I am trying to visualize the result of numerical integration in Wolfram Mathematica using DensityPlot. But dark artifacts appear inside the circle on the graph, this is incorrect. How to fix this?
f[u_, v_]:=(1/1.2)*(1/(3.14159*0.02^2)*E^(-((x-u)^2+(y-v)^2)/0.02^2)+0.2/(3.14159*2^2)*E^(-((x-u)^2+(y-v)^2)/2^2));
i[x_, y_]:=NIntegrate[f[u, v],{u, v} \[Element] Disk[{0,0},2.5], AccuracyGoal -> 30];
DensityPlot[i[x, y], {x, -3, 3}, {y, -3, 3},ColorFunction->"SunsetColors",PlotPoints -> 20, PlotLegends -> Automatic]
Link to the image: https://postimg.cc/47MJTYj7
Well, that is not an artifact, or is it. First I'd like to point out that f is a function of u, v, x and y. So the way it is programmed is not very clean. Now to the main problem. There are two Gaussian peaks. One is very broad, the other very sharp. In many cases the numerical integral just does not get points near the sharp peak, therefore not detecting it. As the rest is smooth and easy, no subset is generated and the integral is just missing this contribution. Luckily, there is an option specifically for this. MinRecursion: NIntegrate may miss sharp peaks of integrands:.... So this works
g[x_, y_, s_] := Exp[ -( x^2 + y^2 ) / s^2] / (s^2 Pi)
f[x_, y_, u_, v_] := g[ x - u, y - v, 0.02 ] / 1.2 + 0.2 g[ x - u, y - v, 2 ]
i[x_, y_] := NIntegrate[ f[x, y, u, v], {u, v} \[Element] Disk[{0, 0}, 2.5], MinRecursion -> 5]
nn = 30;
t = Table[i[x, y], {x, -3, 3, 6/nn}, {y, -3, 3, 6/nn}];
ListDensityPlot[t, ColorFunction -> "SunsetColors", PlotLegends -> Automatic]
Takes quite some time, though. Probably, it would be better to split this up in two integrals and invest some thought on the integration boundaries.
Related
I am trying to model in Mathematica the concentration (density) of an intersection of two graphs, eggs[x,y,z,t] and sperm[x,y,z,t]. The issue arises when one tries to implement an equation describing the graph's boundaries (sperm[x,y,z,t] == eggs[x,y,z,t]) into either DensityPlot3D or say the RegionFunction -> Function[{x,y,z,t},func] specifier. The methodology works for ContourPlot3D, however, DensityPlot3D is much more suitable for the task, as I am trying to animate the solutions and have the color of the graph be the concentration variable (4D output of sperm and egg equations).
Here is my code for the ContourPlot3D process:
In[1]:= sperm[x_,y_,z_,t_]= 1/Sqrt[4*Pi*t]*E^((-x^2-y^2-z^2)/(4*t));
eggs[x_,y_,z_,t_]=1/Sqrt[4*Pi*t]*E^((-(x+3)^2-(y+3)^2-(z+3)^2)/(4*t));
Animate[ContourPlot3D[sperm[x, y, z, t] == eggs[x, y, z, t], {x, -10, 10}, {y, -10, 10}, {z, -10, 10}], {t, 0, 10}]
The output normally crashes the system and thus cannot be uploaded here.
Here is my attempt to apply this method to the DensityPlot3D:
In[2]:= Animate[DensityPlot3D[sperm[x, y, z, t] == eggs[x, y, z, t], {x, -10, 10}, {y, -10, 10}, {z, -10, 10}], {t, 0, 10}]
The output is only a blank 3D plot.
After researching, I had found that RegionFunction (and RegionPlot3D) only seems to use inequalities, and the first argument in DensityPlot3D uses a function that is not in the form of an equation. I had tried to bypass this by doing something like
f = Solve[sperm[x, y, z, t] == eggs[x, y, z, t], {x, y, z, t}, Reals]
and inputting that variable as the first argument, but this did not help, as the result was still considered to be an equation (a complicated one at that).
So, my question is: How can I implement the equation as a boundary for DensityPlot3D?
EDIT: So it turns out the first argument can actually take in equations, as I was able to achieve an animation for s[x, y, t] == e[x, y, t].
Here is the code:
In[1]:= s[x_, y_, t_] = 1/Sqrt[4*Pi*t]*E^((-(x + 0)^2 - (y + 0)^2)/(4*t));
e[x_, y_, t_] = 1/Sqrt[4*Pi*t]*E^((-(x + 3)^2 - (y + 3)^2)/(4*t));
Animate[DensityPlot[s[x, y, t] == e[x, y, t], {x, -10, 10}, {y, -10, 10}, ColorFunction -> "TemperatureMap"], {t, 0, 10}]
I cannot figure out how to save the animation on here, and the picture is not uploading here.
Nevertheless, the problem then is simply the following: Why is DensityPlot3D a blank graph for the "4D" case?
What is the best way of plotting several functions with different domains into the same plot? Is there a way to do this with plot2d, or do I have to use draw2d instead?
I especially like the possibility in plot2d to give several functions in a list, whereas I would have to add the different functions in draw2d as separate parameters, if I understand the documentation correctly.
An example of what I mean:
f(x, a) := sqrt(a) * exp(-(x-a)^2);
fmax(x) := sqrt(x);
In this example I would like to plot f(x, a) for several a (e.g. using makelist(f(x, a), a, [0, 0.5, 1, 2, 5])) from -1 to 10 and fmax from 0 to 5 (to show where the maxima of the f(x, a) family of curves are located).
You can try draw2d
f(x, a) := sqrt(a) * exp(-(x-a)^2);
fmax(x) := sqrt(x);
flist: makelist(f(x, a), a, [0, 0.5, 1, 2, 5]);
par: map(lambda([f], explicit(f, x, -1, 10)), flist);
par: append([explicit(fmax, x, 0, 5), color=red], par);
load(draw);
apply(draw2d, par);
One approach I am not particularly happy with is to declare the functions with smaller domains as parametric curves, with the x axis parameter being simply x:
f(x, a) := sqrt(a) * exp(-(x-a)^2);
fmax(x) := sqrt(x);
plot2d(endcons([parametric, x, fmax(x), [x, 0, 5], [nticks, 80]],
makelist(f(x, a), a, [0, 1/2, 1, 2, 5])),
[x, -1, 10]);
This was frustrating me for hours but I found a way to have multiple differently domained functions on the same graph.
wxplot2d([if x < 0 then -x else sin(x), if x > -1 then x^2],[x,-%pi,%pi],[y,-2,2]);
I am trying to plot 1/r (r-hat) using VectorPlot in Mathematica by decomposing r-hat into x-hat and y-hat.
I am using:
[Chi][x_, y_] := Sqrt[x^2 + y^2]
[Phi][x_, y_] := ArcTan[y/x]
and then plotting the above function in x,y plane.
But, for x<0 and y<0, the vectors do not have the correct sign because the unit vectors have different sign in each of the quadrants. I tried defining a piecewise function to get around this without success.
VectorPlot[ 1/Sqrt[x x + y y] {Cos[ArcTan[x, y]], Sin[ArcTan[x, y]]},
{x, -2, 2}, {y, -2, 2},
RegionFunction -> Function[{x, y, vx, vy, n}, Sqrt[x x + y y] > 1/3],
VectorPoints -> 10]
To address the question we start with the following toy model problem being here just a case study:
Given two circles on a plane (its centers (c1 and c2) and radii (r1 and r2)) as well as a positive number r3, find all circles with radii = r3 (i.e all points c3 being centers of circles with radii = r3) tangent (externally and internally) to given two circles.
In general, depending on Circle[c1,r1], Circle[c2,r2] and r3 there are 0,1,2,...8 possible solutions. A typical case with 8 solutions :
I slightly modified a neat Mathematica implementation by Jaime Rangel-Mondragon on Wolfram Demonstration Project, but its core is similar:
Manipulate[{c1, a, c2, b} = pts;
{r1, r2} = Map[Norm, {a - c1, b - c2}];
w = Table[
Solve[{radius[{x, y} - c1]^2 == (r + k r1)^2,
radius[{x, y} - c2]^2 == (r + l r2)^2}
] // Quiet,
{k, -1, 1, 2}, {l, -1, 1, 2}
];
w = Select[
Cases[Flatten[{{x, y}, r} /. w, 2],
{{_Real, _Real}, _Real}
],
Last[#] > 0 &
];
Graphics[
{{Opacity[0.35], EdgeForm[Thin], Gray,
Disk[c1, r1], Disk[c2, r2]},
{EdgeForm[Thick], Darker[Blue,.5],
Circle[First[#], Last[#]]& /# w}
},
PlotRange -> 8, ImageSize -> {915, 915}
],
"None" -> {{pts, {{-3, 0}, {1, 0}, {3, 0}, {7, 0}}},
{-8, -8}, {8, 8}, Locator},
{{r, 0.3, "r3"}, 0, 8},
TrackedSymbols -> True,
Initialization :> (radius[z_] := Sqrt[z.z])
]
We can easily conclude that in a generic case we have an even number of solutions 0,2,4,6,8 while cases with an odd number of solutions 1,3,5,7 are exceptional - they are of zero measure in terms of control ranges. Thus changing in Manipulate c1, r1, c2, r2, r3 one can observe that it is much more difficult to track cases with an odd number of circles.
One could modify on a basic level the above approach : solving purely symbolically equations for c3 as well as redesignig Manipulate structure with an emphasis on changing number of solutions. If I'm not wrong Solve can work only numerically with Locator in Manipulate, however here Locator seems to be crucial for simplicity of controlling c1, r1, c2, r2 as well as for the whole implementation.
Let's state the questions, :
1. How can we force Manipulate to track seamlessly cases with an odd number of solutions (circles) ?
2. Is there any way to make Solve to find exact solutions of the underlying equations?
( I find the answer by Daniel Lichtblau to be the best approach to the question 2, but it seems in this instance there is still an essential need for sketching of a general technique of emphasizing measure zero sets of solutions while working with Manipulate )
These considerations are of less importance while dealing with exact solutions
For example Solve[x^2 - 3 == 0, x] yields {{x -> -Sqrt[3]}, {x -> Sqrt[3]}}
while in case from the above of slightly more difficult equations extracted from Manipulate setting the following arguments :
c1 = {-Sqrt[3], 0}; a = {1, 0}; c2 = {6 - Sqrt[3], 0}; b = {7, 0};
{r1, r2} = Map[ Norm, {a - c1, b - c2 }];
r = 2.0 - Sqrt[3];
to :
w = Table[Solve[{radius[{x, y} - {x1, y1}]^2 == (r + k r1)^2,
radius[{x, y} - {x2, y2}]^2 == (r + l r2)^2}],
{k, -1, 1, 2}, {l, -1, 1, 2}];
w = Select[ Cases[ Flatten[ {{x, y}, r} /. w, 2], {{_Real, _Real}, _Real}],
Last[#] > 0 &]
we get two solutions :
{{{1.26795, -3.38871*10^-8}, 0.267949}, {{1.26795, 3.38871*10^-8}, 0.267949}}
similarly under the same arguments and equations, putting :
r = 2 - Sqrt[3];
we get no solutions : {}
but in fact there is exactly one solution which we would like to emphasize:
{ {3 - Sqrt[3], 0 }, 2 - Sqrt[3] }
In fact, passing to Graphics such a small difference between two different solutions and the uniqe one is indistinguishable, however working with Manipulate we cannot track carefully with a desired accuracy merging of two circles and usually the last observed configuration when lowering r3 before vanishing all solutions (reminding so-called structural instability) looks like this :
Manipulate is rather a powerful tool, not only a toy, and its mastering could be very useful. The considered issues when appearing in a serious research are frequently critical, for example: in studying solutions of nonlinear differential equations, occurence of singularities in its solutions, qualitative behavior of dynamical systems, bifurcations, phenomena in Catastrophe theory and so on.
As this is a measure zero set, tools that require some granularity will generally have trouble with the concept. Perhaps better is to look for the singularity locus explicitly, where solutions have multiplicity or in other ways depart from the nearby solution behavior(s). It will be a part of the discriminant variety. In particular, you can grab the relevant part by setting your defining polynomials to zero and simultaneously making the Jacobian determinant zero.
Here is your example. I will eventually (wlog) put one center at the origin and the other at (1,0).
centers = Array[c, {2, 2}];
radii = Array[r, 3];
circ[cen_, rad_, x_, y_] := ({x, y} - cen).({x, y} - cen) - rad^2
I'll use your 'k' for both polynomials. Your formulation has pairs (k,l) where each is +-1. We can just use k, arrange by squaring to get a polynomial in k^2, and replace that with 1.
polys =
Table[Expand[
circ[centers[[j]], radii[[3]] + k*radii[[j]], x, y]], {j, 2}]
Out[18]= {x^2 + y^2 - 2 x c[1, 1] + c[1, 1]^2 - 2 y c[1, 2] +
c[1, 2]^2 - k^2 r[1]^2 - 2 k r[1] r[3] - r[3]^2,
x^2 + y^2 - 2 x c[2, 1] + c[2, 1]^2 - 2 y c[2, 2] + c[2, 2]^2 -
k^2 r[2]^2 - 2 k r[2] r[3] - r[3]^2}
We'll remove the part that is linear in k, square the rest, square that removed part, and equate the two. We also then replace k with unity.
p2 = polys - k*Coefficient[polys, k];
polys2 = Expand[p2^2 - (k*Coefficient[polys, k])^2] /. k -> 1;
We now get the determinant of the Jacobian and add that to the brew.
discrim = Det[D[polys2, #] & /# {x, y}];
allrelations = Join[polys2, {discrim}];
Now set the centers as noted earlier (could have done this from the beginning, one would suppose).
ar2 =
allrelations /. {c[1, 1] -> 0, c[1, 2] -> 0, c[2, 1] -> 0,
c[2, 2] -> 0}
Out[38]= {x^4 + 2 x^2 y^2 + y^4 - 2 x^2 r[1]^2 - 2 y^2 r[1]^2 +
r[1]^4 - 2 x^2 r[3]^2 - 2 y^2 r[3]^2 - 2 r[1]^2 r[3]^2 + r[3]^4,
x^4 + 2 x^2 y^2 + y^4 - 2 x^2 r[2]^2 - 2 y^2 r[2]^2 + r[2]^4 -
2 x^2 r[3]^2 - 2 y^2 r[3]^2 - 2 r[2]^2 r[3]^2 + r[3]^4, 0}
We now eliminate x and y to get the locus in r[1],r[2],r[3] parameter space that determines where we'll have multiplicity in our solutions.
gb = GroebnerBasis[ar2, radii, {x, y},
MonomialOrder -> EliminationOrder]
{r[1]^6 - 3 r[1]^4 r[2]^2 + 3 r[1]^2 r[2]^4 - r[2]^6 -
8 r[1]^4 r[3]^2 + 8 r[2]^4 r[3]^2 + 16 r[1]^2 r[3]^4 -
16 r[2]^2 r[3]^4}
If I did this all correctly, then we now have the polynomial defining the locus in parameter space where solution sets can get silly. Off this set they should never have multiplicity, and real counts should always be even. The intersection of this set with real space will be a 2d surface in the 3d space of the radii parameters. It will separate regions that have 0, 2, 4, 6, or 8 real solutions from one another.
Last, I'll point out that in this example the variety in question reduces nicely into a product of planes. I guess from a geometric view this is not too surprising.
Factor[gb[[1]]]
Out[43]= (r[1] - r[2]) (r[1] + r[2]) (r[1] - r[2] - 2 r[3]) (r[1] +
r[2] - 2 r[3]) (r[1] - r[2] + 2 r[3]) (r[1] + r[2] + 2 r[3])
How can I write the code in mathematica to see the result like this:
As you see we have the complex function w=f(z), where z=x+iy and w=u+iv.
In this example w=sin z, and we see the image of vertical line x=c is hyperbola. (Left)
and the image of horizontal line y=c is an elliptic. (Right)
This picture took from the book "Complex Variables and Applications, by James Ward Brown, Ruel Vance Churchill", 8th edition: pages 331 and 333 or third edition pages 96-97
Something like this?
ClearAll[u, v, f];
f[z_] := Sin[z]
u[x_, y_] := Re#f[x + I*y];
v[x_, y_] := Im#f[x + I*y];
EDIT: This just produces the whole thing. If you want to just see what happens for a single path parallel to the imaginary axis, try
ParametricPlot[{u[5, y], v[5, y]}, {y, -3, 3}]
or for the same parallel to the real axis try
ParametricPlot[{u[x, 1], v[x, 1]}, {x, -3, 3}]
EDIT2: Interactive:
ClearAll[u, v, f];
f[z_] := Sin[z]
u[x_, y_] := Re#f[x + I*y];
v[x_, y_] := Im#f[x + I*y];
Manipulate[
Show[
Graphics[{Line[{p1, p2}]}, PlotRange \[Rule] 3, Axes \[Rule] True],
ParametricPlot[
{u[p1[[1]] + t (p2[[1]] - p1[[1]]),
p1[[2]] + t (p2[[2]] - p1[[2]])],
v[p1[[1]] + t (p2[[1]] - p1[[1]]),
p1[[2]] + t (p2[[2]] - p1[[2]])]},
{t, 0, 1},
PlotRange \[Rule] 3]],
{{p1, {0, 1}}, Locator},
{{p2, {1, 2}}, Locator}]
(ugly, yes, but no time to fix it now). Typical output:
or
The idea is that you can vary the line on the left hand side of the figures you give (by clicking around the plot, which amounts to clicking on the Argand diagram...) and see the corresponding images.
Depending on what you want to do with the representations, it might sometimes be helpful to visualize the Riemann surface in 3D. Here's the surface for w=sin(z) in 3D, neatly showing the branch cuts and the different branches (same as acl's first plot, but in 3D).
ParametricPlot3D[
Evaluate[{Re#Sin[z], Im#Sin[z], y} /. z -> x + I y], {x, -2,
2}, {y, -2, 2}]