Coupled differential equations using matrix exponent

[roam]

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.

 

[Orodruin]

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.

If you write out the solutions, you will get your functions as linear combinations of the exponentials $e^{\lambda_i z}$. (Note that the matrix exponential you are looking for is $e^{Mz}$.) The following relation may be of use to you:
$$
e^x = \frac{e^x + e^{-x}}{2} + \frac{e^x - e^{-x}}{2} = \cosh(x) + \sinh(x)
$$

Edit: A more straight-forward way is to directly write your functions as linear combinations of the eigenvectors and use that the ODEs for the eigenvector coefficients separate and are not coupled.

 

[Ray Vickson]

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.

The way I like to evaluate $\exp(A)$ for a matrix such as $A = Mt$ is to use the fact that for any analytic function $f(\cdot)$ we have
$$f(A) = \sum_i E_i \, f(\lambda_i),$$
where the $\lambda_i$ are the eigenvalues of $A$ and the matrices $E_1, E_2, \ldots$ are independent of the function $f(.)$; that is, they are the same for any $f$--see below. (This assumes all eigenvalues are unequal; modifications are needed if you have some repeated eigenvalues.)

Anyway, for your $2 \times 2$ matrix $A$ you can apply that equation to the two functions $f(x) = 1 = x^0$ giving $A^0 = I = E_1 \lambda_1^0 + E_2 \lambda_2^0 = E_1 + E_2$ and $f(x) = x$, giving $A^1 = A =E_1 \lambda_1^1 + E_2 \lambda_2^1 = E_1 \lambda_1 + E_2 \lambda_2$. (In the first equation, $I = A^0$ is the identity matrix.)

After you have solved the two algebraic equations $i = e_1+e_2$ and $a = e_1 \lambda_1 + e_2 \lambda_2$ for $e_1,e_2$ in terms of $i,a$, you can substitute $I$ for $i$ and $A$ for $a$ to get your two matrices $E_1, E_2$.

Now the rest is easy: $\exp(A) = E_1 \exp(\lambda_1) + E_2 \exp(\lambda_2)$.

Why does it work? Well, for an $n \times n$ matrix $A$ with $n$ distinct eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$ we can diagonalize it by a similarity transformation:
$$A = P \pmatrix{\lambda_1 & 0 & \cdots & 0\\ 0 & \lambda_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_n} P^{-1} \equiv P \Lambda P^{-1} $$
We have
$$f(A) = P f(\Lambda) P^{-1}, $$
in which $f(\Lambda)$ is a diagonal matrix with diagonal elements $f(\lambda_i)$. Pulling out each factor $f(\lambda_i)$ separately gives the stated result.

Of course, if you have already found the matrix $P$ you can evaluate the $E_i$ directly, without the need for solving any linear equations to find them. However, you do not need to know $P$ in order to carry out the computations, so you do not need to determine the eigenvectors, etc.

 

[roam]

If you write out the solutions, you will get your functions as linear combinations of the exponentials $e^{\lambda_i z}$. (Note that the matrix exponential you are looking for is $e^{Mz}$.) The following relation may be of use to you:
$$
e^x = \frac{e^x + e^{-x}}{2} + \frac{e^x - e^{-x}}{2} = \cosh(x) + \sinh(x)
$$

Edit: A more straight-forward way is to directly write your functions as linear combinations of the eigenvectors and use that the ODEs for the eigenvector coefficients separate and are not coupled.

Thank you for the relation, this should be helpful in doing the simplification. But how exactly should I write out the solutions to get to the form given in the equations above (1.1-2)?

Since I had already found P, I found this result:

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

$$=\begin{bmatrix}1 & 1\\ \frac{i\sqrt{\kappa^{2}-\delta^{2}}-\delta}{\kappa} & \frac{-i\sqrt{\kappa^{2}-\delta^{2}}-\delta}{\kappa} \end{bmatrix}\begin{bmatrix}e^{\sqrt{\kappa^{2}-\delta^{2}}z} & 0\\ 0 & e^{-\sqrt{\kappa^{2}-\delta^{2}}z} \end{bmatrix}\begin{bmatrix}\frac{i\sqrt{\kappa^{2}-\delta^{2}}+\delta}{2i\sqrt{\kappa^{2}-\delta^{2}}} & -\frac{\kappa}{-2i\sqrt{\kappa^{2}-\delta^{2}}}\\ \frac{i\sqrt{\kappa^{2}-\delta^{2}}+\delta}{2i\sqrt{\kappa^{2}-\delta^{2}}} & \frac{\kappa}{-2i\sqrt{\kappa^{2}-\delta^{2}}}\end{bmatrix}$$

When I carry out all the matrix multiplications I get:

$$\begin{bmatrix}\frac{\left(i\sqrt{\kappa^{2}-\delta^{2}}+\delta\right)\left(e^{\sqrt{\kappa^{2}-\delta^{2}}z}+e^{-\sqrt{\kappa^{2}-\delta^{2}}z}\right)}{2i\sqrt{\kappa^{2}-\delta^{2}}} & \frac{\kappa\left(e^{-\sqrt{\kappa^{2}-\delta^{2}}z}-e^{\sqrt{\kappa^{2}-\delta^{2}}z}\right)}{-2i\sqrt{\kappa^{2}-\delta^{2}}}\\
\frac{\left(-\kappa^{2}\right)e^{\sqrt{\kappa^{2}-\delta^{2}}z}+\left(\kappa^{2}-2\delta^{2}-2\delta i\sqrt{\kappa^{2}-\delta^{2}}\right)e^{-\sqrt{\kappa^{2}-\delta^{2}}z}}{2i\kappa\sqrt{\kappa^{2}-\delta^{2}}} & \frac{\delta\left(e^{-\sqrt{\kappa^{2}-\delta^{2}}z}-e^{\sqrt{\kappa^{2}-\delta^{2}}z}\right)}{i\sqrt{\kappa^{2}-\delta^{2}}}
\end{bmatrix}$$

 

[Orodruin]

You need to multiply the matrix exponential with the initial condition. The matrix exponential by itself is not the solution to the problem.

Edit: I also strongly suggest that you use the $\alpha$ defined in post #1 to tidy up your expressions.

 

[roam]

Thank you very much for the suggestions. Hopefully this is correct:

$$\begin{bmatrix}f(z)\\ b(z) \end{bmatrix}\stackrel{?}{=}\begin{bmatrix}\frac{\left(i\alpha+\delta\right)\left(e^{\alpha z}+e^{-\alpha z}\right)}{2i\alpha} & \frac{\kappa\left(e^{-\alpha z}-e^{\alpha z}\right)}{-2i\alpha}\\ \frac{\left(-\kappa^{2}\right)e^{\alpha z}+\left(\kappa^{2}-2\delta^{2}-2\delta i\alpha\right)e^{-\alpha z}}{2i\kappa\alpha} & \frac{\delta\left(e^{-\alpha z}-e^{\alpha z}\right)}{i\alpha} \end{bmatrix}\begin{bmatrix}f(0)\\ b(0)\end{bmatrix}$$

So, for $f(z)$ this gives:

$$f(z)=\frac{\left(i\alpha+\delta\right)\left(e^{\alpha z}+e^{-\alpha z}\right)}{2i\alpha}f(0)+\frac{\kappa\left(e^{-\alpha z}-e^{\alpha z}\right)}{-2i\alpha}b(0)$$

$$=\frac{1}{i\alpha}\left[\left(i\alpha+\delta\right)\frac{\left(e^{\alpha z}+e^{-\alpha z}\right)}{2}f(0)+\kappa\frac{e^{\alpha z}-e^{-\alpha z}}{2}b(0)\right]$$

$$=\frac{1}{i\alpha}\left[\left(i\alpha+\delta\right)\cosh\left(\alpha z\right)f(0)+\kappa\sinh\left(\alpha z\right)b(0)\right]$$

It is getting closer to the answer but it still doesn't agree with the correct solution posted above. Any ideas what could be wrong here?

Also, how would it be possible to factor $L$ into the solutions? It represents the fiber length, such that $b(L)=0$. I have made a diagram of the geometry of the problem:

 

[Orodruin]

You also have a condition for $b(L)$. You should be able to use that condition to fix $b(0)$ in terms of $f(0)$.

 

[Ray Vickson]

Thank you very much for the suggestions. Hopefully this is correct:

$$\begin{bmatrix}f(z)\\ b(z) \end{bmatrix}\stackrel{?}{=}\begin{bmatrix}\frac{\left(i\alpha+\delta\right)\left(e^{\alpha z}+e^{-\alpha z}\right)}{2i\alpha} & \frac{\kappa\left(e^{-\alpha z}-e^{\alpha z}\right)}{-2i\alpha}\\ \frac{\left(-\kappa^{2}\right)e^{\alpha z}+\left(\kappa^{2}-2\delta^{2}-2\delta i\alpha\right)e^{-\alpha z}}{2i\kappa\alpha} & \frac{\delta\left(e^{-\alpha z}-e^{\alpha z}\right)}{i\alpha} \end{bmatrix}\begin{bmatrix}f(0)\\ b(0)\end{bmatrix}$$
View attachment 213599

As explained in #3, for your matrix
$$A = \pmatrix{i\delta & i \kappa\\-i \kappa & -i \delta}$$
with eigenvalues $\alpha, -\alpha$ (where $\alpha = \sqrt{\kappa^2 - \delta^2}$) the matrices $E_1, E_2$ are obtained from the equations
$$E_1 + E_2 = I, \;\; \alpha E_1 - \alpha E_2 = A$$
so that
$$ E_1 = \frac{1}{2} I + \frac{1}{2 \alpha} A, \;\; E_2 = \frac{1}{2} I - \frac{1}{2 \alpha} A.$$
Thus,
$$\exp(A t) = E_1 e^{\alpha t} + E_2 e^{-\alpha t} = I \cosh(\alpha t) + \frac{1}{\alpha} A \sinh(\alpha t)$$

 

[roam]

$$ E_1 = \frac{1}{2} I + \frac{1}{2 \alpha} A, \;\; E_2 = \frac{1}{2} I - \frac{1}{2 \alpha} A.$$
Thus,
$$\exp(A t) = E_1 e^{\alpha t} + E_2 e^{-\alpha t} = I \cosh(\alpha t) + \frac{1}{\alpha} A \sinh(\alpha t)$$

Thank you. So if I used that last expression I would have:

$$e^{Mz} = \begin{bmatrix}1 & 0\\ 0 & 1 \end{bmatrix}\cosh(\alpha z)+\frac{1}{\alpha}\begin{bmatrix}i\delta & i\kappa\\ -i\kappa & -i\delta \end{bmatrix}\sinh(\alpha z)$$

$$=\begin{bmatrix}\cosh(\alpha z)+\frac{i\delta}{\alpha}\sinh(\alpha z) & \frac{i\kappa}{\alpha}\sinh(\alpha z)\\ \frac{-i\kappa}{\alpha}\sinh(\alpha z) & \cosh(\alpha z)-\frac{i\delta}{\alpha}\sinh(\alpha z) \end{bmatrix}$$

Multiplying this matrix exponential with the initial condition to get the solutions:

$f(z)=\left(\cosh(\alpha z)+\frac{i\delta}{\alpha}\sinh(\alpha z)\right)f(0)+\frac{i\kappa}{\alpha}\sinh(\alpha z)b(0) \tag{i}$

$b(z)=\frac{-i\kappa}{\alpha}\sinh(\alpha z)f(0)+\left(\cosh(\alpha z)-\frac{i\delta}{\alpha}\sinh(\alpha z)\right)b(0) \tag{ii}$

Due to the geometry of the problem $b(L)=0$, so I tried to write $b(0)$ in terms of $f(0)$:

$f(L)=\left(\cosh(\alpha L)+\frac{i\delta}{\alpha}\sinh(\alpha L)\right)f(0)+\frac{i\kappa}{\alpha}\sinh(\alpha L)b(0)$

$\therefore \ \underline{b(0)=\frac{-i\alpha f(L)+\left(i\alpha\cosh(\alpha L)-\delta\sinh(\alpha L)\right)f(0)}{\kappa\sinh(\alpha L)}}$

When I substitute this into (i) I get this expression:

$$f(z)=\frac{\left(\cosh(\alpha z)\sinh(\alpha L)\right)f(0)+\sinh(\alpha z)f(L)-\left(\cosh(\alpha L)\sinh(\alpha z)\right)f(0)}{\sinh(\alpha L)}$$

Is there a mistake somewhere? Or is it still possible to manipulate this expression so that it agrees with equation 1.1?

 

[Orodruin]

I did not try to check your computations explicitly, but looking at your final expression, you should try to eliminate $f(L)$ from it. What you should use to eliminate $b(0)$ is not $f(L)$, it is the expression for $b(L)$, which must be equal to zero.

 

[Ray Vickson]

Thank you. So if I used that last expression I would have:

-------------------------------

$$f(z)=\frac{\left(\cosh(\alpha z)\sinh(\alpha L)\right)f(0)+\sinh(\alpha z)f(L)-\left(\cosh(\alpha L)\sinh(\alpha z)\right)f(0)}{\sinh(\alpha L)}$$

Is there a mistake somewhere? Or is it still possible to manipulate this expression so that it agrees with equation 1.1?

As pointed out in #10, you need to express b(0) in terms of f(0) by setting b(L) = 0. Then, to simplify further, use the "addition" laws
$$\begin{array}{l}\cosh(u-v) = \cosh(u) \cosh(v) - \sinh(u) \sinh(v)\\
\sinh(u-v) = \sinh(u) \cosh(v) - \cosh(u) \sinh(v)
\end{array}
$$
When I did it I got exactly eq (1.1), eventually.

 

[roam]

Thank you very much, Ray Vickson and Orodruin. I really appreciate your time. I got the correct answer too (it makes perfect sense now).