Given the following PDE:


$$-\infty <x<+ \infty , t>0$$ $$u(x,0)=0,u_t(x,0)=\frac{1}{1+x^2}$$

solve it by using the Laplace transform.

Here's my attempt so far:

Let $$\mathcal L \{u(x,t) \} \equiv U(x,s)\equiv\int_0^\infty u(x,t)e^{-st} \,dt .$$

and the result I've calculated is:

$$U_{xx}(x,s)-s^2 U(x,s) = -\frac{1}{1+x^2}-\frac{1}{s+1}=f(x,s)$$

I've concluded that the relation above is a non-homogenous 2nd order ODE in respect to the variable x. I have tried to solve it but so far it hasn't worked.

Which method do you think would be the best to solve it? Could you provide some guiding steps?

  • $\begingroup$ @Aleksas You are right about the initial conditions! Changed it! That method looks useful, but the initial exercise that I'm trying to solve requests that I do it by using the Laplace transform. :( $\endgroup$
    – Tita
    May 19, 2021 at 11:08

2 Answers 2


I think initial conditions is $$u(x,0)=0,u_t(x,0)=\frac{1}{1+x^2}$$ Solve using d’Alembert’s formula http://www.math.usm.edu/lambers/mat606/lecture12.pdf

I get solution $$u(x,t) =\frac{\operatorname{atan}\left( x+t\right) -\operatorname{atan}\left( x-t\right) }{2}+e^{-t}+t-1$$


So, I solved this the way it asked, by using the Laplace transform.

Continuing from where I left off in the question above, I use the Lagrange method to solve the non homogenous 2nd order linear ODE.

The answer will be in the form:


From solving the homogenous part I get the complimentary solution:


But because we need the solution to be finite (since $-\infty<x<\infty$):


Now, taking the particular solution $U_p(x,s)=C(x,s)e^{sx}+D(x,s)e^{-sx}$, we start by finding the Wronskian and continuing up until we find the derivatives of the factors:

$$C'(x,s)=f(x,s)\frac{e^{-sx}}{2s}$$ $$D'(x,s)=f(x,s)\frac{e^{sx}}{-2s}$$

Then we conclude that $U(x,s)=U_p(x,s)$, while leaving the above in their integral form. Therefore, we can write the following:

$$u(x,t)=\mathcal L^{-1}\{\frac{1}{2s}\int_{-\infty,\infty}(-\frac{1}{1+x'^2}-\frac{1}{1+s})e^{-s|x'-x|}dx'\}$$

By calculating each section of the above we get:

$$\mathcal L^{-1}\{\frac{1}{2s}e^{-s|x'-x|}\}=\frac{1}{2}u_{|x'-x|}(t)$$


$$\mathcal L^{-1}\{\frac{1}{2s(s+1)}e^{-s|x'-x|}\}=\frac{1}{2}u_{|x'-x|}(t)-\frac{1}{2}u_{|x'-x|}(t)e^{-t+|x'-x|}$$

where $u_{|x'-x|}(t)$ is the Heaviside function for $c=|x'-x|$. But for $t<|x'-x|=>u_{|x'-x|}(t)=0$, therefore, we limit the solution to where $t\geq|x'-x| => u_{|x'-x|}(t)=1$. That way the new bounds for the integral of $u(x,t)$ are $(x-t,x+t)$.


and the result is:


which is the same as what @Aleksas Domarkas found by using d'Alambert's formula.

I would say this was solved more easily when tried with a Fourier cosine transform, but since the exercise asked for the Laplace transform I wanted to give the answer just in case anyone else wondered. It's really just a solution to an ODE but I thought the last part was interesting.


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