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Green function

  1. Jan 21, 2008 #1
    solve the next differential equation:

    [tex] y´´- a*y= \delta (x-d) [/tex]

    with the boundary conditions:

    [tex] \left.\frac{\partial y}{\partial x} \right|_ {x=0} = 0 [/tex]

    [tex] lim _{x\rightarrow\infty} y = 0[/tex]

    I get the homogeneous solution: [tex] y_H = C_1 exp (\sqrt{a}x) + C_2 exp (-\sqrt{a}x)[/tex]

    and then to obtain the inhomogeneous solution (the particular solution), one should get the Green function. For these case, it is G equal to:

    [tex] \{A exp (\sqrt{a}x) + B exp (-\sqrt{a}x)[/tex]
    [tex] \{C exp (\sqrt{a}x) + D exp (-\sqrt{a}x)[/tex]

    and the A, B, C, D coefficients should be obtain from the boundary conditions. This is my problem, I try to applied this boundary conditions but I have not idea how I can do it. Can someone help me?
  2. jcsd
  3. Jan 21, 2008 #2
    It's easier with variation of parameters. Consider the problem

    [itex]y''(x)-ay(x)=f(x), \quad y'(0)=0, \quad y(x)\rightarrow 0 \hbox{ as } x\rightarrow \infty.[/tex]

    On one side of the boundary (x=0), the solution is

    [tex]u_1(x)=\cosh \sqrt{a} x,[/tex]

    while on the other side ([itex]x=\infty[/itex]), the solution is


    Now, you want a solution of the ode in the form


    where [itex]a(x)\rightarrow 0[/itex] as [itex]x\rightarrow \infty[/itex], and b(0)=0 (so in x=0, [itex]u_1(x)[/itex] is the only term, hence satisfying that boundary condition, and in [itex]x\rightarrow \infty[/itex] the hole solution vanishes, satisfying the other boundary condition).

    Now, using variation of parameters, we require that

    [tex] a'(x)u_1(x)+b'(x)u_2(x)=0,[/tex]
    [tex] a'(x)u_1'(x)+b'(x)u_2'(x)=f(x).[/tex]

    Solving for a'(x) and b'(x)

    [tex]\int_x^\infty a'(\xi)d\xi=-\int_x^\infty \frac{u_2(\xi)}{W[u_1,u_2](\xi)}f(\xi)d\xi,[/tex]

    [tex]\int_0^x b'(\xi)=\int_0^x \frac{u_1(\xi)}{W[u_1,u_2](\xi)}f(\xi)d\xi.[/tex]

    Evaluating, we obtain

    [tex]a(x)=\int_x^\infty \frac{u_2(\xi)}{W[u_1,u_2](\xi)}f(\xi)d\xi,[/tex]

    [tex]b(x)=\int_0^x \frac{u_1(\xi)}{W[u_1,u_2](\xi)}f(\xi)d\xi,[/tex]

    where [itex]W[u_1,u_2](\xi)[/itex] is the Wronskian.

    Substituting in the original solution, we obtain

    [tex]y(x)=\int_x^\infty \frac{u_1(x)u_2(\xi)}{W[u_1,u_2](\xi)}f(\xi)d\xi+\int_0^x \frac{u_1(\xi)u_2(x)}{W[u_1,u_2](\xi)}f(\xi)d\xi,[/tex]

    wich can be written as

    [tex]y(x)=\int_0^\infty G(\xi,x) f(\xi) d\xi,[/tex]

    where [itex]G(\xi,x)[/itex] is the Green function of the problem and it's given by

    [tex]G(\xi,x)=\left\{\begin{array}{ll}\dfrac{u_1(\xi)u_2(x)}{W[u_1,u_2](\xi)} & \hbox{ if } \xi < x, \\ & \\ \dfrac{u_1(x)u_2(\xi)}{W[u_1,u_2](\xi)} & \hbox{ if } x<\xi .\end{array}\right.[/tex]

    The other way to solve the problem is far less intuitive and involves symbolic calculus. You use the fact that in symbolic calculus, [itex]H'(x)=\delta(x)[/itex], where H(x) is the Heaviside function, and [itex]\delta(x)[/itex] is the Dirac Delta function.
    Last edited: Jan 21, 2008
  4. Jan 22, 2008 #3
    first of all thanks for your help, it is very good. I have done the next to obtain [tex]u_1(x) [/tex] and [tex]u_2(x) [/tex] and I am not sure if this is correct:

    1. [tex]y=0 \rightarrow y_H = C_1 + C_2 \rightarrow C_1 = C_2 \rightarrow [/tex]
    [tex]C_1 [exp (\sqrt{a}x) + exp (-\sqrt{a}x)] = cosh (\sqrt{a}x) = u_1(x) [/tex]

    2. [tex]y= \infty \rightarrow y_H = C_2 exp (-\sqrt{a}x)= 0 \rightarrow u_2(x)=exp (-\sqrt{a}x)[/tex]
  5. Jan 22, 2008 #4
    I'll admit I read your post rather quickly, however in general to solve for the coeffients of the Green's function, you have to apply the properties of Green's functions.
    (i.e. those two fcns correspond to x<y and x>y correct? )

    Well at x=y, the Green's fcn must be continuous, and its first derivative differs by unity. This gives you two relations for the coefficients, the other two relations come from your boundary condtions.
  6. Jan 23, 2008 #5
    1. is not correct, as it implies [itex]C_1=-C_2[/itex], hence given [itex]u_1(x)=\sinh \sqrt{a}x[/itex], but the condition of your original post is [itex]y'(0)=0[/itex] wich implies [itex]C_1=C_2[/itex], so [itex]u_1(x)=\cosh \sqrt{a}x[/itex].

    2. is not stated correctly, as it should read

    [tex]\mathop{lim}_{x\rightarrow\infty} y(x)=\mathop{lim}_{x\rightarrow \infty} \{C_3 e^{\sqrt{a}x}+C_4 e^{-\sqrt{a}x}\}=0 \quad \hbox{if and only if}\quad C_3=0[/tex]
    Last edited: Jan 23, 2008
  7. Jan 29, 2008 #6
    Thank you very much AiRAVATA for your help and explanation. You are right with u1, I have also noticed later my error. I have already all the calculations.

    Now, I don't have to calculate anymore but I have a question, maybe it is very stupid. If one consider the same equation but without the boundary conditions, to calculate the Green's function one should use also the variation of parameters method but in this case, u1 and u2 are the solutions of the homogeneous equation. Is it correct?
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