${ u }_{ tt }-{ c }^{ 2 }{ u }_{ xx }=0 ,\ 0<x<\infty ,\ 0\le t <\infty $
Initial conditions: $u(x,0)=0\ $ and ${u}_{t}(x,0)=V$ where $V$ is a positive constant.
Boundary condition: ${u}_{t}(0,t)+a{u}_{x}(0,t)=0$ where $a$ is a constant and $a>c>0$.
I am trying to solve this problem.
My attempt: $ u=f(x+ct)+g(x-ct) $ for some functions $\ f\ and\ g $.
But the problem is that I cannot find a solution that satisfies all the conditions simultaneously. Somebody help. Thanks in advance.
On
Your equation is linear.
Let say $U$ is the global solution. If you have a solution $U_1$ satisfying only the boundary condition. Then you should look for $U_2 = U-U_1$. $U_2$ is subject to homogeneous boundary condition. U2 can be called the asymptotic solution while U1 is the transient one and depends on the initial condition.
Otherwise, Laplace transform in $x$ and $t$ would probably do the job.
Variable separation may also be useful : $u(x,t)=f(x)g(t)$
You need a solution defined in the region $D=\{(x,t):x\ge0,\ t\ge0\}$. If $(x,t)\in D$, then $x+c\,t\ge0$, while $x-c\,t$ can be positive or negative. From the initial conditions you can find in the usual way the values of $f$ and $g$ on $[x,\infty)$: $$ f(x)=\frac{V}{2\,c}\,x,\quad g(x)=-\frac{V}{2\,c}\,x,\qquad x\ge0. $$ Then $$ u(x,t)=V\,t\qquad x\in D,\quad x-c\,t\ge0. $$ In order to define the solution on the region $\{(x,t)\in D:x-c\,t<0\}$ we need to extend $g$ to $(-\infty,0)$. For this we use the boundary condition: $$ u_t(0,t)+a\,u_x(0,t)=(c+a)f'(c\,t)+(a-c)g'(-c\,t)=0. $$ From here we obtain that if $x<0$ then $$ g'(x)=-\frac{a+c}{a-c}\,f'(-x)=-\frac{a+c}{a-c}\,\frac{V}{2\,c}. $$ I hope you can finish now.