Lecture 8: Intro to ODE

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tags: 224a

First order Linear ODE

We begin by proving a basic existence and uniqueness result for first order initial value problems, which will serve as the foundation of such results for second and higher order problems. The notation

$C^k[a,b]$ will denote

$k$ times continuously differentiable functions on

$[a,b]$ (this means they must be defined in a neighborhood of the endpoints), whose co-domain will be apparent from the context.

Theorem. For a finite closed interval

$[a,b]$, the IVP:

$$u^{\prime }(x)=A(x)u(x)+b(x),$$

$$u(x_0)=u_0\in {\mathbb{R}}^d,$$
where

$u\in C^1[a,b]$ and

$A,b\in C[a,b]$ are real vector/matrix/vector valued, has a unique solution

$u$.

Proof. Observe that by the fundamental theorem of calculus, a

$C^1$ function

$u$ satisfies the equation iff it satisfies the integral equation

$$u(x)=u_0+{\int }_{x_0}^x(A(s)u(s)+b(s))ds.$$
Define the operator

$T:C[a,b]\to C[a,b]$ by

$$T\varphi =u_0+{\int }_{x_0}^x(A(s)\varphi (s)+b(s))ds.$$
The fact that

$T\varphi$ is a continuous function follows from continuity of the integrand on

$[a,b]$. Consider

$C[a,b]$ to be a metric space with the sup norm

$|f-g|=\underset{[a,b]}{sup}|f(x)-g(x)|.$

Our goal is to show that

$T$ has a unique fixed point, which will be a solution to our ODE. We will appeal to the Banach fixed point theorem: if

$(M,d)$ is a metric space and

$T:M\to M$ satisfies

$d(T(a),T(b))\le \alpha d(a,b)$ for all

$a,b$ for some

$\alpha <1$, then

$T$ has a unique fixed point.

$T$ itself is not a contraction, but it turns out a high enough power of it is. This method is called Picard iteration.

Claim. There exists

$n$ such that

$|T^{n}f-T^{n}g|\le .9|f-g|$ for every

$f,g\in C[a,b]$.
Proof. To be completed.

Higher order ODE

A general

$n$th order ODE:

$$\sum _{j=0}^n{a}_j(x)u^{(j)}(x)=b(x)$$
with

$a_n(x)\ne 0$ can be written as a system of first order ODE in

$n$ functions constrained by the

$n-1$ equations:

$$u_0=u,u_1=u_0^{\prime },\dots ,u_k=u_{k-1}^{\prime }=u^{(k)},\dots ,u_{n-1}=u_{n-2}^{\prime }.$$ This allows one to write the ODE linearly as

$$a_n(x)u_{n-1}^{\prime }{+}^{\prime }\sum _{j=0}^{n-1}{a}_j(x)u_j(x)=b(x).$$

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Viewing the variables as a single vector valued function

$\hat{u}:[a,b]\to {\mathbb{R}}^n$ and dividing by

$a_n(x)$ the above system is of the form

$${\hat{u}}^{\prime }(x)=A(x)\hat{u}(x)+b(x)e_n,$$
and there is a bijection between solutions of this equation and solutions of the nth order ODE (the matrix

$A(x)$ is just the https://en.wikipedia.org/wiki/Companion_matrix). Thus, by the theorem in the previous section it must also have a unique solution given initial data

$$u(x_0)=u_0,\dots ,u^{(n-1)}(x_0)=u_{n-1}.$$

Dimension of The Solution Space

Let

$L=\sum _{j=0}^n{a}_j{d}^j/d{x}^j$ be an nth order differential operator. For the homogeneous problem

$Lu=0$, the set of solutions is a subspace of

$C^n[a,b]$. For any point

$x_0\in [a,b]$, consider the linear map

$$E_{x_0}(u)=[u(x_0),u^{\prime }(x_0),\dots ,u^{(n-1)}(x_0){]}^T$$
into

${\mathbb{R}}^n$. By existence and uniqueness of solutions to the IVP at

$x_0$, this map must be a bijection. Thus the space of solutions has dimension exactly

$n$ for an

$n$th order ODE.