Why is Green's function method for ODEs only considered for x < z?

[Screwdriver]

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:

$$y''+5y'+6y=sin(x) \; \; \; ; \; \; \; y(0)=y'(0)=0$$

Homework Equations

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

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

The Attempt at a Solution

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

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

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

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

$$\begin{matrix}<br /> C(z)e^{-2z}+D(z)e^{-3z}=0\\ -2C(z)e^{-2z}-3D(z)e^{-3z}=1<br /> <br /> \end{matrix}$$

Solving gives $C(z) = e^{2z}, D(z) = -e^{3z}$. Then:

$$y(x) = \int_{0}^{\infty }G(x,z)sin(z)dz$$

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

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

What I don't get is why that improper integral all of a sudden changed to have limits from 0 to $x$. 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

 

[Screwdriver]

Start spreadin' the news - I'm bumping today.
I want to bump a part of it; this thread, is bumped.

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[vela]

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

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 0^th order derivative to be continuous. Also, the 1^st derivative needs to have a discontinuity of 1 (because the leading coefficient is 1):

$$\begin{matrix}<br /> C(z)e^{-2z}+D(z)e^{-3z}=0\\ -2C(z)e^{-2z}-3D(z)e^{-3z}=1<br /> <br /> \end{matrix}$$

Solving gives $C(z) = e^{2z}, D(z) = -e^{3z}$. Then:

$$y(x) = \int_{0}^{\infty }G(x,z)sin(z)dz$$

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

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

What I don't get is why that improper integral all of a sudden changed to have limits from 0 to $x$. 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

Remember G(x,z) vanishes for x<z, so only when z is between 0 and x is there any contribution to the integral.

 

[Screwdriver]

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.