Bug introduced in 6.0, persisting through 11.1.0 or later, fixed in 12.3.1 or earlier.
I am trying to obtain the general solution, $G(x_1,x_2,x_3,x_4)$ to the very easy system of PDEs: $$ \partial_{x_3} G(x_1,x_2,x_3,x_4) - \partial_{x_1} F(x_1,x_2) = 0, \\ \partial_{x_4} G(x_1,x_2,x_3,x_4) - \partial_{x_2} F(x_1,x_2) = 0 \label{a}\tag{1} $$ and in so doing I am finding the following strange behaviour in Mathematica 10.4.
DSolve[
{D[G[x1, x2, x3, x4], x3] == D[F[x1, x2], x1],
D[G[x1, x2, x3, x4], x4] == D[F[x1, x2], x2]},
G[x1, x2, x3, x4], {x1, x2, x3, x4}
]
returns what I expect:
{{G[x1,x2,x3,x4]->C[1][x1,x2]+x4 (F^(0,1))[x1,x2]+x3 (F^(1,0))[x1,x2]}}
However, if I just reorder my input:
DSolve[
{D[F[x1,x2],x1]==D[G[x1,x2,x3,x4],x3],
D[F[x1,x2],x2]==D[G[x1,x2,x3,x4],x4]},
G[x1,x2,x3,x4],{x1,x2,x3,x4}
]
Mathematica returns nothing! This obviously doesn't cause a problem for this simple example. However, the system I get from some other portion of the code is given in the form of (\ref{a}) which, if plugged into DSolve in that form, gives the same null result. This is because if I have Mathematica Simplify the system, I get a system of the second form which gives a null result. I can do something hacky and have it solve for the derivative of $G$ in terms of everything else, but I would prefer to avoid this.
Update: This is not the only bug of this form in DSolve. It appears the order in which the different equations appear also matters (for 3 or more equations), see e.g. below
For
badPDE = {D[G[x0, x1, y0, y1, z0, z1], z1] == D[F[x0, y0, z0], z0],
D[G[x0, x1, y0, y1, z0, z1], y1] == D[F[x0, y0, z0], y0],
D[G[x0, x1, y0, y1, z0, z1], x1] == D[F[x0, y0, z0], x0]}
we get
DSolve[badPDE, G, {x0,x1,y0,y1,z0,z1}]
failing in the same manner as above, it echos the command. However,
DSolve[Reverse[badPDE], G, {x0,x1,y0,y1,z0,z1}]
gives the correct answer:
{{G->Function[{x0,x1,y0,y1,z0,z1},C[1][x0,y0,z0]+z1 (F^(0,0,1))
[x0,y0,z0]+y1 (F^(0,1,0))[x0,y0,z0]+x1 (F^(1,0,0))[x0,y0,z0]]}}
I am adding this to my original bug report with WRI.
DSolve[SortBy[FreeQ[#, G] &] /@ pde, G[x1, x2, x3, x4], {x1, x2, x3, x4}]
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