- #1
member 428835
Hi PF!
I'm reading my math text and am looking at the heat eq ##u_t = u_{xx}##, where we are are given non-homogenous boundary conditions. We are solving using the method of eigenfunction expansion.
Evidently we begin by finding the eigenfunction ##\phi (x)## related to the homogenous boundary conditions. From here we say the solution takes the form ##u(x,t) = \sum a(t) \phi(x)##. When plugging this result into the heat equation we are not allowed to differentiate w.r.t. x under the sum. The reason evidently is because the boundary conditions for the actual problem and ##\phi(x)## do not agree (the problem has non-homogenous B.C. yet the eigenfunction satisfies homogenous B.C.).
Can anyone tell me why not satisfying the same B.C. conditions implies we cannot differentiate w.r.t. that variable ##x##?
Thanks so much!
I'm reading my math text and am looking at the heat eq ##u_t = u_{xx}##, where we are are given non-homogenous boundary conditions. We are solving using the method of eigenfunction expansion.
Evidently we begin by finding the eigenfunction ##\phi (x)## related to the homogenous boundary conditions. From here we say the solution takes the form ##u(x,t) = \sum a(t) \phi(x)##. When plugging this result into the heat equation we are not allowed to differentiate w.r.t. x under the sum. The reason evidently is because the boundary conditions for the actual problem and ##\phi(x)## do not agree (the problem has non-homogenous B.C. yet the eigenfunction satisfies homogenous B.C.).
Can anyone tell me why not satisfying the same B.C. conditions implies we cannot differentiate w.r.t. that variable ##x##?
Thanks so much!