Use Laplace transforms to solve the boundary value problem $$Y_{xx}(x,t)-2Y_{tx}(x,t)+Y_{tt}(x,t)=0, \quad 0<x<1, t>0$$ $$Y(x,0)=Y_t(x,0)=0, \quad 0<x<1$$ $$Y(0,t)=0, \ Y(t,1)=F(t), \ t>0.$$
I am supposed to solve the PDE using Laplace transform. I know how to solve such PDE's like the wave equation using Laplace, but I dont know how to solve it for this problem where there is a mixed partial of $\frac{\partial^2}{\partial x\partial t}$. I tried searching online but I couldn't find anything.
What I have so far is $$\frac{d^2U}{dx^2}-2s\frac{\partial U}{\partial x}=-s^2U$$ $$U(0,s)=0,\quad U(1,s)=F(s)$$.
Best Answer
Define
$$y(x,s) = \int_0^{\infty} dt \, Y(x,t) \, e^{-s t}$$
Then, integrating by parts:
$$\int_0^{\infty} dt \, Y_t(x,t) \, e^{-s t} = -Y(x,0) + s y(x,s)$$
$$\int_0^{\infty} dt \, Y_{tt}(x,t) \, e^{-s t} = -Y_t(x,0) + s Y(x,0) + s^2 y(x,s)$$
Then using the initial conditions $Y(x,0)=Y_y(x,0)=0$, the PDE becomes the following ODE:
$$y''-2 s y' + s^2 y = 0$$
where $y(0,s)=0$ and $y(1,s)=f(s)$, where the prime represents derivative with respect to $x$, and where
$$f(s) = \int_0^{\infty} dt \, F(t) \, e^{-s t} $$
The general solution of the ODE is (I will not derive here)
$$y(x,s) = (A + B x) \, e^{s x}$$
Using the boundary conditions, we may find $A$ and $B$ and therefore the LT of the solution to the PDE:
$$y(x,s) = x \, f(s) \, e^{-s (1-x)} $$
We may find the inverse LT by convolution, as we know the individual LT's. The ILT of $f(s)$ is obviously $F(t)$ by definition, and the ILT of $e^{-s (1-x)}$ is $\delta(t-(1-x))$. Therefore, the ILT, and the solution to the equation, is
$$y(x,t) = x \int_0^t dt' \, F(t') \delta(t-t'-(1-x)) = x F(t-(1-x)) \theta(t-(1-x))$$
where $\theta$ is the Heaviside step function, which is necessary because the contribution to the integral from the $\delta$ function for $t < 1-x$ is zero (i.e., $t' \gt 0$ in the integral.)