Verifying a (convolution) solution for ODE I'm having trouble in verifying that $f\star \phi$, $\phi = \frac{1}{2} e^{-|x|}$ is a solution for $u -u''=f$. Please help me.
 A: Let's first review what a derivative does to a Fourier transform.  I write the FT of a function $u(x)$ as
$$\hat{u}(k) = \int_{-\infty}^{\infty} dx \: u(x) e^{i k x}$$
The inverse FT is then
$$u(x) =  \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: \hat{u}(k) e^{-i k x}$$
Now, let's consider the FT of the derivative of $u$, $u'(x)$:
$$\begin{align}\int_{-\infty}^{\infty} dx \: u'(x) e^{i k x} &= \int_{-\infty}^{\infty} d(u'(x))  e^{i k x}\\ &= \underbrace{[(e^{i k x}u(x)]_{-\infty}^{\infty}}_{\text{this is zero}} - i k \int_{-\infty}^{\infty} dx \: u(x) e^{i k x} \\ &= -i k \hat{u}(k) \end{align}$$
Each derivative is equivalent to multiplying the FT by $-i k$.  Now we apply this knowledge to the original equation:
$$-u''+u=f(x)$$
Take the FT of both sides.  Note that the 2nd derivative is equivalent to multiplying by $(-i k)^2$:
$$(1+k^2) \hat{u}(k) = \hat{f}(k) \implies \hat{u}(k) = \frac{\hat{f}(k)}{1+k^2}$$
Let $\bar{\phi}(k) = \frac{1}{1+k^2}$. One may show that $\phi(x) = \pi e^{-|x|}$.  The solution $u(x)$ may be written as
$$\begin{align}u(x) &= \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: \hat{u}(k) e^{-i k x}\\ &=\frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: \frac{\hat{f}(k)}{1+k^2} e^{-i k x}\\ &= \int_{-\infty}^{\infty} dx' \: f(x') \phi(x-x') \end{align}$$
by the convolution theorem.
