Green's Function using Laplace Transformation

Click For Summary
SUMMARY

This discussion focuses on the application of Green's Function using Laplace Transformation to solve the differential equation x' + x = f(t) with the initial condition x(t=0,t')=0. The transformation leads to the equation s G(s) + G(s) = e^-st, resulting in G(s) = (1/(s+1)) e^-st'. The final expression for the Green's function is G(t,t') = e^-(t-t') * U(t-t'), where U(t) represents the Heaviside step function. The discussion also raises a question about the system's response when f(t) = U(t-1).

PREREQUISITES
  • Understanding of differential equations and initial value problems
  • Familiarity with Laplace Transform techniques
  • Knowledge of Green's Functions and their applications
  • Basic concepts of the Heaviside step function
NEXT STEPS
  • Study the derivation of Green's Functions for various types of differential equations
  • Learn about the properties and applications of the Laplace Transform in engineering
  • Investigate the response of systems to different forcing functions using Green's Functions
  • Explore advanced topics in functional analysis related to Green's Functions
USEFUL FOR

Mathematicians, physicists, and engineers interested in solving linear differential equations and analyzing system responses using Green's Functions and Laplace Transforms.

dspampi
Messages
16
Reaction score
0
I was wondering if someone could help me go through a simple example in using Green's Function.

Lets say:
x' + x = f(t)
with an initial condition of x(t=0,t')=0;

Step 1 would be to re-write this as:
G(t,t') + G(t,t') = \delta(t-t')

then do you multiply by f(t')\ointdt' ?
which I would believe would give me:

s G(s) + G(s) = e^-st

and G(s) = \frac{1}{s+1} e^-st'
then giving me my G(t,t') = e^-(t-t') * U(t-t') ?

Not sure if that is the expected Green's function or if I screwed up somewhere.

Also, if f(t) = U(t-1), what would be the system's response?
* U fxn is a Heaviside step function
 
Physics news on Phys.org
Please don't double post your questions.
 

Similar threads

Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K