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
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))} *)
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]]}} *)
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