# Math Help - PDE Initial Condition Question

1. ## PDE Initial Condition Question

Hello,
I have the following PDE:

$u_t = k u_xx$

with $u(x,0) = f(x), u_x(0,t) = u_x(L,t) = 0$

which I solve using separation of variables by letting $u = XT$.

Now, for the conditions we have

$u_x(0,t) = X'(0) T(t) = 0$

so $X'(0) = 0$ and

$u_x(L,t) = X'(L) T(t) = 0$

$X'(L) = 0$

My question is why can't $T(t) = 0$ in these cases? This assumption is made in every PDE I've solved using sep of vars, but I would like to know why exactly this is.
Thanks

2. Well, T(t) = 0, when plugged back into your separation guess of u = XT, implies that u = 0, the trivial solution. The trivial solution, I think you'll find, will not work unless f(x) = 0. In addition, most of the time you're interested in a nontrivial solution, which implies that T(t) is not identically zero everywhere. Make sense?