Diffusion equation in polar coordinates

robinegberts
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Homework Statement


I am trying to solve the axisymmetric diffusion equation for vorticity by Fourier transformation.

Homework Equations


$$ \frac{\partial \omega}{\partial t} = \nu \Big( \frac{1}{r}\frac{\partial \omega}{\partial r} + \frac{\partial^2 \omega}{\partial r^2} \Big). $$

The Attempt at a Solution


I know how to solve the 1D analog
$$ \frac{\partial \omega}{\partial t} = \nu \frac{\partial^2 \omega}{\partial x^2} $$
by Fourier transformation. In this case, however, I don't how how to deal with the term
$$ \int_{-\infty}^{\infty}{\frac{\partial \omega}{\partial r} \frac{e^{-i k r}}{r} \mathrm{d} r}, $$
since partial integration only makes matters worse. Any ideas? Thanks.
 
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You cannot solve it by Fourier transform because your domain is not ##r\in (-\infty,\infty)##. If you really want to use transform methods, the transform you would be looking for is the Hankel transform.
 
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Thanks, that makes sense. What would be the easiest way to solve the problem under the initial condition ##\omega(t=0)=\gamma \delta(\vec{r})##?
 
robinegberts said:

Homework Statement


I am trying to solve the axisymmetric diffusion equation for vorticity by Fourier transformation.

Homework Equations


$$ \frac{\partial \omega}{\partial t} = \nu \Big( \frac{1}{r}\frac{\partial \omega}{\partial r} + \frac{\partial^2 \omega}{\partial r^2} \Big). $$

The Attempt at a Solution


I know how to solve the 1D analog
$$ \frac{\partial \omega}{\partial t} = \nu \frac{\partial^2 \omega}{\partial x^2} $$
by Fourier transformation. In this case, however, I don't how how to deal with the term
$$ \int_{-\infty}^{\infty}{\frac{\partial \omega}{\partial r} \frac{e^{-i k r}}{r} \mathrm{d} r}, $$
since partial integration only makes matters worse. Any ideas? Thanks.

As Orodruin points out in post #2, you cannot do a Fourier transform in the ##r## variable. However, you could do a Fourier transform in ##t##. Personally, if I were doing it (in view of the fact that everything starts at time ##t=0##) I would avoid Fourier and go with a Laplace transfer in ##t##.
 
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robinegberts said:
Thanks, that makes sense. What would be the easiest way to solve the problem under the initial condition ##\omega(t=0)=\gamma \delta(\vec{r})##?
Don't do it in polar coordinates. Putting a delta function on a point that is singular in your coordinate system is generally a bad idea if you do not know how to do it properly. Do it in Cartesian coordinates and Fourier transform in both spatial directions.

Edit: What you have here is essentially the computation of the heat kernel in two dimensions (which can easily be related to the Green's function of the heat equation). I have quite some discussion on this in my book and you should find it discussed in any text on mathematical methods that cover Green's functions of the heat equation.
 
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Thanks for the replies. Using Orodruin's approach of doing it in Cartesian coordinates separates the problem nicely into a problem I can solve and is a lot more intuitive than using polar coordinates.
 
robinegberts said:
Thanks for the replies. Using Orodruin's approach of doing it in Cartesian coordinates separates the problem nicely into a problem I can solve and is a lot more intuitive than using polar coordinates.
For future readers of this thread, could you please provide your solution? By forum rules others are not allowed to discuss full solutions in the homework forums until the OP has shown that he or she has solved the problem.
 
Written in Cartensian coordinates, the vorticity diffusion equation for ##\omega(x,y,t)## reads
$$\frac{\partial \omega}{\partial t} = \nu \Big( \frac{\partial^2 \omega}{\partial x^2} + \frac{\partial^2 \omega}{\partial y^2} \Big).$$
Fourier transforming the ##x,y## coordinates to ##k_x,k_y## using
$$\iint{\frac{\partial^2 \omega}{\partial x^2} e^{-2 \pi i (x k_x + y k_y)}\mathrm{d}x\mathrm{d}y} = -4\pi^2 k_x^2 \hat{\omega}(k_x,k_y,t),$$
and similar for the second partial derivative with respect to the y-component yields
$$\frac{\partial \hat{\omega}}{\partial t} = -4 \pi^2 \nu (k_x^2 + k_y^2) \hat{\omega}. $$
The initial condition
$$\omega(x,y,t=0)=\gamma \delta(x,y)$$
can be Fourier transformed to give
$$\omega(k_x,k_y,t=0)=\gamma,$$
and therefore
$$\hat{\omega}=\gamma e^{-4\pi^2 \nu (k_x^2 + k_y^2)t}.$$
Fourier-transforming this back into real space by completing the square in the exponent in the integral then gives the final answer
$$\omega(x,y,t)=\frac{\gamma}{4\pi \nu t} e^{-\frac{x^2 + y^2}{4 \nu t}}.$$
 
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