Consider the Heat Equation： du/dt=k(d2u/dx2), where d is a partial and d2 is the second partial. The B.C.'s are u_x(0,t)=u(0,t) and u_x(L,t)=u(L,t), where u_x is the partial of u with respect to x. The I.C is u(x,0)=f(x)
Now, consider the Boundary Value Problem X''(x)=-lambda*X(x), note the negative sign, with B.C's X'(0)=X(0) and X'(L)=X(L), where L is the length of a 1D rod(at the very end of the rod).
Find the eigenvalues and eigenfunctions for lambda>0, lambda=0, and lambda<0
The Attempt at a Solution
lambda=0 was an easy one, lambda=0, gave an eigenfunction X=0.
lambda<0, wasn't too bad either. Since lambda < 0 I get X=Acosh(sqrt(-lambda)*x)+Bsinh(sqrt(-lambda)*x), where A and B are constants. Also, note that lambda is negative, thus the square root will not give any complex numbers. Using the first initial condition X'(0)=X(0), I get that A=sqrt(-lambda)*B and plugging back in again, using the second BC, I get that lambda=-1 and thus the eigenfunction is X=Bcosh(x)+Bsinh(x).
lambda>0 is the one I'm having trouble with. If do what I did for lambda<0, and use X=Acos(sqrt(lambda)*x)+Bsin(sqrt(lambda)*x), I get that lambda is -1, again, however, lambda is suppose to be greater than 0, and the eigenfunction is complex. Any help would be appreciated. Thanks.