MATH 414 Lecture 24

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Homework Questions

#13

$u''+2u'+2u=3\cos {5t}$

• Solve using method of undetermined coefficients
• pick periodic solution

by MoUC, we have $f=A\,u''+B\,u'+C\,u$

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

• $u'(t)\approx {\frac {u(t)-u(t-h)}{h}}$
• $u''(t)\approx {\frac {u'(t)-u'(t-h)}{h}}={\frac {u(t)-2u(t-h)+u(t-2h)}{h^{2}}}\quad {\overset {\rightarrow }{t=t+h}}\quad {\frac {u(t+h)-2u(t)+u(t-h)}{h^{2}}}$

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

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

Let $t_{k}={\frac {2\pi }{n}}\,k$ and $u_{k}=u(t_{k})$ for $k\in 0,1,\ldots ,n$.

$f(t_{k})=f_{k}\approx A\,\left({\frac {u_{k+1}-2u_{k}+u_{k-1}}{h^{2}}}\right)+B\,\left({\frac {u_{k}-u_{k-1}}{h}}\right)+C\,u_{k}=\alpha u_{k+1}+\beta u_{k}+\gamma u_{k-1}$

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

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

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

From Last Time

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

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

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

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

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

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

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

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 $0\leq j\leq n-1$

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

${\hat {f}}\left({\frac {2\pi }{b-a}}\,k\right)=({\mbox{stuff}})\,\underbrace {{\frac {1}{2\pi }}\,\int _{0}^{2\pi }F(\theta )\,\mathrm {e} ^{-i\,\theta \,k}\,\mathrm {d} \theta } _{c_{k}}$

Observe that since $c_{k}={\frac {1}{n}}\,{\hat {y}}_{k}$, we get

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

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