Solving Diffusion equation with Convection

  • Thread starter PiRho31416
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  • #1
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The problem is as follows:
[tex]\frac{\partial u}{\partial t}=k\frac{\partial^{2}u}{\partial x^{2}}+c\frac{\partial u}{\partial x},[/tex]

[tex] -\infty<x<\infty[/tex]

[tex]u(x,0)=f(x)[/tex]

Fourier Transform is defined as:

[tex]F(\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{i\omega x}dx[/tex]

So, I took the Fourier Transform which brought me to

[tex] \frac{\partial F}{\partial t}=-\omega^{2}F-ci\omega F=-F(\omega^{2}+ci\omega)[/tex]

Solving the first order differential equation brought me to

[tex]F(\omega)=e^{-\frac{1}{6}(3ic+2\omega)\omega^{2}}[/tex]

When I try to integrate using the inverse Fourier transform

[tex]f(x)=\int_{-\infty}^{\infty}F(\omega)e^{-i\omega x}\, d\omega[/tex]

I get stuck. Did I do the steps right?
 
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Answers and Replies

  • #2
798
34
As usual, it is not dificult to find particular solutions and more general solution of the PDE.
The real difficulty is encountered when we have to fit the general solution to the boundary conditions so that the solution of the problem should be derived.
In the present wording, the bondary condition is specified by a function f(x) which is not explicit. So, we cannot say if the problem can be analytically solved.
 

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  • #3
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I had a similar equation which I would like to solve but with variable coefficients; I have produced a solution which I feel is wrong and have posted it - twice - but nobody seemed to be willing to offer their insight.

Anyway, significant progress can be made for your problem with Fourier transforms, since c and k are constants.

Given the original equation

[tex]
\frac{\partial{u(x,t)}}{\partial{t}}=k\frac{\partial^2{u(x,t)}}{\partial{x^2}}+c\frac{\partial{u(x,t)}}{\partial{x}}
[/tex]

taking the f.t. in the space dimension and canceling out any shared factors you obtain

[tex]
\frac{\partial{U(\omega,t)}}{\partial{t}}=-k\omega^2U(\omega,t)+ic\omega U(\omega,t)
[/tex]

which when solved results to

[tex]
U(\omega,t)=U(\omega,0)e^{-(k\omega^2-ic\omega)t}
[/tex]

so, assuming you can determine

[tex]
U(\omega,0)=\int{u(x,0)e^{-i\omega x}dx}
[/tex]

the solution will be given by

[tex]
u(x,t)=\frac{1}{2\pi}\int{U(\omega,0) e^{-kt(\omega^2-\frac{i(c+x)\omega}{k})} d\omega}
[/tex]

I believe this is correct. What is your f(x)? If you can expand U(w,0) in powers of w then you can complete the square in the exponential and solve the resulting gaussian integrals.
If anyone can spot any horrible mistakes in the above, please shout
 
  • #4
798
34
If anyone can spot any horrible mistakes in the above, please shout
Only a very, very small shout for a mistake at the end.
 

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