如何在Mathematica中获得准确的绘图曲线? [英] How to obtain accurate plot curves in Mathematica?

问题描述

在Mathematica中运行以下代码:

Run the following code In Mathematica:

r=6197/3122;
p[k_,w_]:=Sqrt[w^2/r^2-k^2];q[k_,w_]:=Sqrt[w^2-k^2];
a[k_,w_,p_,q_]:=(k^2-q^2)^2 Sin[p]Cos[q]+4k^2 p q Cos[p]Sin[q]
a[k_,w_]:=a[k,w,p[k,w],q[k,w]];
ContourPlot[a[k,w]==0,{w,0,6},{k,0,14}]

这给了我非常不准确的曲线:

This gives me very inaccurate curves:

我尝试将ContourPlot的PlotPoints和WorkingPrecision选项分别设置为30和20，但无济于事.您还将注意到，唯一的数字参数r是精确的有理数.我不知道还能尝试什么.谢谢.

I have tried setting the PlotPoints and WorkingPrecision options of ContourPlot to 30 and 20 respectively, to no avail. You will also notice that the only numerical parameter, r, is an exact rational number. I don't know what else to try. Thanks.

我期望得到的曲线是下图上的三个黑色曲线(分别标记为A1，A2和A3)

The curves I expect to get are the three black ones (marked A1, A2, and A3) on the following picture

推荐答案

您确定a的图片和/或定义吗?根据a的定义，可以得出k==w上的a[k,w]==0，但该曲线未出现在图片中.

Are you sure about the picture and/or the definition for a? From the definition of a it follows that a[k,w]==0 on k==w but that curve doesn't appear in your picture.

无论如何，假设a的定义是正确的，绘制轮廓的问题是在域w^2/r^2-k^2<0中，p[k,w]和Sin[p[k,w]]都变成纯虚构的，这意味着a[k,w]变成纯虚构的虚构的.由于ContourPlot不喜欢复数值函数，因此仅绘制了域w^2/r^2>=k^2中的轮廓部分.

Anyway, assuming that the definition of a is right, the problem with plotting the contours is that in the domain w^2/r^2-k^2<0, both p[k,w] and Sin[p[k,w]] become purely imaginary which means that a[k,w] becomes purely imaginary as well. Since ContourPlot doesn't like complex valued functions only the parts of the contours in the domain w^2/r^2>=k^2 are plotted.

并非Sin[p[k,w]]/p[k,w]对于k和w的所有值都是真实的(并且在限制p[k,w]->0中表现良好).因此，要解决a变得复杂的问题，您可以绘制等高线a[k,w]/p[k,w]==0:

Not that Sin[p[k,w]]/p[k,w] is real for all values of k and w (and it's nicely behaved in the limit p[k,w]->0). Therefore, to get around the problem of a becoming complex you could plot the contours a[k,w]/p[k,w]==0 instead:

ContourPlot[a[k, w]/p[k, w] == 0, {w, 0, 6}, {k, 0, 14}]

结果

的轮廓图

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