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Heat in a Rod Fundamental Solution

  • #1

Homework Statement



Screen_shot_2012_05_23_at_6_35_39_PM.png


The Attempt at a Solution


So I know that I must have boundary conditions u(0,t) = 0 and ux(L,t) = 0. My textbook recommends reducing the given boundary conditions to homogeneous ones by subtracting the steady state solution. But, I thought these were already homogenous boundary conditions (are they?). Is my steady state condition v''(x) = 0, but, then by the boundary conditions I know that this must be a trivial v. Am I thinking about this incorrectly?
 

Answers and Replies

  • #2
So I thought I might have to write something of the form:
Assume the solution can be written [tex]u(x,t) = X(x)T(t)[/tex]. Thus, by the heat equation [tex]u_t = a^2 u_{xx}[/tex], we wind up with two linear differential equations. Namely, [tex]X'' + qX = 0[/tex] and [tex]T' + a^2 q T = 0[/tex]. Now I have to find which values of q make q an eigenvalue of the eigenfunction. We test three cases: q = 0; q > 0; q < 0.

q = 0:
We must have that [tex]X = C_1 x + C_2[/tex], but by the boundary conditions, this forces both of the arbitrary constants to be zero.
q < 0:
[tex]X = C_1 sinh(\sqrt{-q}x) + C_2 cosh(\sqrt{-q}x)[/tex], which implies that C_2 = 0, and, when one takes the derivative, we also have C_1 = 0 by the assumption that q was nonzero.

q > 0:
[tex]X = C_1 cos(\sqrt{q}x) + C_2 sin(\sqrt{q}x)[/tex], which implies that C_1 = 0, and, again, when one takes the derivative, we force the other constant to zero. This means that I am only getting trivial solutions here. What other approaches can I try? Or have I done something wrong?
 

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