PDEs

Can someone please help me with part b of the question? I applied the initial condition but it’s not getting me anywhere. What am I doing wrong? If I apply T(0)=0 then alpha = 0, which means there is a no equation for T thus we don’t have U(y,t).

You sure the question has no typos ? I think there is something erased in front of the time derivative. That coefficient could be important and missing it leads to trivial solution of this problem given the initial condition.

Anyone has any thoughts on how to go about solving part b?

Yes, the x^2 term was with u_t but the lecturer told me it should be with u_xx.

In that case this problem has no meaningful solution. The initial condition cannot be zero. The boundaries are OK, those can be zero.

Look, if your PDE is of the form $u_t = f(u_y,u_{yy})$ then using separation of variables $u(y,t) = Y(y) T(t)$ puts the PDE in the form

$\dfrac{T'}{T} = \tilde{f}(Y,Y',Y'') = -\omega^2$

for some $\omega \in \mathbb{R}$.

However, ignoring the $Y$ entirely, just focus on the $T$ part which gives

$T = Ae^{-\omega^2 t}$

for arbitrary constant $A$.

The fact that the initial data is $u(x,0) = 0 = Y(\ln x) T(0) \iff T(0) = 0$ implies $A = 0$ must be the only way.

But as you saw, this then means $u(y,t) = YT = 0$ (trivial solution).

The initial condition must be something other than 0 for this PDE to have a non-trivial solution.

Check with your lecturer if this is really the initial condition they meant to put down. If the typo was made at PDE level as you say, then it is not unreasonable to suspect this IC to be a typo as well.