Basic concepts in ODEs
Basic Notions
An ODE of the $k$ -th order is in the expression of the form
\[\frac{\mathrm{d}^kx}{\mathrm{d}t^k}=f\left(t,x,\frac{\mathrm{d}x}{\mathrm{d}t},\ldots,\frac{\mathrm{d}^{k-1}x}{\mathrm{d}t^{k-1}}\right)\]A function $\phi (x)$ is called a solution to above ODE on the interval $I=(a, b)$ if this function $k$ times continously differentiable on $I$ or say that $\phi\in C^{k}(I,\mathbb{R})$ .
An ODE plus the initial conditions are called an initial value problem (IVP for short) or Cauchy’s problem.
Malthus population model
We will denote $N (t)$ as the number of individuals in a given populations at time moment $t$ . Then the population number changes during a short time interval $h$ we have
\[N(t+h)=N(t)+bh(t)-dhN(t)\]where $b$ and $d$ here stands for birth rate and death rate, and then we will have
\[\frac{N(t+h)-N(t)}{h}=(b-d)N(t)\]if we postulate that the existence of a derivative, then we will have
\[\frac{\mathrm{d}N}{\mathrm{d}t}=(b-d)N\]if we denote $b-d$ as a parameter $m$ then we will have the population growth model to be an IVP
\[\dot{N}=mN,N(0)=N_0\]where $N(t)$ is the population size at time moment $t$. But we cannot always know the direct form of the model, maybe we can use assume that the law of growth has the general form:
\[\dot{N}=NF(N)\]Where $F$ is some function, which has to be negative for sufficiently large values of $N$, if this function is smooth enough, we have its Taylor formula around $N=0$ as
\[F(N)=F(0)+\frac{F'(0)}{1!}N+\frac{F''(0)}{2!}N^2+o(N^2)\]If we only keep up constant terms and them we will have the equation
\[\dot{N}=mN\]where $m=N(0)$, and if you keep two terms we will have
\[\dot{N}=NF(N)=N(F(0)+F'(0)N)=mN\left(1-\frac{N}{K}\right)\]where $K=-F(0)/F’(0)$ to be another parameter called $K$ in above equation.
Well-posed problems
(Salomon Hadamard) A mathematical problem is well posed if
- its solution exists.
- its solution is unique.
- its solution depends continuously on the initial data.
Theorem(local uniqueness)
Consider the IVP and assume that function $f$ is continous in $t$ and continously differentiable in $x$ for $(t,x)\in (a,b)\times(c,d)$ for some constant $a,b,c,d$, assume that $(t_0,x_0)\in(a,b)\times(c,d)$ then there exists am $\epsilon>0$ such that the solution $\phi(x,t)$ to the equation exists and unique for $t\in(t_0-\epsilon,t_0+\epsilon)$.
However the theorem is local, it only guarantees that the solution exists and is unique on some smaller interval $(t_0-\epsilon,t_0+\epsilon)\subseteq (a,b)$. However the solution of ODE can below up i.e. approach infinity for a finite $t$.
Examples of the uniqueness theorem. Consider the ODE
\[\dot{x}=1+x^2=f(t,x)\]the right hand side is a polynomial for any $(t,x)\in\mathbb{R}$, its solution is given by
\[x(t)=\tan{(t+C)}\]and hence for each fixed $C$ is defined only on the interval $(-\pi/2-C,\pi/2-C)$.
Consider the ODE
\[\dot{x}=\sqrt{x},x(0)=x_0,x\geq0\]One solution can be given as
\[x(t)=\frac{(t+2\sqrt{x_0})^2}{4}\]if $x_0=0$ then the solution is exactly $x(t)=0$, therefore the solution is not unique! A strong theorem on IVP is given below
Theorem
Let IVPs satisfy the local conditions, where two IVPs are defined as
\[\begin{gathered} \dot{x}=f(x,t),x(t_0)=x_0\\ \dot{x}=f(x,t),x(t_0)=x_1 \end{gathered}\]and $t\mapsto x_0(t)$ and $t\mapsto x_1(t)$ be the solution to IVP at the same time $t$ and then we will have
\[|x_1(t)-x_0(t)|\leq |x_1-x_0|\mathrm{e}^{L|t-t_0|}\]where $L$ is a constant that depends on $f$.
Above theorem shows that the solution to a first order ODE depends continously on the initial condition.
Visualizing a ODE system.
Consider the ODE
\[\dot{x}=\sin{x}\]and we plot out its phase diagram.