1
$\begingroup$

I'm studying PDEs from old course material which includes answers to most of the exercises. Given a ODE

$-u''(x)+au(x)=f(x)$

and a function $g(x)=e^{-|x|}$, we should find $u(x)$ by calculating the Fourier transform of $g(x)$ and then applying convolution. I computed $\hat{g}(x)=\int_{-\infty}^{\infty}e^{-|x|}e^{-ikx}\,dx=\frac{1}{1+k^2}$ and hit the wall. The solutions suggest using $h(x)=\frac{g(\sqrt{a}x)}{\sqrt{a}}=\frac{1}{\sqrt{a}}e^{-|\sqrt{a}x|}$ and its Fourier transform $\hat{h}(k)=\frac{1}{a+k^2}$. I took $\hat{h}(k)$ as a typo, because I calculated $\hat{h}(x)=\frac{1}{a+k^2}$ (similarly as $\hat{g}(x)$) but in the next step the solution says that by transforming the ODE, we get

$\hat{u}(k)(a+k^2)=\hat{f}(k)\implies u(x)=\mathcal{F}^{-1}(\hat{h}\hat{f})=h*f=\frac{1}{\sqrt{a}}\int_{-\infty}^{\infty}e^{-\sqrt{a}|s|}f(x-s)\,ds$.

In addition, the solution says that $\hat{h}(k)$ is obtained by the following Lemma:

$\mathcal{F}(f*g)=2\pi\mathcal{F}(f)\diamond\mathcal{F}(g)$,

where $\diamond$ represents Hadamard (or Schur) product. I don't understand why the variable changes from $x$ to $k$ or why would I use the lemma as the calculation seems to be similar to computing $\hat{g}(x)$. I'm also clueless about how am I supposed to get $\hat{u}(k)=\hat{f}(k)/(a+k^2)=\hat{f}(k)\hat{h}(k)$ from applying Fourier transform to the ODE. Where do I get the function $h$? I see the ODE as $u(x)=(f(x)+u''(x))/a$ and can't figure out how would the rhs equal $fh$.

There's also a note about general solution having two free parameters but in this case there's only one solution. Why is that?

$\endgroup$
5
  • 1
    $\begingroup$ $-u''(x)+au(x)=f(x)$ is not a PDE. There is no partial differential in it, only differential. This is an ODE. $\endgroup$
    – JJacquelin
    Jul 7 at 17:17
  • $\begingroup$ Thanks. Question edited. $\endgroup$
    – Jonne
    Jul 7 at 17:19
  • 1
    $\begingroup$ This is the method of the Green's function for solving the ODE. It should be cited somewhere on your notes. In particular $h(x)$ is the Green function. $\endgroup$
    – lcv
    Jul 9 at 9:54
  • $\begingroup$ Good to know. Green's function is not mentioned anywhere in the material. Maybe the lecturer has spoken about it in class. $\endgroup$
    – Jonne
    Jul 9 at 11:33
  • $\begingroup$ ... but there are quite a few books mentioned as sources, probably Green is in some or one of them. $\endgroup$
    – Jonne
    Jul 9 at 11:46

2 Answers 2

1
$\begingroup$

I was writing a solution, when the solution from PC1 came. Nonetheless, I think this might complement it.

Applying the Fourier transform to the equation works like this:

\begin{equation} \begin{aligned} \int_{-\infty}^\infty \left[-u''(x)+au(x)\right]e^{-ikx}dx&=\int_{-\infty}^{\infty}f(x)e^{-ikx}dx\\ -\int_{-\infty}^\infty u''(x)e^{-ikx}dx+a\hat u(k)&=\hat f(k)\\ k^2\hat u(k)+a\hat u(k)=\hat f(k) \end{aligned} \end{equation} where in the last line I have used the fact that \begin{equation} \int_{-\infty}^\infty u''(x)e^{-ikx}dx=-k^2\hat u(k) \end{equation} which can be obtained by applying integration by parts twice.

Then the equation $k^2\hat u(k)+a\hat u(k)=\hat f(k)$ can be easily solved as \begin{equation} \hat u(k)=\frac{\hat f(k)}{a+k^2}=\hat f(k)\hat h(k) \end{equation} where $\hat h(k)=(a+k^2)^{-1}$. Notice that, to my understanding, the inverse transform of $\hat h$ is $$h(x)=\mathcal{F}^{-1}\{\hat h(k)\}=\frac{1}{2\sqrt{a}}e^{-\sqrt{a}|x|}$$

Since a multiplication in Fourier space is a convolution in real space, \begin{equation} u(x)=\mathcal{F}^{-1}\{\hat u(k)\}=\mathcal{F}^{-1}\{(\hat f\hat h)(k)\}=(f\ast h) (x)=\frac{1}{2\sqrt{a}}\int_{-\infty}^{\infty} e^{-\sqrt{a}|s|}f(x-s)ds. \end{equation}

This is very close to what you say that you found in the solutions, but there might be a typo/mistake somewhere. I cannot proceed further without an explicit form for $f$.

The general solution is what PC1 explained, and $C_1$ and $C_2$ are the two parameters.

$\endgroup$
6
  • $\begingroup$ Great! I'll read more on Fourier transform and do the integration also myself. I'll return here to accept this or PCI's answer. Thank you very much! $\endgroup$
    – Jonne
    Jul 7 at 17:37
  • $\begingroup$ I edited the question typos, now it is even closer to your version. The example solution has $1/\sqrt{a}$ and your version has $1/2\sqrt{a}$. $\endgroup$
    – Jonne
    Jul 7 at 17:42
  • $\begingroup$ I don't seem to get the signs right. It looks like it should be $-1/2\sqrt{a}$ (and a $-1/2$ factor for the function $g$ as well). $\endgroup$
    – Jonne
    Jul 8 at 9:29
  • 1
    $\begingroup$ I do not understand where you get this minus sign from. Also, the factor 1/2 should appear when doing the inverse transform of $h$, so I think my answer is still ok, but I could be missing something else. Maybe posting the original exercise from the book could help to clarify things. $\endgroup$
    – pClabdn
    Jul 8 at 22:04
  • $\begingroup$ The exercise is as posted in the question: ODE is given, $g$ is given and the instructions are to calculate Fourier transform of $g$ and apply convolution. It's quite easy to mess up the signs while doing integration, I'm not saying that your answer is wrong. As you can read, I think I don't get the signs right. I'll post my integrals to the question a bit later so you can see/correct them. $\endgroup$
    – Jonne
    Jul 9 at 7:32
1
$\begingroup$

I assume that the question is in the last paragraph.

This is a second order ODE so there are two free parameters. The solution of the homogeneous equation $-u''(x)+au(x)=0$ is $u(x)=C_1e^{\sqrt ax}+C_2e^{-\sqrt ax}$, where $C_1$ and $C_2$ are constants. This linear combination can always be added to your solution.

To get the inhomogeneous part, we take the Fourier transform of both sides. We find: $$\hat u_0(k)=\frac{\hat f(k)}{k^2+a}$$ By defining $\hat h(k)=\frac1{k^2+a}$, we have that $\hat u(x)=\hat f(x)\hat h(x)$. So $u_0(x)$ is the convolution product of $f(x)$ and $h(x)$.

The final solution will be: $$u(x)=C_1e^{\sqrt ax}+C_2e^{-\sqrt ax}+u_0(x)$$ I don't explicitly find $u_0(x)$ here but I think that you can get it from your previous work in the question.

$\endgroup$
5
  • $\begingroup$ @PCI The more important questions are how to get the function $h$ and why does the varible change to $k$. How do you get the result from taking Fourier transform on both sides? It is not clear at all to me. Inside the integral is now an unknown function $u$ and its second derivative. I could calculate $\hat{g}$ because $g$ is defined. $\endgroup$
    – Jonne
    Jul 7 at 16:59
  • $\begingroup$ $h(x)$ is the inverse transform of $\hat h(k)=\frac1{k^2+a}$, depending on your convention. The variable "changes" from $x$ to $k$ as this is how you define a Fourier transform. $k$ here is in the dual space of $x$. This is fundamental, you should read again the definition of the Fourier transform if this is not apparent... $\endgroup$
    – PC1
    Jul 7 at 17:12
  • $\begingroup$ @PCI Ok, I'll read more. But how do you actually calculate the transform of $-u''(x)+au(x)$? As I already wrote, I'm lost because I don't know what $u$ is. $\endgroup$
    – Jonne
    Jul 7 at 17:22
  • 1
    $\begingroup$ The transform is a linear operator so $\mathcal F(au(x))=a\hat u(k)$. Also, by integrating by parts, you can show that $\mathcal F(u^{(n)}(x))=(ik)^n\hat u(k)$, which is why Fourier transforms are so useful to solve differential equations. Here $u^{(n)}$ is the $n$-th derivative of $u$. $\endgroup$
    – PC1
    Jul 7 at 17:40
  • $\begingroup$ Thanks. I just downloaded old course material for Fourier-analysis so I should get some understanding from there. The PDE course material has only some short Fourier-results as an appendix. $\endgroup$
    – Jonne
    Jul 7 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.