Green fuktion on the wave operator in 1 dimension

In summary, Daniel explained to me that I need to solve the differential equation:\hat{O}_{x,t} G(x,x',t,t')=\delta(x-x')\delta(t-t') (1)with \hat{O}_{x,t} G(x,x',t,t')=\theta(R-c\tau) (5)to get phi(0,0). He also told me that my problem is set up wrongly and that I should use the above equation with the delta distribution to solve for G.
  • #1
nobody2100
6
0
I have a Homework here I'm trying to solve like the last 3-4 hours but somehow I'm stuck so I ask you guys for help:

They gave me the Green Function like this:
http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo9/img50.png
with http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo9/img53.png for y >= 0 and http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo9/img55.png for y<0
Now I'm supposed to solve this:
http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo9/img61.png
with
http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo9/img36.png and http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo5/img39.png and http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo9/img62.png

Would be so nice if someone could help me! thanks in advance!
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
So you're basically saying to solve
[tex] box_{1}[I_{G_{1}}(\phi)] [/tex]

That means [itex] box_{1} \int G_{1}(x,t)\phi (x) dx [/itex]
with he "box"/d'alembertian taken wrt to "y" and "t'"("t" prime)
Then u can insert the d'alembertian under the integral and use the differential eq.verified by the propagator.

It think the integration will be immediate...

Daniel.
 
  • #3
thx for your quick reply! But I don't quite following that one, could you explain it to me with more details? and phi is also a function of t ;)
I Expect to get phi(0,0) at the end. I tried some more and got something similar except a minus in front of it but i really don't think that it was mathematically correct what I did there over those 2 pages x_X have to get this one done by tomorrow morning so i would appreciate any kind of help! thx!
What i basically did until now is that I made my integration invervals smaller:
for all t < 0 Teta is 0 anyway so my integral over dt goes from 0 to infinite, now with my t >0 I started to change the interval for dx by only integrating over the support of Teta that means my Interval changed to [-ct , ct]
Now my first problem was that my Integration Interval is a function of t. I did get one differentation to t inside the Integral over dx and got 3 terms and well ... here I'm not sure anymore if that was right x_X
This is the last thing i have where I'm quite sure that it could be right:

[itex] ((1/c^2) \partial_{t}^2 - \partial_{x}^2) \int_{0}^{\infty} dt \int_{-ct}^{ct} dx ( (c/2)\phi (x,t) ) [/itex]
 
Last edited:
  • #4
anyone? still can't solve it x_X
 
  • #5
Okay,first question:I find kind of awkward the shape of the Green's function.
It should always look like this:
[tex] G_{1}(x,x',t,t') [/tex] and should verify identically the differential equation
[tex] \hat{O}_{x,t} G(x,x',t,t')=\delta(x-x')\delta(t-t') [/tex](1)

U can make the substitution
[tex] x-x' \rightarrow R [/tex](2)
[tex] t-t' \rightarrow \tau [/tex] (3)

and the new propagator would be
[tex] G_{1}(R,\tau}) [/tex],which would check the LPDE

[tex] \hat{O}_{R,\tau} G_{1}(R,\tau)=\delta(R)\delta(\tau) [/tex](4)

If u say that your propagator is
[tex] G_{1}(R,\tau)=\theta(R-c\tau) [/tex](5)
and u want to compute
[tex] I_{G_{1}}=\int [O_{R,\tau}G_{1}(R,\tau)]\phi(R,\tau)=\int \delta(R)\delta(\tau)\phi(R,\tau) =\phi(0,0) [/tex]

Daniel.
 
Last edited:
  • #6
BTW,your problem is set up wrongly.
If u were to do exactly what was indicated,u'd end up with a dalembertian applied to a number (the value of the integral) which would be identically zero.

Daniel.
 
  • #7
yes, it should look like G(t,t',x,x') but t' and x' are set 0 we kinda did it in the couse that we subsitute it.
[tex] G_{1}(R,\tau)=\theta(R-c\tau) [/tex]
Is practically the same
and this:
[tex] I_{G_{1}}=\int [O_{R,\tau}G_{1}(R,\tau)]\phi(R,\tau)=\int \delta(R)\delta(\tau)\phi(R,\tau) =\phi(0,0) [/tex]
is exactely what I'm supposed to show
Yes the notation looks like ****, but the "box" operator is kinda a differentiation of a distribution here:
[tex] \partial I_{G_{1}}( \phi(t,x)) = I_{\partial G_{1}}( \phi(t,x)) = - I_{G_{1}}( \partial \phi(t,x)) [/tex]
could you just help me how you got the last line? or how to solve this one (hope it's right this time):
[itex] \int_{0}^{\infty} dt \int_{-ct}^{ct} dx ( ((1/2c) \partial_{t}^2 - (c/2)\partial_{x}^2)\phi (x,t)) [/itex]
 
Last edited:
  • #8
Well,i just I've given a proof.Remember that the Green function/propagator is a solution of this eq.
[tex] \hat{O}G(R,\tau)=\delta(R)\delta\tau [/tex]
which means
[tex] \hat{O}\theta(R-c\tau)=\delta(R)\delta\tau [/tex]
,which is just i used to get the integral involving the product of delta functionals and the 'phi'.

Daniel.
 
  • #9
I know, that this is the solution but in this homework they say I have to show explicit, that this given G actually is a Green function to that Operator by solving that equasion. So I'm not allowed to use that equation above with the delta distribution because that is exactely what i have to proof ^^
I Just cleared up that integral and finally got a [tex] \phi(0,0) [/tex] term. still there are some ugly integrals left which i hope will result to 0 somehow ...
 

1. What is the Green function on the wave operator in 1 dimension?

The Green function on the wave operator in 1 dimension is a function that describes the response of a system to an impulse input. It is used in mathematical physics to solve problems involving wave equations in one dimension.

2. How is the Green function on the wave operator derived?

The Green function on the wave operator is derived by solving the wave equation with a delta function as the initial condition. This results in a solution that can be used to represent any arbitrary input as a sum of delta functions, making it a powerful tool for solving wave equations.

3. What are the properties of the Green function on the wave operator?

The Green function on the wave operator has several important properties, including linearity, time-reversal symmetry, and causality. It is also a fundamental solution of the wave equation, meaning it satisfies the equation and all of its derivatives.

4. How is the Green function on the wave operator used in practical applications?

The Green function on the wave operator is used in a variety of practical applications, such as solving problems in electromagnetism, acoustics, and quantum mechanics. It can also be used to model the behavior of waves in different types of media, making it a versatile tool for understanding and predicting physical phenomena.

5. Are there any limitations to using the Green function on the wave operator?

While the Green function on the wave operator is a powerful tool, it does have limitations. It is only applicable to linear systems, and it assumes that the system is time-invariant and has a well-defined impulse response. Additionally, it may not be suitable for solving problems with complex boundary conditions or non-uniform media.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
4K
  • Calculus and Beyond Homework Help
Replies
1
Views
692
  • Electromagnetism
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
4K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
8
Views
346
  • Introductory Physics Homework Help
Replies
1
Views
12K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top