Differential Equations with Constant Coefficients

Given initial conditions $u(0)=\left[\begin{array}{c}a\text{ }b\end{array}\right]$ a system of linear differential equations such as

$$\begin{array}{c}u\_1^{\prime }=-u\_1+2u\_2\\ u\_2^{\prime \prime }=u\_1-2u\_2\end{array}$$

is rewritten into matrix form $Au=\left[\begin{array}{ccccc}& & & & \end{array}\right]\left[\begin{array}{c}u\_1u\_2\end{array}\right]$. The solution to this system is solved by the general solution $u(t)=c\_1e^{\lambda \_1t}x\_1+c\_2e^{\lambda \_2t}x\_2+\dots$ the $x\_n$ eigenvectors and $\lambda \_n$ eigenvalues of $A$. This solution can be rewritten as a combination of eigenvector columns $u(0)=\left[\begin{array}{ccc}x\_1& \dots & x\_n\end{array}\right]\left[\begin{array}{c}c\_1⋮c\_n\end{array}\right]=Sc$, a property that is central to the following decomposition.

Finding the General Solution

1. All $\lambda$ and $x$ of $A$ are found
2. The solution structure is populated with these values and set equal the initial condition to find the coefficients $c\_1,\dots ,c\_n$: $u(0)=c\_1e^{\lambda \_1t}x\_1+c\_2e^{\lambda \_2t}x\_2+\cdots =\left[\begin{array}{c}ab\end{array}\right]$

Convergence of Solutions

Because the general solution consists of separate exponential factors that can separately converge or diverge. A stable solution only exists if all $\mathrm{Re}(\lambda )<0$ where some of the exponential terms $e^{\lambda t}$ decay to 0 while others ($\lambda =0$) stay constant.

Realizing Diagonalization in the Process

If $A$ is diagonalizable ($\lambda$ all independent) all of the following applies.

The bundled form of any solution is therefore $u(t)=e^{At}u(o)$. Computing $e^{At}=\sum \_0^{\infty }\frac{(At{)}^n}{n!}$ for dense matrices $A$ is unfeasible, thus we diagonalize to use the property $A^n=S{\Lambda }^n{S}^{-1}$ of Eigenmatrices.

We find that $A^n=S{\Lambda }^n{S}^{-1}$ and substitute into the infinite sum out of which we can factor $S$ and $S^{-1}$.

$$e^{At}=S(\sum \_0^{\infty }\frac{{\Lambda }^n{t}^n}{n!})S^{-1}$$

Therefore $e^{At}=S{e}^{\Lambda t}{A}^{-1}$ and solutions $u(t)=e^{At}u(0)=S{e}^{\Lambda t}{A}^{-1}u(0)$. Therefore $u$ is re-expressed as a linear combination of eigenvectors.

Mechanism of this Change-of-Basis

Solutions

Diagonalizing the differential-equation systems breaks the dependency of the equations on one-another apart.

$$e^{\Lambda t}=\left[\begin{array}{ccc}{e}^{\lambda \_1t}& & \\ & \ddots & \\ & & e^{\lambda \_nt}\end{array}\right]$$

Logical Chain

$A=S\Lambda S^{-1}\stackrel{\text{powers}}{\to }{A}^k=S{\Lambda }^k{S}^{-1}\stackrel{\text{series}}{\to }{e}^{At}=S{e}^{\Lambda t}{S}^{-1}\stackrel{u=Sv}{\to }\text{decoupled }{v}^{\prime }=\Lambda v$

The reason your notes’ general solution $u(t)=c\_1e^{\lambda \_1t}x\_1+c\_2e^{\lambda \_2t}x\_2+\cdots$ is equivalent to $S{e}^{\Lambda t}{S}^{-1}u(0)$ is precisely this: the columns of $S$ are the $x\_i$, the diagonal entries of $e^{\Lambda t}$ supply the $e^{\lambda \_it}$ factors, and $S^{-1}u(0)$ extracts the coordinates $c\_i$ in one matrix operation.

Why $S^{-1}u(0)$ extracts the $c$‘s

$S$ is a change-of-basis matrix. It converts eigenbasis coordinates into standard coordinates:

$\underset{\text{eigenbasis}\to \text{standard}}{\underbrace{S}}\cdot \underset{\text{coords in eigenbasis}}{\underbrace{c}}=\underbrace{u(0)}\_\text{standard coords}$

So $S^{-1}$ runs the arrow the other way:

$\underset{\text{standard}\to \text{eigenbasis}}{\underbrace{S^{-1}}}\cdot \underset{\text{standard coords}}{\underbrace{u(0)}}=\underbrace{c}\_\text{coords in eigenbasis}$

Extra: Decomposing Systems of high-dimension Differential Eqs.

An $n$th order system $y^{\prime \prime }+b{y}^{\prime }+ky=0$ can be decomposed into $n\times n$ $1st$ order equations. (Involving $2\times 2$) matrices.

Here, we use $u=\left[\begin{array}{c}{y}^{\prime }y\end{array}\right]$ such that $u^{\prime }=\left[\begin{array}{c}{y}^{\prime \prime }{y}^{\prime }\end{array}\right]=\left[\begin{array}{ccc}-b& -k\text{ }1& 0\end{array}\right]u$