + 4A = 0.00028 $$
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$$ x: 6D + C+ 4B = 1.99086 $$
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$$ x^2: 12E + 1.5D +4C= -0.92814 $$
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$$ x^3: 2E + 4D = 0.43282 $$
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$$ x^4: 4E = -0.10972 $$
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This can be put into a matrix in the form of A c = d
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$$ \begin{bmatrix} 4 &0.5& 2& 0& 0\\ 0& 4 & 1 & 6 & 0\\ 0 & 0 & 4 & 1.5 & 12 \\ 0 & 0 & 0 & 4 & 2\\ 0 & 0 & 0 & 0 & 4 \end{bmatrix} \begin{bmatrix} A \\ B \\ C \\ D \\ E \end{bmatrix} = \begin{bmatrix} 0.00028 \\ 1.99086 \\ -0.92814 \\ 0.43282 \\ -0.10972 \end{bmatrix} $$
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Solving,
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$$ E = -0.02743 $$
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$$ D = 0.12192 $$
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$$ C = -0.19546 $$
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$$ B = 0.36371 $$
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$$ A = 0.05234 $$
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Thus, the particular solution is:
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$$ y_p(t) = 0.05234 + 0.36371t -0.19546t^2+0.12192t^3 -0.02743t^4 $$
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Verifying the correctness of the particular solution:
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$$ 4(0.05234 + 0.36371t -0.19546t^2+0.12192t^3 -0.02743t^4) + 0.5(0.36371 - 0.39092t + 0.36576t^2 - 0.10972t^3) + (-0.39092 +0.73152t - 0.32916t^2) = 0.00028+1.99086t-0.92814t^2+0.43282t^3-0.10972t^4 $$
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We see that the solution checks out:
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$$ 0.000295 + 1.9909t - 0.92812t^2 + 0.43282t^3 -0.10972t^4 \approx 0.00028+1.99086t-0.92814t^2+0.43282t^3-0.10972t^4 $$
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And, the entire solution becomes:
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$$ y(t) = y_h(t)+ y_p(t)= e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)] +0.05234 + 0.36371t -0.19546t^2+0.12192t^3 -0.02743t^4 $$
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Taking the first derivative:
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$$ y'(t) = \frac{-1}{4}e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)]  + e^{-\frac14 t}[-\frac{3 \sqrt7}{4}C_{11}\sin(\frac{3 \sqrt7}{4}t)+ \frac{3 \sqrt7}{4}C_{12}\cos(\frac{3 \sqrt7}{4}t)] + 0.36371 -0.39093t+0.36575t^2 -0.10972t^3 $$
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Plugging in the initial conditions:


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$$ y(0) = C_{11}+0.05234 = 1 $$
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$$ C_{11} = 0.94766 $$
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$$ y'(0) = \frac{-1}{4}C_{11}  +  \frac{3 \sqrt7}{4}C_{12} + 0.36371 =0 $$
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$$ C_{12} = -0.06390 $$
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And the complete solution is:


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$$ y(t) = e^{-\frac14 t}[0.94766\cos(\frac{3 \sqrt7}{4}t)+-0.06390\sin(\frac{3 \sqrt7}{4}t)] +0.05234 + 0.36371t -0.19546t^2+0.12192t^3 -0.02743t^4 $$
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Modal Equation 2
To find the modal displacement, we just need to find the homogenous solution.


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$$ y_2(t) = y_{h2}(t) $$
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To find homogeneous solution we assume trial solution


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$$ y_{h2}(t) = e^{\lambda_{2} t} $$
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in which we retrieve the characteristic equation


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$$ \lambda_2^2 + 6 \,\lambda_2 + 9 = 0 $$
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and the eigenvalues retrieved come out to be


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$$ \lambda_{1,2}=-3 $$
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since the eigenvalues recovered are a double real root the solution takes the form


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$$ y_2=P_{21}e^{-3t}+P{22}te^{-3t} $$
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and thus the modal velocity takes the form


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$$ y_2=-3P_{21}e^{-3t}-3P_{22}te^{-3t}+P_{22}e^{-3t} $$
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to find the undetermined coefficients we must satisfy the modal initial conditions


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$$ y_2(0) = 0 \Rightarrow P_{21} = 0 $$
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$$ y'_2(0) = 1 \Rightarrow -3 P_{21} + P_{22} = 1 \Rightarrow P_{22} = 1 $$
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and thus the modal displacement is


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$$ y_2(t) = t e^{-3 t} $$
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There is no particular solution for modal equation 2. We can check this:

If you were to plug in zero for $$ p_2 $$ and its first and second derivatives

in to modal equation 2, you would see that it solves to zero which is correct.

Complete Solution
Obtain displacements in original coordinate system by multiplying by the eigenvectors


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$$ \mathbf d(t) = \bar{\boldsymbol \Phi} \, \mathbf y(t) $$
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then,


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$$
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\mathbf d(t)=\begin{bmatrix} \frac{1}{\sqrt{5}} & \sqrt{\frac{3}{10}} \\ \sqrt{\frac{6}{5}} & \frac{-2}{\sqrt{5}} \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}

$$
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$$ d_1 = \frac{1}{\sqrt{5}} (e^{-\frac14 t}[0.94766\cos(\frac{3 \sqrt7}{4}t)+-0.06390\sin(\frac{3 \sqrt7}{4}t)] +0.05234 + 0.36371t -0.19546t^2+0.12192t^3 -0.02743t^4) + \sqrt{\frac{3}{10}}(te^{-3 t}) $$
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$$ d_2 =  \sqrt{\frac{6}{5}} (e^{-\frac14 t}[0.94766\cos(\frac{3 \sqrt7}{4}t)+-0.06390\sin(\frac{3 \sqrt7}{4}t)] +0.05234 + 0.36371t -0.19546t^2+0.12192t^3 -0.02743t^4) + \frac{-2}{\sqrt{5}}(t e^{-3 t}) $$
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Plots of the two solutions to the modal equations.



n=6
The basis is


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$$ \{ 1, x, x^2,x^3, x^4, x^5, x^6 \} $$
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so the function f(x) can be approximated as


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$$ f(x) \approx c_0b_0+c_1b_1+c_2b_2+c_3b_3+c_4b_4+c_5b_5+c_6b_6=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5+c_6x^6 $$
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Now, determine the known computable right hand side (using the values determined from the quad function in Matlab):
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$$ \mathbf d = \begin{bmatrix} \langle x^0,\log(1+x) \rangle \\ \langle x^1,\log(1+x)\rangle\\ \langle x^2,\log(1+x) \rangle \\ \langle x^3,\log(1+x)\rangle \\ \langle x^4,\log(1+x)\rangle \\ \langle x^5,\log(1+x)\rangle \\ \langle x^6,\log(1+x)\rangle\end{bmatrix} = \begin{bmatrix} 0.386294334336416\\ 0.250000034521966\\  0.184320336460938\\   0.145833342164217\\   0.120592191814104\\   0.102777809230323\\   0.089538700573044 \end{bmatrix} $$
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Using the Hilbert matrix of order 7 calculated prior, we can plug into the equation (1) and solve for c:


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$$ \boldsymbol \Gamma \mathbf c = \mathbf d $$
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$$ \mathbf c = \begin{bmatrix}      -0.001634682661461\\ 1.064527996641118\\ -1.116675421362743\\   2.712370300665498\\  -4.568075499963015\\   3.854147227481008\\  -1.252616312238388 \end{bmatrix} $$
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Therefore, the function can be approximated as:
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$$ \log(1+x) \approx -0.00163+1.06453x-1.11668x^2+2.71237x^3-4.56808x^4+3.85414x^5-1.25261x^6 $$
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 * }
 * }



As you can see, the approximation follows the graph of the exact function very closely and thus, it is a good approximation. The solution achieved in R4.4 did not show this relationship due to rounding errors. This time, long decimals were kept in matlab in order to achieve a better approximation.

Modal Equation 1
Modal applied forces as given in sec.53f p. 53-48:
 * {| style="width:100%" border="0"

$$ r(t) = \begin{bmatrix} 2\\0 \end{bmatrix} \log(t+1) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Modal equation 1 is given as the L2-ODE with initial conditions:


 * {| style="width:100%" border="0"

$$ y_1'' + \frac12 \, y_1' + 4 \, y_1 = 2log(t+1) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_1(0) = 1, \ y'_1(0) = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solution for modal coordinate $$ y_1(t) $$ is the summation of

the homogeneous and particular solutions


 * {| style="width:100%" border="0"

$$ y_1(t) = y_{h1}(t) + y_{P1}(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

To find homogeneous solution we assume trial solution


 * {| style="width:100%" border="0"

$$ y_{h1}(t) = e^{\lambda_{1} t} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

in which we retrieve the characteristic equation


 * {| style="width:100%" border="0"

$$ \lambda_1^2 + \frac12 \,\lambda_1 + 4 = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

which generates the following complex-conjugate roots


 * {| style="width:100%" border="0"

$$ \lambda_{11} = -\frac14 + i \frac{3 \sqrt7}{4} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \lambda_{12} = -\frac14 - i \frac{3 \sqrt7}{4} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

to which the homogeneous solution then takes the form


 * {| style="width:100%" border="0"

$$ y_{h1}=e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, to find the particular solution $$ y_{P1} $$

The particular solution needs to be subbed in to modal equation 1 and set equal to the new approximation solved for in R4.4


 * {| style="width:100%" border="0"

$$ y_{P1}'' + \frac{1}{2}y_{P1}' + 4y_{P1} = 2log(1+t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sub in the logarithmic approximation from R4.4
 * {| style="width:100%" border="0"

$$ y_{P1}'' + \frac{1}{2}y_{P1}' + 4y_{P1} = 2[-0.00163+1.06453t-1.11668t^2+2.71237t^3-4.56808t^4+3.85414t^5-1.25261t^6] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We know that solving for a 2nd order differential equation requires a 2nd order homogeneous solution in the terms of
 * {| style="width:100%" border="0"

$$ y_p(t)= A + Bt + Ct^2 + Dt^3 +Et^4 + Ft^5 + Gt^6 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

And, differentiating twice,
 * {| style="width:100%" border="0"

$$ y'_p(t)= B + 2Ct + 3Dt^2 +4Et^3 + 5Ft^4 + 6Gt^5 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y''_p(t) = 2C + 6Dt + 12Et^2 + 20Ft^3 + 30Gt^4 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plugging into the original differential equation,
 * {| style="width:100%" border="0"

$$ (2C + 6Dt + 12Et^2 + 20Ft^3 + 30Gt^4) + \frac{1}{2} (B + 2Ct + 3Dt^2 +4Et^3 + 5Ft^4 + 6Gt^5) + 4(A + Bt + Ct^2 + Dt^3 +Et^4 + Ft^5 + Gt^6) = $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ 2[-0.00163+1.06453t-1.11668t^2+2.71237t^3-4.56808t^4+3.85414t^5-1.25261t^6] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Equating the coefficients on both sides,


 * {| style="width:100%" border="0"

$$ 1: 2C + 0.5B + 4A = -0.00326 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ x: 6D + C+ 4B = 2.12906 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ x^2: 12E + 1.5D +4C= -2.23336 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ x^3: 20F + 2E + 4D = 5.42474 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ x^4: 30G + 2.5F + 4E = -9.13616 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ x^5: 3G + 4F = 7.70828 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ x^6: 4G = -2.50522 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

This can be put into a matrix in the form of A c = d
 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 4 & 0.5 & 2 & 0 & 0 & 0 & 0 \\ 0 & 4 & 1 & 6 & 0 & 0 & 0 \\ 0 & 0 & 4 & 1.5 & 12 & 0 & 0 \\ 0 & 0 & 0 & 4 & 2 & 20 & 0 \\ 0 & 0 & 0 & 0 & 4 & 2.5 & 30 \\ 0 & 0 & 0 & 0 & 0 & 4 & 3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 4  \end{bmatrix} \begin{bmatrix} A \\ B \\ C \\ D \\ E \\ F \\ G \end{bmatrix} = \begin{bmatrix} -.00326 \\ 2.12906 \\ -2.23336 \\ 5.42474 \\ -9.13616 \\ 7.70828 \\ -2.50522 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Using reverse substitution we find c
 * {| style="width:100%" border="0"

$$ \begin{bmatrix} A \\ B \\ C \\ D \\ E \\ F \\ G \end{bmatrix} = \begin{bmatrix} -2.545691906005222 \\ 16.947169811320755 \\ 0.8529525279814744 \\ -11.085427135678392 \\ 0.9152542372881356 \\ 2.3967971530249113 \\ -0.6263048016701461 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Recall
 * {| style="width:100%" border="0"

$$ y_p(t)= A + Bt + Ct^2 + Dt^3 +Et^4 + Ft^5 + Gt^6 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Therefore
 * {| style="width:100%" border="0"

$$ y_p(t)= -2.545691906005222 + 16.947169811320755t + 0.8529525279814744^2 - 11.085427135678392^3 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ + 0.9152542372881356t^4 + 2.3967971530249113t^5 - 0.6263048016701461t^6 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Recall
 * {| style="width:100%" border="0"

$$ y_1(t) = y_{h1}(t) + y_{P1}(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Therefore
 * {| style="width:100%" border="0"

$$ y_1=e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)] -2.545691906005222 + 16.947169811320755t + 0.8529525279814744t^2 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ - 11.085427135678392t^3 + 0.9152542372881356t^4 + 2.3967971530249113t^5 - 0.6263048016701461t^6 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plug in initial conditions
 * {| style="width:100%" border="0"

$$ 1 = e^{-\frac14 (0)}[C_{11}cos(\frac{3 \sqrt7}{4}(0))+C_{12}sin(\frac{3 \sqrt7}{4}(0))] - 2.545691906005222 + 16.947169811320755(0) + 0.8529525279814744(0) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ - 11.085427135678392(0) + 0.9152542372881356(0) + 2.3967971530249113(0) - 0.6263048016701461(0) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 1 = (1)C_{11} - 2.545691906005222 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ C_{11} = 3.545691906005222 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Take First derivative to use velocity initial condition
 * {| style="width:100%" border="0"

$$ y'_1 = -\frac{1}{4}e^{-\frac14 t}[C_{12}sin(\frac{3 \sqrt7}{4}t)+C_{11}sin(\frac{3 \sqrt7}{4}t)] + e^{-\frac14 t}[\frac{3 \sqrt7}{4}C_{12}cos(\frac{3 \sqrt7}{4}t)-\frac{3 \sqrt7}{4}C_{11}sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ + 16.947169811320755 + 1.705905056t - 33.25628141t^2 + 3.661016949t^3 + 11.98398557t^4 - 3.75782881t^5 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 0 = -\frac{1}{4}e^{-\frac14 (0)}[C_{12}sin(\frac{3 \sqrt7}{4}(0))+ (3.545691906005222)sin(\frac{3 \sqrt7}{4}(0))] + e^{-\frac14 (0)}[\frac{3 \sqrt7}{4}C_{12}cos(\frac{3 \sqrt7}{4}(0))-\frac{3 \sqrt7}{4}3.545691906005222sin(\frac{3 \sqrt7}{4}(0))] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ + 16.947169811320755 + 1.705905056(0) - 33.25628141(0) + 3.661016949(0) + 11.98398557(0) - 3.75782881(0) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 0 = -.25(3.545691906005222) + 1.98431C_{12} + 16.947169811320755 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ C_{12}= -8.093869184 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Get $$ y_1 $$
 * {| style="width:100%" border="0"

$$ y_1=e^{-\frac14 t}[3.545691906cos(\frac{3 \sqrt7}{4}t)-8.093869184sin(\frac{3 \sqrt7}{4}t)] -2.545691906005222 + 16.947169811320755t $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ + 0.8529525279814744t^2 - 11.085427135678392t^3 + 0.9152542372881356t^4 + 2.3967971530249113t^5 - 0.6263048016701461t^6 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Modal Equation 2
To find the modal displacement, we just need to find the homogenous solution.


 * {| style="width:100%" border="0"

$$ y_2(t) = y_{h2}(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

To find homogeneous solution we assume trial solution


 * {| style="width:100%" border="0"

$$ y_{h2}(t) = e^{\lambda_{2} t} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

in which we retrieve the characteristic equation


 * {| style="width:100%" border="0"

$$ \lambda_2^2 + 6 \,\lambda_2 + 9 = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

and the eigenvalues retrieved come out to be


 * {| style="width:100%" border="0"

$$ \lambda_{1,2}=-3 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

since the eigenvalues recovered are a double real root the solution takes the form


 * {| style="width:100%" border="0"

$$ y_2=P_{21}e^{-3t}+P{22}te^{-3t} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

and thus the modal velocity takes the form


 * {| style="width:100%" border="0"

$$ y_2=-3P_{21}e^{-3t}-3P_{22}te^{-3t}+P_{22}e^{-3t} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

to find the undetermined coefficients we must satisfy the modal initial conditions


 * {| style="width:100%" border="0"

$$ y_2(0) = 0 \Rightarrow P_{21} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y'_2(0) = 1 \Rightarrow -3 P_{21} + P_{22} = 1 \Rightarrow P_{22} = 1 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

and thus the modal displacement is


 * {| style="width:100%" border="0"

$$ y_2(t) = t e^{-3 t} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Complete Solution
Obtain displacements in original coordinate system by multiplying by the eigenvectors


 * {| style="width:100%" border="0"

$$ \mathbf d(t) = \bar{\boldsymbol \Phi} \, \mathbf y(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

then,


 * {| style="width:100%" border="0"

$$
 * style="width:95%" |
 * style="width:95%" |

\mathbf d(t)=\begin{bmatrix} \frac{1}{\sqrt{5}} & \sqrt{\frac{3}{10}} \\ \sqrt{\frac{6}{5}} & \frac{-2}{\sqrt{5}} \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ d_1 = \frac{1}{\sqrt{5}}(e^{-\frac14 t}[3.545691906cos(\frac{3 \sqrt7}{4}t)-8.093869184sin(\frac{3 \sqrt7}{4}t)] -2.545691906005222 + 16.947169811320755t $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ + 0.8529525279814744t^2 - 11.085427135678392t^3 + 0.9152542372881356t^4 + 2.3967971530249113t^5 - 0.6263048016701461t^6 + \sqrt{\frac{3}{10}}(te^{-3 t}) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ d_2 =  \sqrt{\frac{6}{5}}(e^{-\frac14 t}[3.545691906cos(\frac{3 \sqrt7}{4}t)-8.093869184sin(\frac{3 \sqrt7}{4}t)] -2.545691906005222 + 16.947169811320755t $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ + 0.8529525279814744t^2 - 11.085427135678392t^3 + 0.9152542372881356t^4 + 2.3967971530249113t^5 - 0.6263048016701461t^6 + \frac{-2}{\sqrt{5}}(t e^{-3 t}) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plots of the two solutions to the modal equations.

Green = Equation 1

Purple = Equation 2



n=2
The function f(x bar) can be approximated as


 * {| style="width:100%" border="0"

$$ f(x) \approx \bar c_0+ \bar c_1 \bar x+ \bar c_2 \bar x^2 $$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Because we are analyzing the functions on the x bar space, the Hilbert matrix remains the same as calculated in R4.4 for n=2

Therefore, we can use the Matlab function: hilb(3) to determine the Hilbert matrix of order 3.


 * {| style="width:100%" border="0"

$$ \boldsymbol \Gamma = \begin{bmatrix}1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5}\end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Then, using Matlab again, we solve for the inverse of the Hilbert matrix


 * {| style="width:100%" border="0"

$$ \boldsymbol \Gamma^{-1} = \begin{bmatrix}9 & -36 & 30 \\ -36 & 192 & -180 \\ 30 & -180 & 180\end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, we can develop the matrix d bar, the known computable right hand side:
 * {| style="width:100%" border="0"

$$ \mathbf \bar d = \begin{bmatrix} \langle \bar x^0,\log{\left(2 \bar x+1\right)} \rangle \\ \langle \bar x^1, \log{\left(2 \bar x+1\right)} \rangle\\ \langle \bar x^2, \log{\left(2 \bar x+1\right)} \rangle \end{bmatrix} = \begin{bmatrix} \int_{0}^{1} \log{\left(2 \bar x+1\right)} d\bar x \\ \int_{0}^{1} \bar x \log{\left(2 \bar x+1\right)} d\bar x  \\ \int_{0}^{1} \bar x^2 \log{\left(2 \bar x+1\right)}  d\bar x  \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We will need to employ the method of integration by parts to determine the matrix.


 * {| style="width:100%" border="0"

$$ \bar d_0 =\langle \bar x^0,\log{\left(2 \bar x+1\right)} \rangle= \int_{0}^{1} \log{\left(2 \bar x+1\right)} d\bar x = \frac{1}{2}\left(2\bar x+1\right)\left(\log{\left(2\bar x+1\right)}-1\right) |_0^1 = \frac{3\log{3}}{2}-1 = 0.6479 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

http://integrals.wolfram.com/index.jsp?expr=log%282x%2B1%29&random=false


 * {| style="width:100%" border="0"

$$ \bar d_1 = \langle \bar x^1, \log{\left(2 \bar x+1\right)} \rangle=\int_{0}^{1} \bar x \log{\left(2 \bar x+1\right)}  d\bar x = \frac{1}{16}\left(2\bar x+1\right)\left(-2\bar x+\left(4\bar x-2\right)\log{\left(2\bar x+1\right)}+3\right) = \frac{3\log{3}}{8} = 0.4120 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

http://integrals.wolfram.com/index.jsp?expr=x*log%282x%2B1%29&random=false


 * {| style="width:100%" border="0"

$$ \bar d_2 =\langle \bar x^2,\log{\left(2 \bar x+1\right)} \rangle= int_{0}^{1} \bar x^2\log{\left(2 \bar x+1\right)} d\bar x = \frac{1}{72}\left(3\left(8\bar x^3+1\right)\log{\left(2\bar x+1\right)}-2\bar x\left(\bar x\left(4\bar x-3\right)+3\right)\right) |_{0}^{1} = \frac{e\log{3}}{8}-\frac{1}{9} = 0.3009 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

http://integrals.wolfram.com/index.jsp?expr=x%5E2*log%282x%2B1%29&random=false


 * {| style="width:100%" border="0"

$$ \mathbf \bar d=\begin{bmatrix} 0.6479 \\ 0.4120 \\ 0.3009 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, we can solve for c bar,


 * {| style="width:100%" border="0"

$$ \mathbf \bar c= \boldsymbol \Gamma^{-1} \mathbf \bar d $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Performing matrix multiplication in Matlab...
 * {| style="width:100%" border="0"

$$ \mathbf \bar c= \begin{bmatrix}9 & -36 & 30 \\ -36 & 192 & -180 \\ 30 & -180 & 180\end{bmatrix} \begin{bmatrix} 0.6479 \\ 0.4120 \\ 0.3009 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf \bar c \approx \begin{bmatrix} 0.026054914182897\\  1.618691710928587\\  -0.562447009935070\end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, changing back to the original variable x, we sub in for x bar:


 * {| style="width:100%" border="0"

$$ f(\bar x) \approx 0.02605 + 1.6187 \bar x - 0.5624 \bar x^2 \approx f(\frac{x}{2}) \approx 0.026052 + 1.6187 (\frac{x}{2}) - 0.5624 (\frac{x}{2})^2 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

This results in the following plot on the interval [0,2]:



modal equation 1

 * {| style="width:100%" border="0"

$$ y_1'' + \frac12 \, y_1' + 4 \, y_1 = 2log(t+1) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_1(0) = 1, \ y'_1(0) = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solution for modal coordinate $$ y_1(t) $$ is the summation of

the homogeneous and particular solutions


 * {| style="width:100%" border="0"

$$ y_1(t) = y_{h1}(t) + y_{P1}(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

To find homogeneous solution we assume trial solution


 * {| style="width:100%" border="0"

$$ y_{h1}(t) = e^{\lambda_{1} t} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

in which we retrieve the characteristic equation


 * {| style="width:100%" border="0"

$$ \lambda_1^2 + \frac12 \,\lambda_1 + 4 = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

which generates the following complex-conjugate roots


 * {| style="width:100%" border="0"

$$ \lambda_{11} = -\frac14 + i \frac{3 \sqrt7}{4} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \lambda_{12} = -\frac14 - i \frac{3 \sqrt7}{4} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

to which the homogeneous solution then takes the form


 * {| style="width:100%" border="0"

$$ y_{h1}=e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, to find the particular solution $$ y_{P1} $$


 * {| style="width:100%" border="0"

$$ y_{P1}'' + \frac{1}{2}y_{P1}' + 4y_{P1} = 2log(1+t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sub in the logarithmic approximation for n=2
 * {| style="width:100%" border="0"

$$ y_{P1}'' + \frac{1}{2}y_{P1}' + 4y_{P1} = 2[ 0.026052 + 1.6187 (\frac{t}{2}) - 0.5624 (\frac{t}{2})^2] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

To solve for the particular solution, it must be split up in to three

smaller particular solutions. The sum of three equals the total particular solution

by the sum rule.


 * {| style="width:100%" border="0"

$$ y_{P1} = y_{P11} + y_{P12} + y_{P13} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We use this L2-ODE-CC to solve for $$ y_{P11} $$ with a particular

solution guess.


 * {| style="width:100%" border="0"

$$ y_{P11}'' + \frac{1}{2}y_{P11}' + 4y_{P11} = 2(0.026052) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Particular Solution Guess:


 * {| style="width:100%" border="0"

$$ y_{P11} = A $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Take the first and second derivative to plug back in to the modal equation


 * {| style="width:100%" border="0"

$$ y_{P11}' = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_{P11}'' = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plug Back In: :{| style="width:100%" border="0" $$ y_{P11}'' + \frac{1}{2}y_{P11}' + 4y_{P11} = 2(0.026052) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 0 + \frac{1}{2}(0) + 4A = 2(0.026052) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 4A = 2(0.026052) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ A = 0.013026 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Get particular solution $$ y_{P11} $$


 * {| style="width:100%" border="0"

$$ y_{P11} = 0.013026 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, to find the particular solution $$ y_{P12} $$


 * {| style="width:100%" border="0"

$$ y_{P12}'' + \frac{1}{2}y_{P12}' + 4y_{P12} = 2log(1+t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sub in the logarithmic approximation for n=2
 * {| style="width:100%" border="0"

$$ y_{P12}'' + \frac{1}{2}y_{P12}' + 4y_{P12} = 2[ 0.026052 + 1.6187 (\frac{t}{2}) - 0.5624 (\frac{t}{2})^2] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_{P12}'' + \frac{1}{2}y_{P12}' + 4y_{P12} = 2[1.6187(\frac{t}{2})] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Particular Solution Guess
 * {| style="width:100%" border="0"

$$ y_{P12} = K_{1}t + K_{0} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Take First and Second Derivative
 * {| style="width:100%" border="0"

$$ y'_{P12} = K_{1} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y''_{P12} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plug back into logarithmic approximation
 * {| style="width:100%" border="0"

$$ y_{P12}'' + \frac{1}{2}y_{P12}' + 4y_{P12} = 2[1.6187(\frac{t}{2})] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ (0) + \frac{1}{2}(K_{1}) + 4(K_{1}t + K_{0}) = 2[1.6187(\frac{t}{2})] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \frac{1}{2}K_{1} + 4K_{1}t + 4K_{0} = 1.6187t $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solve for coefficients
 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 4t & 0 \\ \frac{1}{2} & 4 \end{bmatrix} \begin{bmatrix} K_{1} \\ K_{0} \end{bmatrix} = \begin{bmatrix} 1.6187t \\ 0 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ K_{1}= 0.404675 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ K_{0}= -0.050584375 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plug coefficients back in to $$ y_{P12} $$
 * {| style="width:100%" border="0"

$$ y_{P12} = 0.404675t - 0.050584575 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, to find the particular solution $$ y_{P13} $$
 * {| style="width:100%" border="0"

$$ y_{P13}'' + \frac{1}{2}y_{P13}' + 4y_{P13} = 2log(1+t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sub in the logarithmic approximation for n=2
 * {| style="width:100%" border="0"

$$ y_{P13}'' + \frac{1}{2}y_{P13}' + 4y_{P13} = 2[ 0.026052 + 1.6187 (\frac{t}{2}) - 0.5624 (\frac{t}{2})^2] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_{P13}'' + \frac{1}{2}y_{P13}' + 4y_{P13} = 2[-0.5624(\frac{t}{2})^2] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Particular Solution Guess
 * {| style="width:100%" border="0"

$$ y_{P13} = F_{2}t^2 + F_{1}t + F_{0} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Take First and Second Derivative
 * {| style="width:100%" border="0"

$$ y'_{P13} = 2F_{2}t + F_{1} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y''_{P13} = 2F_{2} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plug back into logarithmic approximation
 * {| style="width:100%" border="0"

$$ (2F_{2}) + \frac{1}{2}(2F_{2}t + F_{1}) + 4(F_{2}t^2 + F_{1}t + F_{0}) = 2[-0.5624(\frac{t}{2})^2] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solve for Coefficients
 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 4t^2 & 0 & 0 \\ t & 4t & 0 \\ 2 & \frac{1}{2} & 4 \end{bmatrix} \begin{bmatrix} F_{2} \\ F_{1} \\ F_{0} \end{bmatrix} = \begin{bmatrix} -0.2812t^2 \\ 0 \\ 0 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ F_{2} = -0.0703 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ F_{1} = 0.017575 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ F_{0} = 0.032953125 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plug In Coefficients to $$ y_{P13} $$
 * {| style="width:100%" border="0"

$$ y_{P13} = -0.0703t^2 + 0.017575t + 0.032953125 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Find $$ y_{P1} $$
 * {| style="width:100%" border="0"

$$ y_{P1} = y_{P11} + y_{P12} + y_{P13} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_{P1} = (0.013026) + (0.404675t - 0.050584575) + (-0.0703t^2 + 0.017575t + 0.032953125) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_{P1} = -0.0703t^2 + 0.4225t - 0.00460545 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Verify Particular Solution Correctness

Get $$ y_{1} $$
 * {| style="width:100%" border="0"

$$ y_1 = y_{h1} + y_{P1} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_1 = e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)] - 0.0703t^2 + 0.4225t - 0.00460545 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plug in initial conditions
 * {| style="width:100%" border="0"

$$ 1 = e^{-\frac14 (0)}[C_{11}cos(\frac{3 \sqrt7}{4}(0))+C_{12}sin(\frac{3 \sqrt7}{4}(0))] - 0.0703(0) + 0.4225(0) - 0.00460545 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 1 = (1)C_{11} - 0.00460545 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ C_{11} = 1.00460545 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Take First derivative to use velocity initial condition
 * {| style="width:100%" border="0"

$$ y'_1 = -\frac{1}{4}e^{-\frac14 t}[C_{12}sin(\frac{3 \sqrt7}{4}t)+C_{11}sin(\frac{3 \sqrt7}{4}t)] + e^{-\frac14 t}[\frac{3 \sqrt7}{4}C_{12}cos(\frac{3 \sqrt7}{4}t)-\frac{3 \sqrt7}{4}C_{11}sin(\frac{3 \sqrt7}{4}t)] - 0.1406t + 0.4225 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 0 = -\frac{1}{4}e^{-\frac14 (0)}[C_{12}sin(\frac{3 \sqrt7}{4}(0))+ (1.00460545)sin(\frac{3 \sqrt7}{4}(0))] + e^{-\frac14 (0)}[\frac{3 \sqrt7}{4}C_{12}cos(\frac{3 \sqrt7}{4}(0))-\frac{3 \sqrt7}{4}1.00460545sin(\frac{3 \sqrt7}{4}(0))] - 0.1406(0) + 0.4225 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 0 = -.25(1.00460545) + 1.98431C_{12} + .4225 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ C_{12}= -0.0863517482 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Get $$ y_1 $$
 * {| style="width:100%" border="0"

$$ y_1 = e^{-\frac14 t}[1.00460545cos(\frac{3 \sqrt7}{4}t)- 0.0863517482sin(\frac{3 \sqrt7}{4}t)] - 0.0703t^2 + 0.4225t - 0.00460545 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Verify that Modal Displacement Satisfies Initial Conditions

modal equation 2
For modal equation 2, there is no particular solution. The modal displacement is just homogenous solution for modal equation 2.


 * {| style="width:100%" border="0"

$$ y_{h2}=e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solve for Coefficients by using initial conditions


 * {| style="width:100%" border="0"

$$ 1=e^{-\frac14 (0)}[C_{11}cos(\frac{3 \sqrt7}{4}(0))+C_{12}sin(\frac{3 \sqrt7}{4}(0))] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 1=(1)[C_{11}(1)) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ C_{11} = 1 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Take first derivative to solve for $$ C_{12} $$


 * {| style="width:100%" border="0"

$$ y'_{h2}=e^{-\frac14 t}[\frac{3 \sqrt7}{4}C_{12}cos(\frac{3 \sqrt7}{4}t)-\frac{3 \sqrt7}{4}C_{11}sin(\frac{3 \sqrt7}{4}t)] - \frac{1}{4}e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t) + C_{12}sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 0=e^{-\frac14 (0)}[\frac{3 \sqrt7}{4}C_{12}cos(\frac{3 \sqrt7}{4}(0))-\frac{3 \sqrt7}{4}(1)sin(\frac{3 \sqrt7}{4}(0))] - \frac{1}{4}e^{-\frac14 (0)}[(1)cos(\frac{3 \sqrt7}{4}(0)) + C_{12}sin(\frac{3 \sqrt7}{4}(0))] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 0=[\frac{3 \sqrt7}{4}C_{12}] - \frac{1}{4} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ C_{12} = .1259881577 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Get $$ y_2 $$


 * {| style="width:100%" border="0"

$$ y_2 = y_{h2} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_2 = e^{-\frac14 t}[cos(\frac{3 \sqrt7}{4}t)+ .1259881577sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Verify that Modal Displacement Satisfies Initial Conditions

Complete Solution
Obtain displacements in original coordinate system by multiplying by the eigenvectors


 * {| style="width:100%" border="0"

$$ \mathbf d(t) = \bar{\boldsymbol \Phi} \, \mathbf y(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

then,


 * {| style="width:100%" border="0"

$$
 * style="width:95%" |
 * style="width:95%" |

\mathbf d(t)=\begin{bmatrix} \frac{1}{\sqrt{5}} & \sqrt{\frac{3}{10}} \\ \sqrt{\frac{6}{5}} & \frac{-2}{\sqrt{5}} \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_1 = e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)] - 0.0703t^2 + 0.4225t - 0.00460545 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_2 = e^{-\frac14 t}[cos(\frac{3 \sqrt7}{4}t)+ .1259881577sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ d_1 = \frac{1}{\sqrt{5}}(e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)] - 0.0703t^2 + 0.4225t - 0.00460545)+ \sqrt{\frac{3}{10}}e^{-\frac14 t}[cos(\frac{3 \sqrt7}{4}t)+ .1259881577sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ d_2 = \sqrt{\frac{6}{5}}(e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)] - 0.0703t^2 + 0.4225t - 0.00460545)+ \frac{-2}{\sqrt{5}}e^{-\frac14 t}[cos(\frac{3 \sqrt7}{4}t)+ .1259881577sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

n=4
The function f(x bar) can be approximated as


 * {| style="width:100%" border="0"

$$ f(x) \approx \bar c_0+ \bar c_1 \bar x+ \bar c_2 \bar x^2+ \bar c_3 \bar x^3+\bar c_4 \bar x^4 $$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Because we are analyzing the functions on the x bar space, the Hilbert matrix remains the same as calculated in R4.4 for n=4

Therefore, we can use the Matlab function: hilb(5) to determine the Hilbert matrix of order 5.


 * {| style="width:100%" border="0"

$$ \boldsymbol \Gamma = \begin{bmatrix}1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6}\\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} \\ \frac{1}{4} & \frac{1}{5} & \frac{1}{5} & \frac{1}{7} & \frac{1}{8} \\ \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9}\end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Then, using Matlab again, we solve for the inverse of the Hilbert matrix


 * {| style="width:100%" border="0"

$$ \boldsymbol \Gamma^{-1} = 10^5 \begin{bmatrix} 0.0002 & -0.0030 &   0.0105  & -0.0140 &   0.0063\\ -0.0030 &  0.0480  & -0.1890 &   0.2688  & -0.1260\\    0.0105  & -0.1890  &  0.7938  & -1.1760  &  0.5670\\   -0.0140 &   0.2688  & -1.1760  &  1.7920 &  -0.8820\\    0.0063  & -0.1260  &  0.5670  & -0.8820  &  0.4410 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, we can develop the matrix d bar, the known computable right hand side:
 * {| style="width:100%" border="0"

$$ \mathbf \bar d = \begin{bmatrix} \langle \bar x^0,\log{2 \bar x} \rangle \\ \langle \bar x^1, \log{2 \bar x} \rangle\\ \langle \bar x^2, \log{2 \bar x} \\ \langle \bar x^3, \log{2 \bar x} \rangle \\ \langle \bar x^4, \log{2 \bar x} \rangle \rangle \end{bmatrix} = \begin{bmatrix} \int_{0}^{1} \log{2 \bar x} d\bar x \\ \int_{0}^{1} \bar x \log{2 \bar x} d\bar x  \\ \int_{0}^{1} \bar x^2 \log{2 \bar x}  d\bar x  \\ \int_{0}^{1} \bar x^3 \log{2 \bar x}  d\bar x  \\ \int_{0}^{1} \bar x^4 \log{2 \bar x}  d\bar x  \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We will need to employ the method of integration by parts to determine the matrix.


 * {| style="width:100%" border="0"

$$ \bar d_3 = \langle \bar x^3, \log{2 \bar x} \rangle=\int_{0}^{1} \bar x^3 \log{2 \bar x}  d\bar x = \frac{- \bar x^4}{16} + \frac{1}{4} \bar x^4 \log(2 \bar x +1) + \frac{\bar x^3}{24} - \frac{\bar x^2}{32} +\frac{x}{32} - \frac{1}{64} \log(2 \bar x +1) |_{0}^{1}  = \frac{1}{192}(45 \log(3) -4)  = 0.2367 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

https://www.wolframalpha.com/input/?i=integral+from+0+to+1+of+%28x%5E3%29log%282x%2B1%29


 * {| style="width:100%" border="0"

$$ \bar d_4 =\langle \bar x^4,\log{2 \bar x} \rangle= \int_{0}^{1} \bar x^4\log{2 \bar x} d\bar x = \frac{- \bar x^5}{25} + \frac{1}{5} \bar x^5 \log(2 \bar x +1) + \frac{\bar x^4}{4} - \frac{\bar x^3}{60} + \frac{\bar x^2}{80} - \frac{\bar x}{80} + \frac{1}{160}\log(2 \bar x +1) |_{0}^{1} = \frac{495 \log(3) -76}{2400} = 0.1949 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

https://www.wolframalpha.com/input/?i=integral+from+0+to+1+of+%28x%5E4%29log%282x%2B1%29


 * {| style="width:100%" border="0"

$$ \mathbf \bar d=\begin{bmatrix} 0.6479 \\ 0.4120 \\ 0.3009 \\ 0.2367 \\ 0.1949 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, we can solve for c bar,


 * {| style="width:100%" border="0"

$$ \mathbf \bar c= \boldsymbol \Gamma^{-1} \mathbf \bar d $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Performing matrix multiplication in Matlab...


 * {| style="width:100%" border="0"

$$ \mathbf c \approx \begin{bmatrix} 0.001444052549559\\  1.950727199565335\\  -1.576061476482209\\   1.042794468296051\\  -0.321170053975948 \end{bmatrix} $$ Now, changing back to the original variable x, we sub in for x bar:
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f(\bar x) \approx 0.00144 + 1.9507 \bar x - 1.5761 \bar x^2 + 1.0428 \bar x^3 - 0.3212 \bar x^4 \approx 0.00144 + 1.9507 (\frac{x}{2}) - 1.5761 (\frac{x}{2})^2 + 1.0428 (\frac{x}{2})^3 - 0.3212 (\frac{x}{2})^4 $$
 * style="width:95%" |
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 * }
 * }

This results in the following plot on the interval [0,2]:



modal equation 1

 * {| style="width:100%" border="0"

$$ y_1'' + \frac12 \, y_1' + 4 \, y_1 = 2log(t+1) $$
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 * }
 * }


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$$ y_1(0) = 1, \ y'_1(0) = 0 $$
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 * }
 * }

Solution for modal coordinate $$ y_1(t) $$ is the summation of

the homogeneous and particular solutions


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$$ y_1(t) = y_{h1}(t) + y_{P1}(t) $$
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 * }
 * }

To find homogeneous solution we assume trial solution


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$$ y_{h1}(t) = e^{\lambda_{1} x} $$
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 * }
 * }

in which we retrieve the characteristic equation


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$$ \lambda_1^2 + \frac12 \,\lambda_1 + 4 = 0 $$
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 * }
 * }

which generates the following complex-conjugate roots


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$$ \lambda_{11} = -\frac14 + i \frac{3 \sqrt7}{4} $$
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 * }
 * }


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$$ \lambda_{12} = -\frac14 - i \frac{3 \sqrt7}{4} $$
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 * }
 * }

to which the homogeneous solution then takes the form


 * {| style="width:100%" border="0"

$$ y_{h1}=e^{-\frac14 x}[C_{11}cos(\frac{3 \sqrt7}{4}x)+C_{12}sin(\frac{3 \sqrt7}{4}x)] $$
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 * }
 * }

Now, to find the particular solution $$ y_{P1} $$


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$$ y_{P1}'' + \frac{1}{2}y_{P1}' + 4y_{P1} = 2log(1+t) $$
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 * }
 * }

Sub in the logarithmic approximation for n=4
 * {| style="width:100%" border="0"

$$ y_{P1}'' + \frac{1}{2}y_{P1}' + 4y_{P1} = 2[0.00144 + 1.9507 (\frac{x}{2}) - 1.5761 (\frac{x}{2})^2 + 1.0428 (\frac{x}{2})^3 - 0.3212 (\frac{x}{2})^4 ] $$
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 * }
 * }

To solve for the particular solution, it must be split up in to five

smaller particular solutions. The sum of three equals the total particular solution

by the sum rule.


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$$ y_{P1} = y_{P11} + y_{P12} + y_{P13} + y_{P14} + y_{P15} $$
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 * }
 * }

We use this L2-ODE-CC to solve for $$ y_{P11} $$ with a particular

solution guess.


 * {| style="width:100%" border="0"

$$ y_{P11}'' + \frac{1}{2}y_{P11}' + 4y_{P11} = 2(0.026052) $$
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 * }
 * }

Particular Solution Guess:


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$$ y_{P11} = A $$
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 * }
 * }

Take the first and second derivative to plug back in to the modal equation


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$$ y_{P11}' = 0 $$
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 * }
 * }


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$$ y_{P11}'' = 0 $$
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 * }
 * }

Plug Back In:
 * {| style="width:100%" border="0"

$$ y_{P11}'' + \frac{1}{2}y_{P11}' + 4y_{P11} = 2(0.00144) $$
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 * }
 * }


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$$ 0 + 0 + 4A = 2(0.00144) $$
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 * }
 * }


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$$ A = 0.00072 $$
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 * }
 * }

Solve for $$ y_{P12} $$
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$$ y_{P12}'' + \frac{1}{2}y_{P12}' + 4y_{P12} = 2(1.9507 (\frac{x}{2})) $$
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 * }
 * }

Particular Solution Guess
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$$ y_{P12} = Ax + B $$
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 * }

Take First and Second Derivative
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$$ y'_{P12} = A $$
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 * }
 * }


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$$ y''_{P12} = 0 $$
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 * }
 * }

Plug Back in
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$$ (0) + \frac{1}{2}(A) + 4(Ax + B) = 2(1.9507 (\frac{x}{2})) $$
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 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 4x & 0 \\ \frac{1}{2} & 4 \end{bmatrix} \begin{bmatrix} A \\ B \end{bmatrix} = \begin{bmatrix} 1.9507x \\ 0 \end{bmatrix} $$
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 * }
 * }


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$$ A = 0.487675 $$
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 * }
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$$ B = -0.060959375 $$
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Plug in Coefficients
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$$ y_{P12} = .487675x - 0.060959375 $$
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 * }
 * }

Solve for $$ y_{P13} $$
 * {| style="width:100%" border="0"

$$ y_{P13}'' + \frac{1}{2}y_{P13}' + 4y_{P13} = 2(- 1.5761 (\frac{x}{2})^2)) $$
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 * }
 * }

Particular Solution Guess
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$$ y_{P13} = Ax^2 + Bx + C $$
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 * }
 * }

Take First and Second derivative
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$$ y'_{P13} = 2Ax + B $$
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 * }
 * }


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$$ y''_{P13} = 2A $$
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 * }
 * }

Plug Back in
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$$ (2A) + \frac{1}{2}(2Ax + B) + 4(Ax^2 + Bx + C) = -0.78805x^2 $$
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 * }
 * }

Solve
 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 4x^2 & 0 & 0 \\ x & 4x & 0 \\ 2 & \frac12 & 4 \end{bmatrix} \begin{bmatrix} A \\ B \\C \end{bmatrix} = \begin{bmatrix} -0.78805x^2 \\ 0 \\ 0 \end{bmatrix} $$
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 * }
 * }


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$$ A = 0.1970125 $$
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 * }
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$$ B = -0.049253125 $$
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 * }
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$$ C = -0.0923496094 $$
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 * }
 * }

Plug back in to $$ y_{P13} $$
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$$ y_{P13} = .1970125x^2 - 0.049253125x - 0.0923496094 $$
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 * }
 * }

Solve for $$ y_{p14} $$
 * {| style="width:100%" border="0"

$$ y_{P14}'' + \frac{1}{2}y_{P14}' + 4y_{P14} = 2(1.0428 (\frac{x}{2})^3) $$
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 * }
 * }

Particular Solution Guess
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$$ y_{P14} = Ax^3 + Bx^2 + Cx + D $$
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 * }
 * }

Take First and Second Derivative


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$$ y'_{P14} = 3Ax^2 + 2Bx + C $$
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 * }
 * }


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$$ y'_{P14} = 6Ax + 2B $$
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Plug back in
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$$ (6Ax + 2B) + \frac{1}{2}(3Ax^2 + 2Bx + C) + 4(Ax^3 + Bx^2 + Cx + D) = 0.2607x^3 $$
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 * }
 * }

Solve
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$$ \begin{bmatrix} 4x^3 & 0 & 0 & 0 \\ \frac32x^2 & 4x^2 & 0 & 0 \\ 6x & x & 4x & 0 \\ 0 & 2 & \frac12 & 4\end{bmatrix} \begin{bmatrix} A \\ B \\ C \\ D \end{bmatrix} = \begin{bmatrix} 0.2607x^3 \\ 0 \\ 0 \\ 0 \end{bmatrix} $$
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 * }
 * }


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$$ A = 0.065175 $$
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$$ B = -0.024440625 $$
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 * }
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$$ C = -0.0916523438 $$
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 * }
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$$ D = -.0236768555 $$
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Plug back in to $$ y_{P14} $$
 * {| style="width:100%" border="0"

$$ y_{P14} = 0.065175x^3 - 0.024440625x^2 - 0.0916523438x - 0.0236768555 $$
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 * }
 * }

Find $$ y_{P15} $$
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$$ y_{P15}'' + \frac{1}{2}y_{P15}' + 4y_{P15} = 2[- 0.3212 (\frac{x}{2})^4 ] $$
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 * }
 * }

Particular Solution Guess
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$$ y_{P15} = Ax^4 + Bx^3 + Cx^2 + Dx + E $$
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 * }
 * }

Take First and Second Derivative
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$$ y'_{P15} = 4Ax^3 + 3Bx^2 + 2Cx + D $$
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 * }
 * }


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$$ y''_{P15} = 12Ax^2 + 6Bx + 2C $$
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Plug back in
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$$ (12Ax^2 + 6Bx + 2C) + \frac{1}{2}(4Ax^3 + 3Bx^2 + 2Cx + D ) + 4(Ax^4 + Bx^3 + Cx^2 + Dx + E) = - 0.04015x $$
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 * }

Solve
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$$ \begin{bmatrix} 4x^4 & 0 & 0 & 0 & 0 \\ 2x^3 & 4x^3 & 0 & 0 & 0 \\12x^2 & \frac32x^2 & 4x^2 & 0 & 0 \\ 0 & 6x & x & 4x & 0 \\ 0 & 0 & 2 & \frac12 & 4\end{bmatrix} \begin{bmatrix} A \\ B \\ C \\ D \\ E \end{bmatrix} = \begin{bmatrix} -0.04015x^4 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} $$
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 * }
 * }


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$$ A = -.0100375 $$
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$$ B = 0.00501875 $$
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$$ C = 0.0282304688 $$
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 * }
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$$ D = -0.0145857422 $$
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 * }
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$$ E = -0.0122920166 $$
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Plug back in to $$ y_{p15} $$
 * {| style="width:100%" border="0"

$$ y_{P15} = -.0100375x^4 + 0.00501875x^3 + 0.0282304688x^2 - 0.0145857422x - 0.0122920166 $$
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 * }
 * }

Get $$ y_{P1} $$
 * {| style="width:100%" border="0"

$$ y_{P1} = -0.00100375x^4 + .0701938x^3 + 0.200802x^2 + 0.332184x - 0.188558 $$
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 * }
 * }

Verify Correctness of the particular solution

Find $$ y_{1} $$


 * {| style="width:100%" border="0"

$$ y_{1} = y_{h1} + y_{P1} $$
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 * }
 * }


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$$ y_{1} = e^{-\frac14 x}[C_{11}cos(\frac{3 \sqrt7}{4}x)+C_{12}sin(\frac{3 \sqrt7}{4}x)] + (-0.00100375x^4 + .0701938x^3 + 0.200802x^2 + 0.332184x - 0.188558) $$
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 * }
 * }

Find coefficients using initial conditions


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$$ 1 = 1[C_{11}] - 0.188558) $$
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 * }
 * }


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$$ C_{11} = 1.188558 $$
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 * }
 * }


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$$ C_{12} = .1497442327 $$
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 * }
 * }


 * {| style="width:100%" border="0"

$$ y_{1} = e^{-\frac14 x}[1.188558cos(\frac{3 \sqrt7}{4}x)+ .1497442327sin(\frac{3 \sqrt7}{4}x)] + (-0.00100375x^4 + .0701938x^3 + 0.200802x^2 + 0.332184x - 0.188558) $$
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 * }
 * }

modal equation 2
For modal equation 2, there is no particular solution. The modal displacement is just homogenous solution for modal equation 2.


 * {| style="width:100%" border="0"

$$ y_{h2}=e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)] $$
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 * }
 * }

Solve for Coefficients by using initial conditions


 * {| style="width:100%" border="0"

$$ 1=e^{-\frac14 (0)}[C_{11}cos(\frac{3 \sqrt7}{4}(0))+C_{12}sin(\frac{3 \sqrt7}{4}(0))] $$
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 * }
 * }


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$$ 1=(1)[C_{11}(1)) $$
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 * }
 * }


 * {| style="width:100%" border="0"

$$ C_{11} = 1 $$
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 * }
 * }

Take first derivative to solve for $$ C_{12} $$


 * {| style="width:100%" border="0"

$$ y'_{h2}=e^{-\frac14 t}[\frac{3 \sqrt7}{4}C_{12}cos(\frac{3 \sqrt7}{4}t)-\frac{3 \sqrt7}{4}C_{11}sin(\frac{3 \sqrt7}{4}t)] - \frac{1}{4}e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t) + C_{12}sin(\frac{3 \sqrt7}{4}t)] $$
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 * }
 * }


 * {| style="width:100%" border="0"

$$ 0=e^{-\frac14 (0)}[\frac{3 \sqrt7}{4}C_{12}cos(\frac{3 \sqrt7}{4}(0))-\frac{3 \sqrt7}{4}(1)sin(\frac{3 \sqrt7}{4}(0))] - \frac{1}{4}e^{-\frac14 (0)}[(1)cos(\frac{3 \sqrt7}{4}(0)) + C_{12}sin(\frac{3 \sqrt7}{4}(0))] $$
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 * }
 * }


 * {| style="width:100%" border="0"

$$ 0=[\frac{3 \sqrt7}{4}C_{12}] - \frac{1}{4} $$
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 * }
 * }


 * {| style="width:100%" border="0"

$$ C_{12} = .1259881577 $$
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 * }
 * }

Get $$ y_2 $$


 * {| style="width:100%" border="0"

$$ y_2 = y_{h2} $$
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 * }
 * }


 * {| style="width:100%" border="0"

$$ y_2 = e^{-\frac14 t}[cos(\frac{3 \sqrt7}{4}t)+ .1259881577sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Verify that Modal Displacement Satisfies Initial Conditions

n=6
The function f(x bar) can be approximated as


 * {| style="width:100%" border="0"

$$ f(x) \approx \bar c_0+ \bar c_1 \bar x+ \bar c_2 \bar x^2 + \bar c_3 \bar x^3 + \bar c_4 \bar x^4 + \bar c_5 \bar x^5 + \bar c_6 \bar x^6 $$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Because we are analyzing the functions on the x bar space, the Hilbert matrix remains the same as calculated in R4.4 for n=6

Therefore, we can use the Matlab function: hilb(7) to determine the Hilbert matrix of order 7.


 * {| style="width:100%" border="0"

$$ \boldsymbol \Gamma = \begin{bmatrix}1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} &\frac{1}{6} &\frac{1}{7}\\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} &\frac{1}{7} &\frac{1}{8}\\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} &\frac{1}{8} &\frac{1}{9} \\ \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} &\frac{1}{9} &\frac{1}{10}\\ \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9} &\frac{1}{10} &\frac{1}{11} \\ \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9} & \frac{1}{10} &\frac{1}{11} &\frac{1}{12} \\ \frac{1}{7} & \frac{1}{8} & \frac{1}{9} & \frac{1}{10} & \frac{1}{11} &\frac{1}{12} &\frac{1}{13}\end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Then, using Matlab again, we solve for the inverse of the Hilbert matrix


 * {| style="width:100%" border="0"

$$ \boldsymbol \Gamma^{-1} = 10^8 \begin{bmatrix} 0.0000 & -0.0000 &   0.0001 &  -0.0003 &   0.0005 &  -0.0004  &  0.0001\\   -0.0000   & 0.0004  & -0.0032   & 0.0113  & -0.0194   & 0.0160  & -0.0050\\    0.0001  & -0.0032    &0.0286  & -0.1058   & 0.1871  & -0.1572   & 0.0505\\   -0.0003  &  0.0113  & -0.1058   & 0.4032  & -0.7277   & 0.6209  & -0.2018\\    0.0005   &-0.0194   & 0.1871   &-0.7277   & 1.3340   &-1.1526    &0.3784\\   -0.0004   & 0.0160   &-0.1572  &  0.6209  & -1.1526  &  1.0059  & -0.3330\\    0.0001   &-0.0050    &0.0505 &  -0.2018   & 0.3784 &  -0.3330   & 0.1110\\ \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, we can develop the matrix d bar, the known computable right hand side:
 * {| style="width:100%" border="0"

$$ \mathbf \bar d = \begin{bmatrix} \langle \bar x^0,\log{\left(2 \bar x+1\right)} \rangle \\ \langle \bar x^1, \log{\left(2 \bar x+1\right)} \rangle\\ \langle \bar x^2, \log{\left(2 \bar x+1\right)} \rangle \\ \langle \bar x^3, \log{\left(2 \bar x+1\right)} \rangle \\ \langle \bar x^4, \log{\left(2 \bar x+1\right)} \rangle \\ \langle \bar x^5, \log{\left(2 \bar x+1\right)} \rangle \\ \langle \bar x^6, \log{\left(2 \bar x+1\right)} \rangle \end{bmatrix} = \begin{bmatrix} \int_{0}^{1} \log{\left(2 \bar x+1\right)} dx \\ \int_{0}^{1} \bar x \log{\left(2 \bar x+1\right)} dx  \\ \int_{0}^{1} \bar x^2 \log{\left(2 \bar x+1\right)}  dx \\ \int_{0}^{1} \bar x^3 \log{\left(2 \bar x+1\right)}  dx \\ \int_{0}^{1} \bar x^4 \log{\left(2 \bar x+1\right)}  dx \\ \int_{0}^{1} \bar x^5 \log{\left(2 \bar x+1\right)}  dx \\ \int_{0}^{1} \bar x^6 \log{\left(2 \bar x+1\right)}  dx \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We will need to employ the method of integration by parts to determine the matrix.


 * {| style="width:100%" border="0"

$$ \bar d_5 =\langle \bar x^5,\log{\left(2 \bar x+1\right)} \rangle= \int_{0}^{1} \bar x^5 \log{\left(2 \bar x+1\right)} dx =[\frac{-\bar x^6}{36}+\frac{1}{6}\bar x^6 \log{\left(2\bar x+1\right)}+\frac{\bar x^5}{60}-\frac{\bar x^4}{96}+\frac{\bar x^3}{144}-\frac{\bar x^2}{192}+\frac{\bar x}{192}-\frac{1}{384}\log{\left(2\bar x +1\right)}] |_0^1 =\frac{21\log{3}}{128}-\frac{7}{480} =0.1656 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

http://integrals.wolfram.com/index.jsp?expr=x%5E5*log%282*x%2B1%29&random=false


 * {| style="width:100%" border="0"

$$ \bar d_6 = \langle \bar x^6, \log{\left(2 \bar x+1\right)} \rangle =\int_{0}^{1} \bar x^6 \log{\left(2 \bar x+1\right)} dx =[\frac{-\bar x^7}{49}+\frac{1}{7}\bar x^7 \log{\left(2\bar x+1\right)}+\frac{\bar x^6}{84}-\frac{\bar x^5}{140}+\frac{\bar x^4}{224}-\frac{\bar x^3}{336}+\frac{\bar x^2}{48}-\frac{\bar x}{448}+\frac{1}{896}\log{\left(2\bar x +1\right)}] |_0^1 =\frac{129\log{3}}{896}-\frac{111}{7840} =0.1440 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

http://integrals.wolfram.com/index.jsp?expr=x%5E6*log%282*x%2B1%29&random=false


 * {| style="width:100%" border="0"

$$ \mathbf \bar d=\begin{bmatrix} 0.6479 \\ 0.4120 \\ 0.3009 \\ 0.2367 \\ 0.1949 \\ 0.1656 \\ 0.1440 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, we can solve for c bar,


 * {| style="width:100%" border="0"

$$ \mathbf \bar c= \boldsymbol \Gamma^{-1} \mathbf \bar d $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Performing matrix multiplication in Matlab...


 * {| style="width:100%" border="0"

$$ \mathbf c \approx \begin{bmatrix} 0.000087533828719\\  1.994608291424811\\  -1.916677197907120\\   2.099761187098920\\  -1.844738412648439\\   1.010804294608533\\  -0.245285915676504 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, changing back to the original variable x, we sub in for x bar:


 * {| style="width:100%" border="0"

$$ f(\bar x) \approx 0.0001 + 1.9946 \bar x - 1.9167 \bar x^2 + 2.0998 \bar x^3 - 1.8447 \bar x^4 + 1.0108 \bar x^5 - 0.2453 \bar x^6 \approx f(\frac{x}{2}) \approx 0.0001 + 1.9946 \left(\frac{x}{2}\right) - 1.9167 \left(\frac{x}{2}\right)^2 + 2.0998 \left(\frac{x}{2}\right)^3 - 1.8447 \left(\frac{x}{2}\right)^4 + 1.0108 \left(\frac{x}{2}\right)^5 - 0.2453 \left(\frac{x}{2}\right)^6 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

This results in the following plot on the interval [0,2]:



modal equation 1

 * {| style="width:100%" border="0"

$$ y_1'' + \frac12 \, y_1' + 4 \, y_1 = 2log(t+1) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_1(0) = 1, \ y'_1(0) = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solution for modal coordinate $$ y_1(t) $$ is the summation of

the homogeneous and particular solutions


 * {| style="width:100%" border="0"

$$ y_1(t) = y_{h1}(t) + y_{P1}(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

To find homogeneous solution we assume trial solution


 * {| style="width:100%" border="0"

$$ y_{h1}(t) = e^{\lambda_{1} x} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

in which we retrieve the characteristic equation


 * {| style="width:100%" border="0"

$$ \lambda_1^2 + \frac12 \,\lambda_1 + 4 = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

which generates the following complex-conjugate roots


 * {| style="width:100%" border="0"

$$ \lambda_{11} = -\frac14 + i \frac{3 \sqrt7}{4} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \lambda_{12} = -\frac14 - i \frac{3 \sqrt7}{4} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

to which the homogeneous solution then takes the form


 * {| style="width:100%" border="0"

$$ y_{h1}=e^{-\frac14 x}[C_{11}cos(\frac{3 \sqrt7}{4}x)+C_{12}sin(\frac{3 \sqrt7}{4}x)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, to find the particular solution $$ y_{P1} $$


 * {| style="width:100%" border="0"

$$ y_{P1}'' + \frac{1}{2}y_{P1}' + 4y_{P1} = 2log(1+t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sub in the logarithmic approximation for n=6
 * {| style="width:100%" border="0"

$$ y_{P1}'' + \frac{1}{2}y_{P1}' + 4y_{P1} = 2[0.0001 + 1.9946 \left(\frac{x}{2}\right) - 1.9167 \left(\frac{x}{2}\right)^2 + 2.0998 \left(\frac{x}{2}\right)^3 - 1.8447 \left(\frac{x}{2}\right)^4 + 1.0108 \left(\frac{x}{2}\right)^5 - 0.2453 \left(\frac{x}{2}\right)^6 ] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_{P1}'' + \frac{1}{2}y_{P1}' + 4y_{P1} = -0.007665625x^6+0.063175x^5-0.2305875x^4+0.52495x^3-0.95835x^2+1.9964x+0.0002 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We guess the form of the particular solution knowing the form for a polynomial excitation:


 * {| style="width:100%" border="0"

$$ y_{P1}=Ax^6+Bx^5+Cx^4+Dx^3+Ex^2+Fx+G $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We take the first two derivatives of the particular solution


 * {| style="width:100%" border="0"

$$ y_{P1}'=6Ax^5+5Bx^4+4Cx^3+3Dx^2+2Ex+F $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_{P1}''=30Ax^3+20Bx^3+12Cx^2+6Dx+2E $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We substitute the derivatives into the ODE.


 * {| style="width:100%" border="0"

$$ 30Ax^3+20Bx^3+12Cx^2+6Dx+2E+\frac{1}{2}\left(6Ax^5+5Bx^4+4Cx^3+3Dx^2+2Ex+F\right)+4\left(Ax^6+Bx^5+Cx^4+Dx^3+Ex^2+Fx+G\right)= $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 4Ax^6+\left(3A+4B\right) x^5+\left(30A+\frac{5}{2}B+4C\right) x^4+\left(20B+2C+4D\right) x^3+\left(12C+\frac{3}{2}D+4E\right) x^2+\left(6D+E+4F\right) x+\left(2E+\frac{1}{2}F+4G\right) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We can now solve for the coefficients A, B, C, D, E, F, and G by equating the coefficients.


 * {| style="width:100%" border="0"

$$ A=-0.00191640625, B=0.017231054688, C=-0.54043237305, D=0.072103845215, E=-0.104496730042, F=0.417068414688, G=1.64813184743E-4 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So the particular solution is


 * {| style="width:100%" border="0"

$$ y_{P1}=-0.00191640625x^6+0.017231054688x^5-0.54043237305x^4+0.072103845215x^3-0.104496730042x^2+0.417068414688x+1.64813184743E-4 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

And the modal equation is


 * {| style="width:100%" border="0"

$$ y_1(x)=e^{-\frac14 x}[C_{11}cos(\frac{3 \sqrt7}{4}x)+C_{12}sin(\frac{3 \sqrt7}{4}x)]-0.00191640625x^6+0.017231054688x^5-0.54043237305x^4+0.072103845215x^3-0.104496730042x^2+0.417068414688x+1.64813184743E-4 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Substitute in initial conditions of the modal equation and its derivative to solve for C11 and C12.


 * {| style="width:100%" border="0"

$$ y_1(0)=1=C_{11}+1.64813184743E-4 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ C_{11}=0.99835 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_1'(0)=0=\frac{3}{4}\left(C_{12}\sqrt{7}+0.222813\right) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ C_{12}=-0.08425 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_1(x)=e^{-\frac14 x}[0.99835cos(\frac{3 \sqrt7}{4}x)-0.08425sin(\frac{3 \sqrt7}{4}x)]-0.00191640625x^6+0.017231054688x^5-0.54043237305x^4+0.072103845215x^3-0.104496730042x^2+0.417068414688x+1.64813184743E-4 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

modal equation 2
For modal equation 2, there is no particular solution. The modal displacement is just homogenous solution for modal equation 2.


 * {| style="width:100%" border="0"

$$ y_{h2}=e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solve for Coefficients by using initial conditions


 * {| style="width:100%" border="0"

$$ 1=e^{-\frac14 (0)}[C_{11}cos(\frac{3 \sqrt7}{4}(0))+C_{12}sin(\frac{3 \sqrt7}{4}(0))] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 1=(1)[C_{11}(1)) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ C_{11} = 1 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Take first derivative to solve for $$ C_{12} $$


 * {| style="width:100%" border="0"

$$ y'_{h2}=e^{-\frac14 t}[\frac{3 \sqrt7}{4}C_{12}cos(\frac{3 \sqrt7}{4}t)-\frac{3 \sqrt7}{4}C_{11}sin(\frac{3 \sqrt7}{4}t)] - \frac{1}{4}e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t) + C_{12}sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 0=e^{-\frac14 (0)}[\frac{3 \sqrt7}{4}C_{12}cos(\frac{3 \sqrt7}{4}(0))-\frac{3 \sqrt7}{4}(1)sin(\frac{3 \sqrt7}{4}(0))] - \frac{1}{4}e^{-\frac14 (0)}[(1)cos(\frac{3 \sqrt7}{4}(0)) + C_{12}sin(\frac{3 \sqrt7}{4}(0))] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 0=[\frac{3 \sqrt7}{4}C_{12}] - \frac{1}{4} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ C_{12} = .1259881577 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Get $$ y_2 $$


 * {| style="width:100%" border="0"

$$ y_2 = y_{h2} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_2 = e^{-\frac14 t}[cos(\frac{3 \sqrt7}{4}t)+ .1259881577sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Verify that Modal Displacement Satisfies Initial Conditions

Problem Statement
Continuation of R5.3


 * 1) For n = 3, 4, repeat the steps in Part 1 R5.3. Compare the approximated solutions obtained with the Fourier basis to the approximated solutions obtained with the polynomial basis. Compare the results with those in R6.1.

Solution
Modal Equations and initial conditions

Modal Equation #1:
 * {| style="width:100%" border="0"

$$ y_1'' + \frac12 \, y_1' + 4 \, y_1 = 2log(t+1) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_1(0) = 1, \ y'_1(0) = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Modal Equation #2:
 * {| style="width:100%" border="0"

$$ y_2'' + 6 \, y_2' + 9 \, y_2 = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_2(0) = 0, \ y'_2(0) = 1 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

No matter the excitation function or further projection of excitation function, the homogeneous solution will be of the same form.

Since the process and solution was already calculated in R5 there is no need to re-derive.


 * {| style="width:100%" border="0"

$$ y_{h1}=e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

in which coefficients $$ C_{11} and C_{12} $$ are to be determined using initial conditions.

Though the homogeneous solution is of the same form invariant to the excitation the coefficients do

with a different excitation function.

Particular Solution
The Fourier projected logarithm function for n=3 coefficients can be seen below.


 * {| style="width:100%" border="0"

$$ \begin{align} 1.78281828469046-0.684161390102918 cos(\pi x)+0.084898666012778 cos(2 \pi x)\\ -0.082788137760126 cos(3 \pi x) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

thus we can rewrite the first modal equation as


 * {| style="width:100%" border="0"

$$ \begin{align} y_1'' + \frac12 \, y_1' + 4 \, y_1 = 2(1.78281828469046-0.684161390102918 cos(\pi x)\\ +0.084898666012778 cos(2 \pi x)-0.082788137760126 cos(3 \pi x)) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We must now find the particular solution for the given excitation. We assume the particular solution takes the form.


 * {| style="width:100%" border="0"

$$ y_{p1}=A+Bcos(x)+Csin(x)+Dcos(2x)+Esin(2x)+Fcos(3x)+Gsin(3x) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

where


 * {| style="width:100%" border="0"

$$ \begin{align} y_{p1}'= -Bsin(x)+Ccos(x)-2Dsin(2x)+2Ecos(2x)-3Fsin(3x)+3Gcos(3x) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{align} y_{p1}''= -Bcos(x)-Csin(x)-4Dcos(2x)-4Esin(2x)-9Fcos(3x)-9Gsin(3x) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Imputing this into the modal particular equation or


 * {| style="width:100%" border="0"

$$ \begin{align} y_{p1}'' + \frac12 \, y_{p1}' + 4 \, y_{p1} = 2(1.78281828469046-0.684161390102918 cos(\pi x)\\ +0.084898666012778 cos(2 \pi x) -0.082788137760126 cos(3 \pi x)) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

with the choice for particular solution plugged in.


 * {| style="width:100%" border="0"

$$ \begin{align} -Bcos(x)-Csin(x)-4Dcos(2x)-4Esin(2x)-9Fcos(3x)-9Gsin(3x)\\
 * style="width:95%" |
 * style="width:95%" |

+ \frac12 \, [-Bsin(x)+Ccos(x)-2Dsin(2x)+2Ecos(2x)-3Fsin(3x)+3Gcos(3x)]\\ +

4 \, [A+Bcos(x)+Csin(x)+Dcos(2x)+Esin(2x)+Fcos(3x)+Gsin(3x)]\\

= 2(1.78281828469046-0.684161390102918 cos(\pi x)\\ +0.084898666012778 cos(2 \pi x) -0.082788137760126 cos(3 \pi x))

\end{align} $$
 * }
 * }

and combining like terms, we can solve for the coefficients (A-I)


 * {| style="width:100%" border="0"

$$ Constant: 4A = 2(1.718281828459046) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ cos(\pi x): -B + \frac12 C + 4B= 2(-0.684161390102918) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ sin(\pi x): -C - \frac12 B + 4C= 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ cos(2 \pi x): -4D + \frac12 2E + 4D= 2(0.084898666012778) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ sin(2 \pi x): -4E - \frac12 2D + 4E= 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ cos(3 \pi x): -9F+ \frac12 3G + 4F= 2(-0.082788137760126) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ sin(3 \pi x): -9G - \frac12 3F + 4G= 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

we can put this into matrix form to quickly solve for the coefficients.


 * {| style="width:100%" border="0"

$$ C = A^{-1} d $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

where


 * {| style="width:100%" border="0"

$$ \mathbf A= \begin{bmatrix} 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & (-1+4) & \frac12 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 &-\frac12 & (-1+4) & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & (-4+4) & 1 & 0 & 0 & 0 & 0 \\              0 & 0 & 0 & -1 & (-4+4) & 0 & 0 & 0 & 0 \\               0 & 0 & 0 & 0 & 0 & (-9+4) & \frac32 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -\frac32 & (-9+4) & 0 & 0  \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

,


 * {| style="width:100%" border="0"

$$ \mathbf d= \begin{bmatrix} 2(1.718281828459046) \\ 2(-0.684161390102918) \\               0  \\               2(0.084898666012778)  \\               0  \\               2(-0.082788137760126)  \\               0     \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf C= \begin{bmatrix} A \\ B \\ C \\ D \\ E \\ F \\ G   \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

therefore the coefficients A-I become
 * {| style="width:100%" border="0"

$$ \mathbf C= \begin{bmatrix} 0.8591 \\ -0.4438 \\               -0.0740  \\               0  \\               0.1698  \\               0.0304  \\               -0.0091    \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

and thus the particular solution takes the form


 * {| style="width:100%" border="0"

$$ \begin{align} y_{p1}=0.8591-0.4438cos(x)-0.0740sin(x)+0.1698sin(2x)\\+0.0304cos(3x) -0.0091sin(3x) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

n=4
The Fourier projected logarithm function for n=4 coefficients can be seen below.


 * {| style="width:100%" border="0"

$$ \begin{align} 1.78281828469046-0.684161390102918 cos(\pi x)+0.084898666012778 cos(2 \pi x)\\ -0.082788137760126 cos(3 \pi x) + 0.021625349461069 cos(4 \pi x) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Modal Equation 1
We can rewrite the first modal equation as


 * {| style="width:100%" border="0"

$$ \begin{align} y_1'' + \frac12 \, y_1' + 4 \, y_1 = 2(1.78281828469046-0.684161390102918 cos(\pi x)\\ +0.084898666012778 cos(2 \pi x)-0.082788137760126 cos(3 \pi x)\\ + 0.021625349461069 cos(4 \pi x)) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Particular Solution
We must now find the particular solution for the given excitation. We assume the particular solution takes the form.


 * {| style="width:100%" border="0"

$$ \begin{align} y_{p1}=A+Bcos(\pi x)+Csin(\pi x)+Dcos(2 \pi x)+Esin(2 \pi x)+Fcos(3 \pi x)+Gsin(3 \pi x)\\+Hcos(4 \pi x)+Isin(4 \pi x) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

where


 * {| style="width:100%" border="0"

$$ \begin{align} y_{p1}'= -B\pi sin(\pi x)+C\pi cos(\pi x)-2\pi Dsin(2 \pi x)+2\pi Ecos(2 \pi x)-3\pi Fsin(3x)\\+3\pi Gcos(3 \pi x)-4\pi Hsin(4 \pi x)+4\pi Icos(4 \pi x) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{align} y_{p1}''= -\pi^2 Bcos(\pi x)-\pi^2 Csin(\pi x)-4\pi^2 Dcos(2\pi x)-4\pi^2 Esin(2\pi x)-9\pi^2 Fcos(3\pi x)\\-9\pi^2 Gsin(3\pi x)-16\pi^2 Hcos(4\pi x)-16\pi^2 Isin(4\pi x) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Imputing this into the modal particular equation or


 * {| style="width:100%" border="0"

$$ \begin{align} y_{p1}'' + \frac12 \, y_{p1}' + 4 \, y_{p1} = 2(1.78281828469046-0.684161390102918 cos(\pi x)\\ +0.084898666012778 cos(2 \pi x)-0.082788137760126 cos(3 \pi x) + 0.021625349461069 cos(4 \pi x)) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

with the choice for particular solution plugged in.


 * {| style="width:100%" border="0"

$$ \begin{align} -B\pi^2 cos(\pi x)-C\pi^2 sin(\pi x)-4\pi^2 Dcos(2\pi x)-4\pi^2 Esin(2\pi x)-9\pi^2 Fcos(3\pi x)\\ -9\pi^2 Gsin(3\pi x)-16\pi^2 Hcos(4\pi x)-16\pi^2 Isin(4\pi x) + \frac12 \, [-B\pi sin(\pi x)\\ +C\pi cos(\pi x)-2\pi Dsin(2\pi x)+2\pi Ecos(2\pi x)-3\pi Fsin(3\pi x)+3\pi Gcos(3\pi x)\\ -4\pi Hsin(4\pi x)+4\pi Icos(4\pi x)] + 4 \, [A+B\pi ]cos(\pi x)+Csin(\pi x)+Dcos(2\pi x)\\ +Esin(2\pi x)+Fcos(3\pi x)+Gsin(3\pi x)+Hcos(4\pi x)+Isin(4\pi x)]= 2(1.78281828469046\\ -0.684161390102918 cos(\pi x) +0.084898666012778 cos(2 \pi x)-0.082788137760126 cos(3 \pi x)\\ + 0.021625349461069 cos(4 \pi x))
 * style="width:95%" |
 * style="width:95%" |

\end{align} $$
 * }
 * }

and combining like terms, we can solve for the coefficients (A-I)


 * {| style="width:100%" border="0"

$$ Constant: 4A = 2(1.718281828459046) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ cos(\pi x): -\pi^2 B + \frac12 C\pi + 4 B= 2(-0.684161390102918) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ sin(\pi x): -\pi^2 C - \frac12 \pi B + 4 C= 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ cos(2 \pi x): -4\pi^2 D + \frac12 2\pi E + 4 D= 2(0.084898666012778) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ sin(2 \pi x): -4\pi^2 E - \frac12 2\pi D + 4 E= 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ cos(3 \pi x): -9\pi^2 F+ \frac12 3\pi G + 4 F= 2(-0.082788137760126) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ sin(3 \pi x): -9\pi^2 G - \frac12 3\pi F + 4 G= 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ cos(4 \pi x): -16\pi^2 H + \frac12 4\pi I + 4 H= 2(0.021625349461069) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ sin(4 \pi x): -16\pi^2 I - \frac12 4\pi H + 4 I= 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

we can put this into matrix form to quickly solve for the coefficients.


 * {| style="width:100%" border="0"

$$ C = A^{-1} d $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

where


 * {| style="width:100%" border="0"

$$ \mathbf A=\pi \begin{bmatrix} \frac{4}{\pi} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & (-\pi+\frac{4}{\pi}) & \frac12 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 &-\frac12 & (-\pi+\frac{4}{\pi}) & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & (-4\pi+\frac{4}{\pi}) & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & (-4\pi+\frac{4}{\pi}) & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & (-9\pi+\frac{4}{\pi}) & \frac32 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -\frac32 & (-9\pi+\frac{4}{\pi}) & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & (-16\pi+\frac{4}{\pi}) & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -2 & (-16\pi+\frac{4}{\pi}) \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

,


 * {| style="width:100%" border="0"

$$ \mathbf d= \begin{bmatrix} 2(1.718281828459046) \\ 2(-0.684161390102918) \\               0  \\               2(0.084898666012778)  \\               0  \\               2(-0.082788137760126)  \\               0  \\               2(0.021625349461069)  \\               0   \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf C= \begin{bmatrix} A \\ B \\ C \\ D \\ E \\ F \\ G \\ H \\ I  \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

therefore the coefficients A-I become
 * {| style="width:100%" border="0"

$$ \mathbf C= \begin{bmatrix} 0.8591 \\ -0.3789 \\               -0.2207  \\              -0.0062  \\               0.0007  \\               0.0022  \\               -0.0001  \\               -0.0003  \\               0.0000   \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

and thus the particular solution takes the form


 * {| style="width:100%" border="0"

$$ \begin{align} y_{p1}=0.891+0.2175cos(\pi x)-0.0582sin(\pi x)-0.0047cos(2 \pi x)+0.0004sin(2 \pi x)\\ +0.0019cos(3 \pi x)-0.0001sin(3 \pi x)-0.0003cos(4 \pi x) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

we can verify by plugging this into the L2-ODE-CC equation to see if the left hand of

the solution matches the right side. This was done in Desmos in the link below.

https://www.desmos.com/calculator/yet9u2kopm



Modal Equation 1 Solution
The full solution for the first modal equation is the sum of the homogeneous and particular solutions.

therefore


 * {| style="width:100%" border="0"

$$ y_1(t) = y_{h1}(t)+ y_{p1}(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

substituting solved values


 * {| style="width:100%" border="0"

$$ \begin{align} y_1(t) =e^{-\frac14 t}[C_{11}cos(\frac{3 \sqrt7}{4}t)+C_{12}sin(\frac{3 \sqrt7}{4}t)]\\+0.891+0.2175cos(\pi t)-0.0582sin(\pi t)-0.0047cos(2 \pi t)+0.0004sin(2 \pi t)\\+0.0019cos(3 \pi t)-0.0001sin(3 \pi t)-0.0003cos(4 \pi t) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

using the initial conditions for this problem we can solve for the undetermined coefficients.


 * {| style="width:100%" border="0"

$$ y_1(0) = 1, \ y'_1(0) = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_1(0) = 1 = C_{11}+0.891+0.2175-0.0047+0.0019-0.0003 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

therefore


 * {| style="width:100%" border="0"

$$ C_{11} = -0.105732 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

taking the derivative and setting to zero.


 * {| style="width:100%" border="0"

$$ \begin{align} y_1'(0) = 0=3/4 sqrt(7) C_{12}-(C_{11})/4-0.18127 \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

therefore


 * {| style="width:100%" border="0"

$$ \begin{align} C_{12}=-0.0146602 \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

thus the solution for the first modal equation takes the form.


 * {| style="width:100%" border="0"

$$ \begin{align} y_1(t) =e^{-\frac14 t}[-0.105732cos(\frac{3 \sqrt7}{4}t)-0.0146602sin(\frac{3 \sqrt7}{4}t)]\\+0.891+0.2175cos(\pi t)-0.0582sin(\pi t)-0.0047cos(2 \pi t)+0.0004sin(2 \pi t)\\+0.0019cos(3 \pi t)-0.0001sin(3 \pi t)-0.0003cos(4 \pi t) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We then plot to verify the initial conditions



Modal Equation 2
This modal equation is homogeneous and was solved for previously in report 5.


 * {| style="width:100%" border="0"

$$ y_2'' + 6 \, y_2' + 9 \, y_2 = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_2(0) = 0, \ y'_2(0) = 1 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Modal Equation 2 Solution
the modal solution is


 * {| style="width:100%" border="0"

$$ y_2(t) = t e^{-3 t} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Modal Displacements
We have obtained two modal displacement functions


 * {| style="width:100%" border="0"

$$ \begin{align} y_1(t) =e^{-\frac14 t}[-0.105732cos(\frac{3 \sqrt7}{4}t)-0.0146602sin(\frac{3 \sqrt7}{4}t)]\\+0.891+0.2175cos(\pi t)-0.0582sin(\pi t)-0.0047cos(2 \pi t)+0.0004sin(2 \pi t)\\+0.0019cos(3 \pi t)-0.0001sin(3 \pi t)-0.0003cos(4 \pi t) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_2(t) = t e^{-3 t} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

plotting them together



The long term response for the second modal displacement decays to zero. However the first modal displacement decays to a non-zero value.

Complete Solution
We can now project the modal solutions back into the original coordinate system.


 * {| style="width:100%" border="0"

$$ \mathbf d(t) = \bar{\boldsymbol \Phi} \, \mathbf y(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

where


 * {| style="width:100%" border="0"

$$
 * style="width:95%" |
 * style="width:95%" |

\mathbf d(t)=\begin{bmatrix} \frac{1}{\sqrt{5}} & \sqrt{\frac{3}{10}} \\ \frac{1}{\sqrt{5}} \cdot \sqrt{6} & \sqrt{\frac{3}{10}} \cdot \frac{-4}{\sqrt{6}} \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}

$$
 * }
 * }

the first displacement is found to be


 * {| style="width:100%" border="0"

$$ \begin{align} d_{1} = (3^{1/2}10^{1/2}te^{-3t})/10 - (5^{1/2}((47cos(2\pi t))/10000 - (87cos(\pi t))/400\\ - (19cos(3\pi t))/10000 + (3cos(4\pi t))/10000 + (291sin(\pi t))/5000 - sin(2\pi t)/2500\\ + sin(3\pi t)/10000 + e^{-t/4}((7618793532818197cos((37^{1/2}t)/4))/72057594037927936\\ + (8451029920918649sin((37^{1/2}t)/4))/576460752303423488) - 891/1000))/5 \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

the second displacement is found to be


 * {| style="width:100%" border="0"

$$ \begin{align} d_{2} =- (5^{1/2}6^{1/2}((47cos(2\pi t))/10000 - (87cos(\pi t))/400 - (19cos(3\pi t))/10000\\ + (3cos(4\pi t))/10000 + (291sin(\pi t))/5000 - sin(2\pi t)/2500 + sin(3\pi t)/10000\\ + e^{-t/4}((7618793532818197cos((37^{1/2}t)/4))/72057594037927936\\ +(8451029920918649sin((37^{1/2}t)/4))/576460752303423488) - 891/1000))/5\\ - (25^{1/2}te^{-3t})/5 \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

plotting these solutions



we see that as time goes to infinity both displacements approach a non-zero value

Problem Statement
Continuation of R5.6


 * 1) You used Audacity to record the audio signal and run FFT to detect the music notes in the 2nd chord of this beautiful piano classical music piece. Now provide several possible and plausible arrangements of the notes you detected for the 2nd chord of the piano classical music piece for playability on a piano; see my lecture notes sec.25e (rules of thumb) and sec.25g (see the example of note detection of a piano chord and the summary of steps, particularly the elimination of higher harmonics).
 * 2) Now, use your experience with music note detection using FFT to detect the notes of the opening chord of the Beatles’ A Hard Day’s Night in this YouTube recording 2. Use the FFT size of 65536 to generate a note detection spreadsheet to compare with the FFT spectrum in Brown 2004, "Mathematics, Physics and A Hard Day's Night," with my annotation v1.10; confirm or disconfirm whether this YouTube recording is identical to the recording used by Brown 2004.

Describe the rationale of how Brown 2004 arrived at the conclusion that there must be a piano being played alongside with the Beatles’ guitars in the opening chord of A Hard Day’s Night. Below are the points that you need to address in your description:
 * 1) How did Brown 2004 decide that the 3 versions (transcriptions) of the opening chord were wrong?
 * 2) Why did Brown 2004 assign the loudest note D3 to the bass guitar of Paul McCartney? How about the other D3 notes? How many D3 notes do you have in your FFT spectrum of the YouTube recording 2? Which instrument did Brown 2004 assign these extra D3 notes to? What was the rationale?
 * 3) How did Brown 2004 assign the notes to the 12-string guitar of George Harrison, e.g., the A notes? How about the F4 note? Does it exist in your FFT spectrum of the YouTube recording 2? How about the other notes for the 12-string guitar?
 * 4) How many F3 notes do you have in your FFT spectrum of the YouTube recording 2? Which instrument did Brown 2004 assign these F3 notes to? What was the rationale?
 * 5) Which notes did Brown 2004 assign to the piano played by George Martin? What was the rationale?
 * 6) Which note did Brown 2004 assign to the guitar of John Lennon? What was the rationale?
 * 7) How about notes that are within 20 dB from the peak but Brown 2004 did not assign to any instrument? What was the rationale?

Chord possiblities
As a reminder, here is the picture from R5.6 showing the detected notes from the piano chord:



Audacity detected 13 distinct notes within the 20 dB threshold: F2, C3, F3, C4, E♭4, A♭4, A4, C5, E♭5, G5, A5, B♭5, and E♭6. The loudest note was F2 and the pitch classes are F, C, E♭, A♭, A, G, and B♭. We need to look through the harmonics of each detected note to figure out which notes are in the chord being played. The first five harmonics of each of the 13 notes and their resulting notes are shown in the table below. The second, third, fourth, and fifth harmonic frequencies were found by multiplying the first frequency, that of the note itself, by 2, 3, 4, or 5, respectively. The spreadsheet from R5.6 was used with the resulting frequencies to get the note names.

We can now compare the first harmonic frequencies (the notes that Audacity detected) with the other harmonics and eliminate the overlaps. This will let us know which notes Audacity detected, but weren't actually played. We will start with the top note, F2, and move down the table, eliminating every note that appears as a harmonic of another note. The table below shows these modifications.

A strikethrough means that the note was either eliminated by a harmonic or was used to eliminate a note. Italicized text means the harmonic was never used. Either Audacity never detected the note, or it was detected outside the 20 dB threshold. Bolded text represents the notes that are left. We have decided, using this method, that these four notes, F2, C3, E♭4, and A♭4, are the four notes being played in the chord.

Rearranging the notes so they fit in one octave gives the order F A♭ C E♭. We know that F A♭ C is the f minor triad, and E♭ is the seventh note in the f minor scale, so the chord being played is a rearrangement of the fmin7 chord. The provided link confirms this chord. This chord is very easy to play. The left hand will handle F2 and C3 and the right hand will handle E♭4 and A♭4.

A Hard Day's Night FFT
Using the process described and the spreadsheet made in R5.6, the opening chord to the YouTube recording of A Hard Day's Night was analyzed with Audacity's FFT function. The below table contains every frequency with a 20 dB threshold of the loudest frequency. The loudest note detected was D3, and the loudest frequencies for each detected note have been bolded.

A total of 33 unique notes were detected by Audacity: F2, A2, D3, E♭3, F3, G3, A3, C4, D4, F4, G4, A4, C5, D5, F5, F#5, G5, A5, B5, C6, C#6, D6, E6, F6, F#6, G6, A♭6, A6, C#7, D7, E7, F7, and G7.

Brown reported detecting the following 23 notes: A2, D3, F3, G3, A3, C4, D4, G4, A4, C5, D5, G5, B5, C6, D6, E6, F#6, A♭6, A6, D7, E7, F7, G7. The bolded notes from Audacity are the notes that Brown reported. It would appear that the YouTube recording somehow had ten more notes than the recording that Brown used. What if we tried another recording? The table below shows the frequencies from the second recording, as before.

This time, Audacity detected 32 unique notes: F2, A2, D3, E♭3, F3, G3, A3, C4, D4, F4, G4, A4, C5, D5, F5, G5, A5, B5, C6, C#6, D6, E6, F6, F#6, G6, A♭6, A6, C#7, D7, E7, F7, and G7, and the loudest note was D3. The only difference between the two recordings was the first recording had F#5 while the second recording didn't. Again, the bolded notes were also reported by Brown.

Again, it seems that neither recording was the one used by Brown. However, this may not be the case. For his paper, Brown used a Mathematica subroutine to do the FFT of his recording, but we used Audacity both times. Brown also reported using a size of 29,375, when we used a size of 65,536. A larger size means more notes will be detected. It's entirely possible that Brown did use one of the recordings we used, but the "extra" frequencies only showed up due to the slightly different methods. We can't definitively say either way.

Brown 2004
Point 1

Each of the three transcriptions discussed in the article calls for a low G2 being played (indicated by a G3 on the score), but this note does not appear amongst the frequencies detected with an amplitude of 0.02 or larger. This suggests that this G2 was not played, which means these three versions are incorrect.

Point 2

The loudest D3 note was assigned by Brown 2004 to the bass guitar of Paul McCartney because even though George Harrison’s 12 string guitar is evident on the solo, and so assumed to be present in the opening chord, and contains D3 (along with E2, E3, A2, A3, D4, G3, G4, B3, B3, F4, F4), the amplitude of D3 shows that it was much louder than the other notes produced by the 12 string guitar. This extra loud D3 was accounted for by the bass guitar. Despite this finding, there were still three other D3’s left unaccounted for. Two of these can be attributed to the piano, where a single D3 note actually strikes two strings, and the last to the 6 string guitar. In the FFT spectrum of the YouTube recording, five D3 notes were detected. Brown 2004 assigned these extra D3 notes to the piano (with 2 D3’s), bass guitar (1 D3), and one each from the 12 string and 6 string guitar. These combination works because it does not create issues with the numbers of other notes, like the single D4 detected.

Point 3

Brown 2004 assigned the A notes to the 12 string guitar of George Harrison because a 12 string guitar has each string doubled and it’s A strings are A2 and A3, which are both present in the frequencies. The F4 note, which is present on the 12 string guitar and would be produced along with any F3 created on this instrument, must be attributed to the piano because it produces all three F3’s (due to striking 3 stings at once) without making any F4’s, which do not show up on the frequencies in the article. In the FFT spectrum of the YouTube recording 2, two F4’s are present along with three F3’s. Also there are three A2’s, two A3’s, five D3’s, two D4’s, three G3’s, and three G4’s.

Point 4

There are four F3 notes in the FFT spectrum of the YouTube recording 2. Brown assigned all three of these notes to a piano because to produce a single F3 note on a piano three stings are used. This explains why the three F3 notes are of slightly different frequencies, as piano strings are individually tuned and so, may be slightly off from one another.

Point 5

Brown 2004 assigns George Martin’s piano playing to the notes D3, F3, D5, G5, and E6 because each of these notes showed up in the frequency at least three times. Once other instruments were taken into account, each of these notes was shown to occur three times with this exception of D3, which showed up twice. This could all be explained by how a piano strikes one, two, or three strings for a single note, with the ten bottom pitches using single string, C3 to just after D3 using double strings, and the rest using triple strings.

Point 6

John Lennon’s guitar was assigned the note of C5 by Brown 2004 with the rationale that while most of the higher notes present can be attributed to harmonics, C5 has a particularly high amplitude, suggesting it was a note that was played (not just harmonics). John Lennon’s six string guitar had the ability to play such a high note and so could explain this high, loud note.

Point 7

Some of these notes that are within 20dB from the peak are not assigned an instrument by Brown 2004 because they had relatively low amplitudes, like A4 and B5. This suggests that these notes were not specifically played by any instrument but were rather produced by secondary means, such as harmonics. Other notes, like C6 and D6, do not display such low amplitudes, so cannot be explained by the same factors as A4 and B5. Instead, these notes are not assigned due to the fact that they are not detected in threes, which would suggest them being produced by piano, and that they would be particularly inconvenient and difficult to arrange on the 12-string, 6-string, or bass guitars also present.