# ODE solution using convolution and (inverse) Fourier transform

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?

• $-u''(x)+au(x)=f(x)$ is not a PDE. There is no partial differential in it, only differential. This is an ODE. Jul 7 at 17:17
• Thanks. Question edited. Jul 7 at 17:19
• 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.
– lcv
Jul 9 at 9:54
• Good to know. Green's function is not mentioned anywhere in the material. Maybe the lecturer has spoken about it in class. Jul 9 at 11:33
• ... but there are quite a few books mentioned as sources, probably Green is in some or one of them. Jul 9 at 11:46

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.

• 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! Jul 7 at 17:37
• I edited the question typos, now it is even closer to your version. The example solution has $1/\sqrt{a}$ and your version has $1/2\sqrt{a}$. Jul 7 at 17:42
• 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). Jul 8 at 9:29
• 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. Jul 8 at 22:04
• 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. Jul 9 at 7:32

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.

• @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. Jul 7 at 16:59
• $h(x)$ 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...
– PC1
Jul 7 at 17:12
• @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. Jul 7 at 17:22
• 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 $u$.
– PC1
Jul 7 at 17:40
• 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. Jul 7 at 17:44