Introduce the <characteristic coordinate>[null coordinates]
$$
\xi=t-x,\qquad\eta=t+x.
$$
Then $\Box=-4\partial_\xi\partial_\eta$, so the equation becomes
$$
u_{\xi\eta}=-\frac14u.
$$
Write the compatible boundary values as
$$
a(\xi)=u(\xi,0),\qquad b(\eta)=u(0,\eta),\qquad a(0)=b(0)=c.
$$
Twice integrating the equation gives the equivalent <Volterra integral equation>
$$
u(\xi,\eta)
=a(\xi)+b(\eta)-c
-\frac14\int_0^\xi\int_0^\eta u(s,r)\,dr\,ds.
$$

Let $T$ denote the double-integral operator including the factor $-1/4$, and put $g=a+b-c$. Successive approximation gives the <Neumann series>
$$
u=\sum_{m=0}^\infty T^mg.
$$
On a rectangle $|\xi|\leq R$, $|\eta|\leq R$,
$$
\|T^mg\|_\infty
\leq\frac{(R^2/4)^m}{(m!)^2}\|g\|_\infty.
$$
The series and its differentiated series converge locally uniformly. Since $a$ and $b$ are analytic, the sum is analytic and solves the equation and data near the origin.

If two solutions have the same data, their difference $w=Tw$. Iterating and using the same factorial estimate gives $\|w\|_\infty=0$ on every sufficiently small rectangle. This proves uniqueness. The argument is the <Analytic Goursat problem for a Klein--Gordon equation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 105 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-105/1/c/solution