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How to prove [itex]\Phi[/itex] (x,y)

  • Thread starter jhosamelly
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  • #1
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Homework Statement



I need to prove this

<< [itex]\Phi[/itex] (x,y) >> = [itex]\frac{4}{5}[/itex] [itex]< \Phi (x,y)>_{c}[/itex] + [itex]\frac{1}{5}[/itex] [itex]< \Phi (x,y)>_{d}[/itex]


Homework Equations



[itex]< \Phi (x,y)>_{c}[/itex] = [itex]\frac{1}{4}[/itex] [itex]\left[< \Phi (x,y+h)> + < \Phi (x-h,y)> + < \Phi (x,y-h)> \right][/itex]

[itex]< \Phi (x,y)>_{d}[/itex] = [itex]\frac{1}{4}[/itex] [itex]\left[< \Phi (x+h,y+h)> + < \Phi (x-h,y+h)> + < \Phi (x-h,y-h)> + < \Phi (x+h,y-h)> \right][/itex]

The Attempt at a Solution



I tried to substitute the formula for [itex]< \Phi (x,y)>_{c}[/itex] and [itex]< \Phi (x,y)>_{d}[/itex] to no avail. I don't know how to work on this kind of problem. Can somebody please tell me how should I do this? I'm willing to learn. Help much appreciated. Thanks.
 

Answers and Replies

  • #2
Mute
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You're going to have to define what your notation means before anyone can help you.
 
  • #3
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Oww., hmm.. I think [itex]\Phi[/itex] is potential... I dont know about the others. I think this is just a mathematical expression that you need to manipulate to prove the equation.
 
  • #4
Mute
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That's still not enough information to help solve the problem. What do the angular brackets mean? What about the double angular brackets? Are they averages? If it's an average, what is it an average over? h?

If you don't know the answers to these questions, then there's nothing to do. All you've given an expression for ##\langle \langle \Phi \rangle \rangle## in terms of some quantities ##\langle \Phi \rangle_c## and ##\langle \Phi \rangle_d##, which you've also given expressions for, but unless we know what all of those are supposed to be, the best we can do is plug ##\langle \Phi \rangle_c## and ##\langle \Phi \rangle_d## into the expression for ##\langle \langle \Phi \rangle \rangle##.
 
  • #5
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Ic, thanks. I did substitute the equations to the main equation, but the problem is i don't know how to manipulate them. I really need help. Thanks. :))
 
  • #6
vela
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Is your expression for ##\langle \phi(x,y) \rangle_c## missing a term? I'm thinking it's supposed to be
$$\langle \phi(x,y) \rangle_c = \frac{1}{4}[\langle \phi(x+h,y) \rangle + \langle \phi(x-h,y) \rangle + \langle \phi(x,y+h) \rangle + \langle \phi(x,y-h) \rangle]$$ I'm inferring that all of this has to do with solving Laplace's equation numerically. Am I right?
 
  • #7
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Yes yes, correct vela... But the equations that i posted are the ones given by our professor. Im not really sure if there's something wrong with it. I copied it correctly from my notebook. Hmmm.. Is there a site where this thing is discussed? I can't seem to find one. Thanks for all the help.
 
  • #8
vela
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OK, your formula for ##\langle\phi(x,y)\rangle_c## is wrong*. I'll show you how it's derived. The formula for ##\langle\phi(x,y)\rangle_d## can be found in a similar way. I don't know how to show what you've been asked to show, but seeing where the other formulas come from may give you an idea of how to proceed.

The idea is to use a Taylor series to approximate ##\phi##. For example,
\begin{align*}
\phi(x+h,y) &= \phi(x,y) + h\frac{\partial\phi}{\partial x}(x,y) + \frac{1}{2!}h^2\frac{\partial^2\phi}{\partial x^2}(x,y) + \text{ higher-order terms} \\
\phi(x-h,y) &= \phi(x,y) - h\frac{\partial\phi}{\partial x}(x,y) + \frac{1}{2!}h^2\frac{\partial^2\phi}{\partial x^2}(x,y) + \text{ higher-order terms}.
\end{align*} If you add these two equations together, the first-order terms cancel, and you're left with
$$\phi(x+h,y) + \phi(x-h,y) = 2\phi(x,y) + h^2\frac{\partial^2\phi}{\partial x^2}(x,y) + \text{ higher-order terms}.$$ If you do the same thing with y, you end up with
$$\phi(x,y+h) + \phi(x,y-h) = 2\phi(x,y) + h^2\frac{\partial^2\phi}{\partial y^2}(x,y) + \text{ higher-order terms}.$$ Combining the x and y equations and neglecting the higher-order terms, you end up with
$$\phi(x+h,y) + \phi(x-h,y) + \phi(x,y+h) + \phi(x,y-h) = 4\phi(x,y) + h^2\left(\frac{\partial^2\phi}{\partial x^2} + \frac{\partial^2\phi}{\partial y^2}\right).$$ The term in the parentheses vanishes because ##\nabla^2\phi = 0##, so solving for ##\phi(x,y)##, you get
$$\phi(x,y) = \frac{1}{4}[\phi(x+h,y) + \phi(x-h,y) + \phi(x,y+h) + \phi(x,y-h)].$$ You can derive the formula for ##\langle\phi(x,y)\rangle_d## similarly by considering ##\phi(x\pm h, y\pm h)##.


*It might be that your formula for ##\langle\phi(x,y)\rangle_c## is correct, and the missing term gives rise to the expression you're asked to prove. That's another idea you might want to investigate. I think, however, that you simply copied the formula down incorrectly in your notes.
 
Last edited:
  • #9
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Thanks vela. :)) taylor series it is. I'll try to do it now. Thanks :))
 

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