Skip to main content

differential equations - DSolve takes too much time


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

Mathematica graphics


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))} *)

enter image description here





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]]}} *)

Mathematica graphics


Comments

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.