- #1
Gekko
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Homework Statement
du/dt=d2u/dx2, u(0,t)=0, u(pi,t)=0
u(x,0) = sin^2(x) 0<x<pi
Find the solution
Also find the solution to the initial condition:
du/dt u(x,0) = sin^2(x) 0<x<pi
The Attempt at a Solution
From separation of variables I obtain
u(x,t) = B.e^(-L^2t).sin(Lx)
For the boundary condition u(pi,t)=0, u(x,t) = Bm . e^(-m^2t) . sin(mx)
Finally for u(x,0) = sin^2x
u(x,0) = 4(cos(m.pi)-1).e^(-m^2t).sinmx / pi(m^3-4m)
found through superposition and Fourier sum of sines
What I don't see is how to solve the initial condition because when I diff. above wrt t and set t-0 I obtain
du/dt u(x,0) = -m^2 4(cos(m.pi)-1).e^(-m^2t).sinmx / pi(m^3-4m) = sin^2x
How can this be set to sin^2x by choosing a value of m?