User:Egm6321.f11.team1.steele.m/Report3

R*3.7 Question from meeting 15, page 2: Use IFM to show that the solution of (2) below is (3) underneath. $$\dot x(t)= ax(t) + bu(t)$$ $$x(t)=[\exp\{a(t-t_0)\}]x(t_0)+\int^t_{t_0}[\exp\{a(t-\tau)\}]\,b \,u(\tau)\, d \tau$$ From lecture notes [[media:pea1.f11.mtg15.djvu|Mtg 15]]

Solution
The solution requires an understanding of how the Euler Integrating Factor Method (IFM) is based on the 1st and 2nd exactness conditions. Please note that the equations below for the Euler IFM were derived in a previous homework assignment and noted again for clarity. The derivation builds to the solution. First Exactness Condition of N1-ODEs If an N1-ODE is exact then it has a particular form of M(x,y) + N(x,y)y' = 0 where $$ y' = \frac{dy}{dx}$$ The appended derivations to this point are from meeting 8 Second Exactness Condition of N1-ODEs Now, we must derive the 2nd exactness condition. A general ND-ODE is exact if for G the following relations exist: $$\frac{d\phi}{dx} = G$$ $$\phi(x,y) = k$$ $$\frac{d\phi(x,y(x))}{dx} = G(y',y,x) = 0$$ Let $$\phi(x,y)$$ be a smooth function, then for the 2nd exactness condition we have $$M_y(x,y) = \frac {\partial^2\phi(x,y)}{\partial x\partial y} = N_x(x,y) = \frac{\partial^2\phi(x,y)}{\partial y\partial x}$$ The appended derivations to this point are from meeting 9 If an N1-ODE satisfies the 1st exactness condition but not the 2nd, then the N1-ODE can be transformed by using the Euler Integrating Factor Method (IFM). Assume the conditions below. $$M(x,y) + N(x,y)y' = 0 \And M_y(x,y) \ne N_x(x,y)$$ The transformation of the given N1-ODE to a form that fulfills both exactness conditions begins by finding an integrating factor h(x,y). For condition 1, the substitution of h gives $$(hM) + (hN)y' = \bar{M} + \bar{N}y' = 0$$ The key is to derive a solution using $$\bar{M} = hM \And \bar{N} = hN$$ The derivations are as follows: $$\bar{M}_y = h_yM + hM_y = \bar{N}_x = h_xN + hN_x$$ So the Euler IFM gives the following solution using h(x,y) as the transformation. $$h_xN - h_yM + h(N_x - M_y) = 0 $$ The cumulative derivations to this point are from meetings 10-11 But in our case of this problem, the following equation is the direct basis for the derivation. The inhomogeneous form of this equation has a solution based on meeting 11. So the IFM takes a different derivation path. The updated solution process begins here for report 3. $$\underbrace{-a \,x}_{\displaystyle\color{blue}{M(t,x)}} + \underbrace{\color{blue}{1}}_{\displaystyle\color{blue}{N(t,x)}} \cdot dx/dt = bu(t) $$ The 1st order linear differential equation is a function of t with two variables x(t) and u(t). So let us rewrite the equation as $$-ax + x' = bu = M + Nx'$$ Note that M = -ax and N = 1, so that $$\frac{h_t}{h} = \frac{1}{\N}(N_t - M_x) = (0 - a) = n(s)$$ $$\int_{t_0}^t\frac{h_t}{h} = \int_{t_0}^t(-a)$$ The equation reduces to $$\ln h = -a(t-t_0)$$ Which can be exponentiated to give the integrating factor $$ h = \exp ({\ln h}) = \exp -a(t-t_0)$$ Once the integrating factor is found, the following identity is used to integrate for hx. $$\frac{h_t}{h} = -a$$ So rearrangement gives $$h_t = ha = \dot {h}$$ The substitution into the 1st exactness condition for the inhomogenous condition gives $$h\dot {x} + \dot {h} x = \dot{hx} = hbu$$ Integrate both sides and get $$\int_{t_0}^t{\dot {hx}} = \int_{t_0}^t{b*u}*dt$$ The use of the dummy variable tau and rearrangement with respect to the integration constant gives $$x(t)=[\exp\{a(t-t_0)\}]x(t_0)+\int^t_{t_0}[\exp\{a(t-\tau)\}]\,b \,u(\tau)\, d \tau$$ The first term on the left is the homogeneous solution, and the 2nd term on the right is the particular solution.

References for R3.7 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 9) [[media:pea1.f11.mtg9.djvu|Mtg 9]] 8 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 10) [[media:pea1.f11.mtg10.djvu|Mtg 10]] 8 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) [[media:pea1.f11.mtg11.djvu|Mtg 11]] 13 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 15) [[media:pea1.f11.mtg15.djvu|Mtg 15]] 22 Sep 2011

R*3.8 Question from meeting 15, page 4: Generalize (2) p. 15-2 to the case of linear time-variant system (1) p. 14-4. See (1) p. 15-2. Verify that your expression is indeed the solution for (1) p.14-4. $$\mathbf{x}(t) = \left[\exp \{\mathbf{A}(t-t_0) \} \right] \, \mathbf{x}(t_0) + \int^t_{t_0} \left[\exp \{ \mathbf{A} (t-\tau) \} \right] \mathbf{B} \, \mathbf{u}(\tau) \, d\tau$$ $$\mathbf{\dot[x]}(t) = \mathbf{A}(t) \,\mathbf{x}(t) + \mathbf{B}(t)\, \mathbf{u}(t)$$ From lecture notes [[media:pea1.f11.mtg15.djvu|Mtg 15]]

Solution
The equation (2) from p. 15-2 was for the case of n = 1. But the solution can be generalized for all n by simply applying the same derivation to the equation in matrix form. $$\mathbf{\dot[x]}(t) = \mathbf{A}(t) \,\mathbf{x}(t) + \mathbf{B}(t)\, \mathbf{u}(t)$$

The derivation path is similar since the solution of a scalar linear 1st order differential equation is the same for a system of such equations.

The 1st order linear differential equation is a function of t with two matrices x(t) and u(t) and two varying coefficient matrices a(t) and b(t). So let us rewrite the equation as $$-\mathbf{A}\mathbf{X} + \mathbf{X}' = \mathbf{B}\mathbf{U} = \mathbf{M} + \mathbf{N}\mathbf{X}'$$ $$\frac{\mathbf{h_t}}{\mathbf{h}} = \frac{1}{\mathbf{N}}(\mathbf{N_t} - \mathbf{M_x}) = (0 - \mathbf{A}) = \mathbf{n(s)}$$ So rearrangement gives $$ x(t) = \left[\exp\int^t_(t_0)\mathbf{A}(\mathbf{\tau}d\tau\right]\mathbf{x}(t_0) + \int^t_{t_0}\left[exp \int^t_{\tau}\mathbf{A}(s)ds \right]\mathbf{b}(\mathbf{\tau})\mathbf{u(\tau)}\, d\tau $$

References for R3.8 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) [[media:pea1.f11.mtg11.djvu|Mtg 11]] 13 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 14) [[media:pea1.f11.mtg14.djvu|Mtg 14]] 20 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 15) [[media:pea1.f11.mtg15.djvu|Mtg 15]] 22 Sep 2011.

R*3.9 Question from meeting 15, page 5: Verify that (1) satisfies (2)-(3). The equations (1) to (3) are shown from top to bottom below. $$\mathbf{\Phi}(t,t_0) = \exp \{\mathbf{A}(t-t_0) \}$$ $$\displaystyle\frac{d}{dt} \mathbf{\Phi}(t,t_0) = \mathbf{A} \, \mathbf{\Phi}(t,t_0)$$ $$\mathbf \Phi (t_0,t_0) = \mathbf I$$ From lecture notes [[media:pea1.f11.mtg15.djvu|Mtg 15]]

Solution
The derivation of a matrix is a direct derivation of a linear system. So it follows the general rules of derivation and matrix algebra. The derivative of an exponential function as a scalar is the same as that for a linear system in matrix algebraic notation. Therefore, the simple rules of derivation apply to prove that (1) satisfies (2) $$\frac {d}{dt} {\mathbf{\Phi}(t,t_0)} = \frac {d}{dt} \exp \{\mathbf{A}(t-t_0) \}$$ $$\frac {d}{dt} {\mathbf{\Phi}(t,t_0)} = \frac {d}{dt} [\mathbf{A}(t-t_0)] * \exp \{\mathbf{A} (t-t_0) \}$$ This simply reduces to show that (1) satisfies (2). $$\displaystyle\frac{d}{dt} \mathbf{\Phi}(t,t_0) = \mathbf{A} \, \mathbf{\Phi}(t,t_0)$$ In addition, the identity matrix is derived from general matrix algebra. $${\mathbf{\Phi}(t_0,t_0)} = \exp \{\mathbf{A}(t_0-t_0) \} = \exp \{\mathbf{A}(0) \} $$ $${\mathbf{\Phi}(t_0,t_0)} = \exp {\begin{bmatrix} 1     & \cdots & 0      \\ \vdots & \ddots & \vdots \\ 0     & \cdots & 1 \end{bmatrix}} $$ Therefore, the substitutions give the identity matrix, which is a matrix with values of 1 along the diagonal and 0 elsewhere.

References for R3.9 King, A.C., J. Billingham, and S.R. Otto. Differential Equations: Linear, Nonlinear, Ordinary, Partial. New York: Cambridge University Press, 2003. Oulia, Masoud and David E. Stevens. Fundamentals of Engineering Exam, 2nd Edition. Boston: Barron's Educational Services Inc., 2008. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 15) [[media:pea1.f11.mtg15.djvu|Mtg 15]] 22 Sep 2011.