Question

Let u(x, t) be a solution of the one- dimensional wave equation and A, B, C and D be any vertices of any parallelogram whose sides are characteristics wave (i.e the line parallel to x+ct=0, x-ct=0)

Then show that u(A)+u(C)=u(B)+u(D)

Answer 1

A general solution:

$u(x,t)=F(x-ct)+G(x+ct)$

Consider any parallelogram of vertices A, B, C, D with sides parallel to the

characteristics $x=\pm ct+\xi$

Let

$A=(x,t)$

$B=(x+cs, t+s)$

$C=(x+cs-c\tau,t+s+\tau)$

$D=(x-c\tau,t+\tau)$

be the coordinates of the vertices of a characteristic parallelogram, where $s$ and $\tau$ are

positive parameters.

Then:

$u(A)=F(x-ct)+G(x+ct)$

$u(C)=F(x-2c\tau-ct)+G(x+2cs+ct)$

$u(B)=F(x-ct)+G(x+2cs+ct)$

$u(D)=F(x-2c\tau-ct)+G(x+ct)$

Therefore,

$u(A)+u(C)=u(B)+u(D)$