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Math Help - Green's function for a BVP

  1. #1
    Senior Member bkarpuz's Avatar
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    Green's function for a BVP

    Dear MHF members,

    I need help in finding the Green's function for the BVP
    y^{\prime\prime}+\frac{1}{x}y^{\prime}=-f(x) with \lim_{x\to0^{+}}|y(x)|<\infty and y(1)=0.

    Many thanks.
    bkarpuz

    Note. Typo is corrected after TheEmptySet's warning.
    Last edited by bkarpuz; May 11th 2011 at 01:51 PM.
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    Quote Originally Posted by bkarpuz View Post
    Dear MHF members,

    I need help in finding the Green's function for the BVP
    y^{\prime\prime}-\frac{1}{x}y^{\prime}=-f(x) with \lim_{x\to0^{+}}|y(x)|<\infty and y(1)=0.

    Many thanks.
    bkarpuz
    I am going to assume that there is a typo in the statement of the problem otherwise the boundary condition at zero isn't needed as the solution to the homogenous problem is

    y=c_1+c_2x^2

    Which is bounded no matter what the constants are but if we solve the problem

    y''+\frac{1}{x}y'=-f(x) \iff -\frac{d}{dx}\left( xy' \right)=xf(x)

    Now this is in its self adjoint form.

    Solving the homogeneous equation gives

    y_c=c_1+c_2\ln(x)

    Now we need to find two solutions each satisfying one of the boundary conditions.

    y_0=1

    other wise the log function is unbounded at zero and

    y_1=\ln(x) to give zero at 1.

    So the Green's function has the form

    g(x,s)=\begin{cases}A\ln(x), \text{ if } s \le x \\  B\ln(s), \text{ if } s \ge x \end{cases}

    Now if we take the Wronskian we get

    \begin{vmatrix}A\ln(x) & B \\  \frac{A}{x} & 0 \end{vmatrix}=\frac{-AB}{x}

    Now we take the function off of the derivative of y in the self adjoint form and multiply it by the Wronskian and set it equal to negative 1 to get

    -AB=-1

    So we can pick any real numbers for A and B that satisfy this equation so let

    A=B=1

    So the Green's function is

    g(x,s)=\begin{cases}\ln(x), \text{ if } s \le x \\  \ln(s), \text{ if } s \ge x \end{cases}

    So the solution to the ODE is

    y=\ln(x)\int_{0}^{x}sf(s)ds+\int_{x}^{1}s\ln(s)f(s  )ds

    Notice this satisfies the boundary conditions and that

    y'=\frac{1}{x}\int_{0}^{x}sf(s)ds+x\ln(x)f(x)-x\ln(x)f(x)=\frac{1}{x}\int_{0}^{x}sf(s)ds

    and

    y''=-\frac{1}{x^2}\int_{0}^{x}sf(s)ds-f(x)

    If you plug this back into the ODE you get

    -\frac{1}{x^2}\int_{0}^{x}sf(s)ds-f(x)+\frac{1}{x}\left( \frac{1}{x}\int_{0}^{x}sf(s)ds\right)=-f(x)
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  3. #3
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    I am going to assume that there is a typo in the statement of the problem otherwise the boundary condition at zero isn't needed as the solution to the homogenous problem is ...
    Thanks a lot TheEmptySet. Let me tell you what I have done. The problem was actually the following equation
    y^{\prime\prime}+\dfrac{1}{x}y^{\prime}-\dfrac{\lambda^{2}}{x^{2}}y=-f(x) with |y(0^{+})|<\infty and y(1)=0,
    whose self-adjoint form is
    (xy^{\prime})^{\prime}-\dfrac{\lambda^{2}}{x}y=-xf(x)
    The functions \phi_{1}=2\lambda x^{\lambda} and \phi_{2}=(x^{\lambda}-x^{-\lambda})/(2\lambda) are respective solutions of the homogeneous equations
    (x y^{\prime})^{\prime}-\dfrac{\lambda^{2}}{x}y=-xf(x) with |y(0^{+})|<\infty
    and
    (x y^{\prime})^{\prime}-\dfrac{\lambda^{2}}{x}y=-xf(x) with y(1)=0.
    Then, xW(\phi_{1},\phi_{2})=2\lambda, which yields
    G(x,t)=\dfrac{1}{2\lambda}\begin{cases}(x/t)^{\lambda}-(tx)^{\lambda}, &x\leq t\\ (t/x)^{\lambda}-(tx)^{\lambda}, &x\geq t\end{cases}
    ............ =\dfrac{1}{2\lambda}\big[(t/x)^{\lambda\mathrm{sgn}(x-t)}-(tx)^{\lambda}\big].
    Thus the solution is given by
    \phi=\frac{1}{2\lambda}\int_{0}^{1}\big[(s/x)^{\lambda\mathrm{sgn}(x-s)}-(sx)^{\lambda}\big]sf(s)\mathrm{d}s.

    Letting here \lambda\to0 (assuming that it is possible), I get
    \phi=\frac{1}{2}\int_{0}^{1}\big[\mathrm{sgn}(x-s)\ln(s/x)-\ln(sx)\big]sf(s)\mathrm{d}s
    ..... =\int_{0}^{1}\left\{\begin{array}{c}\ln(s),\ x\leq s\\ \ln(x),\ x\geq s\end{array}\right\}sf(s)\mathrm{d}s, which is exactly the same expression you have obtained. But the problem here is that in my case, the Wronskian tends to 0. What is the reason that the method I have applied above does not work (by considering the homogeneous equation with different boundary conditions separately)?

    Thanks.
    bkarpuz
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    Here is what I think is going on.

    When we compare the two equations and their solutions.

    y''+\frac{1}{x}y'+\frac{\lambda^2}{x^2}y=0

    y_c=c_1x^\lambda+c_2x^{-\lambda}

    and

    y''+\frac{1}{x}y'=0

    y_c=d_1+d_2\ln(x)

    Now if we take the limit as

    \lambda \to 0

    The first ODE goes to the 2nd but

    \lim_{\lambda \to 0}c_1x^\lambda+c_2x^{-\lambda}=c_1+c_2

    This does not go to the complimentary solution of the 2nd equation.
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  5. #5
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Here is what I think is going on.

    When we compare the two equations and their solutions.

    y''+\frac{1}{x}y'+\frac{\lambda^2}{x^2}y=0

    y_c=c_1x^\lambda+c_2x^{-\lambda}

    and

    y''+\frac{1}{x}y'=0

    y_c=d_1+d_2\ln(x)

    Now if we take the limit as

    \lambda \to 0

    The first ODE goes to the 2nd but

    \lim_{\lambda \to 0}c_1x^\lambda+c_2x^{-\lambda}=c_1+c_2

    This does not go to the complimentary solution of the 2nd equation.
    But if we write the solution of the first one as y=c_{1}x^{\lambda}+c_{2}(x^{\lambda}-x^{-\lambda})/(2\lambda) it does.
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    Quote Originally Posted by bkarpuz View Post
    But if we write the solution of the first one as y=c_{1}x^{\lambda}+c_{2}(x^{\lambda}-x^{-\lambda})/(2\lambda) it does.
    But the Wronskian of the above function is independent of lambda and non zero.

    \begin{vmatrix} x^\lambda & \frac{x^{\lambda} - x^{-\lambda}}{2\lambda}  \\ \lambda x^{\lambda -1} & \frac{x^{\lambda-1}+x^{-\lambda-1}}{2} \end{vmatrix} = \frac{x^{2\lambda-1}+x^{-1}}{2} - \frac{ x^{2\lambda -1}- x^{-1}}{2} =\frac{1}{x}  \ne 0
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