Reflection on the solution method

Systems of differential equations: Linear systems of differential equations

Reflection on the solution method

The differential equation of exponential growth $\frac{\dd y}{\dd t}=r\cdot y$ has the solution $y(t)=C\cdot e^{r\cdot t}$ with constant $C$ . The exponential function can be written as a power series: $e^{x}=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots=\sum_{k=0}^{\infty}\frac{1}{k!}x^k$

We consider again a pair of coupled homogeneous linear first-order differential equations with constant coefficients of the form $\left\{\begin{aligned} \frac{\dd x}{\dd t} &= a\, x+ b\, y\\[0.25cm] \frac{\dd y}{\dd t} &= c\, x + d\, y\end{aligned}\right.$ and write it in the matrix-vector form $\frac{\dd}{\dd t}\cv{x\\y}=\matrix{a & b\\ c & d}\cv{x\\y }$ Name the matrix $A=\matrix{a & b\\ c & d}$ There is thus an equation of the form $\frac{\dd}{\dd t}\vec{v}=A\vec{v}$ It is tempting to write the solution as $\vec{v}=e^{tA}\vec{C}$ with a certain vector $\vec{C}$ of constants. But what is meant by the exponential function applied to a matrix and how do you calculate this?

The answer to the first question comes from the series expansion of the exponential function:

For a square matrix $A$ we can define $\exp(A)=e^A$ as $e^A=I+\sum_{k=1}^{\infty}\frac{1}{k!}A^k$

For the calculation of $e^A$ we first consider once again the case $A$ is a $2\times 2$ diagonal matrix, say $A=\matrix{\lambda_1 & 0\\ 0 & \lambda_2}$ . Then $\begin{aligned}e^A &=I + \sum_{k=1}^{\infty}\frac{1}{k!}\matrix{\lambda_1 & 0\\ 0 & \lambda_2}^k\\ \\ &= I + \sum_{k=1}^{\infty}\frac{1}{k!}\matrix{\lambda_1^k & 0\\ 0 & \lambda_2^k} \\ \\ &= \matrix{\sum_{k=0}^{\infty}\frac{1}{k!}\lambda_1^k & 0\\ 0 & \sum_{k=0}^{\infty}\frac{1}{k!}\lambda_1^k } \\ \\ &= \matrix{e^{\lambda_1} & 0\\ 0 & e^{\lambda_2}}\end{aligned}$ When $A$ can brought via a similarity transformation $T$ into the diagonal form $\Lambda$ , say $T^{-1}AT=\Lambda$ , then you can use $(T^{-1}AT)^k = T^{-1}A^k T$ and understand that: $T^{-1}e^AT=e^{T^{-1}e^AT}=e^{\Lambda}$ Thus: $e^A=Te^{\Lambda}T^{-1}$ From linear algebra we already how to determine $T$ when two eigenvalues are different, or if there is only one eigenvalue with a 2-dimensional eigenspace: write the eigenvectors as columns in the matrix $T$ . We do not discuss here the case of one eigenvalue with a one-dimensional eigenspace, but the computational work is then doable, too.