19.1.1.2 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 $\vert\xi\vert$.) Find out what function $\tilde{u}$ is relative to some given second anti-derivative $u_0$ of $f$.

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. The most general solution is this particular solution plus the general solution of the homogeneous equation. Use your knowledge of ordinary differential equations to show that the most general solution is

$$u_0 + A + B x$$

It must therefore be shown that the Green’s fuction solution $\tilde{u}$ is of this form.

Now restrict the region of integration of $\tilde{u}$ to $-R$ $\raisebox{-.3pt}{$\leqslant$}$ $\xi$ $\raisebox{-.3pt}{$\leqslant$}$ ${R}$ where $R$ is some large number. Take $R$ $\raisebox{.3pt}{$>$}$ $\xi_{\rm {max}}$ so that the integral is zero beyond $R$. Also, take $R$ large enough that the particular $x$ at which you want to find $u$ is in the range $-R$ $\raisebox{.3pt}{$<$}$ $x$ $\raisebox{.3pt}{$<$}$ $R$. Show that that since $f$ is zero for $\xi$ $\raisebox{.3pt}{$>$}$ $\xi_{\rm {max}}$,

$$u_0(\xi) = C_1 + C_2 \xi\qquad\mbox{for}\qquad\xi >\xi_{\rm {max}}$$

Also, since $f$ is zero for $\xi$ $\raisebox{.3pt}{$<$}$ $-\xi_{\rm {max}}$,

$$u_0(\xi) = D_1 + D_2 \xi\qquad\mbox{for}\qquad\xi <-\xi_{\rm {max}}$$

According to the above, your value of $R$ is large enough that $R$ is in the first range and $-R$ is in the second.

Split the integral into two parts $\xi$ $\raisebox{.3pt}{$<$}$ $x$ and $\xi$ $\raisebox{.3pt}{$>$}$ $x$ 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 $\pm{R}$ fall in the mentioned ranges.

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