I am looking to solve the following PDE:

$$\begin{align*}&u_{tt} - c^2 u_{xx} = 0, &0 < x < \infty, t>0\\&u(x,0)=0,\quad u_t (x,0)=0 &0 < x < \infty\\&u_x(0,t)=h(t) \qquad &t > 0\end{align*}$$

There is another question on here which solves this by assuming a solution in the form of $u(x,t) = f(x+ct) - g(x-ct)$ and I am looking to solve this equation by other methods. I am only concerned for when $x \leq ct$.

The other question can be found here.

My current attempt consists of trying to make a substitution to zero the initial condition but it has not led me anywhere.


We can use the Laplace transform to solve the PDE. The rules are pretty much the same as solving an ODE with the Laplace transform. \begin{align} \mathcal{L}\{u_{tt}(x,t)\} &= s^2U(x,s) - su(x,0) - u_t(x,0)\\ \mathcal{L}\{u_{xx}(x,t)\} &= U_{xx}(x,s) \end{align} So the Laplace transform of the PDE is then $$ c^2U_{xx}(x,s) = s^2U(x,s) - su(x,0) - u_t(x,0)\Rightarrow U_{xx}(x,s) = \frac{s^2}{c^2}U(x,s)\tag{1} $$ Then the general solution to $(1)$ is $$ U(x,s) = A(s)\exp[sx/c] + B(s)\exp[-sx/c]\tag{2} $$ Now this wasn't explicitly stated but I am assuming we require $\lim_{x\to\infty}\lvert u(x,t)\rvert < \infty$. Therefore, equation $(2)$ becomes $$ U(x,s) = B(s)\exp[-sx/c]\tag{3} $$ Now we need to take the Laplace transform of $u_x(0,t) = h(t)$ but this is simply $U_x(0,s) = H(s)$. Using our transformed boundary condition, we have $$ U_x(0,s) = \frac{-s}{c}B(s) = H(s)\Rightarrow B(s) = \frac{-c}{s}H(s) $$ Now, we have our transformed solution to the PDE. \begin{align} U(x,s) &= \frac{-c}{s}H(s)\exp[-sx/c]\\ \mathcal{L}^{-1}\{U(x,s)\} &= -c\mathcal{L}^{-1}\{H(s)\exp[-sx/c]/s\}\tag{4}\\ u(x,t) &= -ch(t-x/c)\mathcal{U}(t-x/c)\tag{5} \end{align} Equation $(4)$ reduces to $(5)$ by Laplace shifting theorems and $\mathcal{U}(t-x/c)$ is the shifted unit step function.


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