Find the general solution of the inhomogeneous wave equation $$u_{tt} − c^2u_{xx} = \sin t$$ What is the solution to the initial-value problem for this equation with $u(x, 0) = 0$ and $u_t(x, 0) = 1$?
I know that the general solution to the homogenous equation is $u(x,t) = f(x+ct) + g(x-ct)$ and I know how to solve the initial value problem from there. It's the inhomogenous part of this equation that is confusing me.
Thanks in advance!
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$$u_{tt} − c^2u_{xx} = \sin t$$ You know that the general solution to the homogenous equation is $u(x,t) = f(x+ct) + g(x-ct)$
Then you have to find a particular solution of the inhomogeneous equation (any one). Only by inspection it is easy to see that $u_p=-\sin(t)$ is a solution.
If you don't see it, try any fonction of $t$ only, that is on the form $u_p(t)$. So, you solve the ODE $u_p''(t)=\sin(t)$ . No need for all solutions. Any one among them is sufficient, for example $u_p=-\sin(t)$ .
Thus the general solution of the PDE is : $$u(x,t)=f(x+ct)+g(x-ct)-\sin(t)$$ $u_t=cf'(x+ct)-cg'(x-ct)-\cos(t)$
Conditions :
$\begin{cases} u(x,0)=f(x)+g(x)=0\quad\implies\quad g(x)=-f(x) \\ u-t(x,0)=cf'(x)-cg'(x)-1=1 \quad\implies\quad 2cf'(x)=2 \end{cases}$
Integrating the second equation leads to $\quad f(x)=\frac{x}{c}+C \quad$ and $\quad g(x)=-\frac{x}{c}-C$ .
Putting them into the above general solution gives : $$u(x,t)=(\frac{x+ct}{c}+C)+(-\frac{x-ct}{c}-C)-\sin(t)$$ Finally, after simplification the solution is : $$u(x,t)=2t-\sin(t)$$
First, observe that a particular solution is $u_p(t) = -\sin t$.
Then, find the remaining homogeneous solution $v(x,t)$ such that $u = u_p + v$. This gives the BVP
\begin{cases} v_{tt} - c^2 v_{xx} = 0 \\ v(x,0) = u(x,0) - u_p(0) = 0 \\ v_t(x,0) = u_t(x,0) - u_p'(0) = 2 \end{cases}
d'Alembert's formula gives
$$v(x,t) = \frac{1}{2c}\int_{x-ct}^{x+ct} 2 ds = \frac{1}{c}\big[(x+ct) - (x-ct) \big] = \frac{2ct}{c} = 2t$$