# How to solve these coupled differential equation?

I came across the following coupled differential equation while studying a coupled split-ring resonator as two lumped RLC circuits. Using Kirchoff's voltage law we get the following coupled equations:

$$\frac{1}{C}\int i_1dt+R \ i_i + L \ \frac{di_1}{dt}=L_{12}\frac{di_2}{dt}+gE_{in},$$ $$\frac{1}{C}\int i_2dt+R \ i_2 + L \ \frac{di_2}{dt}=L_{12}\frac{di_i}{dt}.$$

where $L$ is inductance, $R$ is resistance, $C$ is capacitance, $E_{in}$ is the incident electric field, $L_{12}$ mutual inductance and $g$ is a constant gap size. How can we solve this differential equation?

I was thinking to get rid of the integrals by differentiating with respect to $t$ and proceed further, but could not do it. Solutions are:

$$i_1(\omega)=gE_{in}\frac{\frac{1}{i\omega C}+R+i\omega L}{\left(\frac{1}{i\omega C}+R+i\omega L\right)^2-(i \omega L_{12})^2},$$

$$i_2(\omega)=gE_{in}\frac{i \omega L_{12}}{\left(\frac{1}{i\omega C}+R+i\omega L\right)^2-(i \omega L_{12})^2}.$$

• try taking the laplace transform – Amara Dec 11 '17 at 15:06
• hint: replace $\int$ with $\frac {1}{i\omega}$ and $d/dt$ with $i\omega$ then solve the linear system of equations for two variables. – hyportnex Dec 11 '17 at 15:18
• @hyportnex that's interesting. But, how could we do that? Could you please explain it a little bit. – Pradip Kattel Dec 11 '17 at 16:35
• assume that $E_{in}(t) = \Re {[E_0 e^{i \omega t} ]}$, $i_1(t)=\Re{[I_1 e^{i\omega t}]}$, etc., then integrate and/or differentiate the exponentials. – hyportnex Dec 11 '17 at 16:43

The given solutions clearly assume that the voltage source is sinusoidal with frequency $\omega$. To make the math easier, use complex exponentials and then take the real part 'at the end'. Assume

$$v_{in}(t) = E_{in}e^{i\omega t}$$

where $E_{in}$ is a complex constant. It follows that the currents are of the form $i(t) = Ie^{i\omega t}$ where $I$ is a complex constant. Then, the coupled integro-differential equations become

$$\frac{1}{i\omega C}i_1 + Ri_1 + i\omega L i_1 = i\omega L_{12}i_2 + gv_{in}$$

$$\frac{1}{i\omega C}i_2 + Ri_2 + i\omega L i_2 = i\omega L_{12}i_1$$

Now see that, e.g., the second equation can be written as

$$i_2 = \frac{i\omega L_{12}}{\frac{1}{i\omega C} + R + i\omega L}i_1$$

Can you take it from here?