- #1
eckiller
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I am to reduce the following PDE to 2 ODEs and find only the particular solutions:
u_tt - u_xx - u = 0; u_t(x,0) = 0; u(0,t) = u(1,t) = 0
I guess u = X(x)T(t), and plug u_tt, u_xx into PDE and divide by u to get:
T''/T = X''/X + 1 = K
I solve X'' + (1-K)X = 0 first.
From characteristic equation, r = +- sqrt(1-K)
Becausse of boundary conditions, we must have 1-K < 0
So r = +- i sqrt(K-1)
(I think I make an error somewhere around here...)
==> u = c1 cos(sqrt(K-1)x) + c2 sin(sqrt(K-1)x)
From boundary conditions, c1 =0, and sin(sqrt(K-1)) = 0 => sqrt(K-1) = n*pi
=> K-1 = n^2*pi^2
=> K = n^2*pi^2 + 1
Correct so far?
Now for T:
T'' - KT = 0
T'' - (n^2*pi^2 + 1)T = 0
r = +- sqrt(n^2*pi^2+1)
I think my K = n^2*pi^2 + 1 is wrong because it is strictly positive and I don't think an e^t solution will satisfy the initial condition.
Please help.
The book's answer is cos( sqrt(n^2*pi^2-1)t) * sin(n*pi*x)
u_tt - u_xx - u = 0; u_t(x,0) = 0; u(0,t) = u(1,t) = 0
I guess u = X(x)T(t), and plug u_tt, u_xx into PDE and divide by u to get:
T''/T = X''/X + 1 = K
I solve X'' + (1-K)X = 0 first.
From characteristic equation, r = +- sqrt(1-K)
Becausse of boundary conditions, we must have 1-K < 0
So r = +- i sqrt(K-1)
(I think I make an error somewhere around here...)
==> u = c1 cos(sqrt(K-1)x) + c2 sin(sqrt(K-1)x)
From boundary conditions, c1 =0, and sin(sqrt(K-1)) = 0 => sqrt(K-1) = n*pi
=> K-1 = n^2*pi^2
=> K = n^2*pi^2 + 1
Correct so far?
Now for T:
T'' - KT = 0
T'' - (n^2*pi^2 + 1)T = 0
r = +- sqrt(n^2*pi^2+1)
I think my K = n^2*pi^2 + 1 is wrong because it is strictly positive and I don't think an e^t solution will satisfy the initial condition.
Please help.
The book's answer is cos( sqrt(n^2*pi^2-1)t) * sin(n*pi*x)