DSolve takes too much time

Question

I have the following problem:

DSolve[D[l[w1, w2], w1] a w2 - D[l[w1, w2], w2] a w1 == 
  l[w1, w2] + w1 + a^2 w2^2, l[w1, w2], {w1, w2}]

I expect DSolve to return a complete polynomial of second degree as a solution to this differential equation. Yet it is taking an awful lot of time to solve this. Why is it?

EDIT

I expect a solution like:

l = a1 w1 + a2 w2 + a11 w1^2 + a12 w1 w2 + a22 w2^2; 

Since:

h = D[l, w1] a w2 - D[l, w2] a w1 - l - w1 - a^2 w2^2; 
Solve[Table[CoefficientRules[h, {w1, w2}][[i]][[2]] == 0, {i, 1, 3}]] 

Outputs:

{{a1 -> -1 - a*a2, a12 -> -(a11/a), a22 -> ((1 + 2*a^2)*a11)/(2*a^2)}, {a -> 0, a1 -> -1, a11 -> 0, a12 -> 0}}

EDIT 2

In the past edit I didn't add the assumption a>0. Indeed the verification turns a trivial solution. I've corrected this, and now we have a real solution with the following code.

Solve[-a11 - a*a12 == 0 && 
  2*a*a11 - a12 - 2*a*a22 == 0 && -1 - a1 - a*a2 == 0 && 
  a > 0  , {a11, a12, a1, a2, a22}, Reals]

Answer 1

Probably because DSolve is looking for a general solution, while a solution like l = a1 w1 + a2 w2 + a11 w1^2 + a12 w1 w2 + a22 w2^2 is far beyond general. For example, with the following code we can find another part of the general solution (The definition of DChange can be found here.):

neweqn = DChange[D[l[w1, w2], w1]*a*w2 - D[l[w1, w2], w2]*a*w1 == 
       l[w1, w2] + w1 + a^2*w2^2, l[w1, w2] == L[w1*w2]]

(* (-a)*w1^2*Derivative[1][L][w1*w2] + a*w2^2*Derivative[1][L][w1*w2] == 
   w1 + a^2*w2^2 + L[w1*w2] *)  

DSolve[neweqn /. w1 -> W/w2, L@W, W] /. {L@W -> l[w1, w2], W -> w1 w2} // Simplify

Notice this solution is still incomplete, it only represents solutions that satisfy $l(w_1,w_2)=L(w_1 w_2)$, yet it's already much more complicated than a polynomial. One can expect the complete solution for the PDE is even more complicated and hard to obtain at least for Mathematica.

Finally, I hate to admit it, but Maple does a better job on this PDE:

pdsolve([diff(l(w1,w2),w1)*a*w2-diff(l(w1,w2),w2)*a*w1 = l(w1,w2)+w1+a^2*w2^2],l(w1,w2))

(* {l(w1,w2) = (Intat(exp(-1/a*arctan(_a/(-_a^2+w1^2+w2^2)^(1/2)))*(-_a^2*a^2+(w1^2+w2^2)*a^2+_a)/a/(-_a^2+w1^2+w2^2)^(1/2),_a = w1)+_F1(w1^2+w2^2))*exp(1/a*arctan(w1/w2))} *)

---

Update

Inspired by the form of the general solution given by Maple, I figured out how to obtain it fast with DSolve. We just need to transform to polar coordinate!:

neweqn = Assuming[{r > 0, -Pi < th < Pi}, 
  DChange[D[l[w1, w2], w1] a w2 - D[l[w1, w2], w2] a w1 == 
    l[w1, w2] + w1 + a^2 w2^2, {Sqrt[w1^2 + w2^2] == r, th == ArcTan[w1, w2]}, {w1, 
    w2}, {r, th}, l[w1, w2]]]

(* l[r, th] + r*(Cos[th] + a^2*r*Sin[th]^2) + 
     a*Derivative[0, 1][l][r, th] == 0 *)

DSolve[neweqn, l[r, th], {r, th}] /. {l[r, th] -> l[w1, w2], r -> Sqrt[w1^2 + w2^2], 
   th -> ArcTan[w1, w2]} // Simplify

(* {{l[w1, w2] -> (1/(
    2 + 10 a^2 + 
     8 a^4))(4 a^3 (1 + a^2) w1 w2 - (1 + 4 a^2) (2 w1 + (a^2 + a^4) w1^2 + 
         a w2 (2 + a w2 + a^3 w2)) + a^2 (1 + a^2) (w1^2 + w2^2) Cos[2 ArcTan[w1, w2]]) +
     E^(-(ArcTan[w1, w2]/a)) C[1][Sqrt[w1^2 + w2^2]]}} *)

Answer 2

Clear[l];

The differential equation is

deqn = D[l[w1, w2], w1] a w2 - D[l[w1, w2], w2] a w1 == 
   l[w1, w2] + w1 + a^2 w2^2;

The assumed solution is

aSoln = l -> (a1 #1 + a2 #2 + a11 #1^2 + a12 #1 #2 + a22 #2^2 &);

Substituting the assumed solution into the differential equation

eqn = deqn /. aSoln

(*  a w2 (a1 + 2 a11 w1 + a12 w2) - a w1 (a2 + a12 w1 + 2 a22 w2) == 
 w1 + a1 w1 + a11 w1^2 + a2 w2 + a12 w1 w2 + a^2 w2^2 + a22 w2^2  *)

Equating the coefficients on the LHS of eqn with those on the RHS of eqn

coefEqn = Thread[
  (CoefficientList[#, {w1, w2}] // Flatten) & /@
   eqn]

(*  {True, a a1 == a2, a a12 == a^2 + a22, -a a2 == 1 + a1, 
 2 a a11 - 2 a a22 == a12, True, -a a12 == a11, True, True}  *)

Solving for the coefficients of the assumed solution

coefSoln = Solve[
    coefEqn, {a1, a2, a11, a12, a22},
    Reals][[1]] // Simplify

(*  {a1 -> -(1/(1 + a^2)), 
   a2 -> -(a/(1 + a^2)), 
   a11 -> -((2*a^4)/(1 + 4*a^2)), 
   a12 -> (2*a^3)/(1 + 4*a^2), 
   a22 -> -((a^2 + 2*a^4)/
          (1 + 4*a^2))}  *)

Substituting these coefficients into the assumed solution

soln = aSoln /. coefSoln

(*  l -> (-(#1/(1 + a^2)) - 
        (a*#2)/(1 + a^2) - 
        ((2*a^4)*#1^2)/(1 + 4*a^2) + 
        ((2*a^3)*#1*#2)/(1 + 4*a^2) - 
        ((a^2 + 2*a^4)*#2^2)/
          (1 + 4*a^2) & )  *)

Verifying that soln satisfies deqn

deqn /. soln // Simplify

(*  True  *)

So the solution is

l[w1_, w2_] = l[w1, w2] /. soln

(*  -(w1/(1 + a^2)) - (2*a^4*w1^2)/
     (1 + 4*a^2) - (a*w2)/(1 + a^2) + 
   (2*a^3*w1*w2)/(1 + 4*a^2) - 
   ((a^2 + 2*a^4)*w2^2)/(1 + 4*a^2)  *)