The wave equation for a semi-infinite string with boundary conditions: $$\begin{cases}u_{xx}=u_{tt} &x,t>0 \\ u_{x}(0,t)=au(0,t)+bu_{t}(0,t)+h(t) &t\ge 0 \\ u(x,0)=f(x) &x>0 \\ u_{t}(x,0)=g(x) & x>0\end{cases}$$
First, I wonder what the physical interpretation might be at $x=0$. Is it a motor $h(t)$, a spring with springconstant $a$ and vertical friction with the motor with friction constant $b$?
Second, I have to solve this problem using the general solution for the wave equation: $$u(x,t)=F(x-ct)+G(x+ct)$$ and $h=g=a=0$. I cannot figure out how the wave equation is solved at $x=0$ using the general solution and the method of characteristics.
Thank you!
$$u(x,t)=\frac{1}{2}(f(x-t)+f(x+t))+\frac{1}{2}\int_{x-t}^{x+t}g(x_{0})dx_{0}$$
The standard solution for $u(x,t)$ for $c=1$ and $x_{0}$ a dummy variable. The derivative of $u(x,t)$ with respect to $x$ and $t$ are:
$$u_{x}(x,t)=\frac{1}{2}(f'(x+t)+f'(x-t))$$ and $$u_{t}(x,t)=\frac{1}{2}(f'(x+t)-f'(x-t))$$ now $$u_{x}(0,t)=b*u_{t}(0,t)$$ $$\frac{1}{2}(f'(t)+f'(-t))=\frac{b}{2}(f'(t)-f'(-t))$$ The function $f(t)$ is known and therefore also the derivative $f'(t)$. The extension of $f(t)$ has to be determined. $$f'(-t)=\frac{b-1}{b+1}f'(t)$$ To meet the boundary condition at $x=0$ the derivative of $f(t)$ has to be extended even around $x=0$ and multiplied with $\frac{b-1}{b+1}$, therefore $f(t)$ has to be extended uneven around $x=0$ and multiplied with $\frac{b-1}{b+1}$. The solution in the general form $u(x,t)=F(x-ct)+G(x+ct)$ is: $$u(x,t)=\frac{1}{2}f_{ext}(x-t)+\frac{1}{2}f_{ext}(x+t)$$ with $f_{ext}$ the uneven extension around $x=0$ mulitplied by $\frac{b-1}{b+1}$ of $f(t)$.