Does any one know a trick to make DSolve
find solution to this coupled linear first order PDE system: (these are Cauchy-Riemann PDE equations, but with one of them having one of the dependent variables as well).
ClearAll[F1,F2,x,y];
ode1 = D[F1[x,y],y]-D[F2[x,y],x] == 0
ode2 = D[F1[x,y],x]+D[F2[x,y],y] == y (*y here causes the problem*)
DSolve[{ode1,ode2},{F1[x,y],F2[x,y]},{x,y}]
This can be solved in Maple:
restart;
#infolevel[pdsolve]:=3;
eq1:= diff(F1(x,y),y)-diff(F2(x,y),x) = 0;
eq2:= diff(F1(x,y),x)+diff(F2(x,y),y) = y;
pdsolve({eq1,eq2},{F1(x,y),F2(x,y)});
Solution it gives is
F1(x, y) = _F1(y-I*x)+_F2(y+I*x)
F2(x, y) = I*_F1(y-I*x)-I*_F2(y+I*x)+(1/2)*y^2+_C1
Screen shot:
If the RHS of the second equation is not y
but a constant or some other parameter, then Mathematica can now solve it:
ClearAll[F1,F2,x,y,m];
ode1 = D[F1[x,y],y]-D[F2[x,y],x] == 0
ode2 = D[F1[x,y],x]+D[F2[x,y],y] == m
DSolve[{ode1,ode2},{F1[x,y],F2[x,y]},{x,y}]
Is this a known limitation of DSolve
or is there a trick or some other method to get the same solution as in Maple?
Using version 11.2 on windows 7.
Simplify[Unevaluated[D[F1[x, y], x] + D[F2[x, y], y] == y] /. F2[x, y] -> G2[x, y] + y^2/2]
as cheating? $\endgroup$ – bbgodfrey Jan 11 '18 at 4:39ClearAll[F1,F2,G2,x,y,m]; ode1=D[F1[x,y],y]-D[F2[x,y],x]==0; ode2=Simplify[Unevaluated[D[F1[x,y],x]+D[F2[x,y],y]==y]/.F2[x,y]->G2[x,y]+y^2/2]; DSolve[{ode1,ode2},{F1[x,y],G2[x,y]},{x,y}]
but it still does not solve it. $\endgroup$ – Nasser Jan 11 '18 at 4:51