Consider the Cauchy problem
$$\left\{ \begin{array}{l l} u_{tt} - u_{xx} = e^{-t}\sin x & \quad \mbox{$x \in \mathbb{R}, t>0$,} \\ \quad u(x,0) = 0, \\ \quad u_t(x,0) = 0. \end{array} \right. $$
(a) Find the solution $u(x,t)$ using Duhamel's principle.
(b) Compute the limit $\lim_{t\rightarrow\infty}{u(x,t)}$
I think the general solution to this looks like this: $$\int_{0}^{t}u(x,t) = U(x,t-s,s)ds$$ but I am not sure of how to get there.
Or I am also aware that the solution might take the form of this: $$v(x,t) = \frac{1}{2}[\phi(x+ct)+\phi(x-ct)] +\frac{1}{2c} \int_{x-ct}^{x+ct} \psi(s) ds$$
In this case, it this were the general solution, the the entire left side of the addition would disappear because of the zero from the initial conditions?
Also, I am not sure of where to even begin with computing the limit since I am already lost here...
Thank you for your help.
The general solution that you gave is for the homogeneous problem (right-hand-side equal to zero), however, it is still important to know this to derive a formula for the inhomogeneous problem. The essence behind Duhamel's principle is to consider the homogenous problem with $f$ as part of your initial condition at some given time slice, viz.,
$$u_{tt} - u_{xx} = 0, \\ u(x,s) = 0,\\ u_t(x,s) = f(x,s) \,ds.$$
By the general solution you already derived, this is solved by
$$u(x,t) = \frac{1}{2}\left(\int^{x+(t-s)}_{x-(t-s)} f(y,s) \,dy \right) ds$$
Duhamel's principle then tells us that we can find the solution to the inhomogeneous problem by integrating all of the time slices and adding it to the original general solution. Thus, the solution to the inhomogeneous problem is given by
$$u(x,t) = \frac{1}{2}\left(\phi(x+t)+\phi(x-t) \right) + \frac{1}{2}\int^{x+t}_{x-t} \psi(s) \,ds +\frac{1}{2} \int_0^t \left(\int^{x+(t-s)}_{x-(t-s)} f(y,s) \,dy \right)ds$$