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Math Help - Heat Equation trouble

  1. #1
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    Question Heat Equation trouble

    I'm going through the following heat equation problem, and I understand how the solution is constructed except for one algebraic re-arrangement that I can't make sense of (and its probably really obvious) which is right at the start of the solution. I was hoping someone could enlighten me.

    Solve u_t = ku_{xx} on 0 \leq x \leq L with IC's u(x,0) = f(x) and BC's u_x(0,t) = u_x(L,t) = 0.

    The example says:

    Let u = XT.

    Confused from here:
    Sub in XT' = kX''

    \frac{X''}{X} = \frac{T'}{kT} = - \lambda

    Then X'' + \lambda X = 0.
    To the line above

    From here I have no problems understanding and solving this problem.

    This is probably embarrassingly simple, but I keep getting

    \frac{X''}{X} = \frac{T'}{k}

    and why is this equal to -\lambda
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  2. #2
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    Quote Originally Posted by CooperCripps View Post
    I'm going through the following heat equation problem, and I understand how the solution is constructed except for one algebraic re-arrangement that I can't make sense of (and its probably really obvious) which is right at the start of the solution. I was hoping someone could enlighten me.

    Solve u_t = ku_{xx} on 0 \leq x \leq L with IC's u(x,0) = f(x) and BC's u_x(0,t) = u_x(L,t) = 0.

    The example says:

    Let u = XT.

    Confused from here:
    Sub in XT' = kX''

    \frac{X''}{X} = \frac{T'}{kT} = - \lambda

    Then X'' + \lambda X = 0.
    To the line above

    From here I have no problems understanding and solving this problem.

    This is probably embarrassingly simple, but I keep getting

    \frac{X''}{X} = \frac{T'}{k}

    and why is this equal to -\lambda
    You are assuming the production from of the solutions

    \displaystyle u(x,t)=X(x)T(t) this gives that

    \displaystyle u_t(x,t)=X(x)T'(t) and

    \displaystyle u_{xx}(x,t)=X''(x)T(t)

    So the PDE gives us

    \displaystyle u_{t}=ku_{xx} \iff X(x)T'(t)=kX''(x)T(t)

    Now isolating the T's and X's gives

    \displaystyle \frac{T'(t)}{kT(t)}=\frac{X''(x)}{X(x)}

    Now since this equality must hold for all values of x and t each side of the equation must be constant this gives


    \displaystyle \frac{T'(t)}{kT(t)}=\frac{X''(x)}{X(x)}=c or the 2 ODE's

    \displaystyle \frac{X''(x)}{X(x)}=c \text{ and } \frac{T'(t)}{kT(t)}=c

    Now to answer you second question:

    First note the the boundary conditions

    u_x(0,t)=u_x(L,t)=0 \implies \text{ that } X'(0)=X'(L)=0

    Using the X equation above gives

    X''(x)-cX(x)=0 Now if we multiply this by X(x) and integrate we get

    \displaystyle \int_{0}^{L}X(x)X''(x)dx -c\int_{0}^{L}[X(x)]^2dx=0

    If we integrate the first integral by parts with

    u=X(x) \implies du=X'(x) \text{ and } dv=X''(x) \implies v=X'(x) we get

    \displaystyle X(x)X'(x)\bigg|_{0}^{L}-\int_{0}^{L}[X'(x)]^2dx -c\int_{0}^{L}[X(x)]^2dx=0

    Now using the boundary conditions we get

    \displaystyle X(L)X'(L)-X(0)X'(0)-\int_{0}^{L}[X'(x)]^2dx -c\int_{0}^{L}[X(x)]^2dx=0

    \displaystyle -\int_{0}^{L}[X'(x)]^2dx -c\int_{0}^{L}[X(x)]^2dx=0

    Notice that both of the integrands are nonnegative, So the only way this equation can be true is if c < 0. This is why the separation constant is -\lambda
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post

    Now to answer you second question:

    First note the the boundary conditions

    u_x(0,t)=u_x(L,t)=0 \implies \text{ that } X'(0)=X'(L)=0

    Using the X equation above gives

    X''(x)-cX(x)=0 Now if we multiply this by X(x) and integrate we get

    \displaystyle \int_{0}^{L}X(x)X''(x)dx -c\int_{0}^{L}[X(x)]^2dx=0

    If we integrate the first integral by parts with

    u=X(x) \implies du=X'(x) \text{ and } dv=X''(x) \implies v=X'(x) we get

    \displaystyle X(x)X'(x)\bigg|_{0}^{L}-\int_{0}^{L}[X'(x)]^2dx -c\int_{0}^{L}[X(x)]^2dx=0

    Now using the boundary conditions we get

    \displaystyle X(L)X'(L)-X(0)X'(0)-\int_{0}^{L}[X'(x)]^2dx -c\int_{0}^{L}[X(x)]^2dx=0

    \displaystyle -\int_{0}^{L}[X'(x)]^2dx -c\int_{0}^{L}[X(x)]^2dx=0

    Notice that both of the integrands are nonnegative, So the only way this equation can be true is if c < 0. This is why the separation constant is -\lambda
    Thanks EmptySet. This may be a dumb question, but apart from using integration as you have a done above, is their an faster method of determining the separation constant? For example, if in an exam I don't think I would like to use this method because it may take too long.

    Would you know the separation constant to be - \lambda by intuition?
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  4. #4
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    You can look at cases

    c = \omega^2, c = 0, c = -\omega^2 (i.e. c > 0, = 0, < 0),

    and solve X'' - c X = 0 subject to X'(0) =0, X'(L)=0.

    The only case that gives a nonzero answer is c = -\omega^2
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