# Thread: Exponential simultaneous Equations

1. ## 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!

2. ## Re: Exponential simultaneous Equations

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

$[1]: \ (a+y)e^{-ikx} + (b+y)e^{-imx} + (c+y)e^{ikx} + (d+y)e^{imx} = Ey$

$[2]: \ (f+y)e^{-ikx} + (g+y)e^{-imx} + (h+y)e^{ikx} + (j+y)e^{imx} = Ly$

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

$[1] - [2]: \ (a-f)e^{-ikx} + (b-g)e^{-imx} + (c-h)e^{ikx} + (d-j)e^{imx} = (E-L)y$

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

$0=(Ae^{-ikx}+Be^{ikx})+(Ce^{-imx}+De^{imx})+(Fe^{-i2kx}+Ge^{i2kx})+(He^{-i2mx}+Ie^{i2mx})+(Je^{-i(k+m)x}+Ke^{i(k+m)x})+(Me^{-i(k-m)x}+Ne^{i(k-m)x})+P$

$A=fE-aL, \ B=hE-cL, \ C=gE-bL, \ D=jE-dL, \ F=a-f, \ G=c-h, \ H=b-g, \ I=d-j, J=F+H, \ K=G+I, \ M=F+I, \ N=G+H, \ P=J+K$

This can be simplified using the following two identities.

$e^{\pm u} = \cos{(u)} \pm \sin{(u)}$

$\sin{(v)} \cos{(w)} \pm \cos{(v)} \sin{(w)} = \sin{(v \pm w)}$

$\Rightarrow 0=Q_{1} \sin{(kx + \phi_{1})} + Q_{2} \sin{(mx + \phi_{2})} + Q_{3} \sin{(2kx + \phi_{3})} + Q_{4} \sin{(2mx + \phi_{4})} + Q_{5} \sin{[(k+m)x + \phi_{5}]} + Q_{6} \sin{[(k-m)x + \phi_{6}]} + P$

$Q_{1} = 2 \sqrt{AB}, \ \phi_{1} = \arcsin{ \left( \frac{A+B}{2 \sqrt{AB}} \right) },...$

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.

3. ## 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?

4. ## 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.

$[1]: \ (a_{1} + a_{2}y)e^{-ikx} + (b_{1} + b_{2}y)e^{-imx} = Ey$

$[2]: \ (f_{1} + f_{2}y)e^{-ikx} + (g_{1} + g_{2}y)e^{-imx} = Ly$

$[1] - [2]: \ (a_{1} - f_{1})e^{-ikx} + (a_{2} - f_{2})ye^{-ikx} + (b_{1} - g_{1})e^{-imx} + (b_{2} - g_{2})ye^{-imx} = (E-L)y$

$y = \frac{(a_{1} - f_{1})e^{-ikx} + (b_{1} - g_{1})e^{-imx}}{(E-L) + (f_{2} - a_{2})e^{-ikx} + (g_{2} - b_{2})e^{-imx}}$

Plugging this into [1] or [2] would not give just linear terms in $e^{i \xi x}$, where $\xi$ 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}.