Confused about Laplace Wave Equation Transformation

Question

$$\frac{\partial^2}{\partial t^2} u(x,t) -\frac{\partial^2}{\partial x^2} u(x,t) =f(x), \quad 0<x<1, \quad t>0\\ u(x,0)=0, \quad \frac{\partial}{\partial t} u(x,0)=0\\ u(0,t)=0, \quad u(1,t)=0$$ I am supposed to take the Laplace transform of the wave equation that yields a non-homogeneous ordinary differential equation in terms of $\mathcal{L} \{f(x)\}$ and $\mathcal{L} \{u(x,t)\}$ and its $x$-derivatives.

Can someone please explain the relationship between the two functions, and how it's $x$-derivatives changes will be reflected in the function? This question was posed as a challenge question, and I am seeking some guidance.

How does one setup this Laplace transformation? Why is there $t$ and $x$ in the same system?

Answer 1

I've not used Laplace transform a lot but I think that it would act similarly as the Fourier transform (the only difference is that Fourier transform acts on position variable, so $x,y,z,\cdots$ and Laplace transform acts on the time variable). With that in mind if you take the Laplace transform on both side of the equation you would get: $$\mathcal{L}\{\partial^2_{tt}u\}-\mathcal{L}\{\partial^2_{xx}u\} = \mathcal{L}\{f(x)\}\\ \color{red}{\mathcal{L}\{\partial^2_{tt}u\}}-\underbrace{\partial^2_{xx}\mathcal{L}\{u\} = f(x)\color{blue}{\mathcal{L}\{1\}}}_{\text{the transform acts only}\\ \text{on time so it's independent}\\ \text{of the }x}\\ \color{red}{s^2U(x,s)-su(x,0^+)-\partial_tu(x,t)|_{t=0^+}} - \partial^2_{xx}U(x,s)=\frac{f(x)}{\color{blue}{s}}$$ The red one follows from the property of derivation for the Laplace transform and the blue from the Laplace transform of $1$.

As you can see you began with an ode with two partial derivation and you arrived at a second order ode in the function $U(x,s)$. This means that one you've found the solution $U(x,s)$ you can directly find $u(x,t)$ by anti-transforming! If this isn't the most beautiful thing you've ever seen, I don't know what it would be!

Answer 2

Assuming $\mathcal{L}$ transforms the $t$ variable into the $s$ variable. E.g. $u(x, t)$ into $U(x, s)$.

I added the $t$ subscript to the operator to indicate it acts on $t$.

The subscripts $xx$ mean second order partial derivative regarding $x$, $tt$ mean second order partial derivative regarding $t$.

My guess would be: $$ \mathcal{L}_t \{ u_{tt}(x, t) - u_{xx}(x, t) \} = \mathcal{L}_t \{ f(x) \} \iff \\ s^2 \, U(x, s) - s \, u(x, 0+) - u_t(x, 0+) - U_{xx}(x, s) = f(x) \mathcal{L}_t \{ 1 \} $$

The idea is to apply the initial conditions here, then get an ordinary differential equation for $U$, solve for $U$ and then use the inverse Laplace transform to get a solution $u$.