Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

In this problem is very difficult to analyze the roots s. Firstly I thought that the problem is in precision, but I am not sure. The criterion which I want to obtain is that one of these roots has a positive real part. Characteristic equation has the following form:

$$\frac{2 \pi }{\int_{-\infty }^{\infty } \frac{1}{\text{D1}(k,s)} \, dk}+\text{a6} s^2+\text{k0}=0$$

The imaginary axis of the complex s-plane can be parametrized by conidering s=i*r (i=sqrt(-1)). So the equation has new form for consideration

  k0 = a6 r^2 - (1/(2 Pi)
  NIntegrate[1/(
  a1 a3 k^4 - a2 a4 (-k a5 + r)^2), {k, -\[Infinity], \[Infinity]}])^-1

where

a1 = 2*10^11;
a2 = 7.687*10^-3;
a3 = 3.055*10^-5;
a4 = 7849;
a5 = 500;
a6 = 70;
\[Alpha] = Sqrt[(a1 a3)/(a4 a2)];
c = Sqrt[k0/a6];

a5 > < Sqrt[4 \[Alpha] c];

The mapping rule follows from the characteristic equation, which should be rewritten to express the chosen parameter explicitly. Once the mapping is accomplished, a mapped line is obtained, which divides the parameter plane into domains with different number of roots s* possessing a positive real part. Every line is normally shaded from the side, which is related to the right-hand side of the imaginary axis of the (s)-plane now. Obviously, crossing a decomposition curve in the direction of the shading, one extra root with a positive real part is gained. This curve (rule) should be obtained for parameter r from -Infinity to +Infinity for discuss. So I need two separately diagrams for a5 > Sqrt[4 [Alpha] c] or a5 < Sqrt[4 [Alpha] c].

I am completely not sure how to obtain curves from the equation

  k0 = a6 r^2 - (1/(2 Pi)
  NIntegrate[1/(
  a1 a3 k^4 - a2 a4 (-k a5 + r)^2), {k, -\[Infinity], \[Infinity]}])^-1

presented here

enter image description here

where N is number of roots in the region

share|improve this question
    
Would RootLocusPlot be useful? –  bill s Mar 11 at 15:10
    
@bill s how to implement RootLocusPlot on this equation? –  Pipe Mar 11 at 15:12
    
@bill s I tried RootLocusPlot[a6 r^2-a1 a3 k^4 - a2 a4 (-k a5 + r)^2], {k, 0, 0.01000}] but it is not that diagram –  Pipe Mar 11 at 20:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.