Explore BrainMass

Explore BrainMass

    Differential equations

    Not what you're looking for? Search our solutions OR ask your own Custom question.

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

    Consider the following problem for u = u(x,t):
    ,
    ,

    a) Seeking a solution of this problem of the form
    ,

    show that f and g satisfy the coupled system
    ; where and ;
    , ; , .

    b) Eliminating g between the differential equations in a), show that f satisfies

    c) Assuming that

    and noting that

    conclude that

    d) Applying the boundary conditions for f in a), deduce that
    where

    e) Now use a) to show that
    .

    f) Applying the boundary conditions for g in a), finally conclude that
    and hence obtain the result.

    since

    © BrainMass Inc. brainmass.com December 24, 2021, 7:17 pm ad1c9bdddf
    https://brainmass.com/math/calculus-and-analysis/differential-equations-boundary-conditions-177146

    Attachments

    SOLUTION This solution is FREE courtesy of BrainMass!

    Please see the attached file.

    Consider the following problem for u = u(x,t):
    ,
    ,

    a) Seeking a solution of this problem of the form
    ,

    show that f and g satisfy the coupled system
    ; where and ;
    , ; , .

    Now,

    Dividing throughout by 

    =
    Comparing the cos terms, and comparing the sin terms,
    For u(0, t) = cos(t), f(0) = 1 and g(0) = 0

    b) Eliminating g between the differential equations in a), show that f satisfies

    Given that f'' = 2k2g,
    Differentiating with respect to x on both sides, f''' = 2k2g'
    Differentiating again with respect to x on both sides, f(4) = 2k2g''
    But g'' = -2k2f, so f(4) = 2k2(-2k2f) = -4k4f
    Therefore,

    c) Assuming that

    and noting that

    conclude that

    f'(x) = r erx
    f''(x) = r2 erx
    f'''(x) = r3 erx
    f(4)(x) = r4 erx
    Plugging in the equation , we get r4 erx + 4k4 erx = 0
    Or (r4 + 4k4) erx = 0, which gives (r4 + 4k4) = 0
    But
    Therefore,
    Taking
    Or
    Or (r + k)2 = -k2
    Or r + k = -ki, where i =
    Or r = -k - ki
    Therefore, f(x) = e-k-ki
    By Euler's formula,
    Similarly, by taking , it can be shown that r = k - ki
    Or f(x) = ek-ki =
    Combining the two,

    d) Applying the boundary conditions for f in a), deduce that
    where
    Given that ,
    so
    or
    or c1 + c3 = 1
    Similarly, c3 = c4= 0
    Plugging in c3 = 0 in c1 + c3 = 1 gives c1 = 1
    Therefore,
    =
    =
    If , we get

    e) Now use a) to show that
    .
    f'' = 2k2 g(x)  g(x) = f"/2k2
    But f = e-kx h(x)
    Therefore, f'(x) = -ke-kx h(x) + h'(x) e-kx = e-kx(-kh + h') = -kf + e-kx h'(x)
    Or f' + kf = e-kx h'(x)
    But g(x) = -(f' + kf)/k, therefore,

    f) Applying the boundary conditions for g in a), finally conclude that
    and hence obtain the result.

    since

    Given the boundary condition g(0) = 0, 0 = , which means that
    h'(0) = 0
    But , therefore, h'(x) = -k sin(kx) + c2 cos(kx)
    At x = 0, h'(0) = -k*sin(0) + c2*cos(0)
    Or 0 = 0 + c2, which gives c2 = 0
    Therefore,
    Therefore, =
    And =
    Now, = cos(t) + sin(t) Or .

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

    © BrainMass Inc. brainmass.com December 24, 2021, 7:17 pm ad1c9bdddf>
    https://brainmass.com/math/calculus-and-analysis/differential-equations-boundary-conditions-177146

    Attachments

    ADVERTISEMENT