- #1

roam

- 1,271

- 12

## Homework Statement

Solve the following coupled differential equations by finding the eigenvectors and eigenvalues of the matrix and using it to calculate the matrix exponent:

$$\frac{df}{dz}=i\delta f(z)+i\kappa b(z)$$

$$\frac{db}{dz}=-i\delta b(z)-i\kappa f(z)$$

In matrix form:

$$\frac{d}{dz}\begin{pmatrix}f\\b\end{pmatrix}=-i\begin{pmatrix}-\delta & -\kappa\\ \kappa &\delta\end{pmatrix}\begin{pmatrix}f\\b\end{pmatrix}$$

## Homework Equations

These are known as the fiber mode equations in photonics, and I know that the analytic solutions to the above equations must be:

$$f(z)=f(0)\frac{\alpha\cosh(\alpha(L-z))-i\delta\sinh(\alpha(L-z))}{\alpha\cosh(\alpha L)-i\delta\sinh(\alpha L)} \tag{1.1}$$

$$b(z)=if(0)\frac{\kappa\sinh(\alpha(L-z))}{\alpha\cosh(\alpha L)-i\delta\sinh(\alpha L)} \tag{1.2}$$

where ##\alpha=\sqrt{\kappa^{2}-\delta^{2}}##. And ##L## is the fiber length (there is a boundary condition that ##b(z=L)=0##).

## The Attempt at a Solution

So, I have found the eigenvalues by solving ##\text{det}|M-\lambda I|=0##:

$$\text{det}\begin{bmatrix}i\delta-\lambda & i\kappa\\ -i\kappa & -i\delta-\lambda \end{bmatrix}=\delta^{2}+\lambda^{2}-\kappa^{2}=0$$

$$\underline{\lambda=\pm \sqrt{\kappa^{2}-\delta^{2}}}$$

Now, to find the eigenvectors we find a basis for ##\text{Nul}(M-\lambda I)##:

We want ##\begin{bmatrix}x_{1}\\ x_{2} \end{bmatrix}## such that ##\begin{bmatrix}i\delta \pm \sqrt{\kappa^{2}-\delta^{2}} & i\kappa\\ -i\kappa & -i\delta \pm \sqrt{\kappa^{2}-\delta^{2}} \end{bmatrix}\begin{bmatrix}x_{1}\\ x_{2} \end{bmatrix}=\begin{bmatrix}0\\ 0 \end{bmatrix}##, so this gives

$$x_{2}=\frac{i(i\delta\pm\sqrt{\kappa^{2}-\delta^{2}})}{\kappa}x_{1},\ \therefore\text{eigenvectors}=\begin{bmatrix}1\\ \frac{\pm i\sqrt{\kappa^{2}-\delta^{2}}-\delta}{\kappa} \end{bmatrix}.$$

If this is correct so far, ##M## can be diagonalized as follows:

$$P=\begin{bmatrix}1 & 1\\ \frac{i\sqrt{\kappa^{2}-\delta^{2}}-\delta}{\kappa} & \frac{-i\sqrt{\kappa^{2}-\delta^{2}}-\delta}{\kappa} \end{bmatrix},\ \text{and}\ M=P\begin{pmatrix}\sqrt{\kappa^{2}-\delta^{2}} & 0\\ 0 & -\sqrt{\kappa^{2}-\delta^{2}} \end{pmatrix}P^{-1}$$

I believe the matrix exponential would be given by:

$$e^{M}=P\begin{bmatrix}e^{\lambda_{1}} & 0\\ 0 & e^{\lambda_{2}} \end{bmatrix}P^{-1} \tag{2}$$

So how does finding ##e^M## lead to the final answer? How do we get to the final answer as given in 1.1-2? I am very confused about where the hyperbolic functions come from.

Any explanation would be greatly appreciated.