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. 9f48a11b019613d1dcbd25486a5f5fa0.png