# Solve non linear system of equations

I am trying to solve a system of two equations with two unknowns. In these equations I have, a part from constants:

• Unknown nr 1, $$D_{\perp}$$
• Unknown nr 2, $$\omega_C$$
• Known function of r: $$\mu(r)$$

The full system looks like:

equation1: $$D_{||}= \frac 1 {2 \omega_0}(-\alpha-1)\sqrt{(-\omega_C+\mu(r)D_{||})^2+(\mu(r)D_{\perp})^2}+ \frac{\alpha}{2\omega_0}\sqrt{(-2\omega_0-\omega_C+\mu(r)D_{||})^2+(\mu(r)D_{\perp})^2}+ \frac{1}{2\omega_0}\sqrt{(2\omega_0-\omega_C+\mu(r)D_{||})^2+(\mu(r)D_{\perp})^2}$$ and

equation2: $$1= \mu(r) \frac{\alpha}{2\omega_0} \ln{\left[\frac{-2\omega_0-\omega_C+\mu(r)D_{||}+\sqrt{(-2\omega_0-\omega_C+\mu(r)D_{||})^2+(\mu(r)D_{\perp})^2}}{-\omega_C+\mu(r)D_{||}+\sqrt{(-\omega_C+\mu(r)D_{||})^2+(\mu(r)D_{\perp})^2}}\right]}+ \frac{\mu(r)}{2\omega_0} \ln{\left[\frac{2\omega_0-\omega_C+\mu(r)D_{||}+\sqrt{(2\omega_0-\omega_C+\mu(r)D_{||})^2+(\mu(r)D_{\perp})^2}} {-\omega_C+\mu(r)D_{||}+\sqrt{(-\omega_C+\mu(r)D_{||})^2+(\mu(r)D_{\perp})^2}}\right]}$$

So the solution of the system of equations will be $$D_{\perp}(r),\;\omega_C(r)$$.

What I've tried to do is simply

    Solve[{equation1, equation2},{Dorthogonal, omegaC}]


but Mathematica keeps on running forever without any output. I have also tried:

    DorthFun[r_]:=Solve[{equation1, equation2},{Dorthogonal, omegaC}][[1,1]]
omegaCFun[r_]:=Solve[{equation1, equation2},{Dorthogonal, omegaC}][[1,2]]


and it just keeps on running... It doesn't return any errors. Just...eternal running. Forrest Gump Syndrome...

I have also tried to solve the system putting $$\mu(r)=1$$ without any change.

I have given Mathematica about 20 minutes. Should I give it more time or does this mean that Mathematica cannot solve this? Or is there something I could do differently?

My code looks like this:

    NSolve[{
Dparallel==1/(2 omega0) Sqrt[(-omegaC + Dparallel)^2 + (Dorth)^2] (-alpha-1)+alpha/(2omega0)Sqrt[(-2omega0-omegaC+Dparallel)^2+(Dorth)^2]+1/(2omega0)Sqrt[(2omega0-omegaC+Dparallel)^2+(Dorth)^2],
1 == alpha/(2omega0)Log[(-2omega0-omegaC+Dparallel+Sqrt[(-2omega0-omegaC+Dparallel)^2+(Dorth)^2])/(-omegaC+Dparallel+Sqrt[(-omegaC+Dparallel)^2+(Dorth)^2])]+1/(2omega0)Log[(2omega0-omegaC+Dparallel+Sqrt[(2omega0-omegaC+Dparallel)^2+(Dorth)^2])/(-omegaC+Dparallel+Sqrt[(-omegaC+Dparallel)^2+Sqrt[(-omegaC+Dparallel)^2+(Dorth)^2]])]
}, {Dorth, omegaC}]

-
I think your problem are the "_". You can't use them as variables, those are used to define types of expressions. Check it out. Just change D_orthFun to DorthFun and D_orthogonal to Dorthogonal –  Arcotick Mar 9 at 19:56
I seem to have slipped when writing, my actual code has no underscores in the variable names. Thank you for the input though! I edited my question to include my code. I also tried NSolve as you can see but still with no end to the iterations. –  user12291 Mar 9 at 20:05
What is the definition/value of omega0? –  m_goldberg Mar 9 at 20:46
omega0=0.03. Why? –  user12291 Mar 9 at 21:11

I'm not sure what the correct way of showing an answer to a question I myself asked. But here is how I did anyway:

I used ContourPlot3D

Manipulate[ContourPlot[{
(*29*)
Dparallel==1/(2 omega0) Sqrt[(-omegaC + mu[r] Dparallel)^2 + (mu[r] Dorth)^2] (-alpha - 1) +
alpha/(2 omega0) Sqrt[(-2 omega0 - omegaC + mu[r] Dparallel)^2 + (mu[r] Dorth)^2] +
1/(2 omega0) Sqrt[(2 omega0 - omegaC + mu[r] Dparallel)^2 + (mu[r] Dorth)^2],
(*30*)
1 ==
alpha mu[r]/(2 omega0)*
Log[(-2 omega0 - omegaC + mu[r] Dparallel + Sqrt[(-2 omega0 - omegaC + mu[r] Dparallel)^2 + (mu[r] Dorth)^2])/
(-omegaC + mu[r] Dparallel + Sqrt[(-omegaC + mu[r] Dparallel)^2 + (mu[r] Dorth)^2])] +
mu[r]/(2 omega0)*
Log[(2 omega0 - omegaC + mu[r] Dparallel + Sqrt[(2 omega0 - omegaC + mu[r] Dparallel)^2 + (mu[r] Dorth)^2])/
(-omegaC + mu[r] Dparallel + Sqrt[(-omegaC + mu[r] Dparallel)^2 + Sqrt[(-omegaC + mu[r] Dparallel)^2 + (mu[r] Dorth)^2]])]
}, {Dorth, -5, 5}, {omegaC, -5, 5}], {r, 0, 150}]


Where the solution to the system of equations would be where the surfaces intersect.

Unfortunately mine do not intersect... But now I know where to start!

-