Mastering PDEs: Solving the Non-Constant Coefficient d^2G/dxdy Equation

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Discussion Overview

The discussion revolves around solving the partial differential equation (PDE) given by d²G/dxdy + (a-1)dG/dx*dG/dy = 0, where G is a function of x and y. Participants explore various methods and assumptions for finding solutions, particularly considering cases where the coefficient a may not be constant.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes the solution form G(x,y) = X(x)Y(y) but another participant disagrees, suggesting that this assumption may not lead to a valid solution.
  • Another participant claims to have found a specific solution G(x,y) = -ln([1+(a-1)(X(x)+Y(y))]^[1/(a-1)]) and expresses a desire to understand how to derive this solution.
  • There is a suggestion that assuming G(x,y) = X(x) + Y(y) could simplify the problem, leading to the conclusion that either X=0 or Y=0 or a=1, with the latter being deemed uninteresting.
  • One participant mentions the possibility of using a transformation (z = x + y, w = x - y) to potentially simplify the PDE.
  • A reference to the method of characteristics is made, indicating a connection to established mathematical techniques for solving PDEs.
  • Another participant questions the validity of the solutions derived from assuming G(x,y) = X(x) + Y(y), suggesting that these cases may not provide meaningful solutions.
  • There is a mention of the problem resembling a Goursat problem, indicating a specific type of boundary value problem in PDEs.
  • A later post asks about numerical solutions for the PDE using MATLAB, indicating interest in computational approaches.

Areas of Agreement / Disagreement

Participants express differing views on the validity of certain assumptions and proposed solution forms. There is no consensus on the best approach to solve the PDE, and multiple competing methods are discussed.

Contextual Notes

Some assumptions about the nature of the function a and its implications for the solution are not fully explored. The discussion includes various proposed methods without resolving the effectiveness or correctness of each approach.

Who May Find This Useful

Readers interested in advanced mathematics, particularly those studying partial differential equations and their solutions, may find this discussion relevant.

toptrial
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d^2G/dxdy+(a-1)*dG/dx*dG/dy=0
where G is a function of x and y.

Moreover, what if a is not a constant, but instead a function of x and y?
 
Last edited:
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Can't you assume a solution of the form G(x,y) = X(x)Y(y)
 
I am afraid not.
In fact, I know G(x,y)=-ln([1+(a-1){X(x)+Y(y)}]^[1/(a-1)]) is the solution. I am just trying to figure out how to slove this pde.
Thanks
 
If you assume G(x,y)=X(x)+Y(y)
will that do the trick?

You'll get:
(a-1)XY=0, then either X=0 or Y=0 or a=1, the last one is uninterseting, so you have two solutions here.

There not much to go out here, you use multiplication or addition, division and substraction are defined apostriori by them.

Edit: or any other composition of the elementary functions.

I am quite sure the way your textbook or teacher made this question, that they knew already the answer, and then found which equation it satisifies, you know the solution so take the derivative and see his process of devising the question.
 
Last edited:
For some problems like this one can sometimes use the transformation

z = x + y
w = x - y

Maybe some terms drop off and a new simpler PDE is achieved? It's like a rotation.
 
What RedBranchKnight refers to is a special case of the http://en.wikipedia.org/wiki/Method_of_characteristics" .
 
Last edited by a moderator:
loop quantum gravity said:
If you assume G(x,y)=X(x)+Y(y)
will that do the trick?

You'll get:
(a-1)XY=0, then either X=0 or Y=0 or a=1, the last one is uninterseting, so you have two solutions here.

Um, aren't all of those uninteresting cases? This is saying that any G which is a function of just one of the variables (x or y, but not both) is a solution. In that case, one of the 1st-derivatives will be zero, as will the mixed 2nd-derivative. The original equation becomes 0=0.
 
Well, I understand now, the solution's of toptrial is the answer itself, because you don't know what is X(x), Y(y), you need to take the derivative of G(x,y), and find what are X(x) and Y(y).
Or so I think.
 
  • #10
Hi guys, how would I solve this pde numerically in matlab?

d^2G/dxdy+A*dG/dx*dG/dy=0
where G and A are both functions of x and y.
 

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