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Is there any general solution to the linear differential system?

$i\begin{pmatrix}\dot{c_1}(t) \\ \ \dot{c_2}(t) \\ \ \dot{c_3}(t) \end{pmatrix} = \begin{pmatrix} e_1 & J & J\\ J & e_2 & J \\ J & J & e_3 \\ \end{pmatrix} \begin{pmatrix} c_1(t) \\ c_2(t) \\ c_3(t) \end{pmatrix} $.

I can find the eigenvalues when $e_1=e_2$ (or in general when two values of the diagonal terms are the same, but I cannot find a way to find the solution to the general system, as I cannot find the eigenvalues with analytical techniques.

The initial conditions are $c_1(0)=1,c_2(0)=c_3(0)=0$.

Is there a way to solve it? Or at least in the regime $e_2=-e_3$?

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    $\begingroup$ Are the $e_{i}$ terms just constants? What is $J$? Is that really an imaginary on the LHS? $\endgroup$ – Moo Apr 17 '18 at 14:20
  • $\begingroup$ Maple can compute the eigenvalues when the $e_i$ are all distinct. When printed as a PDF, the three eigenvalues and eigenvectors require 173 pages. $\endgroup$ – Kyle Apr 17 '18 at 15:16
  • $\begingroup$ J and e_j are real constants, I tried to solve the system in Mathematica but without success $\endgroup$ – Galuoises Apr 18 '18 at 11:49
  • $\begingroup$ @kyle is, in Maple, the solution more simple if $e_2=-e_3$? $\endgroup$ – Galuoises Apr 19 '18 at 16:07
  • $\begingroup$ It is a bit less complicated... 59 instead of 173 pages. It might be possible to find "sub-expressions" that are common among the eigenvalues $\lambda_1, \lambda_2, \lambda_3$ and eigenvector $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$, that would allow one to simplify further... I'm not really sure. $\endgroup$ – Kyle Apr 19 '18 at 23:12
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Let $A$ be your $3\times 3$ matrix. You ask for the first column of $e^{-itA}$ in a formal form; of course, it's practically impossible as shown by @Kyle (I wonder how long the calculation lasted!). That shows, on the one hand,the extraordinary power of the formal calculus and, on the other hand, its uselessness in this extreme case.

Note that the eigenvalues of $e^{-itA}$ are in the form $(e^{-it\lambda_j})_j$ where $\lambda_j\in\mathbb{R}$; then your $(c_j(t))$ are linear combinations of many $\sin(),\cos()$ and there is not any dominant term when $t\rightarrow\infty$. Thus the solution is a tangle of oscillating terms and, indeed, it's difficult to have an idea about the behavior of the solutions; in conclusion, it's a difficult problem and that's why one asks students to work on the subject; otherwise you would be useless.

Special cases happen when $\lambda_j/\lambda_k$ is closed to a rational number $p/q$ with not too large $p,q$.

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  • $\begingroup$ Thank you for your help! $\endgroup$ – Galuoises Apr 28 '18 at 12:57
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Here's a solution. Given

$$i\mathbf{\dot c}=\mathbf{M}\mathbf{c}$$

rearrange to get

$$\mathbf{\dot c}=-iJ\mathbf{A}\mathbf{c}$$

where

$$\mathbf{A} \equiv \frac{1}{J} \mathbf{M}= \begin{bmatrix} e_1/J & 1 & 1 \\ 1 & e_2/J & 1 \\ 1 & 1 & e_3/J \end{bmatrix}= \begin{bmatrix} f_1 & 1 & 1 \\ 1 & f_2 & 1 \\ 1 & 1 & f_3 \end{bmatrix}$$

for $J\ne 0$. (For $J=0$ the equations are uncoupled and the solution is $c_1=\exp(-i e_1 t)$, $c_2=c_3=0$.)

Now we'll seek solutions of the form

$$\mathbf{c}=\mathbf{S}\exp\left(-iJ\mathbf{\Lambda}t\right)\mathbf{S^{-1}}\mathbf{c_0}$$

where $\mathbf{S}$ is the matrix whose columns are the eigenvectors of $\mathbf{A}$ and $\mathbf{\Lambda}$ is the matrix whose diagonal elements are the eigenvalues of $\mathbf{A}$.

The characteristic equation for $\mathbf{A}$ is

$$\lambda^3-(f_1+f_2+f_3)\lambda^2+(f_1f_2+f_2f_3+f_3f_1-3)\lambda-(f_1f_2f_3+2-(f_1+f_2+f_3))=0$$

or

$$\lambda^3+a_2\lambda^2+a_1 \lambda+a_0=0$$

The roots can be found by standard methods for solving a cubic polynomial (see here for example):

$$\lambda_k = \left(2/3\right)\sqrt{a_2^2-3a_1}\cos((\theta+2k\pi)/3)-a_2/3,\,\,\,\,k=1,2,3$$

where

$$\theta = \cos^{-1}\left(\frac{9a_1a_2-27a_0-2a_2^3}{2\sqrt{(a_2^2-3a_1)^3}}\right)$$

Since $\mathbf{A}$ is symmetric we are guaranteed 3 real roots (eigenvalues) and we are also guaranteed that the eigenvectors will be mutually orthogonal.

The eigenvectors $\mathbf{u_k}$ are determined by

$$\mathbf{A}\mathbf{u_k}=\lambda_k\mathbf{u_k}$$

or

$$(\mathbf{A}-\lambda_k\mathbf{I})\mathbf{u_k}=\mathbf{0}$$

or

$$ \begin{bmatrix} f_1-\lambda_k & 1 & 1 \\ 1 & f_2-\lambda_k & 1 \\ 1 & 1 & f_3-\lambda_k \end{bmatrix} \begin{bmatrix} u_{1k} \\ u_{2k} \\ u_{3k} \end{bmatrix}=\mathbf{0} $$

With some algebra we can solve this to get $$ \mathbf{u_k} = \begin{bmatrix} (f_2-\lambda_k-1)(f_3-\lambda_k-1) \\ (f_3-\lambda_k-1)(f_1-\lambda_k-1) \\ (f_1-\lambda_k-1)(f_2-\lambda_k-1) \end{bmatrix} $$

For the case where two of the $e_k$ are equal, the formulation for the eigenvectors given above fails. This is because, when (say) $e_2=e_3$ (and therefore $f_2=f_3$), then one of the eigenvalues will be $\lambda_k=f_2 -1$. Then $\mathbf{u_k}=\mathbf{0}$ will not be an eigenvector. To address this case, we assign $\lambda_3=f_2 -1$. Then we factor this out of the characteristic equation

$$\lambda^3-(f_1+2f_2)\lambda^2+(2 f_1 f_2 + f_2^2)\lambda-(f_1 f_2^2 +2-(f_1+2 f_2))=0$$

by synthetic division to get

$$(\lambda_3-f_2 +1)(\lambda^2-(f_1 + f_2 +1)\lambda + (f_1 f_2 + f_1 -2)) = 0$$

Then the two remaining eigenvalues are

$$\lambda_{1,2}=\frac{(f_1 + f_2 +1) \pm \sqrt{(f_1 + f_2 +1)^2 - 4(f_1 f_2 + f_1 -2)}}{2}$$

With some algebra we get the corresponding eigenvectors

$$\mathbf{u_k}= \begin{bmatrix} -\frac{2}{f_1 - \lambda_k} \\ 1 \\ 1 \end{bmatrix}\,\,\,k=1,2 $$

We get $\mathbf{u_3}$ by making it orthogonal to $\mathbf{u_1}$ and $\mathbf{u_2}$

$$\mathbf{u_3}= \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} $$

For the case where $e_1 = e_2 = e_3$ (so $f_1 = f_2 = f_3 \equiv f$), a similar analysis gives $\lambda_1 = f + 2$, $\lambda_{2,3} = f - 1$

and

$$\mathbf{S}= \begin{bmatrix} 1 & -2 & 0\\ 1 & 1 & 1 \\ 1 & 1 & -1 \end{bmatrix} $$

Once the eigenvectors are known, the next step is to normalize them

$$N_k = \sqrt{u_{1k}^2 + u_{2k}^2 + u_{3k}^2}$$ $$\mathbf{U_k} = \frac{1}{N_k} \mathbf{u_k}$$

The matrix of eigenvectors is $$ \mathbf{S} = \begin{bmatrix} \mathbf{U_1} & \mathbf{U_2} & \mathbf{U_3} \end{bmatrix} $$

Since the columns of $\mathbf{S}$ are orthonormal we have

$$\mathbf{S^{-1}}=\mathbf{S^T}$$

We now have all of the terms of

$$\mathbf{c}=\mathbf{S}\exp\left(-iJ\mathbf{\Lambda}t\right)\mathbf{S^{-1}}\mathbf{c_0}$$

Carrying out the algebra for the case of distinct $e$'s gives

$$c_1(t) = \left[\frac{(f_2-\lambda_1-1)(f_3-\lambda_1-1)}{N_1}\right]^2 e^{-i J \lambda_1 t} + \left[\frac{(f_2-\lambda_2-1)(f_3-\lambda_2-1)}{N_2}\right]^2 e^{-i J \lambda_2 t} + \left[\frac{(f_2-\lambda_3-1)(f_3-\lambda_3-1)}{N_3}\right]^2 e^{-i J \lambda_3 t} $$

with similar expressions for $c_2$ and $c_3$.

Hope that helps.

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