# 1D heat equation

see attached

Consider the following problem

ut = 4uxx 0<x<Pi, t>0

u(0,t)=a(t), u(Pi,t)=b(t) t>0

u(x,0)=f(x) 0<x<Pi

(a) Show that the solution (which exists and is unique for reasonably nice functions f,a,b) u(x,t) is of the formâ€¨

U(x,t) = v(x,t)+(1-x/Pi) a(t)+x/Pi b(t)

where v solves a heat equation of the form vt = 4vxx + q(x, t) with homogeneous boundary conditions:

v(0,t) = v(Ï€,t) = 0 for t > 0. Determine q(x,t).

. (b) Assume a(t) â‰¡ a0,b(t) â‰¡ b0 are constant. Determine the steady state solution uE. How does this â€¨solution depend on the initial value f(x)? â€¨

. (c) Show that for large t one has u(x, t) â‰ˆ uE (t) + C eâˆ’4t sin x, for some constant C . Determine C . â€¨

https://brainmass.com/math/fourier-analysis/1d-heat-equation-624657

## SOLUTION This solution is **FREE** courtesy of BrainMass!

The equation is:

(1.1)

The boundary conditions are

(1.2)

And the initial condition is

(1.3)

If we set

(1.4)

We see that

(1.5)

And:

(1.6)

Plugging (1.4) into (1.1) we obtain:

(1.7)

â€ƒ

Then we define

(1.8)

Therefore

(1.9)

Is a solution to the original equation if

(1.10)

And

(1.11)

Now the boundary functions are constants

(1.12)

For the steady state we require and the boundary conditions still hold

(1.13)

This gives

(1.14)

â€ƒ

Applying boundary conditions

(1.15)

This solution is completely independent of the initial condition.

If the boundary functions are constants then

(1.16)

Which means that we can write the solution as:

(1.17)

And the equation system for is:

(1.18)

Boundary conditions:

(1.19)

And the initial condition is

(1.20)

We solve the homogenous equation using separation of variables. We set:

(1.21)

And plugging it into (1.18) we get:

(1.22)

The equation is now separated, so each side must be equal the same constant, specifically:

(1.23)

with the boundary conditions

(1.24)

Case 1:

The equation becomes

(1.25)

The solution is

(1.26)

Applying boundary conditions

So this is the trivial solution.

Case 2:

The equation becomes

(1.27)

The solution is

(1.28)

Applying boundary conditions

Again, this is the trivial solution.

Case 3:

The equation becomes

(1.29)

The solution is

(1.30)

Applying boundary conditions

The eigenfunctions are therefore:

(1.31)

Going back to the time-dependent equation:

(1.32)

The solution is

(1.33)

Therefore:

(1.34)

And the general solution is a linear combination of all possible solutions:

(1.35)

And from (1.17):

(1.36)

The time dependent exponents decay quite rapidly, so asymptotically we need to keep only the first term in the sum :

(1.37)

https://brainmass.com/math/fourier-analysis/1d-heat-equation-624657