Use the Fourier transform directly to solve the heat equation

[richyw]

Homework Statement

Use the Fourier transform directly to solve the heat equation with a convection term
[tex]u_t =ku_{xx} +\mu u_x,\quad −infty<x<\infty,\: u(x,0)=\phi(x),
assuming that u is bounded and k > 0.

Homework Equations

fourier transform
inverse Fourier transform
convolution thm

The Attempt at a Solution

taking the FT of both sides i get
[tex]U_t=-k w^2U-iw\mu U[/tex]
[tex]U(0,t)=\Phi(w,0)[/tex]
I solved the ode and got
[tex]U(w)=e^{(\mu i w- w^2k)t}[/tex]
but now I am a bit confused on the next step, is this where I want to get my initial condition involved, or do I want to try and get it back as u(x,t) using inverse FT. I can see that my solution is a gaussian multiplied by another function of F, so I think I might be able to use convolution thm?

 

[vela]

Homework Statement

Use the Fourier transform directly to solve the heat equation with a convection term
[tex]u_t =ku_{xx} +\mu u_x,\quad −infty<x<\infty,\: u(x,0)=\phi(x),
assuming that u is bounded and k > 0.

Homework Equations

fourier transform
inverse Fourier transform
convolution thm

The Attempt at a Solution

taking the FT of both sides i get
[tex]U_t=-k w^2U-iw\mu U[/tex]
[tex]U(0,t)=\Phi(w,0)[/tex]

Don't you mean ##U(\omega,0) = \Phi(\omega,0)##?

I solved the ode and got
[tex]U(w)=e^{(\mu i w- w^2k)t}[/tex]
but now I am a bit confused on the next step, is this where I want to get my initial condition involved, or do I want to try and get it back as u(x,t) using inverse FT. I can see that my solution is a gaussian multiplied by another function of F, so I think I might be able to use convolution thm?

You left out the arbitrary constant when you solved for ##U(\omega,t)##. You should have ##U(\omega,t) = A(\omega) e^{(i\mu\omega-k\omega^2)t}.##