Exponential simultaneous Equations

I have two coupled equations of the form

(a+ y)*exp(-i*k*x) + (b + y)*(-i*m*x) + (c + y)*exp(i*k*x) +( d + y)*exp(i*m*x) = e*y

(f+ y)*exp(-i*k*x) + (g + y)*(-i*m*x) + (h + y)*exp(i*k*x) +(j + y)*exp(i*m*x) = l*y

and need to solve for x,y, where a-f are real. Actually the coefficients are very long with many terms!

What is the best way to go about this?

Re: Exponential simultaneous Equations

I'm assuming that the second terms in each equation should involve an exp,

Subtracting [2] from [1] gives an expression for y in terms of x:

Subbing this expression for y into [1] and reorganizing gives

This can be simplified using the following two identities.

Unfortunately, this equation is transcendental in x, which means it cannot be solved analitically. The solution for x (for which there will be an infinite number due to the periodic nature of the trig functions, as long as P isn't too large compared to the Qs) can be found either graphically or numerically. For a graphical solution, you plot the right-hand side of the equation and see where it crosses the x-axis (these will be the points where the function on the right is equal to zero, satisfying the equation). For a numerical solution check out root finding methods like the bisection method. Once you have values for x you can plug them into the above expression for y.

Re: Exponential simultaneous Equations

Hi Brilliant thanks for that, I suppose if the y terms in equation 1 and 2 each had different coefficients a similar method would also hold?

Re: Exponential simultaneous Equations

It wouldn't be so simple. You would get a more complicated denominator in your expression for y. When you subtract the two equations all those y's on the left don't cancel like they did above. The example below is a truncated version of your equations, but still demonstrates what I mean.

Plugging this into [1] or [2] would not give just linear terms in , where are the different combinations of k and m. Although, I suppose if you could eliminate the problem by giving the whole expression a common denominator. Since the expression is equal to 0, you can multiply both sides by this common denominator and get the 0 = {linear expression}.