# The non-homogeneous heat equation

Concerning heat flow

I am confused about turning a non homogeneous equation (heat generation) into a homogeneous equation; could this process be explained in detail with an example....i unfortunately need this by noon on Thursday (EDT)

Thank You.

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

Any homogenous heat equation with non-homogenous DirichletNeumann boundary conditions can be easily converted to a non-homogenous heat equation with homogeneous DirichletNeumann boundary conditions.

So we shall look at the general non-homogenous heat equation with Dirichlet BC.

In this case, the system looks like:

(1.1)

The associated homogenous system is

(1.2)

We know, using separation of variables, that the homogenous system gives rise to a set of orthogonal eigenfunctions

(1.3)

For brevity and generality sake we will just note that the eigenfunctions satisfy

(1.4)

Where is a constant.

Furthermore

(1.5)

For example, in the context of homogenous Dirichlet BC we see that and

And usually at this point we would find the expansion of the initial condition in terms of the eigenfunctions and equate coefficients.

However, in this case we have to look at the non-homogenous system.

We now assume that the non-homogenous solution is in the form:

(1.6)

Note that now the coefficient is yet-to-be defined function of t.

When substituted back into the non-homogenous system we obtain:

(1.7)

Since equation (1.7) can be written as:

(1.8)

Now, if we expand in terms of the eigenfunctions:

(1.9)

â€ƒ

Where

(1.10)

Equation (1.8) becomes:

(1.11)

So if we equate coefficients on both sides we get a set of n first order ordinary differential equations for :

(1.12)

Since this is a first order equation we need an initial condition. This can only comes from the initial condition of the original heat equation system:

(1.13)

This gives:

(1.14)

So if we expand the initial condition in terms of the eigenfunctions

(1.15)

Where the constants are:

(1.16)

â€ƒ

Equating the coefficients of (1.14) to that of (1.16) gives

(1.17)

And these are the initial conditions to the set of the ordinary differential equations (1.12) that can be solved using integration factor:

(1.18)

Where A is an integration constant.

And when we apply the initial conditions (1.17) we get

(1.19)

â€ƒ

So the general solution of the non-homogenous equation with homogenous boundary conditions

(1.20)

is:

(1.21)

Where

(1.22)

And:

(1.23)

And

(1.24)

â€ƒ

Recap

The system is with homogenous (could be mixed) boundary conditions and initial condition

â€¢ Find the eigenfunctions of the associate homogenous system that satisfy the boundary conditions. Define

â€¢ Expand the non-homogenous forcing term into a series of the eigenfunctions where

â€¢ Expand the initial condition in terms of the eigenfunctions where

â€¢ Solve the set of differential equations with the initial condition The solution of such a differential equation is

â€¢ The solution is

â€ƒ

Special case: Time independent forcing term

If , that is the forcing term is time independent, then is a constant.

This gives:

(1.25)

And

(1.26)

We see that indeed the initial condition is satisfied:

(1.27)

And as we get the steady state:

(1.28)

â€ƒ

Recall that the steady state must satisfy

(1.29)

And from (1.28) we get

(1.30)

Which is the definition of the expansion of the forcing term. Also since satisfies the boundary conditions, so is the steady state.

â€ƒ

Converting a non-homogenous equation with time-independent forcing term to a homogenous equation.

The most general case with homogenous boundary conditions is:

(1.31)

where are constants, and may be time dependent functions.

We solve for the steady state solution

(1.32)

That satisfies the above boundary conditions:

(1.33)

Then we define

(1.34)

When we apply it to the equation we get:

(1.35)

But from (1.32) we get

(1.36)

Thus:

(1.37)

The first boundary condition becomes

(1.38)

But since satisfies the original boundary conditions then we get:

(1.39)

And by the same token:

(1.40)

The initial condition is now:

(1.41)

So the system now becomes homogenous:

(1.42)

To Recap:

To convert the system

To a homogeneous system:

â€¢ Find the steady state that solves and satisfies the boundary conditions and

â€¢ Define

â€¢ Define

The system becomes

The solution to the original system is

Example:

the system

(1.43)

First, we need to solve for the steady state:

(1.44)

Integrating once:

Integrating twice:

(1.45)

Applying boundary conditions:

(1.46)

Or:

(1.47)

So if we set

(1.48)

We get a homogenous equation with respect to

(1.49)

The boundary conditions for are

(1.50)

And:

(1.51)

So far we have

(1.52)

All that is left is to transform the initial condition:

(1.53)

The system is now:

(1.54)

Which is completely homogeneous.

And the solution to the original system is

(1.55)

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