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Math Help - Separation of variables for boundary value problems

  1. #1
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    Separation of variables for boundary value problems

    Quick pre-exam question..

    Say we have some sort of heat equation with boundary conditions:

    u(0,t) = 0,
    u(L,t) = 0

    where u = u(x,t)

    And using separation of variables we get to;
    X''/X = (1/a^2) T'/T

    where X = X(x), T = T(t), a is the alpha from the heat equation.

    So both sides of the equation must be constant;
    (A): X''/X = k
    (B): (1/a^2) T'/T = k

    Now according to my textbook k must be negative, so that T goes to zero as t --> infinity.

    I can't see why this is the case. What determines whether k is negative or positive? Is it determined by the boundary conditions - if the boundary conditions were derivatives (ie u'(some x, some t) = 0) would that produce a positive constant?

    I'm confused because as far as I understand, a -ve constant would leads to a basis involving cosines and sines, whereas a +ve constant would lead to a basis involving hyperbolic sines/cosines. I don't know when to use which!

    thanks,
    Squigles
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  2. #2
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    Quote Originally Posted by Squiggles View Post
    Quick pre-exam question..

    Say we have some sort of heat equation with boundary conditions:

    u(0,t) = 0,
    u(L,t) = 0

    where u = u(x,t)

    And using separation of variables we get to;
    X''/X = (1/a^2) T'/T

    where X = X(x), T = T(t), a is the alpha from the heat equation.

    So both sides of the equation must be constant;
    (A): X''/X = k
    (B): (1/a^2) T'/T = k

    Now according to my textbook k must be negative, so that T goes to zero as t --> infinity.

    I can't see why this is the case. What determines whether k is negative or positive? Is it determined by the boundary conditions - if the boundary conditions were derivatives (ie u'(some x, some t) = 0) would that produce a positive constant?

    I'm confused because as far as I understand, a -ve constant would leads to a basis involving cosines and sines, whereas a +ve constant would lead to a basis involving hyperbolic sines/cosines. I don't know when to use which!

    thanks,
    Squigles
    Let's look at it from a different point of view, from \frac{X''}{X} = k.

    Now your BCs give u(0,t) = T(t)X(0) = 0\;\;\Rightarrow\;\;X(0) = 0 and u(L,t) = T(t)X(L) = 0\;\;\Rightarrow\;\;X(L) = 0

    Now the three possible choices for k are  k = - \omega^2, k = 0, k = \omega^2 (negative, zero or positive)

    If k = \omega^2 then X=c_1e^{\omega x} + c_2 e^{-\omega x}. The BC's X(0) = 0,\;\;X(L)=0 gives

    c_1e^{0} + c_2 e^{0}=0, and c_1e^{\omega L} + c_2 e^{-\omega L} = 0

    which gives c_1 = 0,\;\;c_2 = 0, thus, X(x) = 0\;\;\;\Rightarrow u = 0, the trival solution. Similarly for k = 0 you'll get the trival solution so the only possibility is k = -\omega^2
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  3. #3
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    Hey thanks sorry I didn't reply earlier - yeah I think I remember being shown that in class. So, are there any boundary conditions that would produce a positive k? Since it seems I'm not expected to go through checking for k positive/negative/zero for every problem I do, is it safe to assume it's always going to be negative..??
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    Quote Originally Posted by Squiggles View Post
    Hey thanks sorry I didn't reply earlier - yeah I think I remember being shown that in class. So, are there any boundary conditions that would produce a positive k? Since it seems I'm not expected to go through checking for k positive/negative/zero for every problem I do, is it safe to assume it's always going to be negative..??
    Yes, for your BC's, k will be negative so no, you don't need to check every time.
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    But the question was whether there exist other boundary conditions for which k is not negative.


    Squiggles, when the boundary conditions are u(0,t)= u(L,t)= 0, then you can say X(0)T(t)= 0, X(L)T(t)= 0 so X(0)= 0, X(L)= 0 no matter what t is. The solution for X must NOT be one to one and so cannot be linear or exponential as would be the case for k zero or positive. The solution must be periodic and k must be negative.

    If the boundary conditions were, say, u(0,t)= f(t), u(L,t)= g(t), then we would have X(0)T(t)= f(t), X(L)T(0)= g(t) so we cannot "separate" X from t.

    In that case, I would recommend defining v(x,t)= u(x,t)- f(t)(L-x)/L- g(t)x/L so that v(0,t)= u(0,t)- f(t)= f(t)- f(t)= 0 and v(L,t)= u(L,t)- g(t)= g(t)- g(t)= 0.
    Of course, although v_xx= u_xx, v_t= u_t- f'(t) \frac{L - x}{L} - g'(t) \frac{x}{L} so that the differential equation becomes v_{xx}= \frac{1}{a^2} v_t - \frac{1}{a^2} f'(t) \frac{L - x}{L} - \frac{1}{a^2}g'(t) \frac{x}{L} which might not be "separable".
    Last edited by mr fantastic; March 28th 2009 at 02:15 PM. Reason: Fixed some latex
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