2.1.1.2 Solution gf1d-b

Question:

Show that

$$\tilde u(x) = \int_{\xi =-\infty}^\infty{\textstyle\frac{1}{2}}\vert x-\xi\vert f(\xi){ \rm d}\xi$$

is a solution to

$$u_{xx} = f(x) \quad -\infty < x < \infty$$

You can assume that function $f(\xi)$ becomes zero rapidly at large $\xi$. (If you want, you can assume it is zero beyond some value $\xi_{\rm {max}}$ of $\vert\xi\vert$.)

Answer:

First of all, define a second anti-derivative of $f$ to be $u_0$. That allows you from now on to write $f$ as $u_0''$. Note also that $u_0$ is one possible solution to the Poisson equation.

Now restrict the region of integration of $\tilde{u}$ to $\vert\xi\vert\le{R}$ where $R$ is some large number. (You can take the limit $R\to\infty$ at the end of the story.)

Split the integral into two parts $\xi <x$ and $\xi >x$ because the absolute value in the integral is different in these two cases. You get

$$\tilde u = \int_{\xi =-R}^x {\textstyle\frac{1}{2}}(x-\xi) u... ...i =x}^R {\textstyle\frac{1}{2}}(\xi -x) u_0''(\xi){ \rm d}\xi$$

Integrate each term by parts and clean up. Use the fact that if $R>\xi_{\rm {max}}$,

$$u_0(R) = u_0'(R) R + \mbox{some constant} \qquad u_0(-R) = u_0'(-R) (-R) + \mbox{some constant}$$

because $f=u_0''$ is zero beyond $\xi_{\rm {max}}$. That also makes $u_0'$ some constant at $R$, and some constant at $-R$.

You then find that $\tilde{u}$ is indeed a solution to the Poisson equation.