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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.

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user157227
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1 Answers1

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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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