Question #93157
Solve the following boundary value problem.

uₙ = uₓₓ , 0 < x < L , n > 0

u(0,n) = u(L,n) = 0

u(x,0) = x(L-x) , 0 ≤ x ≤ L
1
Expert's answer
2019-08-28T05:14:08-0400

Let's assume that the solution will take the form

u=φ(x)G(t)u=\varphi(x)G(t)

Substituting this from in the equation, will lead us to

φ(x)G(t)=φ(x)G(t)\varphi(x)G'(t)=\varphi''(x)G(t)

G(t)G(t)=φ(x)φ(x)\frac{G'(t)}{G(t)}=\frac{\varphi''(x)}{\varphi(x)}

Since the left hand side depends only of t and the right hand side only of x, the both sides can be equal only if they are equal to some constant.

Let's denote this constant as λ-\lambda

Then

G(t)G(t)=φ(x)φ(x)=λ\frac{G'(t)}{G(t)}=\frac{\varphi''(x)}{\varphi(x)}=-\lambda

Therefore we obtain two ordinary differential equations

G+λG=0G'+\lambda G=0

φ+λφ=0\varphi''+\lambda\varphi=0

Using boundary condition will lead us to

φ(0)=φ(L)=0\varphi(0)=\varphi(L)=0

The general solution of the equation is

φ=c1cos(λx)+c2sin(λx)\varphi=c_1\cos(\sqrt{\lambda}x)+c_2\sin(\sqrt{\lambda}x)

Applying the first boundary condition gives us

φ(0)=c1=0\varphi(0)=c_1=0

Applying the second boundary condition, and using the above result, gives us

φ(L)=c2sin(λL)=0\varphi(L)=c_2\sin(\sqrt{\lambda}L)=0

We will only have a non-trivial solution, if sin(λL)=0\sin(\sqrt{\lambda}L)=0 , λ0.\lambda \not =0.

Hence

λn=(πnL)2,nZ,n0.\lambda_n=\left(\frac{\pi n}{L}\right)^2,\quad n\in \Zeta, n \not =0.

Then for G(t) we will have

G=λnGG'=-\lambda_nG

This is a linear first order differential equation and its solution is

G=BneλntG=B_ne^{-\lambda_nt }

Hence the solution is

u=n=1Bneλntsinλxu=\sum\limits_{n=1}^{\infty}B_ne^{-\lambda_nt }\sin\sqrt{\lambda}x

The initial condition will lead us to

u(x,0)=n=1Bnsinλx=x(Lx)u(x,0)=\sum\limits_{n=1}^{\infty}B_n\sin\sqrt{\lambda}x=x(L-x)

which is the sine Fourier series of x(Lx)x(L-x)


It's coefficients can be obtained using the formula

Bn=2L0Lx(Lx)sin(πnL)xdx=2L2(2+2cos(nπ)+nπsin(nπ))n3π3=4L2((1)n1)n3π3B_n= \frac{2}{L}\int\limits_{0}^{L}x(L-x)sin(\frac{\pi n}{L})xdx=-\frac{2 L^2 (-2 + 2 cos(n π) + n π sin(n π))}{n^3 π^3}=-\frac{4 L^2((-1)^n-1)}{n^3 π^3}

Hence

B2n=0B_{2n}=0

B2n1=8L2(2n1)3π3B_{2n-1}=\frac{8 L^2}{(2n-1)^3 π^3}

Therefore we obtain the final form of the function

u=n=18L2(2n1)3π3eλ2n1tsinλ2n1xu=\sum\limits_{n=1}^{\infty}\frac{8 L^2}{(2n-1)^3 π^3}e^{-\lambda_{2n-1}t }\sin\sqrt{\lambda_{2n-1}}x




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