Cauchy problems for beginners in mathematics

cauchy-problems-for-beginners-in-mathematics


In this article, we introduce Cauchy problems for beginners in mathematics. To simplify our exposition, we will talk only about linear Cauchy problems. First, we study such problems in the set of real numbers. Second, generalize the result to finite-dimensional spaces. Finally, treat the case of general Banach spaces, partial differential equations. At the end of this article, we shall give some known Cauchy problems. The background of this post can be useful for engineers as well as for graduate students.


The origin of Cauchy’s problems


In one of his lectures at the École Polytechnique de Paris, The French mathematican Austin-Louis Cauchy asked his student the following question: is there a continuous function $ u: \mathbb{R}\to \mathbb{R} $ such that \begin{align*}\tag{Eq} u (t + s) = u (t ) u (s)\end{align*} for all $t, s\in\mathbb{R}$. In fact, this question dated two centuries ago. Of course, the null function is a solution to this problem. On the other hand, if $u(0)$ and if $u$ satisfies the above functional equation (Eq), then by replacing $s$ by zero we get $f(t)=0$, which is the null function. This case is not interesting, we then assume that $u(0)\neq 0$. In this case, as $u(0)=u(0+0)=u(0)u(0),$ we deduce that $u(0)=u(0)^2$, so $u(0)=1$.


The Differential equation


The equation (Eq) implies that $u$ is differentiable on $\mathbb{R}$. On one the hand , let the set be $a=\dot{u}(0)$ the derivative of $u$ at $0$. Then the function $u$ satisfies the following differential equation \begin{align*}(CP)\qquad \begin{cases} \dot{u}(t)=a u(t),& t\in \mathbb{R},\cr u(0)=1.\end{cases}\end{align*} We say that (CP) is a Cauchy problem in $\mathbb{R}$ and its solution if exaclty the following exponential \begin{align*} u(t)=e^{t a},\qquad t\in\mathbb{R}.\end{align*} So, the name of Cauchy problem stems from a differential equation with an intial condition.


We recall that the real exponential function is given by the following series. \begin{align*}e^{at}=\sum_{n=0}^{+\infty} \frac{t^n}{n!}a^n\end{align*}where $n!=1\times 2\times \cdots\times n$ with the convention $0!=1$.


Linear systems on spaces of finite dimension


In this section, we slightly generalize the result of the previous section to a space of finite dimension. In fact, instead of working on $\mathbb{R}$ (which is one-dimensional space) to $n$-dimension space $\mathbb{R^n}$.


The exponential of a matrix


We ask the following question: Can we define the exponential of a square matrix $A$ of order $n$?. By analogy to the real case, we define \begin{align*}e^{tA}:=\sum_{n=0}^{\infty} \frac{t^n}{n!} A^n,\quad t\in\mathbb{R}.\end{align*}


This definition makes sense only if the above series converges. In fact, by using the matrices norms we have $\|A^n\|\le \|A\|^n,$ so that \begin{align*} \left\| \frac{t^n}{n!} A^n\right\|\le \frac{\left(|t|\|A\|\right)^n}{n!}.\end{align*} Also, using series properties it follows that the quantity $e^{t A}$ is well-defined which we call exponential of a matrix.


An important remark on the exponential of matrices


In general, let $A$ and $B$ be arbitrary square matrices. Then \begin{align*} e^{A+B}\neq e^{A}e^{B}.\end{align*} This is because $AB\neq BA$, and then we can not use the Binomial formula. However, if $A$ and $B$ commute, $AB=BA$. Then by using a series product and the Binomial formula, we prove that $ e^{A+B}\neq e^{A}e^{B} $.


The solution is defined by the exponential of a matrix


For instance, consider the Cauchy problem \begin{align*}\begin{cases} \dot{u}(t)=A u(t),& t\in\mathbb{R},\cr u(0)=x\in \mathbb{R}^n.\end{cases}\end{align*} The solution of this problem satisfies following exponential of matrix \begin{align*} u(t)=e^{tA}x,\quad t\in\mathbb{R}.\end{align*} Although, to solve linear systems of the form (CP) it suffices to calculate the exponential of matrices. The latter is simple to compute if $A$ is diagonal or $A$ in nilpotent, that is, $A^p=0$ for some natural number $p$. We can also use reduction of matrices (diagonalization and trigonalization by Jordan blocs) to calculate the above exponential.


Partial differential equations


The case if the partial differential equations (PDE) is quite different. In fact, contrary to finite systems, the solution of a PDF $u(t,x)$ follows two variations, the variation of the time $t$ and the variation of the space coordinates. As an example the following equation \begin{align*} \frac{\partial u}{\partial t}(t,x) = \frac{\partial u}{\partial x}(t,x) ,\quad t\ge 0,\quad x\in [0,r].\end{align*}.


This equation represents the deplacement a line of lent $r$. Here, for each $t$, the solution $u(t,\cdot)$ is not a vector, but it is a function $u(t,\cdot):[0,r]\to \mathbb{R}$. The analogical $A$ of the previous section is \begin{align*} A =\frac{d}{dx}\end{align*} is not a matrix bu it is a differential operator. Moreover, the quantity $Au$ is well-defined if $u$ is diffrential, due to the definition of $A$. We then remark that $A$ is defined on the natural domain ${\rm Dom}(A)=C^1([0,r])$ the space of continuousely differential functions, and the state space is $X=L^2([0,r])$ the space of square integrable functions. An other example the eveolution of a population.


The general case


Let us define a general Cauchy problem. To this end, let $X$ be a complete normed space (Banach space), and a linear operator $A: {\rm Dom}(A) \subset X\to X$. Define \begin{align*} \begin{cases}\dot{u}(t)=Au(t),& t\ge 0,\cr u(0)=f.\end{cases}\end{align*} In the case of heat equation we have $A=\frac{d^2}{dx^2}$ with domaine the set of functions $f$ such that the second derivative of $f$ exists and $Af\in L^2$.


The Cauchy problem (CP) has a solution if and only if $A$ satisfies the condition of the Hille-Yosida theorem. That is the set (graph) $\{(f,Af):f\in {\rm Dom}(A) \}$ is closed in $X \times X$; $(\lambda I-A)^{-1}$ exists for any $\lambda\in \mathbb{C}$ with $Re\lambda >\omega$ (for some $\omega\in\mathbb{R}$), and \begin{align*} \|(\lambda I-A)^{-1}\|\le \frac{M}{( Re\lambda -\omega)^n}\end{align*} for all $n\in\mathbb{N}$ and some real number $M\ge 1$. The problem (CP) can also, in some cases, solved by Laplace transform or Fourier transform.


If (CP) has a solution $X$, there exists a family of linear bounded operators $T(t):X\to X$ for $t\ge 0,$ satisfying $T(0)=I,$ $T(t+s)=T(t)T(s)$ and $T(t)f-f\to 0$ as $t\to 0,$ for any $f\in X$. Thus $T(\cdot)$ satisfies the exponential properties. The solution of (CP) is: \begin{align*}u(t)=T(t)f,\quad t\ge 0.\end{align*}. The familly $(T(t))_{t\ge 0}$ is called strongly continuous semigroups.


A Cauchy problem describing a transport equation


We consider a function defined by \begin{align*} u(t,x)=f(t+x),\quad t\in \mathbb{R}^+,\quad x\in [-r,0]\end{align*} for $f\in L^2([-r,0])$. If we assume that $f$ is differentiable, then by compostion of differential function on can be that \begin{align*} \frac{\partial u}{\partial t}(t,x)=Df(t+x)= \frac{\partial u}{\partial x}(t,x),\end{align*} where $Df(z)$ is the differntial function of $f$ at point $z$.


The translation semigroup


Now, if we set \begin{align*} A g=g’,\quad {\rm Dom}(A)=\{g:[-r,0]\to \mathbb{R},\; {\rm diff},\; g’\in L^2:g(0)=0\};\end{align*} the the function $u$ satisfies the Cauchy problem \begin{align*} u'(t)=Au(t),\quad t\ge 0,\quad u(0)=f.\end{align*}. We derive then the solution from: \begin{align*} u(t)=T(t)f,\qquad t\ge 0,\end{align*} where \begin{align*} (T(t)f)(s)=f(t+s),\qquad t,s\ge 0,\quad f\in L^2([-r,0]).\end{align*}.

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