Solve the following boundary value problem
Ut=Uxx,0<x<l,t>0
U(0,t)=U(l,t)=0
U(x,0)=x(l-x),0<=x<=1
1
Expert's answer
2019-04-22T10:22:20-0400
Separate variables. Let
U(x,t)=X(x)T(t)
Then
Ut=XT′,Uxx=X′′T
Substitute
XT′=X′′T
Dividing this equation by XT, we have
TT′=XX′′=−λ
for some constant λ. Therefore, if there exists a solution U(x,t)=X(x)T(t) of the heat equation, then X(x) and T(t) must satisfy the equations
TT′=−λ
XX′′=−λ
Solve the eigenvalue problem
X′′+λX=0,X(0)=X(1)=0
We have to find nontrivial solutions X of the eigenvalue problem. If λ=0, then X(x)=αx+β and 0=X(0)=0+β implies that β=0 and 0=X(1)=α implies that α=0. If λ<0 then 0=X(0)=αcosh(0)+βsinh(0)=α and 0=X(1)=0+βsinh(1)=βsinh(1) shows that also β=0. We conclude that λ≤0 is not an eigenvalue of the problem. Finally, consider λ>0. Then 0=X(0)=αcos(λ⋅0)+0=α and 0=X(1)=βsin(λ⋅1).
Since X is nontrivial solution, β=0 and hence sin(λ⋅1)=0. Consequently,
λ=(1nπ)2,n≥1ππ
and the corresponding eigenfunction is given by
Xn(x)=sin(1nπx)
After substituting λ=(1nπ)2 , we get the family of solutions
Tn(t)=Bne−k(1nπ)2t
Thus we have obtained the following sequence of solutions
Un(x,t)=Xn(x)Tn(t)=Bnsin(1nπx)e−k(1nπ)2t
We obtain more solutions by taking linear combinations of the Un ’s ( recall the superposition principle)
Comments