# Use the Fourier transform directly to solve the heat equation

## Homework Statement

Use the Fourier transform directly to solve the heat equation with a convection term
$$u_t =ku_{xx} +\mu u_x,\quad −infty<x<\infty,\: u(x,0)=\phi(x), assuming that u is bounded and k > 0. ## Homework Equations fourier transform inverse fourier transform convolution thm ## The Attempt at a Solution taking the FT of both sides i get [tex]U_t=-k w^2U-iw\mu U$$
$$U(0,t)=\Phi(w,0)$$
I solved the ode and got
$$U(w)=e^{(\mu i w- w^2k)t}$$
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?

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## Homework Statement

Use the Fourier transform directly to solve the heat equation with a convection term
$$u_t =ku_{xx} +\mu u_x,\quad −infty<x<\infty,\: u(x,0)=\phi(x), assuming that u is bounded and k > 0. ## Homework Equations fourier transform inverse fourier transform convolution thm ## The Attempt at a Solution taking the FT of both sides i get [tex]U_t=-k w^2U-iw\mu U$$
$$U(0,t)=\Phi(w,0)$$
Don't you mean $U(\omega,0) = \Phi(\omega,0)$?

I solved the ode and got
$$U(w)=e^{(\mu i w- w^2k)t}$$
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 $U(\omega,t) = A(\omega) e^{(i\mu\omega-k\omega^2)t}.$