# Thread: di ffusion equation

1. ## di ffusion equation

Good morning all, I am stuck on this question.

Given u(x,t)=f(x-at) is a solution to the diffusion equation, find f and show that the speed is arbitrary.
Thanks for any help at all guys and girls.

2. ## Re: di

Did you mean wave equation? If so, I'm not sure you can determine f uniquely. You're actually exhibiting one of the two d'Alembert solutions, where f is arbitrary, but a is not. The other d'Alembert solution looks like f(x+at).

3. ## Re: di

Originally Posted by Ackbeet
Did you mean wave equation? If so, I'm not sure you can determine f uniquely. You're actually exhibiting one of the two d'Alembert solutions, where f is arbitrary, but a is not. The other d'Alembert solution looks like f(x+at).
No, I mean the diffusion equation $\displaystyle u_t-ku_{xx}=0$. That's why this question confuses me.

4. ## Re: di

Hmm. Well, I'm not sure I see my way through to the end of the problem, but plug in your f into the diffusion equation. What do you get?

5. ## Re: di

I get $\displaystyle -a \cdot f'(x-at) -k \cdot f''(x-at)=0$,
so $\displaystyle f'(x-at)= - \frac{a}{k}$

I think I am confusing myself.

6. ## Re: di

So now suppose s = x - at. You have an ODE for f, right? I agree with your first equation, but not the second. How did you get that?

7. ## Re: di

Sorry, that second was a mistake, makes no sense.

-kf''(s)-af'(s)=0
kf''(s)+af'(s)=0

So the characteristic equation is kz^2+az=0
which gives z=0 or z=-a/k

Is this correct?

8. ## Re: di

Originally Posted by kittycat485
Sorry, that second was a mistake, makes no sense.

-kf''(s)-af'(s)=0
kf''(s)+af'(s)=0

So the characteristic equation is kz^2+az=0
which gives z=0 or z=-a/k

Is this correct?
Yes, so far.

So the solution is...

9. ## Re: di

Originally Posted by Ackbeet
Yes, so far.

So the solution is...
got it. thanks.

10. ## Re: di

You're very welcome. Have a good one!