There is a problem to Find the bounded solution of the wave equation $$ \dfrac{\partial^2 u}{\partial t^2}= 4\dfrac{\partial^2 u}{\partial x^2} \; \; \;\; $$
Given boundary condition $$u(0,t) = \sin t,\;\; t>0;$$
And initial condition $$ u(x,0) = 0\;\; x>0\quad\text{and}\quad u_t(x,0) = e^{-x},$$
My problem When I saw this problem I thought of using the solution $$u(x,t)= \sum \left( a_n \cos n \frac{\pi}{L} t + b_n\sin n \frac{\pi}{L}t \right)\sin(n \frac{\pi}{L} x) $$.
I have been put back by the boundary value condition.
May someone supply me the solution to this problem. I thank you in advance
Follow up question I am very thankful to Martin and Jim. Their contribution has made me suggest that the solution to this is $$u(x,t)= \sin(x+2t)+\sin(x-2t)+\frac{1}{2} \sinh(x-2t)$$.
Is this answer correct? If correct, Please help me know how I can move to this one using Martins method.
The formula you wrote works for the case where the domain is bounded, which is not the case here.
As stated, it seems as if the problem has no solution.
In general, as Jim mentioned, $u(x,t)=f(x+2t)+g(x-2t)$. From the initial conditions, $$ 0=u(x,0)=f(x)+g(x),\ \ x>0. $$ This tells us that $g=-f$. We have $$ u_t(x,t)=2f'(x+2t)+2f'(x-2t) $$From the second initial condition, $$ e^{-x}=4f'(x). $$ So $f(x)=-\frac14\,e^{-x}+c$.
If we look at the boundary condition, $$ \sin t=f(2t)+g(-2t),\ \ t>0, $$ so now we would need $$\sin t=f(2t)-f(-2t)=\frac14\,(e^{2t}-e^{-2t}),\ \ t>0.$$
The formula on the right is fairly close to $\sinh t$. Are you sure that was not the statement of the problem?