Bug introduced in 8.0.4 or earlier, persisting in part through 11.3.

I was willing to solve this nonlinear PDE:

DSolve[D[l[w1, w2], w1] 0.5 w2 - D[l[w1, w2], w2] 0.5 w1 - w1 == 0, 
 l[w1, w2], {w1, w2}]

Any general solution is ok, so I didn't include boundary conditions. My goal is only to find out if there is a l[w1,w2] that solves the equation. Mathematica isn't outputting anything. What is it I'm getting wrong?


I would like to point out that such a solution may easily be found by hand. For instance this already is a possible solution:

$l=-2*w2+A(w1^2 + w2^2)$

with A being any constant.

  • $\begingroup$ When plotting numerically I get the warning: "No DirichletCondition or Robin-type NeumannValue was specified for \ {l}; the result is not unique up to a constant" $\endgroup$
    – Feyre
    Nov 7, 2016 at 12:57
  • $\begingroup$ Sorry, would you paste your bits of code. I don't receive any result at all. I don't expect the result to be unique as I have not specified any sort of boundary condition. I'm just interested in verifying that \{l} exists. @Feyre $\endgroup$ Nov 7, 2016 at 13:30
  • $\begingroup$ Like I said, I could only solve it numerically: s=NDSolve[D[l[w1, w2], w1] 0.5 w2 - D[l[w1, w2], w2] 0.5 w1 - w1 == 0, l[w1, w2], {w1, 0, 3}, {w2, 0, 3}][[1,1,2]]. Plot with: Plot3D[s, {w1, 0., 3.}, {w2, 0., 3.}] $\endgroup$
    – Feyre
    Nov 7, 2016 at 13:44

1 Answer 1


A solution can be obtained easily using DSolve, if 0.5 is replaced by 1/2.

Flatten@DSolve[D[l[w1, w2], w1]  w2/2 - D[l[w1, w2], w2] w1/2 - w1 == 0, 
    l[w1, w2], {w1, w2}]
(* {l[w1, w2] -> -2 Sqrt[w2^2] + C[1][1/2 (w1^2 + w2^2)], 
    l[w1, w2] -> 2 Sqrt[w2^2] + C[1][1/2 (w1^2 + w2^2)]} *)

It is not obvious (to me) why this change matters, because

0.5 == 1/2
(* True *)


As noted below by xzczd, one of the two solutions derived here is spurious due to a bug in DSolve. To show this, plug the two solutions in turn into the original PDE:

FullSimplify[Unevaluated[D[l[w1, w2], w1]  w2/2 - D[l[w1, w2], w2] w1/2 - w1] /. #] & /@ %%
(* {w1 (-1 + w2/Sqrt[w2^2]), -((w1 (w2 + Sqrt[w2^2]))/w2)} *)

Thus, only the first solution is valid for w2 > 0, and only the second for w2 < 0. This error is related to the one discussed in more detail in 130857.

Update for Version 11.1.0

With version 11.1,

Flatten@DSolve[D[l[w1, w2], w1] 0.5 w2 - D[l[w1, w2], w2] w1/2 - w1 == 0, 
    l[w1, w2], {w1, w2}] // Simplify
(* {l[w1, w2] -> -2 Sqrt[w2^2] + C[1][1/2 (w1^2 + w2^2)], 
    l[w1, w2] -> 2 Sqrt[w2^2] + C[1][1/2 (w1^2 + w2^2)]} *)

which is the machine-precision equivalent of the first result in this answer. This certainly is a step in the right direction. However, the problem remains only one of the two solutions is correct.

  • 1
    $\begingroup$ It might be the old "Rationalize" thing. I often bump into this sort of problem. I never get why some times it works and othertimes it doesn't. $\endgroup$ Nov 7, 2016 at 20:41
  • 1
    $\begingroup$ @MirkoAveta Your suggestion seems reasonable and was, in fact, my first thought. However, the fact that 0.5 == 1/2 is True causes me to wonder. I ran Trace[] on both versions but obtained no useful insight. Trace[] with TraceInternal->True might be more informative, but would take a while to assess, because the output probably would be huge. $\endgroup$
    – bbgodfrey
    Nov 7, 2016 at 20:49
  • 1
    $\begingroup$ I just want to point out this solution has the same issue as the one you mentioned in this post, the correct solution is actually l -> Function[{w1, w2}, -2 w2 + C[1][1/2 (w1^2 + w2^2)]] $\endgroup$
    – xzczd
    Nov 12, 2016 at 3:48
  • $\begingroup$ @xzczd Yes, I somehow overlooked this connection. Thanks. $\endgroup$
    – bbgodfrey
    Nov 12, 2016 at 3:53

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