User:Egm6321.f12.team2/Report5

=Introduction=

Summary
This report explores various topics involving differential equations, matrices and geometry. Specific topics include matrix exponentiation and decomposition utilizing Eigenvalues, exactness of N2-ODEs, Taylor series expansions with respect to matrices and N2-ODEs, and proofs of hypergeometric functions.

Team Members
Zhe Wang Yitien Yan Ranjit Alexander Karl Van Wyk Anubhav Mohan

=R*5.1 - Show proof of Matrix Exponentiation=

Solution
The general expression for matrix exponentiation is as follows,

where $$ \mathbf A $$ is a generic matrix and $$ \mathbf I $$ is an identity matrix with the same dimensions as $$ \mathbf A $$. Using ($$) we can find the LHS, $$ \exp [ \mathbf A^T ] $$,

where $$ \mathbf I^T = \mathbf I $$, so the statement can be simplified to,

Now we can find the RHS, $$ \exp [ \mathbf A ]^T $$. Using the general expression in ($$), we already have $$\exp \mathbf A $$, so $$ \exp [ \mathbf A ]^T $$ can be written as,

Utilizing the following rule,

where, $$ \mathbf A $$ and $$ \mathbf B $$ are separate matrices,

we can simplify ($$) to,

And so, ($$) is true.

Show That
where,

Solution
Making use of the general expression for matrix exponentiation as stated in ($$), we can substitute in matrix $$ \mathbf D $$,

where $$ \mathbf D^2 $$ can be written as follows,

Since a diagonal matrix has no off-diagonal terms, the only terms that can be multiplied are within the diagonal.

Furthermore, the following is true for a diagonal matrix raised to the power $$k$$:

Therefore, for every element $$ d_{i}, \, i=1, \ldots, n $$ on the diagonal in the calculated matrix $$\exp[ \mathbf D ]$$ in ($$), the expansion is as follows (all off diagonal elements are all zero):

which, by Taylor expansion, can be approximated to:

In this case, the entire exponentiation of $$ \mathbf D $$ can then be written as: $$ \exp [ \mathbf D ] = \text{Diag} [e^{d_1},...,e^{d_n}] \in \mathbb C^{n \times n}$$

so, ($$) is true.

Given
Given a diagonalizable matrix $$A \in \mathbb R^{n \times n}$$.

Find
Show that,

Where $$\Phi=[\phi_1, \ldots, \phi_n]$$, is a matrix of eigenvectors of $$A$$, and $$\lambda_1, \ldots, \lambda_n$$ are the eigenvalues of $$A$$.

Solution
Since $$A$$ is diagonalizable, it can be decomposed as follows,

where $$\mathbf \Lambda$$ is a diagonal matrix of the eigenvalues of $$\mathbf A $$. Therefore,

where ($$) is derived by the exponentiation of a matrix. In order to further simplify this equation, the following property will be proven by induction:

Proof:

Assume, the case is true for the power of $$k$$,

Then, show true for the case for the power of $$k+1$$,

Therefore, ($$) must be true. Now, ($$) can be simplified as follows,

At this point, (insert equation in here from problem R5.2) can be used to bring ($$) into its simplest form:

Given
The following matrix,

Part 1
Show that,

Part 2
Then show that,

Part 1
Make the substitution,

Now the following steps are taken to decompose $$A$$ into its eigenvalues, $$\lambda_i$$, and eigenvectors, $$\phi_i$$.

Not taking the trivial solution where $$\phi=0$$ implies that $$|\mathbf A - \lambda \mathbf I | = 0$$

Therefore,

Now we find the eigenvector corresponding to each eignvalue by plugging in the eigenvalue into ($$):

For $$\lambda_1 = it$$,

From ($$), the following two equations can be established:

Solving ($$) for $$\phi_2$$ yields,

Substituting ($$) into ($$) yields,

This means that $$\phi_{11}$$ can take any value and so we choose $$\phi_{11}=1$$. Substituting this result into ($$) yields $$\phi_{12} = -i$$. Therefore, the eigenvector corresponding to $$\lambda_1=it$$ is $$\phi_{1}=\begin{bmatrix} 1 \\ -i \end{bmatrix}$$

The same process is continued for the second eigenvalue, $$\lambda = -it $$, and the corresponding eigenvector is found to be $$\phi_2 = \begin{bmatrix} 1 \\ i \end{bmatrix}$$

Combining the eigenvalues and eigenvectors into matrices for the eigenvalue problem of a matrix, the following relationship is established,

where $$\mathbf \Phi = [\phi_1, \ldots, \phi_n]$$ and $$\mathbf \Lambda = \text{Diag}[\lambda_1,\ldots,\lambda_n] $$

In this particular case,

Rearranging ($$) yields the decomposition for $$A$$,

Taking the inverse of $$ \mathbf \Phi$$ yields,

Substituting ($$), ($$), and ($$) into ($$) yields,

Part 2
Using the property in ($$) and ($$), the following relationship is established,

Multiplying ($$) out yields,

Using the following exponential forms,

($$) reduces to,

Obviously,

Given
The following N2-ODE,

Prove exactness
Prove that equation ($$) is exact.

Find L1-ODE-VC
Additionally, Find the first integral, $$\phi(x,y,p)$$ which is a L1-ODE-VC with $$p(x):= y'(x)$$.

Find y(x)
Solve ($$) for $$y(x)$$.

Proof of exactness
The second exactness condition for N2-ODEs:

where,

Start by finding every element of the 2nd exactness condition equations. Derivation of $$f_{xx}$$

Note here that

For $$g_{y}$$

For $$g_{xp}$$

Inject the terms in equations ($$)-($$) into ($$) and ($$) to test for exactness gives

Where both the statements are true. Hence the equation is exact.

Find L1-ODE-VC
Recalling the form of a second order ODE,

From equations ($$) and ($$), it is seen that,

We start by solving for $$\phi_p$$ where $$p=y'$$,

Next,taking the total derivative with respect to x,

Set ($$) and ($$) equal to each other yields,

In order to find h(x,y), the most obvious case is chosen first where,

From ($$),

Taking the partial derivative with respect to x of ($$) and setting it equal to ($$) yields,

Therefore, combining ($$),($$) and ($$) yields $$\phi$$,

Find y(x)
In order to solve for y(x), we determine whether or not equation ($$) is exact. First exactness condition,

Thus, equation ($$) satisfies the first exactness condition.

Thus, the second exactness condition is not satisfied and equation ($$) is not exact. First, we rewrite equation ($$) as,

We find an integrating factor, $$h(x,y)$$ that will make the equation ($$) exact.

Multiplying equation ($$) through by the integrating factor, and using the fact that $$h'=hx^2sec\;x$$, the following is obtained,

Given
$$ \begin{align} \\& \displaystyle g_{0}=\frac{\partial}{\partial y}\left(\frac{d\phi}{dx}\right)=\phi_{xy} \\& \frac{d}{dx}g_{1}=\frac{d}{dx}\left(\phi_{xy'}+\phi_{yy'}y'+\phi_{y}+\phi_{y'y'}y''\right) \\& \frac{d}{dx}g_{2}=\frac{d}{dx}\left(\phi_{y'x}+\phi_{y'y}y'+\phi_{y'y'}y''\right) \end{align} $$

Find
Show the equivalence of 2nd exactness condition.

Solution
From the lecture notes (1)-(2) p.21-8, the second exactness condition is:

After Substituting the equations given in this problem into ($$):

Collecting each term according to its order of derivatives of $$x$$ and $$y$$:

Since the coefficients $$ C_{0}, C_{1}, C_{2}, C_{3} $$ are linearly independent, we can conclude that $$ C_{0}=C_{1}=C_{2}=C_{3}=0 $$. This yields the symmetry of the mixed 2nd partial derivatives of the first integral $$ \phi $$: {| style="width:100%" border="0" $$ \begin{align} \\& \phi_{xy}=\frac{d}{dx}\left(\phi_{y}\right)=\phi_{yx} \\& \phi_{xy'}=\phi_{y'x} \\& \phi_{yy'}=\phi_{y'y} \end{align} $$
 * style="width:95%" |
 * style="width:95%" |

Given
When $$\displaystyle n=2$$, there exists:

Find
Use ($$) to show the symmetry equivalence:

Solution
Since:

So ($$)becomes:

We get:

The LHS of($$) is:

And the RHS of($$) is:

From equations($$) ($$) ($$) ($$) we can tell that:

Reference
[http://upload.wikimedia.org/wikiversity/en/9/94/Pea1.f12.sec22.djvu 1. Dr. Loc Vu-Quoc, Mtg 28, Oct. 18, 2012]

=R*5.8 Equivalence of Two Forms of the Second Exactness Conditions for N2-ODEs, by Method 2=

Given
The second exactness condition for N2-ODEs is given by

where

Find
Show that the form above is equivalent to the form of the second exactness condition for N2-ODEs given by

Solution
Let

Then,

yields

As noted in the lecture notes, since 1 and q are linearly independent, the only way for the above equation to be true is for their coefficients to be equal to zero.

Thus, ($$) have been shown to be true.

Find
Use the Taylor series at $$ x = 0 $$(MacLaurin series) to derive ($$) and ($$).

Solution
Deine $$ f_{1}\left(x\right)=\left(1-x\right)^{-a} $$, and take the derivative of $$ f_{1}\left(x\right) $$ with respect to $$ x $$ in order to derive the Maclaurin Series of $$ f\left(x\right)=f_{1}\left(x\right) $$

Substitute the results into ($$), ($$) can be derived as followed:

Unlike the derivation of ($$), deriving ($$) in the same fashion would be significantly longer and more difficult. Deriving Maclaurin Series of $$ \arctan \left( 1+x \right) $$ first and take advantage of dividing Maclaurin Series of $$ \arctan \left( 1+x \right) $$ by $$ x $$ would be a better way to derive ($$). Define $$ g\left(x\right)=\arctan \left( x \right) $$ for conveniency, and do the same processes as that in deriving ($$):

Substitute each term into ($$): By dividing ($$) by $$ x $$, ($$) can then be obtained:

Find
Show that in the region near x = 0, ($$) is true.

Solution
As we can see, the contours of $$ F(5,-10;1;x) $$ and $$ \left(1-x\right)^{6}\left(1001x^{4}-1144x^{3}+396x^{2}-44x+1\right) $$ are almost the same. The deviations between two functions are extremely small near $$ x = 0 $$ region.

Given
The governing equation for a particle moving with air resistance with initial velocity,$$z(0)=0$$ is

This relation comes from equating the unbalanced forces on the particle.ie$$mz'+kz^n=mg$$, where $$m$$ is mass, $$z$$ is the verticle velocity,$$v_y$$,$$k$$, the proportionality constant for air resistance and $$g$$ is the acceleration due to gravity.

Also the solution to the integral is given as

Where

In this particular problem,

Find
For each value of $$t$$, solve for ,$$z(t)$$ and then plot $$z(t)$$ versus $$t$$. Find the time when the projectile hits the ground.

Solution
From the values of $$a=2, b=10, n=3$$ the solution for $$z(t)$$ is

Then,

This approximates to,

The equation ($$), is an implicit equation in z(t) and the velocity of the particle when it hits the ground is got by solving for $$z(t)$$ from the equation by inputing the time at which it hits the ground. For some values of t, the velocity,z(t) is calculated numerically(from wolfram alpha). The results are given below. As can be seen from the values, the rate of increase in velocity decreases as time increases. This is illustrated in the figure below by plotting $$z(t)$$ versus time $$t$$.



Given
The hypergeometric differential equation is given as

Also given is the Hypergeometric function,$$ \,_2F_1(a,b;c;x)$$ as the solution to the hypergeomtric differential equation, where

Find
1. Is ($$) exact? 2. Is it in the power form? 3. Verify whether the hypergeometric function $$\;_2F_1(a,b;c;x)$$ is a solution to ($$).

Proof for exactness
The second exactness condition for N2-ODEs:

where,

Start by finding every element of the 2nd exactness condition equations. Derivation of $$f_{xx}$$

Note here that

For $$g_{y}$$

For $$g_{xp}$$

Inject the terms in equations ($$)-($$) into ($$) and ($$) to test for exactness gives

Since the second exactness condition is not satisfied, the hypergeometric differential equation is not exact.

Converting into the power form
For the equation to be converted into the power form it has be in form.

The hypergemetric differential equation cannot be expressed in this form and hence cannot be converted into the exact form using the Integrating Factor Method(IFM).

Checking if the 2F1(a,b,c,x) is a solution to the Hypergeometric differential equation
Given Now, taking the first differential of $$}} gives

Taking the second differential yields us $$y''$$.

Now substitute this in the hypergeometric differential equation gives,

This when expanded gives the following equation,

In order to show that ($$) does indeed sum to 0, we have to look at the coefficients of each power of x and show that these coefficients are 0. Then, we need to show that any remaining terms equal 0 as well.

Coeffecients for $$x^0$$
For coeffecients of $$x^0$$ gives

Coeffecients for $$x$$
For coeffecients of $$x$$ gives Which reduces to zero. Hence this is true with co-effecient of x.

Coefficients for $$x^i$$
The coefficients for $$x^i, i=0, \ldots, k-1$$ sum to zero as well by induction.

Coefficients for $$x^k$$
The remaining terms are those for coefficients of $$x^k$$ which yields This implies that the left side of ($$) is indeed equal to 0. Thus, the hypergeometric function is a solution to the differential equation.

Given
Given the governing L2-ODEs-VC for some classical special functions.

Legendre

Bessel

Hermite

Find
1. Verify the exactness of 5.13.1 - 5.13.3. For this use two methods.Method 1: Use the 2nd exactness condition given by

For Method 2, use

where

2. If the Hermite function given in ($$) is not exact, check whether it can be made exact by using the power form and whether it can be made exact using the IFM, $$h(x,y)=x^my^n$$ 3. The first few Hermite polynomials are given as

Verify that ($$)-($$) are solutions to the homogeneous Hermite differential equation, given by ($$)

Legendre Differential equation
The legendre differential equation is given as The second exactness condition for N2-ODEs:

where,

Start by finding every element of the 2nd exactness condition equations. Derivation of $$f_{xx}$$

Note here that

For $$g_{y}$$

For $$g_{xp}$$

Inject the terms in equations ($$)-($$) into ($$) and ($$) to test for exactness gives

Since only the second exactness condition is satisfied, the Legendre differential equation is not exact.

Bessel Differential equation
The Bessel differential equation is given as The second exactness condition for N2-ODEs:

where,

Start by finding every element of the 2nd exactness condition equations. Derivation of $$f_{xx}$$

Note here that

For $$g_{y}$$

For $$g_{xp}$$

Inject the terms in equations ($$)-($$) into ($$) and ($$) to test for exactness gives

Since only the second exactness condition is satisfied, the Bessel differential equation is not exact.

Hermite Differential equation
The Hermite differential equation is given as The second exactness condition for N2-ODEs:

where,

Start by finding every element of the 2nd exactness condition equations. Note here that

For $$g_{y}$$

For $$g_{xp}$$

Inject the terms in equations ($$)-($$) into ($$) and ($$) to test for exactness gives

Since both second exactness conditions are not satisfied, the Hermite differential equation is not exact.

Legendre Differential equation
The legendre differential equation is given as The second exactness condition for N2-ODEs: Here

Hence,

Inject the terms in equations ($$)-($$) into ($$) to test for exactness gives

Since the second exactness condition is not satisfied, the Legendre differential equation is not exact.

Bessel Differential equation
The Bessel differential equation is given as The second exactness condition for N2-ODEs: Here

Hence,

Inject the terms in equations ($$)-($$) into ($$) to test for exactness gives

Since the second exactness condition is not satisfied, the Legendre differential equation is not exact.

Hermite Differential equation
The Hermite differential equation is given as The second exactness condition for N2-ODEs: Here

Hence,

Inject the terms in equations ($$)-($$) into ($$) to test for exactness gives

Which is an invalid statement. Hence the differential equation is not exact

Converting the Hermite differential equation into the power form
For the equation to be converted into the power form it has be in form.

The hermite differential equation is given as This particular equation is in the power form with The hermite differential equation is given as Hence this can be made exact by using the Integrating factor $$x^my^n$$. Multiplying the differential equation with the integrating factor gives, Here we assign the coefficient of 2y in the differential equation as $$n'$$ to differentiate it from the $$n$$ used in the integrating factor. To find $$m,n$$ ,we use second exactness condition for N2-ODEs:

where,

Start by finding every element of the 2nd exactness condition equations.

Inject the terms in equations ($$) into ($$) to test for exactness gives Since the 0nly way equation ($$), can be valid is if $$n=0$$. Now, substituting in the equation ($$) to test for exactness gives, Substituting the value $$n=0$$ in the above equation gives the relation Since in the LHS of equation,$$x$$ is raised to $$m-2$$ and in the RHS of the equation, x is raised to ,$$m$$, the only way LHS=RHS is if the coefficients on both sides are zero.Hence, from ($$) we get Hence, m is either 0 or 1. Since both m and n cannot be zero here to make the equation exact, m=1. Substituting,m=1 in the coeffecient on the RHS gives, Hence for the General Hermite Differential equation to be exact, $$m=1,n=0,n'=-2$$. Rewriting the equation in its exact form gives,

Hermite Ploynomials as Homogeneous solution to the Hermite differential equation
The Hermite differential equation is given as The first three Hermite polynomials are given as

Checking to see whether they are indeed the solution to the Hermite differential equation. Substituing $$H_0(x)$$ as $$y(x)$$ with ,$$n=0$$ in ($$) gives Hence $$H_0(x)$$ is indeed a homogenous solution to ($$).

Now,Checking to see for $$H_1(x)$$ as $$y(x)$$ with ,$$n=1$$ in ($$) gives Hence $$H_2(x)$$ is also a homogenous solution to ($$).

Now,Checking to see for $$H_2(x)$$ as $$y(x)$$ with ,$$n=2$$ in eqn34 gives Hence $$H_1(x)$$ is also a homogenous solution to eqn 34.

Hence the Hermite polynomials $$H_0(x),H_1(x)$$ and $$H_2(x)$$ are homogenous solutions to the Hermite differential equation.

Given
The solution, as the solution to the following L4-ODE-CC, and the following properties,

Find
The expression for $$X(x)$$ in terms of $$\cos K x,\sin K x, \cosh K x, \sinh K x$$.

Solution
Since ($$) is fourth order, the root of the characteristic equation must have four solutions:

Therefore, ($$) and ($$) yield,

Substituting ($$)-($$) into ($$) where appropriate yields the following,

Thus, $$X(x)$$ has been written in terms of $$\cos K x,\sin K x, \cosh K x, \sinh K x$$