2

How to solve this pde problem : $u_{ttx}-u_{xxx}=0$ with $u_{x}(x,0)=0$ & $u_{tx}(x,0)=\sin(x)$

I worked on this problem by changing variable and I could find $ u_{x}(x,t)=A(x+t)+B(t-x)$ with the help of the answer of wave equation. Now how can I find $A$ and $B$ ?

M.L.M
  • 127

1 Answers1

1

I will use the slightly different notation

$$u_x(x,t) = f(x+t)+g(x-t)$$

You have the system of equations

$$\begin{cases}u_x(x,0) = f(x)+g(x) = 0\\ u_{xt}(x,0) = f'(x)-g'(x) = \sin x\end{cases} \implies f(x) = -g(x) = C-\frac{1}{2}\cos x$$

for $C\in \Bbb{R}$. Where does it go next?

Ninad Munshi
  • 34,407
  • So according to boundary condition $C=0$ and consequently, $u( x ,t)= +1/2 \sin (t-x)-1/2 \sin ( t+ x) +K$ – M.L.M Jan 16 '22 at 02:14