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 $$\frac{d^2U}{dx^2}-2s\frac{\partial U}{\partial x}=-s^2U$$ $$U(0,s)=0,\quad U(1,s)=F(s)$$.

  • $\begingroup$ I haven't read the paper but here is one on the topic. It is a method called the Laplace substitution method. $\endgroup$
    – dustin
    Commented Nov 24, 2014 at 22:48
  • $\begingroup$ When I transform $Y_{tx}$, I get $\int_{0}^{\infty}e^{-st}Y_{tx}\,dt = -Y_{x}(x,0)+sU(x,s)$. There are evaluation terms when I transform $Y_{tt}$, too. $\endgroup$ Commented Nov 26, 2014 at 0:13
  • $\begingroup$ When you're integrating in $t$, just write $Y_{xt}=Y_{tx}=\frac{\partial}{\partial t}Y_{x}$, and think of holding the $x$ variables constant while you work in $t$. $\endgroup$ Commented Nov 26, 2014 at 0:33
  • $\begingroup$ If $U$ is the transform of $Y$, then the transform of $Y_{t}$ is $sU+\cdots$ and the transform of $Y_{tx}$ is $s\frac{\partial}{\partial x}U+\cdots$ (where ... indicates evaluation terms.) So you transform without the derivatives in $x$, and then take those derivatives at the end. There are lots of regularity assumptions, but there always are anyway. $\endgroup$ Commented Nov 26, 2014 at 1:25
  • $\begingroup$ @T.A.E. So then will it just be $$\frac{d^2U}{dx^2}-2s\frac{\partial U}{\partial x}+s^2U=0?$$ Which is just an ODE? Hm, I don't think I am getting it still. I should place a bounty on the question. $\endgroup$
    – Robben
    Commented Nov 26, 2014 at 1:44

1 Answer 1



$$y(x,s) = \int_0^{\infty} dt \, Y(x,t) \, e^{-s t}$$

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.)

  • $\begingroup$ Thank you very much!! It is much more clear to me now. $\endgroup$
    – Robben
    Commented Nov 26, 2014 at 3:28

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