Green's Function of Linear ODE

So, in summary, to find the Green's function for ##u''+u=f##, we make an ansatz of the form ##G(x,x_0) = \theta(x-x_0)g_+(x) + \theta(x_0 - x) g_-(x)## or integrate the differential equation for the Green's function, excluding ##u##, from ##x_{0-}## to ##x_{0+}##, and set it equal to 0 in the limit. We also require four boundary conditions, typically given as 1 at each end point, continuity of ##G##, and a jump discontinuity at ##x=x_0##.
  • #1
member 428835

Homework Statement


Find Green's function of ##u''+u=f##.

Homework Equations


What we all know.

The Attempt at a Solution


Let Greens function be ##G##. Then ##G''+G=\delta(x-x_0)##. This admits solutions superimposed of sine and cosine. Let's split the function at ##x=x_0##. Then we require four boundary conditions. We are typically given 1 at each end point (2 total), we know ##G## is continuous, and lastly there is a jump discontinuity at ##x=x_0##. This is where I'm stuck. In determining this fourth BC, would we integrate the ODE from ##x_0-## to ##x_0+##? This would imply ##u'_+-u'_- + \int_-^+G = 1##.

Is there an easier way?
 
Physics news on Phys.org
  • #2
joshmccraney said:

Homework Statement


Find Green's function of ##u''+u=f##.

Homework Equations


What we all know.

The Attempt at a Solution


Let Greens function be ##G##. Then ##G''+G=\delta(x-x_0)##. This admits solutions superimposed of sine and cosine. Let's split the function at ##x=x_0##. Then we require four boundary conditions. We are typically given 1 at each end point (2 total), we know ##G## is continuous, and lastly there is a jump discontinuity at ##x=x_0##. This is where I'm stuck. In determining this fourth BC, would we integrate the ODE from ##x_0-## to ##x_0+##? This would imply ##u'_+-u'_- + \int_-^+G = 1##.

Is there an easier way?

That is as easy as it gets: ##\int_{-x_0}^{+x_0} G \, dx = 0##, so ##G'(+x_0) - G'(-x_0) = 1##.
 
  • #3
joshmccraney said:
we know ##G## is continuous,

This is often stated without motivation. It may or may not be true depending on the differential equation.

and lastly there is a jump discontinuity at ##x=x_0##. This is where I'm stuck. In determining this fourth BC, would we integrate the ODE from ##x_0-## to ##x_0+##? This would imply ##u'_+-u'_- + \int_-^+G = 1##.

Is there an easier way?

I never liked the "integrate from ##x_{0-}## to ##x_{0+}## argument. I find it much more instructive and transparent to make an ansatz of the form ##G(x,x_0) = \theta(x-x_0)g_+(x) + \theta(x_0 - x) g_-(x)## and insert it into the differential equation and then start identifying terms.

However, if you do want to use the integration approach, it is the differential equation for the Green's function that you want to integrate. This does not contain any u (exchange your u+ and u- for G+ and G-). Also the integral of the function itself vanishes in the limit.
 
  • Like
Likes vanhees71 and Enelafoxa
  • #4
Thanks everyone! Yea, I made a few typos and had a mind blank.
 

1. What is the Green's Function of a Linear ODE?

The Green's Function of a Linear ODE is a mathematical tool used to solve a linear ordinary differential equation (ODE) with a specified boundary condition. It represents the solution of the ODE at a given point in terms of a source term or forcing function.

2. How is the Green's Function used to solve a Linear ODE?

The Green's Function is used in a convolution integral to solve a linear ODE. The convolution integral involves multiplying the Green's Function by the source term and integrating over the domain of the ODE. The result is the solution to the ODE with the specified boundary condition.

3. What is the advantage of using the Green's Function to solve a Linear ODE?

The Green's Function provides a general solution to a linear ODE, which can then be applied to any specific case by incorporating the source term and boundary condition. This eliminates the need to solve the ODE for each individual case, saving time and effort.

4. Can the Green's Function be used for non-linear ODEs?

No, the Green's Function can only be used for linear ODEs. Non-linear ODEs do not have a unique solution that can be represented by a Green's Function.

5. How is the Green's Function related to the Impulse Response Function?

The Green's Function is closely related to the Impulse Response Function, which represents the output of a system in response to an impulse input. The Impulse Response Function is the Green's Function for a specific type of linear ODE known as a homogeneous ODE.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
908
  • Calculus and Beyond Homework Help
Replies
1
Views
708
  • Calculus and Beyond Homework Help
Replies
16
Views
570
  • Calculus and Beyond Homework Help
Replies
1
Views
297
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
17
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
0
Views
173
Back
Top