Green's Functions (for ODE)

  • #1
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Homework Statement



I'm teaching myself the Green's function method for ODEs, because it looks relevant to my interests. This is a (slightly contrived) problem I just came up with arbitrarily:

[tex]y''+5y'+6y=sin(x) \; \; \; ; \; \; \; y(0)=y'(0)=0[/tex]

Homework Equations



i) When considered as a function of [itex]x[/itex] alone, [itex]G(x, z)[/itex] must obey the homogeneous initial conditions.

ii) The derivatives of [itex]G(x, z)[/itex] with respect to [itex]x[/itex] up to order [itex]n - 2[/itex] are continuous at [itex]x = z[/itex], but the [itex](n - 1)^{th}[/itex] order derivative has a discontinuity of [itex]\large{\frac{1}{a_{n}(z)}}[/itex] at this point.

The Attempt at a Solution



Proceeding with the complementary solution (following the procedure outlined in my text):

[tex]G(x, z)=\left\{\begin{matrix}
A(z)e^{-2x} + B(z)e^{-3x}\Rightarrow & x<z \\
C(z)e^{-2x} + D(z)e^{-3x}\Rightarrow & x>z
\end{matrix}\right.[/tex]

We require [itex]G(0, z) = G'(0, z) = 0[/itex]. For some reason this implies [itex]A(z) = B(z) = 0[/itex]. I mean, it's obvious that this is true if [itex]G(x, z) = A(z)e^{-2x} + B(z)e^{-3z}[/itex], but it's a piecewise function, so why do we only consider [itex]x < z [/itex]?

Anyways, this equation is of order 2, so we need the 0th order derivative to be continuous. Also, the 1st derivative needs to have a discontinuity of 1 (because the leading coefficient is 1):

[tex]\begin{matrix}
C(z)e^{-2z}+D(z)e^{-3z}=0\\ -2C(z)e^{-2z}-3D(z)e^{-3z}=1

\end{matrix}[/tex]

Solving gives [itex]C(z) = e^{2z}, D(z) = -e^{3z}[/itex]. Then:

[tex]y(x) = \int_{0}^{\infty }G(x,z)sin(z)dz [/tex]

[tex]y(x)=\int_{0}^{x }[e^{2z-2x}-e^{3z-3x}]sin(z)dz[/tex]

[tex]y(x)=\frac{1}{10}(e^{-3x} (2e^{x}-1) - cos(x) + sin(x)) \leftarrow \text{Lazily computed with Mathematica.}[/tex]

What I don't get is why that improper integral all of a sudden changed to have limits from 0 to [itex]x[/itex]. Furthermore, why is it even an improper integral at all? I suspect that it's because the initial conditions don't specify a boundary, but then why shouldn't the lower limit be negative infinity?

Thanks so much, and sorry about the wall of Tex :redface:
 
Last edited:

Answers and Replies

  • #2
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  • #3
vela
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We require [itex]G(0, z) = G'(0, z) = 0[/itex]. For some reason this implies [itex]A(z) = B(z) = 0[/itex]. I mean, it's obvious that this is true if [itex]G(x, z) = A(z)e^{-2x} + B(z)e^{-3z}[/itex], but it's a piecewise function, so why do we only consider [itex]x < z [/itex]?
The way you solved the problem, you've found the solution on the interval [0, ∞), which makes sense because you have the initial conditions at x=0 and you want to know how the system responds subsequently. Consequently, z will lie somewhere in that interval and split the interval into two pieces. The initial conditions apply to only the x<z part of the Green's function because x=0 is in the x<z part of the interval.
Anyways, this equation is of order 2, so we need the 0th order derivative to be continuous. Also, the 1st derivative needs to have a discontinuity of 1 (because the leading coefficient is 1):

[tex]\begin{matrix}
C(z)e^{-2z}+D(z)e^{-3z}=0\\ -2C(z)e^{-2z}-3D(z)e^{-3z}=1

\end{matrix}[/tex]

Solving gives [itex]C(z) = e^{2z}, D(z) = -e^{3z}[/itex]. Then:

[tex]y(x) = \int_{0}^{\infty }G(x,z)sin(z)dz [/tex]

[tex]y(x)=\int_{0}^{x }[e^{2z-2x}-e^{3z-3x}]sin(z)dz[/tex]

[tex]y(x)=\frac{1}{10}(e^{-3x} (2e^{x}-1) - cos(x) + sin(x)) \leftarrow \text{Lazily computed with Mathematica.}[/tex]

What I don't get is why that improper integral all of a sudden changed to have limits from 0 to [itex]x[/itex]. Furthermore, why is it even an improper integral at all? I suspect that it's because the initial conditions don't specify a boundary, but then why shouldn't the lower limit be negative infinity?

Thanks so much, and sorry about the wall of Tex :redface:
Remember G(x,z) vanishes for x<z, so only when z is between 0 and x is there any contribution to the integral.
 
  • #4
129
0
Okay totally gotcha, so it really just comes down to the initial conditions. Thanks for clearing that up - I was beginning to lose hope! This method seems like more trouble than it's worth, but I'm sure it will come in handy some day.
 

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