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?
 A: 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.
A: 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.
