---
title: "SU-MATh53 JAN262023"
source: https://www.jemoka.com/posts/kbhsu_math53_jan262023/
date: 2023-01-26
---

Underdetermined ODEs Complex ODE System Matrix Exponentiation Finding eigenvectors \(A = n \times n\) matrix, the task of finding eigenvalues and eigenvectors is a linear algebra problem:
\begin{equation} A v = \lambda v \end{equation}
Finding specific solutions to IVPs with special substitution For some:
\begin{equation} \begin{cases} x&rsquo; = Ax \\ x(t_0) = x_0 \end{cases} \end{equation}
we can leverage the first task:
find \(v\), \(\lambda\) for \(A\) guess \(x = u(t)v\), this is &ldquo;magical substitution&rdquo; and now, we can see that \(x&rsquo; = u&rsquo;v = A(uv) = \lambda u v\) meaning \(u&rsquo; = \lambda u\) finaly, \(u(t) = ce^{\lambda} t\) Eigenbasis case Suppose \(A\) has a basis of eigenvectors, and real eigenvalues. We can write its entire solution set in terms of these basis eigenvectors:
\begin{equation} x(t) = u_1(t) v_1 + \dots + u_{n}(t) v_{n} \end{equation}
this means:
\begin{equation} x&rsquo;(t) = Ax = u_1&rsquo; v_1 + \dots +u_{n} &rsquo; v_{n} = \lambda_{1} u_{1} v_1 + \dots + \lambda_{n} u_{n} v_{n} \end{equation}
Because \(v\) forms a basis, each \(u_j&rsquo; = \lambda_{j} u_{j}\).
We thereby decomposed our entangled expression seperably by changing into eigenbasis.
After solving each \(u\), we obtain:
\begin{equation} x(t) = c_1 e^{\lambda_{1}} v_1 + \dots + c_{n} e^{\lambda_{n}} v_{n} \end{equation}
We can identify \(c_{j}\) by noting, that \(x(0)\) resolves to:
\begin{equation} x(0) = c_1v_1 + \dots + c_{n}v_{n} \end{equation}
Finally, we can write this as:
\begin{equation} x(0) = x_0 = \mqty[v_1 &amp; \dots &amp; v_{n}] c \end{equation}
Meaning, we can solve for initial conditions as:
\begin{equation} \mqty[v_1 &amp; \dots &amp; v_{n}]^{-1} x_0 = c \end{equation}
Practice Solving Let:
\begin{equation} A = \mqty(0 &amp; 1 &amp; 1 \\ 1 &amp; 0 &amp; 1 \\ 1 &amp; 1 &amp; 0) \end{equation}
We have two eigenspaces:
\begin{equation} \lambda = -1, v = \left\{\mqty(-1 \\ 1 \\ 0), \mqty(0 \\ 1 \\ -1)\right\} \end{equation}
and
\begin{equation} \lambda = 2, v = \left\{\mqty(1 \\ 1 \\ 1)\right\} \end{equation}
This gives rise to a basis of eigenvectors with all three vectors. We obtain:
\begin{equation} x(t) = c_1 e^{-t} \mqty(-1 \\ 1\\0) + c_2 \mqty(0 \\ 1 \\ -1) e^{-t} + c_3 \mqty(1 \\ 1 \\ 1) e^{2t} \end{equation}
