User:Stellaliu718

=Introduction=

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

=R5.3 - Show a property of exponentiation of a matrix with regards to its decomposition=

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 $$\lambda = it$$,

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)$$.

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

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

where,


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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
{| style="width:100%" border="0" $$ \begin{align} \\& \displaystyle g_{0}=\frac{\partial}{\partial y}\left(\frac{d\phi}{dx}\right)=\phi_{xy} \\& \frac{d}{dx}g_{1}=\frac{d}{dx}\left(\phi_{xy'}+\phi_{yy'}y'+\phi_{y}+\phi_{y'y'}y''\right) \\& \frac{d}{dx}g_{2}=\frac{d}{dx}\left(\phi_{y'x}+\phi_{y'y}y'+\phi_{y'y'}y''\right) \end{align} $$
 * style="width:95%" |
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Find
Show the equivalence of 2nd exactness condition.

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. 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" $$ \begin{align} \\& \displaystyle g_{0}=\frac{\partial}{\partial y}\left(\frac{d\phi}{dx}\right)=\phi_{xy} \\& \frac{d}{dx}g_{1}=\frac{d}{dx}\left(\phi_{xy'}+\phi_{yy'}y'+\phi_{y}+\phi_{y'y'}y''\right) \\& \frac{d}{dx}g_{2}=\frac{d}{dx}\left(\phi_{y'x}+\phi_{y'y}y'+\phi_{y'y'}y''\right) \end{align} $$
 * style="width:95%" |
 * style="width:95%" |

}