Verify that convolution with $\exp(-|z|)$ gives a solution of $y''-y=f$

Question

I am currently trying to find a solution to the ODE $y''(x)-y(x)=f(x)$. Now, I calculated the solution, which was given to be: $$y(x)=-\frac{1}{2}\int_{-\infty}^\infty f(x-z)e^{-|z|}dz$$ I am trying to now show that the integral in fact solves this ODE. First I must find $y''(x)$. Moving this into the integral yields: $$y''(x) = -\frac{1}{2}\int_{-\infty}^\infty \frac{\partial^2 f(x-z)}{\partial x^2} e^{-|z|} dz$$ Now a hint says to divide this integral up into parts where $y\geq x$ and $y<x$, and integrate twice over each region. How does this help, and how am I supposed to proceed? I don't really see why this hint is there.

Answer 1

First, observe that $$ \frac{\partial^2 f(x-z)}{\partial x^2} = \frac{\partial^2 f(x-z)}{\partial z^2} $$ Then integrate by parts twice, throwing the derivative onto the exponential term. The latter is not differentiable at $z=0$, so the integral should be split: $$-2y''(x) = \int_{-\infty}^0 \frac{\partial^2 f(x-z)}{\partial z^2} e^{z} dz + \int_{-0}^\infty \frac{\partial^2 f(x-z)}{\partial z^2} e^{-z} dz \\ = \frac{\partial f(x-z)}{\partial z} e^{z} \bigg|_{-\infty}^0 - \int_{-\infty}^0 \frac{\partial f(x-z)}{\partial z} e^{z} dz + \frac{\partial f(x-z)}{\partial z} e^{-z} \bigg|_{0}^\infty + \int_{-0}^\infty \frac{\partial f(x-z)}{\partial z} e^{-z} dz $$ The boundary values at $\infty$ are zero, and at $0$ they cancel out. Integrate by parts again: $$ - f(x-z)e^z\bigg|_{-\infty}^0 + \int_{-\infty}^0 f(x-z) e^{z} dz + f(x-z)e^{-z}\bigg|_0^{\infty} + \int_{-0}^\infty f(x-z) e^{-z} dz $$ This simplifies to $$ -2f(x) + \int_{-0}^\infty f(x-z) e^{-|z|} dz = -2f(x) - 2y(x) $$ as expected.

Answer 2

You can also use directly the property of the convolution product under differentiation $(f*g)'=f'*g=f*g'$ to conclude that $y''=(f*g)''=f*g''$ and with $g(x)=-\frac12e^{-|x|}$ \begin{align} g'(x)&=(u(x)-\frac12)e^{-|x|}\\ g''(x)&=\delta(x)e^{-|x|}-\frac12(2u(x)-1)^2e^{-|x|}\\ &=\delta(x)+g(x) \end{align} where $u(x)$ is the unit jump at $x=0$, $u(x)=0$ for $x<0$, $u(x)=1$ for $x\ge 0$ so that $(\frac{d}{dx}|x|)^2=(2u(x)-1)^2=1$ (a.e.).

Now the claim follows as $\delta * f=f$.