What does this blog do

We are treating linear ODE problems as an example of quotient space in this blog post. You are assumed to be able to solve linear ODEs without theoretical problems, and have some basic understanding of normal subgroups.

General theories

Quotient Space

Let $X$ be a vector space, and $N$ a subspace of it. Naturally $N$ is normal in $X$ since $X$ is abelian. Define

$\pi(x)=x+N$

for $x \in X$, then the collection of sets ${\pi(x):x \in X}$ is the quotient space of $X$ modulo $N$, in which case we write $X/N$. Addition and scalar multiplication are defined by

$\pi(x+y)=\pi(x)+\pi(y)\quad \pi(\alpha x)=\alpha\pi(x)$

This is well-defined since $N$ is a vector space. The kernel of $\pi$ or the origin of $X/N$ has to be $N=0+N$. People call $\pi$ the canonical map.

Linear ODE

If one solves a linear ODE problem of order $n$ on an interval $(a,b)$, namely

$\frac{d\mathbf{y}}{dx}=\mathbf{Ay}+\mathbf{f} \quad a<x<b$

where $\mathbf{A}$ is a constant matrix, they will find that the solution can be

$\mathbf{y}=\mathbf{\Phi}(x)\mathbf{c}+\mathbf{\Phi}(x)\int_0^x\mathbf{\Phi^{-1}}(s)\mathbf{f}(s)ds$

where $\mathbf{c}$ is a given constant vector and $\mathbf{\Phi}(x)=\begin{bmatrix}\mathbf{\varphi_1},\mathbf{\varphi_2},\cdots,\mathbf{\varphi_n}\end{bmatrix}$ and $\{\mathbf{\varphi_i}\}$ are the fundamental solutions of $\frac{d\mathbf{y}}{dx}=\mathbf{Ay}$. We’ll translate this into the language of quotient space. We have to show that the solutions of the equation forms an element of $X/N$.

Steps to quotient space

So where is the $X$? It suffices to pick $\mathcal{C}^n$, the space of all functions $\mathbf{y}=(y_1,y_2,\cdots,y_n)^T$ such that $y_k$ is $n$-time differentiable. The crux therefore becomes finding $N$. And we’ll show that it’s denoted by $\mathbf{\Phi}(x)\mathbf{c}$.

The solutions of $\frac{d\mathbf{y}}{dx}=\mathbf{Ay}$ form a vector space

Indeed, it’s trivially verified since $\frac{d}{dx}$ and matrix multiplication are linear. The question is, how does this vector space look like? Why the fundamental solutions to this equation has and only has $n$ elements? Does that mean this space (denoted by $N$), has dimension $n$? Fortunately, the following isomorphism answers this question in the affirmative.

Theorem The vector space $N$ is isomorphic to $\mathbb{R}^n$

Pick and fix $x_0 \in (a,b)$. Picard’s existence and uniqueness theorem ensures that, the initial value problem

$\begin{cases}\frac{d\mathbf{y}}{dx}=\mathbf{Ay}\\\mathbf{y}(x_0)=\mathbf{y}_0\end{cases}$

has a unique solution. Therefore we have a bijection

$\begin{aligned}H: N &\mapsto \mathbb{R}^n \\\mathbf{y}(x)&\to \mathbf{y}_0\end{aligned}$

It suffices to prove that $H$ is linear. Namely, we need to show that

$H(C_1\mathbf{y}_1+C_2\mathbf{y}_2)=C_1\mathbf{y}_1(x_0)+C_2\mathbf{y}_2(x_0)$

which is trivial, since

$(C_1\mathbf{y}_1+C_2\mathbf{y}_2)(x_0)=C_1\mathbf{y}_1(x_0)+C_2\mathbf{y}_2(x_0)$

Quotient space and coset are there

Let’s see this solution again

$\mathbf{y}=\mathbf{\Phi}(x)\mathbf{c}+\mathbf{\Phi}(x)\int_{x_0}^x\mathbf{\Phi^{-1}}(s)\mathbf{f}(s)ds$

For $\mathbf{c}=(c_1,c_2,\cdots,c_n)^T$, we have

$\mathbf{\Phi}(x)\mathbf{c}=\sum_{k=1}^{n}c_k\mathbf{\varphi}_k$

Notice that $\{\mathbf{\varphi_k}\}$ is a basis of $N$ and there we have it. $\mathbf{c}$ can be any element of $\mathbb{R}^n$, hence $\mathbf{\Phi}(x)\mathbf{c}$ goes through all elements of $N$. Also we know, the function

$\mathbf{z}=\mathbf{\Phi}(x)\int_{x_0}^x\mathbf{\Phi^{-1}}(s)\mathbf{f}(s)ds$

is a special solution to $\frac{d\mathbf{y}}{dx}=\mathbf{Ay}+\mathbf{f}$. Thus we have the coset

$\mathbf{z}+N=\pi(\mathbf{z})$

to be the set of all solutions where $\pi(\mathbf{z}) \in X/N$.