# Solving PDE with Laplace transform

Please help me to solve the following PDE with Laplace transform. In class we got less than 10 mins to study the topic (the semester ended and we were behind the schedule) and I am totally confused about how to handle/approach the problem. My texts do not show any relevant stuff for the problem. If fact I have only one sample and it confuses me a lot and explains nothing.

I can solve ODEs and compute Laplace/inverse Laplace transforms well, so do not bother with it. Just please show me how to reduce the PDE problem to solving ODE and/or computations of (inverse)Laplace transform and I will be able to handle the rest.

The problem:

Find a (just one) solution of the equation: $$u_{xx}=u_{tt}, t>0, -\infty <x<+ \infty, u|_{t=0}=3x, u_t|_{t=0}=\sin 2x.$$

Thanks a lot for your help!

• Use laplace transformation on the equation and use the properties of the Lp-trafo to get rid of the differentions in respect to x on the left side of the equation. This gives another equation which is hopefully easier to solve, then you need to transform it back. (Cant eloberate further since i am on my moblie atm, sorry, maybe this helps) Dec 15, 2017 at 21:50
• Thanks for the comment, but it can not helps me since I know that in theory, but never saw how to do that. Dec 15, 2017 at 21:58

We're considering the problem $$u_{xx}=u_{tt} ,\qquad t>0, \quad x\in \mathbb R$$ with $u(x,0) = 3x$ and $u_t(x,0) = \sin 2x$. The Laplace transform works almost the same way on PDEs as it does with ODEs, the main difference is that you have to pick which variable to perform the transform with. Since the conditions we're given are at $t=0$, take our transform with respect to the variable $t$ as the Laplace transform of a derivative can be expressed in terms of the functions initial value. I'll use the variable $\mathcal L$ to denote the Laplace transform operator with respect to the variable $t$, and I'll call our transformed variable $s$. Finally, I'll use capital letters to denote Laplace transform of it's lowercase counterpart. i.e. $$\mathcal L \{f(x,t) \} \equiv F(x,s)\equiv\int_0^\infty f(x,t)e^{-st} \,dt .$$
With our notation in order, let's apply the transform to the wave equation. We have that $$\mathcal L \{ u_{tt}(x,t) \} = s^2 \mathcal L \{u(x,t) \} -s u(x,0)- u_t(x,0)$$ which is the standard expansion of the Laplace transform of a derivative. Hopefully it is clear now why we are transforming with respect to $t$ as the functions $u(x,0)$ and $u_t(x,0)$ are already given to us. Plugging these in we have $$\mathcal L \{ u_{tt}(x,t) \} = s^2 U(x,s) - 3sx- \sin(2x).$$ On the other hand, we have that $$\mathcal L \{ u_{xx}(x,t) \} = \frac{\partial^2}{\partial x^2}L \{ u(x,t) \} = \frac{\partial^2 U}{\partial x^2}.$$ We can assume that the partial in $x$ commutes with the transform as this variable is independent to $t$. The wave equation then becomes $$\frac{\partial^2 U}{\partial x^2}=s^2 U - 3sx- \sin(2x)$$ which is effectively an ODE as the only derivatives are with respect to $x$. You can use your favourite method to solve this (variation of parameters, undetermined coefficients, etc..). I get that $$U(x,s) = c_1 e^{sx} + c_2 e^{-sx} + \frac{\sin(2x)}{s^2+4} + \frac{3x}{s}$$ for some constants $c_1,c_2$. Next we impose that our solution must be bounded as $x \to \pm \infty$, in particular that tells us that $c_1=c_2=0$ since the exponentials will be unbounded. So we have $$U(x,s) = \frac{\sin(2x)}{s^2+4} + \frac{3x}{s}.$$ All that we have to do now is invert the Laplace transform with respect to the variable $s$ $$u(x,t) = \mathcal L^{-1} \{ U(x,s) \} = \mathcal L^{-1} \left\{ \frac{\sin(2x)}{s^2+4} + \frac{3x}{s} \right\} = \sin(2x) \mathcal L^{-1} \left\{ \frac{1}{s^2+4} \right\} + 3x \mathcal L^{-1} \left\{ \frac 1s \right\} = \sin(2x) \left( \frac 12 \sin(2t) \right) + 3x (1).$$ Therefore $$u(x,t) = \frac 12 \sin(2x) \sin(2t)+3x.$$
• if $u(x,0)=3x$ how can we assume that $u$ is bounded? Feb 5, 2021 at 8:22