0
$\begingroup$

The given equations are: $$ \dot{y}_1=-iky^*_2\\\dot{y}_2=-iky^*_1 $$ $$ \begin{bmatrix} \dot{y}_1 \\ \dot{y}_2\end{bmatrix}=\begin{bmatrix} 0 & -ik \\ -ik & 0\end{bmatrix}\begin{bmatrix} {y}_1^* \\ {y}_2^*\end{bmatrix} $$ The solutions are given as: $$y_1(t)=y_1(0)\cos(hkt)-iy_2^*(0)\sin(hkt)$$ and $$y_2(t)=y_2^*(0)\cos(hkt)+iy_1(0)\sin(hkt)$$

If we had $y_1$ and $y_2$ instead of $y_1^*$ and $y_2^*$, I could do it by finding the eigenvalue and eigenvectors of the matrix.

How do I proceed further and obtain the desired answer ?

1 Answers 1

0

Transform this $2 \times 2$ differential system in $\mathbb{C}$ into a $4 \times 4$ differential system with coefficient in $\mathbb{R}$ by setting:

$$y_1=X_1+iY_1 \ \ \ \text{and} \ \ y_2=X_2+iY_2$$

yielding:

$$\frac{d}{dt}\left(\begin{array}{c}X_1\\X_2\\Y_1\\Y_2 \end{array}\right)=-k\underbrace{\left(\begin{array}{rrrr}&&&1\\&&1&\\&1&&\\1&&& \end{array}\right)}_{A}\left(\begin{array}{c}X_1\\X_2\\Y_1\\Y_2 \end{array}\right)$$

It remains to diagonalize matrix $kA$.

It is easy to check that $kA$ has 2 double roots $+ k, +k, -k, -k$ and has associated eigenvectors the columns of matrix

$$P=\sqrt{2}\left(\begin{array}{rrrr}0&1&1&0\\1&0&0&1\\1&0&0&-1\\0&1&-1&0 \end{array}\right)$$

(in the same order). As $P^{-1}=P$, the diagonalization identity is:

$$A=P.diag(k,k,-k,-k).P$$

Thus:

$$exp(tA)=P.diag(e^{kt},e^{kt},e^{-kt},e^{-kt}).P$$

Application of formula $X(t)=exp(tA).X(0)$ gives:

$$\left(\begin{array}{c}X_1(t)\\X_2(t)\\Y_1(t)\\Y_2(t) \end{array}\right)=2\left(\begin{array}{rrrr}0&1&1&0\\1&0&0&1\\1&0&0&-1\\0&1&-1&0 \end{array}\right)\left(\begin{array}{rrrr}e^{kt}&&&\\&e^{kt}&&\\&&e^{-kt}&\\&&&e^{-kt} \end{array}\right)\left(\begin{array}{rrrr}0&1&1&0\\1&0&0&1\\1&0&0&-1\\0&1&-1&0 \end{array}\right)\left(\begin{array}{c}a\\b\\c\\d \end{array}\right)$$

where $a=X_1(0), b=X_2(0), c=Y_1(0), d=Y_2(0)$.

It remains to do the matrix multiplications (block multiplications can be useful) to obtain:

$$\left(\begin{array}{c}X_1(t)\\X_2(t)\\Y_1(t)\\Y_2(t) \end{array}\right)=2\left(\begin{array}{rrrr}e^{t}&e^{-t}&e^{t}&-e^{-t}\\e^{-t}&e^{t}&-e^{-t}&e^{t}\\-e^{-t}&e^{t}&e^{-t}&e^{t}\\e^{t}&-e^{-t}&e^{t}&e^{-t} \end{array}\right)\left(\begin{array}{c}a\\b\\c\\d \end{array}\right)$$

and check that the good results are found, by expressing for example:

$$y_1(t)=X_1(t)+iY_1(t)=2(e^t(a+c)+e^{-t}(b-d)+i(e^t(b+d)+e^{-t}(-a+c)))$$