Question

If I have the solution u(x,t)=cos(t) sin(x), how would I show that this solution also has the form u(x,t) = p(x-ct) + q(x+ct)?

Answer 1

It looks like you can reduce the order of the equation using
\( \left\{
\begin{array}{11}
u_t + cu_x = w & 1)\\
\ w_t - c w_x = 0 & 2)
\end{array}
\right.\)

When you solve this, you get one part of the solution.:

\(
\begin{align*}
\frac{dt}{dx} &= \frac{1}{-c} \\
-cdt &= dx \\
0 &= cdt + dx \\
k &= ct + x \\
f(k) &= f(ct +x), & f(k) \mbox{ is a solution to 2) } \\
u_t + cu_x &= \color{green} {w = f(ct+x) }\\
f(ct+x) &= -\sin t \sin x + c \cos t \cos x
\end{align*}
\)

Answer 2

Then you do the same for
\(
\left\{
\begin{array}{11}
u_t - cu_x = v & 1)\\
\ v_t + c v_x = 0 & 2)
\end{array}
\right.
\)

And add the two parts together.

I get

\( u(x,t) = -\sin t \sin x + c \cos t \cos x -\sin t \sin x -c \cos t \cos x\)

Answer 3

Whoops, looks like that's wrong.

Answer 4

You need to isolate \(u_t\) and \(u_x\) by adding and subtracting \(f(x+ct)\) and \( g(x−ct) \).

Then integrate, then add the results together.