Not many chances

Preben Alsholm — Fri, 25 Feb 2011 17:46:08 Z

Using that the two mixed partials w.r.t. x and y are equal you can write down a condition for solving the system:

restart;
pde1 := diff(f(x,y),x) + A(x) * ( p11* f(x,y) + p12 + p13*x )
= p14 * B(y)/B(x) * diff(f(x,y),x);

pde2 := diff(f(x,y),y) + A(y) * ( p21* f(x,y) + p22 + p23*y )
= p24 * B(x)/B(y) * diff(f(x,y),y);

diff(isolate(pde1,diff(f(x,y),x)),y);
diff(isolate(pde2,diff(f(x,y),y)),x);
factor(rhs(%%-%));
Nm:=numer(%):
#Your first special case:
collect(eval(Nm,{p11=0,p21=0}),diff,factor); #This has to be zero
#Your second special case:
factor(eval(Nm,{p11=0,p21=0,p24=0}));#This has to be zero

In both cases you get requirements on the coefficients for solutions to exist.

In the latter special case the requirement is that

A(x)*B(y)^2*B(x)*(diff(B(y), y))*p14*(p12+p13*x)=0 for all (x,y) in some open non-empty set.

Unless p14 = 0 or p12 = p13 = 0 then the only interesting case would be B = constant.

sad

PatrickT — Sat, 26 Feb 2011 11:41:42 Z

> Using that the two mixed partials w.r.t. x and y are equal you can write down a condition for solving the system

Oh that's very useful, thanks A LOT Preben.

On the other hand I'm a little sad: I don't see any "solvability" condition that matches my modeling constraints...

MaplePrimes - answers and comments on Question, Is this a standard PDE ?

Not many chances

sad