# PDE Initial Condition Question

• Nov 30th 2010, 02:46 AM
MelodicRhapsody
PDE Initial Condition Question
Hello,
I have the following PDE:

\$\displaystyle u_t = k u_xx\$

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

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

Now, for the conditions we have

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

so \$\displaystyle X'(0) = 0\$ and

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

\$\displaystyle X'(L) = 0\$

My question is why can't \$\displaystyle 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
• Nov 30th 2010, 02:57 AM
Ackbeet
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?