# Non homogeneous 1D heat equation

ut = 3uxx + 2, 0 < x < 4, t > 0,

u(0,t) = 0, u(4,t) = 0, t = 0

u(x,0) = 5sin2Ï€x,0 < x < 4.

(a) Find the steady state solution uE(x)

(b) Find an expression for the solution.

(c) Verify, from the expression of the solution, that limtâ†’âˆž u(x, t) = uE (x)

for all x, 0 < x < 4.

https://brainmass.com/math/fourier-analysis/non-homogeneous-heat-equation-624667

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

The eqaution is:

(1.1)

With boundary condition

(1.2)

And initial conditions

(1.3)

The steady state is, by definition:

(1.4)

The function is now only x-dependent so the partial derivatives become full derivatives and the ordinary differential equation is

(1.5)

And the boundary conditions must still hold:

(1.6)

Integrating (1.5) twice we obtain:

(1.7)

â€ƒ

Applying boundary conditions we get the steady state solution.

(1.8)

We would like to turn the system into a homogenous system.

So we write:

(1.9)

When we apply it to the original equation we get:

(1.10)

If we set

(1.11)

the equation becomes homogenous.

For the boundary conditions we get:

If we set

(1.12)

The boundary conditions become

(1.13)

So if we combine (1.11) and (1.130 we see that the equation for is the same as that of the steady state, hence:

(1.14)

So now the initial conditions are:

(1.15)

Or:

(1.16)

To find the general solution we use separation of variables, but first we find the eigenfunctions of the homogenous system:

(1.17)

We set

(1.18)

And plugging it into (1.9) we get:

(1.19)

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

(1.20)

with the boundary conditions

(1.21)

Case 1:

The equation becomes

(1.22)

The solution is

(1.23)

Applying boundary conditions

So this is the trivial solution.

Case 2:

The equation becomes

(1.24)

The solution is

(1.25)

â€ƒ

Applying boundary conditions

Again, this is the trivial solution.

Case 3:

The equation becomes

(1.26)

The solution is

(1.27)

Applying boundary conditions

The eigenfunctions are therefore:

(1.28)

Going back to the time-dependent equation:

(1.29)

where

The solution is

(1.30)

Therefore:

(1.31)

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

(1.32)

When we apply the initial condition we get:

(1.33)

To find we expand in terms of the eigenfunctions , which according to Fourier sine series we can write:

(1.34)

Where

(1.35)

Integration by parts

(1.36)

Hence:

And:

therefore:

(1.37)

So we have:

(1.38)

Note that

(1.39)

Therefore only the odd indexed terms survive:

(1.40)

Going back to (1.33):

(1.41)

Since the series expansion is unique, we must have the coefficients of the eigenfunctions on both sides to be identical. All the even-indexed terms except that that corresponds to must vanish. This gives:

(1.42)

And when we bring it into the general expression for

(1.43)

As all the time dependent exponentials disappear and we have

(1.44)

Going back to (1.9)

(1.45)

(1.46)

And we see that indeed:

(1.47)

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