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Bug introduced in 8.0 or earlier and persisting through 11.1.0 or later


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.

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  • $\begingroup$ A workaround: DSolve[SortBy[FreeQ[#, G] &] /@ pde, G[x1, x2, x3, x4], {x1, x2, x3, x4}]. $\endgroup$ – Michael E2 Apr 16 '17 at 17:00
  • $\begingroup$ Thanks for the workaround @MichaelE2. I really have the expressions in a list with expressions of the form F + G == 0 or F-G==0. If I simplify this I get F+G==0 and F==G respectively. Your workaround works in the later case but not in the former. Is there a way to always get this in the form G == something? $\endgroup$ – Jeremy Upsal Apr 16 '17 at 18:02
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    $\begingroup$ I see this bug even in Mathematica 8.0.4 I have at hand currently. $\endgroup$ – innaiz Apr 16 '17 at 18:36
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This is a workaround, as requested in a comment and extended to handle the updated problem. The idea is to solve the pde system for the derivatives so that we can put the derivatives of G on the left-hand side of the equation. There is a hard-coded internal pattern that assumes the problem will be set up this way. Solve returns these in the form of a Rule. Replacing Rule by Equal converts them back to equations, but with the derivatives on the LHS. Update: Additionally, the hard-coded pattern requires the derivatives of G to be in a the same order as the arguments' order, that is, D[.., x0] ==.., D[.., y0] ==.., D[.., z0] ==... This can be fixed by sorting the derivatives of G.

normal[U_] = Function[pde, 
   Reverse@Sort@First@Solve[pde, 
        Cases[Variables[pde /. Equal -> List], _?(! FreeQ[#, U] &)]] /. 
    Rule -> Equal];

DSolve[normal[G]@{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}]

Mathematica graphics

N.B. We're assuming we've got a linear first-order pde system here. (A similar approach could work on higher-order systems, with some work.)

Example in the update

DSolve[normal[G]@badPDE, G, {x0, x1, y0, y1, z0, z1}]

Mathematica graphics

It seems the internal code assumes in the 3D case that the equations have been set up in the order

Grad[G, {x1, y1, z1}] == V /; Curl[V, {x1, y1, z1}] == {0, 0, 0}

This suggests this change to normal:

normal[u_, vars_] = Function[pde, First@Solve[pde, Grad[u, vars]] /. Rule -> Equal];

And this change in its usage, with the arguments being specified:

normal[G[x0, x1, y0, y1, z0, z1], {x1, y1, z1}]@badPDE

Here is another way to use normal to fix DSolve:

ClearAll[gradify];
SetAttributes[gradify, HoldAll];
gradify[code_DSolve] := Internal`InheritedBlock[{DSolve`DSolvePDEs},
  Unprotect[DSolve`DSolvePDEs];
  DSolve`DSolvePDEs[eqs_, {u_}, v : {x_, y_, z_}, c_, i_] := 
   With[{gradeqs = normal[u @@ v, v]@eqs},
    DSolve`DSolvePDEs[gradeqs, {u}, {x, y, z}, c, i] /; gradeqs =!= eqs];
  Protect[DSolve`DSolvePDEs];
  code
  ]

Example:

gradify@DSolve[badPDE, G, {x0, x1, y0, y1, z0, z1}]

Mathematica graphics

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  • $\begingroup$ Is DSolve`DSolvePDEs documented anywhere? $\endgroup$ – bbgodfrey Apr 18 '17 at 19:30
  • $\begingroup$ @bbgodfrey No, I doubt it. DSolve has several special utility solvers for different types of problems, judging from the DSolve` context. $\endgroup$ – Michael E2 Apr 18 '17 at 20:48
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This looks like a bug. Clearly the order should not make a difference. Here is another version of the problem. Putting both on the same side

lhs1 = D[g[x1, x2, x3, x4], x3];
rhs1 = D[f[x1, x2], x1];
lhs2 = D[g[x1, x2, x3, x4], x4];
rhs2 = D[f[x1, x2], x2];

Now

 DSolve[{lhs1==rhs1,lhs2==rhs2},g[x1,x2,x3,x4],{x4,x3,x2,x1}]

Mathematica graphics

But

 DSolve[{lhs1-rhs1==0,lhs2-rhs2==0},g[x1,x2,x3,x4],{x1,x2,x3,x4}]

Mathematica graphics

In Maple, the order does not affect the result

Mathematica graphics

I remember seeing issue once where order of variables made difference in result, but this I remember was in integrate not dsolve? may be someone knows about this more.

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  • 1
    $\begingroup$ I agree it is a bug. I wonder if there's any clever ways around it... Maybe the most clever solution is to use maple for this. $\endgroup$ – Jeremy Upsal Apr 16 '17 at 15:40
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    $\begingroup$ @JeremyUpsal You should report it to WRI. It appears your PDE system belongs to a special case. The solver is set up for the equations to be in the order LHS = derivative of unknown, RHS is free of unknown. A pre-processor ought to be able to rewrite the system in the proper order, but either there isn't one or it fails. I wouldn't think it would hard for them to fix. $\endgroup$ – Michael E2 Apr 16 '17 at 16:52
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    $\begingroup$ Nasser and @MichaelE2 see new edit for another similar bug. Again this doesn't have any issues in Maple, although the "bad" arrangement for Mathematica does take longer for Maple to solve than the "good" arrangement, so it is also doing some rearrangement. $\endgroup$ – Jeremy Upsal Apr 17 '17 at 14:33
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    $\begingroup$ There was an issue with the order of dependent variables in DAE (see here), but that has been fixed. Not sure, whether that issue is related. $\endgroup$ – gwr Apr 17 '17 at 16:20

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