How can i solve P(x,y) with 2 equations of diff

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esat gulhan
esat gulhan le 12 Sep 2020
Commenté : esat gulhan le 12 Sep 2020
Hi guys, .
My problem is to find P(x,y), how can i find P(x,y) with this equations. I can accept with analtical as well as numerical solutions. I know dP/dx, and dP/dy but i dont know how to solve P(x,y). My code is below
clc; clear;
syms a b p x y u v c P
u=a*x+b;
v=-a*y+c*x;
eq1=diff(P,x)==-p*(u*diff(u,x)+v*diff(u,y))
eq2=diff(P,y)==-p*(u*diff(v,x)+v*diff(v,y))

Réponse acceptée

BOB MATHEW SYJI
BOB MATHEW SYJI le 12 Sep 2020
If initial conditions are given, then this code might help.
clc; clear;
syms a b p x y u v c P t1 t2
u=(a*x)+b;
v=(-a*y)+(c*x);
s1=-p*(u*diff(u,x)+v*diff(u,y));
s2=-p*(u*diff(v,x)+v*diff(v,y));
s3(x)=int(s1,x)+t1;
% s3 and s4 are P(x,y) itself. t1 and t2 represents
% constants containing variables y and x respectively after integration
% t1 can be solved by applying the initial conditions to s3(x)
% t1=s3(x)-int(s1,x); for given x value and corresponding value of P(x,y)
s4(y)=int(s2,y)+t2;
% t2 can be solved by applying the initial conditions to s4(x)
% t2=s4(y)-int(s2,y); for given y value and corresponding value of P(x,y)
P(x,y)=(s3(x)+s4(y))/2
  1 commentaire
esat gulhan
esat gulhan le 12 Sep 2020
Hey thanks,
That solution is very easy and clear, how can not i think to integrate them, my was with dsolve but it does not give exact solution. It is worth to accept.

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