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I have three equations of the form:

$$\frac{i_1^3}{P_1}+i_1(Z_1+Z_2)+(i_2+i_3)Z_2-U_1=0$$ $$\frac{i_2^3}{P_2}+i_2(Z_1+Z_2)+(i_1+i_3)Z_2-U_2=0$$ $$\frac{i_3^3}{P_3}+i_3(Z_1+Z_2)+(i_1+i_2)Z_2-U_3=0$$

where $P_1,P_2,P_3,K,U_1,U_2,U_3,Z_1$ and $Z_2$ are all known constant complex numbers.

I am attempting to solve these equations to find $i_1$, $i_2 $and $i_3$. To help, I would like to visualise the functions.

Can anyone recommend a way of visualising these three functions?

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There are commercial software packages that can "graph" these equations for you. –  Gerry Myerson Jun 12 '13 at 9:46

1 Answer 1

Let me try to raise a quite unconventional method, since your question was on a way of visualisation. Your equations are quite symmetric and I cannot resist the temptation to suggest this. However, having an elegant analytical approach for a visualisation will be a calculation challenge and you would need to dig into the theory required for this.

Your equations:

$$i_1^3/P_1 + i_1(Z_1+Z_2) + (i_2+i_3)Z_2 - U_1= 0$$ $$i_2^3/P_2 + i_2(Z_1+Z_2) + (i_1+i_3)Z_2 - U_2 = 0$$ $$i_3^3/P_3 + i_3(Z_1+Z_2) + (i_1+i_2)Z_2 - U_3= 0$$

can easily be reformed to the right hand side of a system of non-linear differential equations, to which your equations seek the roots/zeros:

$$\frac{dq_1}{dt}=a_1 \;q_1^3+b \;q_1+c\;q_2+c\;q_3+d_1$$ $$\frac{dq_2}{dt}=a_2 \;q_2^3+c\;q_1+b \;q_2+c\;q_3+d_2$$ $$\frac{dq_3}{dt}=a_3 \;q_3^3+c\;q_1+c\;q_2+b \;q_3+d_3$$

I changed your variable $i$ with $q$ so it fits directly into the next reference. The parameters are just collecting your parameters and easily extractable.

There is a very elegant routine proposed by Hermann Haken (commonly used in laser physics and also applied to the so called Ginzburg-Landua equations) see here pages 24 upwards, that can be elegantly applied, because your equations already fit perfectly the shape of that procedure. It will help you to shift your system (visually about $d$) to a symmetric point, then distinguish between linear and non-linear parts, transform to the eigenspace (because you are only interested in the roots/zeros it is possible to take care of the non-linear parts of exponent $3$), and reform the complete system with eigenvectors/eigenvalues of the linear part. Then you will obtain an elegant visulisation based on an analytic equation system (of eigenmodes), moreover understand your system near the roots in the eigenspace by deriving elegantly a Lyapunov function, which you can plot and for which also scetch phase digramms.

Mainly why I suggest this is, because you would need a powerful analytical and comprehensive mathematical tool to be able to vary your control parameters (which are complex) and see what happens to near environment of your roots.

I know the way is quite unconventional but I am sure you will receive a comprehensive and beautiful approach through this. The power of this method is in its ability to happen in the eigenspace and that you keep the control parameters variable. I would not suggest this if the shape of your equations were not that elegant fit.

The rather ugly but fast alternative might be to feed a software package to solve your equations nummerically or simulate the environment around the roots, while you wont be able to intuitively vary your control parameters.

I hope this answers your question.

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I don't understand how did you convert a system of algebraic equations into a differential system. –  metacompactness Jul 1 '13 at 19:46
    
The algebraic equation here can be regarded as a stationary case of the differential system. –  al-Hwarizmi Jul 1 '13 at 19:51
    
I got it but I don't think atomh33ls had studied differential systems and Lyapunov functions; I think he expects an answer at the level of first year calculus, linear algebra and precalculus. –  metacompactness Jul 1 '13 at 20:01
1  
there is no need that he solves the equation, the process suggested just considers the right hand side and uses linear algebra methods. The Lyapunov function is an outcome if wanted. Dont judge the method becasue a DGL has a left side, the right side is identical with the algebraic equation system. –  al-Hwarizmi Jul 1 '13 at 20:08

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