Question:- Determine the solution of the initial boundary-value problem $$u_{tt}=c^2 u_{xx},~~~~~~0<x<\infty,~~t>0,~~~~~(1)$$ $$u(x,0)=f(x),~~~~~~~0\leq x<\infty,~~~~~(2)$$ $$u_t(x,0)=0,~~~~~~~0\leq x<\infty,~~~~~(3)$$ $$u_x(0,t)+h u(0,t)=0~~~~~~~t\geq 0,~~~h=\text{constant}~~~~~(4)$$ State the compatibility condition of $f$.
Answer:- Using characteristic curves, $\xi=x+ct,~\eta=x-ct$, we obtain $$u(x,t)=\phi(x+ct)+\psi(x-ct)~~~~~(5)$$ from $(2)$, $f(x)=u(x,0)=\phi(x)+\psi(x)~~~~~(6)$
from $(3)$, $0=u_t(x,0)=c\phi'(x)-c \psi '(x)~~~~~(7)$
Using $(5), (6),$ and $(7)$, for $x>ct$, the solution to $(1)$ is given by $$u(x,t)=\frac{1}{2}\left(f(x+ct)+f(x-ct)\right).~~~~~(8)$$ Now, for $0<x<ct$, we use boundary condition $(4)$, so we have $$0=u_x(0,t)+h~u(0,t)=\phi'(ct)+\psi'(-ct)+h(\phi'(ct)+\psi'(-ct))$$ $$-h=\frac{d(\phi(ct)+\psi(-ct))}{\phi(ct)+\psi(-ct)}.$$ Integrate $0$ to $ct$, $$-h(ct)+k_1=ln(\phi(ct)+\psi(-ct))~~~~~~~\text{where}~k_1=ln(\phi(0)+\psi(0))$$ $$\phi(ct)+\psi(-ct)=k~e^{-h(ct)}~~~~~~~\text{where}~k=\phi(0)+\psi(0)$$ Let $\alpha=-ct$. $$\psi(\alpha)=k~e^{-h\alpha}-\phi(-\alpha)$$ Replacing $\alpha$ by $x-ct$, $$\psi(x-ct)=k~e^{(ct-x)h}-\phi(ct-x)$$. Hence, $$u(x,t)=\frac{1}{2}\left(f(x+ct)-f(ct-x)\right)+ke^{(ct-x)}h.~~~~~(9)$$
Equation $(8)$ and $(9)$ together form solution.
State the compatibility condition of $f$. What to show for compatibility.