SolutionSuppose
u(x,t)=X(x)T(t)∂x∂u∂x2∂2u=X′T∂t∂u=X′′T=XT′ XT′−161X′′TXT′16XX′′=0=161X′′T=TT′=−λ X′′+16λXT′+λT=0=0 X′′=0⇒X(x)=C1x+C2u(0,t)=0,X(0)=0=C2u(1,t)=0,X(1)=C1=0 ⇒ This is a trivial solution.
Case 2: λ=−α2<0 X′′−16α2X=0X(x)=C1cosh(4αx)+C2sinh(4αx)u(0,t)=0,u(1,t)=0,C1=0C2sinh4α=0 Since
sinh4α=0,C2=0⇒ This is a trivial solution
Case 3:
X′′+16α2X=0X(x)=C1cos(4αx)+C2sin(4αx)u(0,t)=0,u(1,t)=0,C1=0C2=sin4α=0 To obtain non-trivial solution,
sin4α=04α=nπ,α=4nπ,n=1,2,3,...n=1,2,3,... Eigenvalue:
λ=α2=16n2π2,n=1,2,3,... Eigenfunction:
Xn(x)=C2sin((4nπ)x)=C2sin(nπx) T′+α2T=0T(t)Tn(t)=C3e−α2t=C3e−16n2π2t un(x,t)=XnTn=Bne−16n2π2tsin(nπx),n=1,2,3,... By superposition principle,
u(x,t)=n=1∑∞Bne−16n2π2tsin(nπx) Given
u(x,0)=2sin(2πx)2sin(2πx)Bn=n=1∑∞Bnsin(nπx)=2∫012sin(2πx)sin(nπx)dx ∫012sin(2πx)sin(nπx)dx={1,0,n=2otherwise Bn={2,0,n=2otherwise ∴u(x,t)=B2e−4π2tsin(2πx)=2e−4π2tsin(2πx)