# Very evil diffusion equation

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1. Sep 6, 2016

### Remixex

1. The problem statement, all variables and given/known data
$$\frac{\partial U}{\partial t}=\nu \frac{\partial^{2} U}{\partial y^{2}}$$
$$U(0,t)=U_0 \quad for \quad t>0$$
$$U(y,0)=0 \quad for \quad y>0$$
$$U(y,t) \rightarrow {0} \quad \forall t \quad and \quad y \rightarrow \infty$$
2. Relevant equations
This is a diffusion problem on fluid mechanics, but it's more of a math problem so i posted it here.

3. The attempt at a solution
I'm trying to solve this via separation of variables (the textbook uses a "similarity" method i've never seen before, and concludes the function U must be erf) is it even possible to reach an analytic result via SV?
The first boundary condition is what gets me, I tried
$$U_{0} e^{{k^{2}t}} e^{{-\frac{k}{\sqrt{\nu}}y}}$$
But it clearly doesn't work for any boundary condition except the last.
I don't think sinusoidal is the answer here either because it must eventually converge to zero, for every t.
Is there really no analytic answer?

2. Sep 7, 2016

### Orodruin

Staff Emeritus

3. Sep 7, 2016

### Ray Vickson

Use Laplace transforms with respect to $t$. Let
$$W(y,s) = \int_0^{\infty} e^{-st} U(y,t) \, dt$$
be the Laplace transform. Then, using standard properties of Laplace transforms, we get the DE
$$\nu W_{yy}(y,s) = s W(y,s) - U(y,0) = s W(y,s),$$
where $W_{yy} = \partial^2 W / \partial y^2$.
Also: $U(0,t) = U_0$ implies that
$$W(0,s) = \frac{U_0}{s}$$
Finally, the initial value theorem requires that $\lim_{s \to \infty} s W(y,s) = 0$ for $y > 0$.

These are enough to determine $W(y,s)$. Then it is just a matter of taking the inverse Laplace transform of $W(y,s)$ to get $U(y,t)$.

Separation of variables will never work in this example, simply because it leads to the wrong kind of function.

Last edited: Sep 7, 2016
4. Sep 7, 2016

THANK YOU :D