Question

Solve

u_t = u_xx 0<x<l, t>0

u(0, t)= u(l, t)= 0

u(x, 0)= x(l-x) , 0<=x<=l

Answer 1

Solution

Solution can be found in the form  u(x,t) = X(x)*T(t)

X’’/X = T’/T  => X’’/X = T’/T = -λ²  

X’’+ λ²X = 0  =>  X = Asin(λ*x) + Bcos(λ*x) . Here A,B are arbitrary constants.

From boundary condition u(0, t) = 0  => B = 0  

From boundary condition u(l, t) = 0  => sin(λ*x) = 0  =>  λ = π*n/l (n = 1,2, …)  

So  X = Asin(π*n *x/l)

Equation on T gives now  T’ + λ² T = 0  =>  T = exp(- λ² t) = exp(- (π*n/l)² t)   

And u_n = A_n* exp(- (π*n/l)² t)* sin(π*n *x/l)

$u(x,t)=\displaystyle\sum_{n=1}^\infty A_n e^{-(\pi n/l)^2t} sin(\pi n \frac{x}{l})$

From initial condition u(x, 0)= x(l-x) , 0<=x<=l

$x(l-x)=\displaystyle\sum_{n=1}^\infty A_n sin(\pi n \frac{x}{l})$

$A_n=\frac{\displaystyle\int_0^l x(l-x) sin(\pi n x/l) dx}{\displaystyle\int_0^l sin^2(\pi n x/l) dx}$

$A_n= \frac{2}{l} \displaystyle\int_0^l x(l-x) sin(\pi n x/l) dx$

Let y = x/l => $A_n= 2l^2 \displaystyle\int_0^1 y(1-y) sin(\pi n y) dy$

Let z = π*n *y  => $A_n= \frac{2l^2}{(\pi n)^2} \displaystyle\int_0^{\pi n} z(1-\frac{z}{\pi n}) sin(z) dz$

$A_n=\frac{2l^2}{(\pi n)^2} [sin(z)-z cos(z) -\frac{1}{\pi n} (2z sin(z)-(z^2-2) cos(z))] |_0^{\pi n}$

$A_n​=\frac{2l^2}{(\pi n)^2} [\pi n (-)^{n+1}+\frac{1}{\pi n}((\pi n)^2-2)(-)^n+\frac{2}{\pi n}]$

$A_n​=\frac{4l^2}{(\pi n)^3} [(-)^{n+1}+1]$

A_2m = 0, A_2m+1 = 8l²/(π*(2m+1))³   (m = 0,1, …)

$u(x,t)=8l^2\displaystyle\sum_{m=0}^\infty \frac{1}{(\pi (2m+1))^3} e^{-(\pi (2m+1)/l)^2t} sin(\pi (2m+1) \frac{x}{l})$

Answer

$u(x,t)=8l^2\displaystyle\sum_{m=0}^\infty \frac{1}{(\pi (2m+1))^3} e^{-(\pi (2m+1)/l)^2t} sin(\pi (2m+1) \frac{x}{l})$