Need help. What is the answer? Thank you sir.

Question

Find the solution the following heat equation ∂u/dt = 4 ∂²u/dx², 0<x<1, t>0

with boundary conditions u(0,t)=0,u(1,t)=0, t>0

and initial condition u(x,0)=x(1-x), 0<x<1

Partial Differential Equations

Answer 1

Note: Make sure to double check my algebra on this. There are also some tricky steps that you should review in detail.

Using the technique separation of variables we can let u = T(t)X(x).

∂/∂t T X = 4 ∂²/∂x² T X

X ∂/∂t T = 4 T ∂²/∂x² X

1/T ∂/∂t T = 4/X ∂²/∂x² X

1/T ∂/∂t T = -4k², 4/X ∂²/∂x² X = -4k²

The boundary conditions for position are periodic so I chose the constant to make the solution easier.

This gives the solutions

T = A exp(-4k²t)

X = B sin(k x) + C cos(kx).

u = T X

If u(0,t) = 0, then

X(0) = C = 0

If u(1,t) = 0, then

X(1) = B sin(k), so k = nπ.

This gives the series solution

u = ∑ B_n exp(-4 n² π² t) sin( n π x).

To find B_n we can use the Fourier Trick.

∫₀¹ sin(n π x)sin(m π x) dx = 1/2 if n=m and 0 if n≠m

Given u(x,0) = =x(1-x)

∑ B_n sin( n π x) = x(1-x)

∫₀¹ ∑ B_n sin( n π x) sin(m π x) dx = ∫₀¹ x(1-x) sin(m π x) dx

B_m = ∫₀¹ x(1-x) sin(m π x) dx = 4(1-(-1)^m)/(π m)³.

Giving a solution,

u = ∑_m=1..∞ 4(1-(-1)^m)/(π m)³ exp(-4 n² π² t) sin( n π x)