MATH 414 Lecture 24

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Homework Questions

#13

$u^{″}+2u^{\prime }+2u=3\cos 5t$

• Solve using method of undetermined coefficients
• pick periodic solution

by MoUC, we have $f=A{u}^{″}+B{u}^{\prime }+Cu$

Convert to a "discrete" version of the equation - get difference equation: (backward difference)

• $u^{\prime }(t)\approx \frac{u(t)-u(t-h)}{h}$
• $u^{″}(t)\approx \frac{u^{\prime }(t)-u^{\prime }(t-h)}{h}=\frac{u(t)-2u(t-h)+u(t-2h)}{h^2}\overrightarrow{t=t+h}\frac{u(t+h)-2u(t)+u(t-h)}{h^2}$

Let's assume that the solution $u$ describes a function $f$ is a $2\pi$-periodic function ($f(t+2\pi )=f(t)$)

$f(t)\approx A\left(\frac{u(t+h)-2u(t)+u(t-h)}{h^2}\right)+B\left(\frac{u(t)-u(t-h)}{h}\right)+Cu(t)$

Let $t_k=\frac{2\pi }{n}k$ and $u_k=u(t_k)$ for $k\in 0,1,\dots ,n$.

$f(t_k)=f_k\approx A\left(\frac{u_{k+1}-2u_k+u_{k-1}}{h^2}\right)+B\left(\frac{u_k-u_{k-1}}{h}\right)+C{u}_k=\alpha u_{k+1}+\beta u_k+\gamma u_{k-1}$

Since $f$ is $2\pi$-periodic, then $u_k$ is $n$-periodic, so ${{ℱ}}_n[u_{k+1}{]}_j={\omega }^j{{ℱ}}_n[u_k{]}_j={\omega }^j{\hat{u}}_j$, and ${{ℱ}}_n[u_{k-1}{]}_j={\omega }^{-j}{{ℱ}}_n[u_k{]}_j={\omega }^{-j}{\hat{u}}_j$.

Now we can solve the following for ${\hat{u}}_j$ and take the inverse discrete fourier transform:

${\hat{f}}_j=\frac{1}{h^2}(\alpha {\omega }^j+\beta +\gamma {\omega }^{-j}){\hat{u}}_j$

From Last Time

$\hat{f}(\lambda )=\frac{1}{\sqrt{2\pi }}\underset{a}{\overset{b}{\int }}f(t){\mathrm{e}}^{-i\lambda t}\mathrm{d}t$.

We change variables to go to $[0,2\pi ]$ instead of $[a,b]$: let $\theta =\frac{(t-a)\cdot 2\pi }{b-a}$, then

$\hat{f}(\lambda )=\frac{b-a}{2\pi }\cdot \frac{1}{\sqrt{2\pi }}\underset{0}{\overset{2\pi }{\int }}f(a+\frac{b-a}{2\pi }\theta ){\mathrm{e}}^{-i\lambda (a+\frac{b-a}{2\pi }\theta )}\mathrm{d}\theta$

Let $F(\theta )=f(a+\frac{b-a}{2\pi }\theta )$

$\hat{f}(\lambda )=\frac{b-a}{\sqrt{2\pi }}\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}F(\theta ){\mathrm{e}}^{-i\lambda a}{\mathrm{e}}^{-i\left(\frac{b-a}{2\pi }\right)\lambda \theta }\mathrm{d}\theta$

Let $\eta =\frac{b-a}{2\pi }\lambda$ so $\lambda =\frac{2\pi }{b-a}\eta$. Then

$\begin{array}{rl}\hat{f}\left(\frac{2\pi }{b-a}\eta \right)& =\frac{b-a}{\sqrt{2\pi }}\frac{{\mathrm{e}}^{-i\left(\frac{2\pi }{b-a}\eta \right)a}}{2\pi }\underset{0}{\overset{2\pi }{\int }}F(\theta ){\mathrm{e}}^{-i\eta \theta }\mathrm{d}\theta \\ & =\frac{1}{\sqrt{2\pi }}(b-a){\mathrm{e}}^{-i\frac{2\pi a}{b-a}\eta }\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}F(\theta ){\mathrm{e}}^{-i\eta \theta }\mathrm{d}\theta \end{array}$

Let $t_j=a+\frac{b-a}{n}j$ and ${\theta }_j=\frac{(t_j-a)2\pi }{b-a}=\frac{2\pi }{n}j$

We have $y_i=f(t_i)=F\left(\frac{2\pi }{n}j\right)$. We know $0\le j\le n-1$

Take ${\lambda }_k=\frac{2\pi }{b-a}k$

$\hat{f}\left(\frac{2\pi }{b-a}k\right)=(\text{stuff})\underset{c_k}{\underbrace{\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}F(\theta ){\mathrm{e}}^{-i\theta k}\mathrm{d}\theta }}$

Observe that since $c_k=\frac{1}{n}{\hat{y}}_k$, we get

$\hat{f}\left(\frac{2\pi }{b-a}k\right)=\frac{b-a}{n\sqrt{2\pi }}{\mathrm{e}}^{-\frac{2\pi ika}{b-a}}{\hat{y}}_k$

This is an approximation to $\hat{f}(\lambda )$ at ${\lambda }_k=\frac{2\pi }{b-a}k$