

Subject:
Differential Equation : Coupled, Linear, First Order
Category: Science > Math Asked by: johnpinckneyga List Price: $25.00 
Posted:
25 May 2006 17:42 PDT
Expires: 24 Jun 2006 17:42 PDT Question ID: 732451 
Analytical solution for this problem: f=f(x,y) g=g(x,y) (d/dx)f + f = g (d/dy)g + g = f under conditions: f(0,y)= 1 g(x,0)= 0 

There is no answer at this time. 

Subject:
Re: Differential Equation : Coupled, Linear, First Order
From: sabertoothed123ga on 30 May 2006 11:20 PDT 
HI u know cant understand ur question .............. 
Subject:
Re: Differential Equation : Coupled, Linear, First Order
From: berkeleychocolatega on 06 Jun 2006 17:37 PDT 
Here's some information suggesting that there is no solution with your given boundary conditions. df/dx = g  f =  dg/dy. Take the derivative of your first equation with respect to y and substitute for dg/dy to get: d^2f/dxdy + df/dx + df/dy = 0. You probably knew that. Now try a solution of the form: f(x,y) = u(x)*v(y). After some work one gets that f(x,y) = A*exp(cx  cy/(c+1)) and g(x,y) = A(c+1)*exp(cx  cy/(c+1)). Noticing that if f and g are solutions, then so are f + k and g + k, we get solutions with three parameters: the above f + k and the above g + k. None of these can fulfill your boundary conditions. The pde in f is obviously linear homogeneous with solutions exp((cx  cy/(c+1)) and 1. This makes it infinite dimensional since c can vary. I guess that's what happens in pde's. There may be other solutions to the equations, but I don't know what they would be. 
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