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I have need to solve for a variable in this particular equation at work, and I have been unable to find a closed form solution for this despite several shots at it.

The system is as follows: enter image description here equation.
I'm trying to solve for $\Phi$, which is the only variable here. The constants are all arbitrary real numbers. I am grateful for your help with this.

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  • $\begingroup$ You have two equations with one variable and 8 parameters...perhaps someone else can make something with this, but it looks pretty hopeless $\endgroup$
    – DonAntonio
    Apr 18, 2017 at 8:49
  • $\begingroup$ I've clarified in my edit, $\Phi$ is the only variable. c1, c2, k, etc. are all constants. $\endgroup$
    – CaladanBrood
    Apr 18, 2017 at 8:52
  • $\begingroup$ I think that's precisely what I wrote... $\endgroup$
    – DonAntonio
    Apr 18, 2017 at 8:52
  • $\begingroup$ My apologies for the redundancy. I was trying to make the question as clear as possible. $\endgroup$
    – CaladanBrood
    Apr 18, 2017 at 8:54
  • $\begingroup$ This reduces to an algebraic equation for the unknown $z=e^{i\Phi}$ but, unless you say something more about the constants, there is no hope to find a solution in a closed form. $\endgroup$
    – Jon
    Apr 18, 2017 at 9:22

2 Answers 2

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The idea is as follows. Let us consider the set of equations $$ \sin(k-k'\Phi)=c_1\cos\Phi+c_2\sin\Phi+c_3 $$ $$ \cos(k-k'\Phi)=c_4\cos\Phi+c_5\sin\Phi+c_6. $$ Multiply by $i$ the first equation and sum it to the second. You will get $$ e^{i(k-k'\Phi)}=Ae^{i\Phi}+Be^{-i\Phi}+C $$ where now $A,\ B,\ C$ are complex constants. Take as an unknown the complex variable $z=e^{-i\Phi}$ you will arrive at the algebraic equation $$ e^{ik}z^{k'}=Az^{-1}+Bz+C. $$ As already suggested, the best choice in the general case is to use a numerical algorithm.

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The first step is to do the matrix product. This gives 2 equations.

I write p for $\Phi$.

Then use the addition formula for sine to get the equations in terms of sin and cost of p and k'p.

Then use $\cos^2 =1-\sin^2$ to eliminate cos.

If k' is an arbitrary real, the resulting equations can only be solved numerically. If k' is an integer, you can use the formula for sin(k'p) in terms of sin p amd cos p and then eliminate cos p as before.

Numerically is probably best in all cases.

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