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    Fourier methods in one dimension

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    Using the method of separation of variables, solve the partial differential equation
    u subscript(tt)+2(pi)u subscript(t)-u subscript(xx)=-3sin(3(pi)x)
    for 0 less than or equal to x less than or equal to 1
    with boundary conditions u(0,t)=u(1,t)=0
    and initial conditions u(x,0)=u subscript (t)(x,0)=0

    © BrainMass Inc. brainmass.com October 5, 2022, 6:26 pm ad1c9bdddf
    https://brainmass.com/math/fourier-analysis/fourier-methods-one-dimension-477689

    SOLUTION This solution is FREE courtesy of BrainMass!

    Expansion of a function into a series of orthogonal eigenfunctions.
    We define a basis of orthogonal functions with respect to a weight function over the interval as:
    (1.1)
    Where A is some constant.

    We can expand the function into a series of the orthogonal functions:
    (1.2)
    The question is to find the set of expansion coefficients
    For this we multiply both sides by :
    (1.3)
    Since the summation is over n and not m, we can bring the term into the sum:
    (1.4)
    Now we integrate both sides:
    (1.5)
    Integration of a sum is the same as sum of integrals, and since are constants:
    (1.6)
    According to the orthogonal relations (1.1) we see that the only term that is not zero on the right hand side sum is the term where and we are left with:

    (1.7)
    And this completes the expansion (1.2).
    It is important to note that the expansion is unique. That is, there is only one set of coefficients that forms the expansion.
    We prove this by contradiction.
    Set
    (1.8)
    And for at least one n we have
    Then:
    (1.9)

    Or:
    (1.10)
    Define then:
    (1.11)
    But according to (1.7) this means that for all n we have , and therefore for all n we have
    (1.12)
    Which contradicts our assumption that there at least one value of n where .
    Hence the expansion is unique.

    The non-homogenous damped wave equation with periodic boundary condition.
    The problem is:
    (2.1)
    With the boundary conditions:
    (2.2)
    And the initial conditions:
    (2.3)
    We start by finding the spatial orthogonal eigenfunctions and eigenvalues of the homogenous wave equation:
    (2.4)
    Subjected to the same periodic boundary conditions.
    We assume that is a product of two independent univariate functions:
    (2.5)
    Then:
    (2.6)

    Plugging this back into equation (2.4) we get:

    (2.7)
    Dividing both sides by we get:
    (2.8)
    This is a separated equation. Both sides are completely independent of each other, and since it is true for any , this means that the only way this can be true is if both sides equal the same constant:
    (2.9)
    The spatial equation becomes:
    (2.10)
    The boundary conditions are:
    (2.11)
    We take into account three cases.

    Case 1:
    The equation is:
    (2.12)
    The solution is:
    (2.13)
    Where A,B are arbitrary constants
    Applying the boundary conditions:

    (2.14)
    We get the trivial solution which is not very interesting.

    Case 2:
    Equation (2.10) is now:
    (2.15)
    The solution is:
    (2.16)

    Applying the boundary conditions:

    (2.17)
    Again, we get the trivial solution.

    Case 3:
    In this case, equation (2.10) becomes the harmonic equation:
    (2.18)
    And its solutions are:
    (2.19)
    Applying the boundary conditions:

    The spatial eigenfunctions are
    (2.20)
    And the eigenvalues are:
    (2.21)

    Note that the spatial eigenfunctions form an orthogonal set:
    (2.22)
    Now we can go back to the non homogenous equation
    (2.23)
    And we are going to write the solution as:
    (2.24)
    And we shall expand into a series of the eigenfunctions:
    (2.25)
    According to (1.7) and (2.25) we have:
    (2.26)
    We also note that:
    (2.27)
    Where, again:
    (2.28)

    We plug all this into (2.23):

    (2.29)
    Since the eigenfunctions expansion is unique (as we have shown) we must have:
    (2.30)
    This is a set of n second order non-homogenous differential equations (damped harmonic oscillator) with the initial conditions as the original wave equation:
    (2.31)
    To solve the homogenous equation we "guess" a solution in the form:
    (2.32)
    Plugging it back into the homogenous equation we get:

    (2.33)

    The solutions are:
    (2.34)
    Since we have two possible solutions:
    For n=1 we a have a single root with a multiplicity of 2, hence:
    (2.35)
    (2.36)
    And for :
    (2.37)
    So we have complex roots and therefore an oscillatory damped solution:
    (2.38)
    Where
    (2.39)
    While for the particular solution we use the method of undetermined coefficients and "guess" a solution in the form:
    (2.40)

    Plugging this back into equation (2.32):

    (2.41)
    Equating the coefficients of the polynomials on both sides we get:
    (2.42)
    Hence the particular solution is:
    (2.43)
    And the general solution to equation (2.33) is:
    (2.44)
    Applying initial conditions:

    And the general solution for is:
    (2.45)
    While:

    The general solution for is:
    (2.46)
    So finally we can write the general solution to the non-homogenous wave equation:

    (2.47)

    Recalling that:
    (2.48)
    This becomes:
    (2.49)
    And:
    (2.50)
    In our specific case:
    (2.51)
    Note that is already given in an eigenfunction expansion form.
    Since the expansion is unique, we can immediately say that
    (2.52)
    So the solution is very simple:

    (2.53)

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 5, 2022, 6:26 pm ad1c9bdddf>
    https://brainmass.com/math/fourier-analysis/fourier-methods-one-dimension-477689

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