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    The non-homogeneous heat equation

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    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.

    © BrainMass Inc. brainmass.com December 24, 2021, 11:57 pm ad1c9bdddf

    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:
    The associated homogenous system is
    We know, using separation of variables, that the homogenous system gives rise to a set of orthogonal eigenfunctions
    For brevity and generality sake we will just note that the eigenfunctions satisfy
    Where is a constant.

    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:
    Note that now the coefficient is yet-to-be defined function of t.
    When substituted back into the non-homogenous system we obtain:

    Since equation (1.7) can be written as:
    Now, if we expand in terms of the eigenfunctions:

    Equation (1.8) becomes:
    So if we equate coefficients on both sides we get a set of n first order ordinary differential equations for :
    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:
    This gives:
    So if we expand the initial condition in terms of the eigenfunctions
    Where the constants are:

    Equating the coefficients of (1.14) to that of (1.16) gives
    And these are the initial conditions to the set of the ordinary differential equations (1.12) that can be solved using integration factor:

    Where A is an integration constant.
    And when we apply the initial conditions (1.17) we get

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


    The system is with homogenous (could be mixed) boundary conditions and initial condition
    • Find the eigenfunctions of the associate homogenous system that satisfy the boundary conditions. Define

    • Expand the non-homogenous forcing term into a series of the eigenfunctions where

    • Expand the initial condition in terms of the eigenfunctions where

    • Solve the set of differential equations with the initial condition The solution of such a differential equation is

    • The solution is

    Special case: Time independent forcing term
    If , that is the forcing term is time independent, then is a constant.
    This gives:



    We see that indeed the initial condition is satisfied:
    And as we get the steady state:

    Recall that the steady state must satisfy
    And from (1.28) we get

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

    Converting a non-homogenous equation with time-independent forcing term to a homogenous equation.
    The most general case with homogenous boundary conditions is:
    where are constants, and may be time dependent functions.
    We solve for the steady state solution
    That satisfies the above boundary conditions:
    Then we define
    When we apply it to the equation we get:
    But from (1.32) we get


    The first boundary condition becomes

    But since satisfies the original boundary conditions then we get:
    And by the same token:
    The initial condition is now:

    So the system now becomes homogenous:

    To Recap:
    To convert the system

    To a homogeneous system:
    • Find the steady state that solves and satisfies the boundary conditions and
    • Define
    • Define
    The system becomes

    The solution to the original system is

    the system
    First, we need to solve for the steady state:
    Integrating once:

    Integrating twice:
    Applying boundary conditions:

    So if we set
    We get a homogenous equation with respect to
    The boundary conditions for are


    So far we have
    All that is left is to transform the initial condition:

    The system is now:
    Which is completely homogeneous.
    And the solution to the original system is

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