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Math Help - Solve the given differential equation by using Green's function method

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    Member roshanhero's Avatar
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    Solve the given differential equation by using Green's function method

    I am really struggling with the concept and handling of the Green's function. I have to solve the given differential equation using Green's function method
    \frac{d^{2}y}{dx^{2}}+k^{2}y=\delta (x-x');\qquad y(0)=y(L)=0
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    Re: Solve the given differential equation by using Green's function method

    The "Green's function" for a given differential operator, D(f), is a function, G(x,x'), satisfying D(G)= 0 as long as x\ne x' and such that \lim_{x'\to x^+} G(x,x')- \lim_{x'\to x^-} G(x, x')= 1.

    Functions satisfying y''+ k^2y= 0 are all of the form y(x)= C cos(kx)+ D sin(kx). So we must have G(x, x')= C_1cos(kx)+ D_1(kx) for some constants (with respect to x, they may depend on x') for x\le x' and G(x, x')= C_2 cos(kx)+ D_2 sin(kx) for x\ge x'. So the problem is to determine C_1, C_2, D_1, and D_2.

    If x= 0 then x\le x' because x' must also be between 0 and L so we must have y(0)= C_1 cos(0)+ D_1 sin(0)= C_1= 0.

    Similarly of x= L then x\ge x' so we must have y(L)= C_2 cos(L)+ D_2 sin(L)= 0. That gives C_2= tan(L) D_2

    So far, we have G(x, x')= D_1sin(x) if x\le x' and G(x, x')= D_2(tan(L)cos(x)+ sin(x)).

    At x= x' those must both be true: G(x', x')= D_1 sin(x')= D_2(tan(L)cos(x')+ sin(x')).

    The derivative, from the left, at x= x', is D_1 cos(x') and, from the right, D_2(cos(x')- tan(L)sin(x')) so we must have D_1 cos(x')- D_2(cos(x')- tan(L)sin(x'))= 1.

    That gives two equation to solve for D_1 and D_2 in terms of x'.
    Last edited by HallsofIvy; July 25th 2014 at 07:50 AM.
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    Member roshanhero's Avatar
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    Re: Solve the given differential equation by using Green's function method

    I think C_{2}=-D_{2}tan(L)
    What is the role of delta function in the right hand side then ?
    Last edited by roshanhero; July 25th 2014 at 09:28 AM.
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