Solving a 2D PDE using the Fourier Transform

[CGandC]

Homework Statement

Solve the following partial differential equation , using Fourier Transform:

Given the following:

And a initial condition:

Homework Equations

The Attempt at a Solution

First , i associate spectral variables to the x and t variables:
$k$ is the spectral variable corresponding to $x$
$\omega$ is the spectral variable corresponding to $t$

Using Fourier transform on the PDE , i get:

$(i\omega)(ik)\widetilde{u(\omega,k)}=(ik)^2 \widetilde{u(\omega,k)}$

After simplifying , I get : $\omega = k$

How am I supposed to proceed from here? ( I didn't find $\widetilde{u(\omega,k)}$ )

 

[Ray Vickson]

Homework Statement

Solve the following partial differential equation , using Fourier Transform:
View attachment 234811

Given the following:
View attachment 234812

And a initial condition:
View attachment 234813

Homework Equations

The Attempt at a Solution

First , i associate spectral variables to the x and t variables:
$k$ is the spectral variable corresponding to $x$
$\omega$ is the spectral variable corresponding to $t$

Using Fourier transform on the PDE , i get:

$(i\omega)(ik)\widetilde{u(\omega,k)}=(ik)^2 \widetilde{u(\omega,k)}$

After simplifying , I get : $\omega = k$

How am I supposed to proceed from here? ( I didn't find $\widetilde{u(\omega,k)}$ )

Your method fails because it has no straightforward way to incorporate the initial condition on $u(x,0).$ If I were doing the problem I would just take the $x$-transform, and write
$$u(x,t) = \int_R \tilde{u}(k,t) e^{-ik x} \, dk$$ and so transform the PDE to
$$(-ik) \frac{\partial \tilde{u}(k,t)}{\partial t} = (-ik)^2 \tilde{u}(k,t), $$
with $\tilde{u}(k,0) =$ Fourier transform of the given function $u(x,0).$

However, if you really insist on transforming both variables $x$ and $t$ you should recognize that the Fourier transform does not fit well with the restriction $t \geq 0.$ It would be better to use the Laplace transform
$$g(s) =( {\cal L} f)(s) = \int_0^\infty e^{-st} f(t) \, dt,$$
and to recognize the possibility of acknowledging the initial condition through the standard property
$$({\cal L}\, dt/dt)(s) = s ({\cal L} f)(s) - f(0).$$ You will get a solvable transformed system: try it and see.

 

[CGandC]

Your method fails because it has no straightforward way to incorporate the initial condition on $u(x,0).$ If I were doing the problem I would just take the $x$-transform, and write
$$u(x,t) = \int_R \tilde{u}(k,t) e^{-ik x} \, dk$$ and so transform the PDE to
$$(-ik) \frac{\partial \tilde{u}(k,t)}{\partial t} = (-ik)^2 \tilde{u}(k,t), $$
with $\tilde{u}(k,0) =$ Fourier transform of the given function $u(x,0).$

However, if you really insist on transforming both variables $x$ and $t$ you should recognize that the Fourier transform does not fit well with the restriction $t \geq 0.$ It would be better to use the Laplace transform
$$g(s) =( {\cal L} f)(s) = \int_0^\infty e^{-st} f(t) \, dt,$$
and to recognize the possibility of acknowledging the initial condition through the standard property
$$({\cal L}\, dt/dt)(s) = s ({\cal L} f)(s) - f(0).$$ You will get a solvable transformed system: try it and see.

I managed to solve the question , like you meant - using the restriction that the time variable $t$ is not Fourier transformable ( meaning: I only use Fourier transform on the x variable ) .

But I don't fully understand why the time variable $t$ is not Fourier transformable , can you please elaborate?

 

[Ray Vickson]

I managed to solve the question , like you meant - using the restriction that the time variable $t$ is not Fourier transformable ( meaning: I only use Fourier transform on the x variable ) .

But I don't fully understand why the time variable $t$ is not Fourier transformable , can you please elaborate?

I did not say it was not Fourier transformable; I said that the Fourier transform does not fit well in that case.

Of course, you can always take the F.T. of a function of the form
$$F(t) = \begin{cases} 0 & t < 0\\
f(t) & t \geq 0
\end{cases} $$
because you can perform the Fourier integral. However, such problems fit more nicely into the area of Laplace transforms (which are, essentially, Fourier transforms at a complex argument $k$). But go ahead and do the Fourier transform if you want to; just remember that the formula for the transform of $df/dt$ will not be what you think it is. You will need extra terms corresponding to the value of $f(0)$. I will let you work it out for yourself, or look it up somewhere.