Question #88293
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)U(x, t)=X(x)T(t)

Then


Ut=XT,Uxx=XTU_t=XT', U_{xx}=X''T

Substitute


XT=XTXT'=X''T

Dividing this equation by XTXT, we have


TT=XX=λ\frac{T'}{T}=\frac{X''}{X}=-\lambda

for some constant λ\lambda. Therefore, if there exists a solution U(x,t)=X(x)T(t)U(x, t)=X(x)T(t) of the heat equation, then X(x)X(x) and T(t)T(t) must satisfy the equations 

TT=λ\frac{T'}{T}=-\lambda

XX=λ\frac{X''}{X}=-\lambda

Solve the eigenvalue problem


X+λX=0,X(0)=X(1)=0X''+\lambda X=0, X(0)=X(1)=0

We have to find nontrivial solutions X of the eigenvalue problem. If λ=0, then X(x)=αx+βX(x)=\alpha x+\beta and 0=X(0)=0+β0=X(0)=0+\beta implies that β=0\beta=0 and 0=X(1)=α0=X(1)=\alpha implies that α=0\alpha=0. If λ<0\lambda<0 then 0=X(0)=αcosh(0)+βsinh(0)=α0=X(0)=\alpha\cosh(0)+\beta\sinh(0)=\alpha and 0=X(1)=0+βsinh(1)=βsinh(1)0=X(1)=0+\beta\sinh(1)=\beta\sinh(1) shows that also β=0\beta=0. We conclude that λ0\lambda\leq0  is not an eigenvalue of the problem. Finally, consider λ>0.\lambda>0. Then 0=X(0)=αcos(λ0)+0=α0=X(0)=\alpha\cos(\sqrt{\lambda}\cdot0)+0=\alpha and 0=X(1)=βsin(λ1).0=X(1)=\beta\sin(\sqrt{\lambda}\cdot1).

Since XX is nontrivial solution, β=\beta=\not0 and hence sin(λ1)=0.\sin(\sqrt{\lambda}\cdot1)=0. Consequently,


λ=(nπ1)2,n1ππ\lambda=(\frac{n\pi}{1})^2, n\geq1\pi\pi

and the corresponding eigenfunction is given by


Xn(x)=sin(nπx1)X_n(x)=\sin(\frac{n\pi x}{1})

After substituting λ=(nπ1)2\lambda=(\dfrac{n\pi }{1})^2 , we get the family of solutions


Tn(t)=Bnek(nπ1)2tT_n(t)=B_ne^{-k({n\pi \over 1})^2t}

Thus we have obtained the following sequence of solutions


Un(x,t)=Xn(x)Tn(t)=Bnsin(nπx1)ek(nπ1)2tU_n(x, t)=X_n(x)T_n(t)=B_n\sin(\frac{n\pi x}{1})e^{-k({n\pi \over 1})^2t}

We obtain more solutions by taking linear combinations of the UnU_n ’s ( recall the superposition principle) 


U(x,t)=n=1NXn(x)Tn(t)=n=1NBnsin(nπx1)ek(nπ1)2tU(x, t)=\displaystyle\sum_{n=1}^NX_n(x)T_n(t)=\displaystyle\sum_{n=1}^NB_n\sin(\frac{n\pi x}{1})e^{-k({n\pi \over 1})^2t}

and then by passing to the limit N,N\to \infin,


U(x,t)=n=1Xn(x)Tn(t)=n=1Bnsin(nπx1)ek(nπ1)2tU(x, t)=\displaystyle\sum_{n=1}^{\infin}X_n(x)T_n(t)=\displaystyle\sum_{n=1}^{\infin}B_n\sin(\frac{n\pi x}{1})e^{-k({n\pi \over 1})^2t}

Finally, we consider the initial condition. At t=0t=0, we must have


U(x,0)=n=1Bnsin(nπx1)=x(1x)U(x, 0)=\displaystyle\sum_{n=1}^{\infin}B_n\sin(\frac{n\pi x}{1})=x(1-x)

Bn=2101x(1x)sin(nπx1)dxB_n={2 \over 1}\displaystyle\int_{0}^1x(1-x)\sin(\frac{n\pi x}{1})dx

xsin(nπx)dx=1πnxcos(nπx)+1πncos(nπx)dx=\int xsin(n\pi x)dx=-{1 \over \pi n}xcos(n\pi x)+{1 \over \pi n}\int cos(n\pi x)dx=

=1πnxcos(nπx)+1π2n2sin(nπx)+C1=-{1 \over \pi n}xcos(n\pi x)+{1 \over \pi^2 n^2}sin(n\pi x)+C_1

x2sin(nπx)dx=1πnx2cos(nπx)21πnxcos(nπx)dx-\int x^2sin(n\pi x)dx={1 \over \pi n}x^2cos(n\pi x)-2{1 \over \pi n}\int xcos(n\pi x)dx

21πnxcos(nπx)dx=21π2n2xsin(nπx)+21π2n2sin(nπx)dx=-2{1 \over \pi n}\int xcos(n\pi x)dx=-2{1 \over \pi^2 n^2}xsin(n\pi x)+2{1 \over \pi^2 n^2}\int sin(n\pi x)dx=

=21π2n2xsin(nπx)21π3n3cos(nπx)+C2=-2{1 \over \pi^2 n^2}xsin(n\pi x)-2{1 \over \pi^3 n^3}cos(n\pi x)+C_2

Bn=2[1πnxcos(nπx)+1π2n2sin(nπx)]10+B_n=2[-{1 \over \pi n}xcos(n\pi x)+{1 \over \pi^2 n^2}sin(n\pi x)]\begin{matrix} 1 \\ 0 \end{matrix}+

+2[1πnx2cos(nπx)2π2n2xsin(nπx)2π3n3cos(nπx)]10=+2[{1 \over \pi n}x^2cos(n\pi x)-{2 \over \pi^2 n^2}xsin(n\pi x)-{2 \over \pi^3 n^3}cos(n\pi x)]\begin{matrix} 1 \\ 0 \end{matrix}=

=2πncos(nπ)+2πncos(nπ)4π3n3cos(nπ)+4π3n3==-{2 \over \pi n}cos(n\pi )+{2 \over \pi n}cos(n\pi )-{4 \over \pi^3 n^3}cos(n\pi )+{4 \over \pi^3 n^3}=


=4(1(1)n)π3n3={4(1-(-1)^n) \over \pi^3 n^3}

We arrive at the solution


U(x,t)=n=1Xn(x)Tn(t)=n=14(1(1)n)π3n3sin(nπx1)ek(nπ1)2tU(x, t)=\displaystyle\sum_{n=1}^{\infin}X_n(x)T_n(t)=\displaystyle\sum_{n=1}^{\infin}{4(1-(-1)^n) \over \pi^3 n^3}\sin(\frac{n\pi x}{1})e^{-k({n\pi \over 1})^2t}

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