Hints

Start from the coupled differential equations for the circuits. Then switch to frequency space (phasor representation) and solve the algebraic system of equations.

You may need \(\sqrt{1-x}\approx1-x/2+\mathcal{O}\left(x^{2}\right)\) and \(1/\sqrt{1-x^{2}}\approx1+x^{2}/2\).

For the eigenfrequency analysis short the voltage supply and find an equation for \(\omega\), the characteristic equation.

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Determining the Currents

We start, as usual with the integro-differential equations governing the system:\[\begin{eqnarray*}  U_{\mathrm{ext}}\left(t\right)&=&L_{1}\dot{I}_{1}\left(t\right)+R_{1}I_{1}\left(t\right)+\frac{1}{C_{1}}\int_{-\infty}^{t}I_{1}\left(\tau\right)d\tau-L_{c}\dot{I}_{2}\left(t\right)\\0&=&L_{2}\dot{I}_{2}\left(t\right)+R_{2}I_{2}\left(t\right)+\frac{1}{C_{2}}\int_{-\infty}^{t}I_{2}\left(\tau\right)d\tau-L_{c}\dot{I}_{1}\left(t\right)\ .  \end{eqnarray*}\]The most important here is the sign of the coupling term which represents an external voltage source.

To get rid of the integral, we may derive the equations once more with respect to time. Then we have have two linear coupled differential equations. In this case it is always conventient to go to a Fourier representation to find the algebraic equations\[\begin{eqnarray*}  -\mathrm{i}\omega U_{\mathrm{ext}}\left(\omega\right)&=&\omega^{2}L_{1}I_{1}\left(\omega\right)-\mathrm{i}\omega R_{1}I_{1}\left(\omega\right)+\frac{1}{C_{1}}I_{1}\left(\omega\right)+\omega^{2}L_{c}I_{2}\left(\omega\right)\ \mathrm{and}\\0&=&\omega^{2}L_{2}I_{2}\left(\omega\right)-\mathrm{i}\omega R_{2}I_{2}\left(\omega\right)+\frac{1}{C_{2}}I_{2}\left(\omega\right)+\omega^{2}L_{c}I_{1}\left(\omega\right)\ .\end{eqnarray*}\]Devision by \(-\mathrm{i}\omega\) results in\[\begin{eqnarray*}  U_{\mathrm{ext}}\left(\omega\right)&=&Z_{1}\left(\omega\right)I_{1}\left(\omega\right)+\mathrm{i}\omega L_{c}I_{2}\left(\omega\right)\\0&=&Z_{2}\left(\omega\right)I_{2}\left(\omega\right)+\mathrm{i}\omega L_{c}I_{1}\left(\omega\right)\ \mathrm{with}\\Z_{i}\left(\omega\right)&=&R_{i}-\mathrm{i}\omega L_{i}+\frac{\mathrm{i}}{\omega C_{i}}\ . \end{eqnarray*}\]We can read-off\[\begin{eqnarray*}  I_{1}\left(\omega\right)&=&-\frac{Z_{2}\left(\omega\right)}{\mathrm{i}\omega L_{c}}I_{2}\left(\omega\right)\ ,\ \mathrm{so}\\U_{\mathrm{ext}}\left(\omega\right)&=&-\frac{Z_{1}\left(\omega\right)Z_{2}\left(\omega\right)}{\mathrm{i}\omega L_{c}}I_{2}\left(\omega\right)+\mathrm{i}\omega L_{c}I_{2}\left(\omega\right)  \end{eqnarray*}\]and find the solution to the first task of the problem:\[\begin{eqnarray*}  I_{2}\left(\omega\right)&=&\frac{\mathrm{i}\omega L_{c}}{\left[\left(\mathrm{i}\omega L_{c}\right)^{2}-Z_{1}\left(\omega\right)Z_{2}\left(\omega\right)\right]}U_{\mathrm{ext}}\left(\omega\right)\ ,\\I_{1}\left(\omega\right)&=&-\frac{Z_{2}\left(\omega\right)}{\mathrm{i}\omega L_{c}}I_{2}\left(\omega\right)=-\frac{Z_{2}\left(\omega\right)}{\left[\left(\mathrm{i}\omega L_{c}\right)^{2}-Z_{1}\left(\omega\right)Z_{2}\left(\omega\right)\right]}U_{\mathrm{ext}}\left(\omega\right)\ .  \end{eqnarray*}\]

Detuning

Now we can look when both terms can be infinite. This can be achieved if the denominator for both currents vanishes. However, since \(\left(\mathrm{i}\omega L_{c}\right)^{2}=-\omega^{2}L_{c}^{2}\) is a real negative number, one may only find that the currents go to infinity if the resistances \(R_{i}\) vanish. Otherwise, we would obtain mixed imaginary terms. Then,\[\begin{eqnarray*}  Z_{1}\left(\omega\right)Z_{2}\left(\omega\right)&=&\left(-\mathrm{i}\omega L_{1}+\frac{\mathrm{i}}{\omega C_{1}}\right)\left(-\mathrm{i}\omega L_{2}+\frac{\mathrm{i}}{\omega C_{2}}\right)\\&=&-\omega^{2}L_{1}L_{2}-\frac{1}{\omega^{2}C_{1}C_{2}}+\frac{L_{1}}{C_{2}}+\frac{L_{2}}{C_{1}}\\&\overset{!}{=}&\omega^{2}L_{c}^{2}\ . \end{eqnarray*}\]We may devide the whole equation by \(L_{1}L_{2}\) and multiply by \(\omega^{2}\) to find\[\begin{eqnarray*}  0&=&-\omega^{4}\left(1+\frac{L_{c}^{2}}{L_{1}L_{2}}\right)-\frac{1}{L_{1}C_{1}L_{2}C_{2}}+\omega^{2}\left(\frac{1}{L_{2}C_{2}}+\frac{1}{L_{1}C_{1}}\right)\\&\equiv&-\omega^{4}\left(1+\kappa_{1}\kappa_{2}\right)-\omega_{1}^{2}\omega_{2}^{2}+\omega^{2}\left(\omega_{2}^{2}+\omega_{1}^{2}\right)\\&=&-\left(\omega^{2}-\omega_{1}^{2}\right)\left(\omega^{2}-\omega_{2}^{2}\right)-\omega^{4}\kappa_{1}\kappa_{2}  \end{eqnarray*}\]using \(\omega_{i}^{2}=1/L_{i}C_{i}\) and \(\kappa_{i}=L_{c}/L_{i}\).

The latter equation is quadratic in \(\omega^{2}\) and we can find two solutions, namely\[\begin{eqnarray*}  \omega_{\pm}^{2}&=&\frac{1}{2}\frac{1}{1-\kappa_{1}\kappa_{2}}\left[\omega_{1}^{2}+\omega_{2}^{2}\pm\sqrt{\left(\omega_{1}^{2}-\omega_{2}^{2}\right)^{2}+4\kappa_{1}\kappa_{2}\omega_{1}^{2}\omega_{2}^{2}}\right]\ .  \end{eqnarray*}\] We can re-obtain the eigenfrequencies of the sole circuits if we switch the coupling off, \(L_{c}=0\) (or \(\kappa_{i}=0\)). If the coupling is present, however, this is not the case. Especially if \(\omega_{1}=\omega_{2}=\omega_{1,2}\), we find with \(\kappa^{2}=\kappa_{1}\kappa_{2}\) the detuning \[\begin{eqnarray*}  \omega_{+}-\omega_{-}&=&\frac{1}{\sqrt{2}}\frac{1}{\sqrt{1-\kappa^{2}}}\left[\sqrt{2\omega_{1,2}^{2}+\sqrt{0+4\kappa^{2}\omega_{1,2}^{4}}}-\sqrt{2\omega_{1,2}^{2}-\sqrt{0+4\kappa^{2}\omega_{1,2}^{4}}}\right]\\&=&\frac{\sqrt{1+\kappa}-\sqrt{1-\kappa}}{\sqrt{1-\kappa^{2}}}\omega_{1,2}\approx\kappa\omega_{1,2}+\mathcal{O}\left(\kappa^{2}\right)\ .  \end{eqnarray*}\]This phenomenon is called detuning or energy splitting and very general throughout the natural sciences. In the figure below you can see a graphical representation for different detunings between the eigenfrequencies. Can you see that \(\kappa/\omega_{1,2}=0.01\) has been chosen?

Eigenfrequency analysis

The eigenfrequency analysis can by done by shortening the voltage supply. We directly arrive at the equations\[\begin{eqnarray*} 0&=&\left(-\mathrm{i}\omega L_{1}+\frac{\mathrm{i}}{\omega C_{1}}\right)I_{1}\left(\omega\right)-\mathrm{i}\omega L_{c}I_{2}\left(\omega\right)\\0&=&\left(-\mathrm{i}\omega L_{2}+\frac{\mathrm{i}}{\omega C_{2}}\right)I_{2}\left(\omega\right)-\mathrm{i}\omega L_{c}I_{1}\left(\omega\right)  \end{eqnarray*}\]Now, multiply by \(\mathrm{i}\omega\) and each line with \(1/L_{i}\), we find in a matrix formulation\[\begin{eqnarray*}  0&=&\left(\begin{array}{cc}\omega^{2}-\omega_{1}^{2} & \omega^{2}\kappa_{1}\\\omega^{2}\kappa_{2} & \omega^{2}-\omega_{2}^{2}\end{array}\right)\left(\begin{array}{c}
I_{1}\left(\omega\right)\\I_{2}\left(\omega\right)\end{array}\right)\ .  \end{eqnarray*}\]This yields the characteristic equation\[\begin{eqnarray*}  \left(\omega^{2}-\omega_{1}^{2}\right)\left(\omega^{2}-\omega_{2}^{2}\right)-\omega^{4}\kappa_{1}\kappa_{2}&=&0 \end{eqnarray*}\]which is identical with our result for the infinite currents. Note that we could in the same way arrive at a characteristic equation with losses in the circuits. We would find out that then the eigenfrequencies are indeed complex numbers accounting for the time-harmonic losses.