In this post, we propose two results in the instability of solutions to nonlinear systems. Here, we study ordinary differential equations. Well-know stability theorems are due to Liapunov, based on the linearization of the vector field.

Consider a continuous function $f:\Omega\subset \mathbb{R}^d\to\mathbb{R}^d$, and $x_0\in \Omega$. Let the Cauchy problem\begin{align*}\tag{Eq}\dot{u}(t)=f(u(t)),\quad u(0)=x_0.\end{align*}

The flow of an autonomous system

According to Peano’s theorem, the maximal solution of the differential equation $({\rm Eq})$ exists. We denote by $J_{x_0}$ the interval in which the maximal solution is well defined. We denote \begin{align*}D(f)=\bigcup_{x\in\Omega}\left(J_x\times\{x\}\right).\end{align*} The flow associated to the equation $({\rm Eq})$ is the following application \begin{align*} \Phi: D(f)\to \Omega,\quad (t,x)\mapsto \Phi(t,x)=u(t),\end{align*}where $u$ is the maximal solution associated to the initial condition $u(0)=x$. Also, we denote $\Phi_t(x)=\Phi(t,x)$.

We mention that $J_{\Phi_t(x)}=J_x-t$, and if $t_1+t_2\in J_x$, we have\begin{align*}\Phi_{t_2}\circ \Phi_{t_1} (x)= \Phi_{t_1+t_2}(x).\end{align*}

Assume that $f$ is locally Lipschitz on $\Omega$. Then $D(f)$ is an open set of $\mathbb{R}\times \Omega$. Moreover, the flow $\Phi$ is locally Lipschitz, in particular, it is continuous, on $D(f)$.

 Instability of solutions to nonlinear systems

An equilibrium point of $f:\Omega\to \mathbb{R}^d$ is an element $x_0\in \Omega$ such that $f(x_0)=0$. An immediate consequence is that the constant function equal to $x_0$ satisfies the differential equation $({\rm Eq})$.

The equilibrium point $x_0$ is stable if and only if for any neighborhood $W$ of $x_0$ in $\Omega,$ there exists a neighborhood $U$ of $x_0$ in $\Omega$ such that for all $x\in U,$ $\Phi_t(x)$ is defined for any $t\ge 0,$ i.e. global solution, and takes value in $W$.

The equilibrium $x_0$ is unstable if and only if it is not stable.

The equilibrium $x_0$ is asymptotically stable if and only if it is stable and there exists a neighborhood $W$ of $x_0$ in $\Omega$ such that for any $x\in W,$ $\Phi_t(x)$ is defined for any $t\ge 0$ and $\Phi_t(x)\to x_0$ as $t\to +\infty$.

In this article, we give an application of the open mapping theorem in functional analysis. This fundamental theorem in functional analysis plays a key role in the study of evolution equations.

In the sequel, we propose a nice functional analysis worksheet on some deep applications in different domains of mathematical analysis.

An application of the open mapping theorem

Exercise: Let $f$ be an application from $\mathbb{R}^n$ to $\mathbb{R}^n$. Suppose that there exists $C>0$ such that \begin{align*}\forall (x,y)\in \mathbb{R}^n\times \mathbb{R}^n ,\quad \langle f(x)-f(y),x-y\rangle\ge C\|x-y\|^2.\end{align*}We propose to demonstrate that $ f $ is a homeomorphism from $ \mathbb{R}^n $ on $ \mathbb{R}^n $. To do so, consider the differential equation \begin{align*}\tag{Eq} \dot{x}(t)=-f(x(t)).\end{align*}

• Let $x$ and $y$ two solutions of $(Eq)$ defined on $[0,T]$ with $T>0$. If we put $\varphi=\|x-y\|^2,$ prove that \begin{align*} \varphi(t)\le \varphi(0)e^{-2Ct},\quad \forall t\in [0,T].\end{align*} Deduce that \begin{align*} \|\dot{x}(t)\|\le \|\dot{x}(0)\| e^{-Ct}.\end{align*}
• Let $x$ be a maximal solution of the Cauchy problem $\dot{x}=-f(x)$ and $x(0)=0$. prove that $x$ is defined on $[0,+\infty)$, global solution, and $x$ admits a finite limit $\ell$ as $t\to +\infty$ such that $f(\ell)=0$.
• Give a conclusion

Solution: 1) As the application $x-y$ is differentiable, then $\varphi$ is differentiable and \begin{align*}\varphi'(t)&=2\langle x(t)-y(t),x'(t)-y'(t)\rangle\cr & \le -2C\varphi(t).\end{align*} This implies that \begin{align*} \varphi(t)\le \varphi(0)e^{-2Ct}.\end{align*}

On the other hand, let we fix $s\in [0,T[$. It is not difficult to see that the application $x_s:[0,T-s]\to \mathbb{R}^n$ such that $t\mapsto x_s(t)=x(t+s)$ is a solution of the equation $(Eq)$. Then by taking $y=x_s$ in the above estimate, we get \begin{align*}\forall t\in [0,T-s],\quad \|x(t+s)-x(t)\|\le \|x(t)-x(0)\| e^{-C t}.\end{align*} This inequality is true for any $(s,t)$ with $0\le s<T$ and $0\le t\le T-s$. Taking $s\in (0,T-t)$, so \begin{align*}\left\|\frac{x(t+s)-x(t)}{s}\right\|\le \left\|\frac{x(t)-x(0)}{s}\right\| e^{-C t}.\end{align*} Now by letting $s\to 0,$ we obten the result.

2) Let $x:[0,T[\to \mathbb{R}^n$ the maximal solution of the Cauchy problem with initial condition $x(0)=0$. If $T<+\infty$, then $\|x(t)\|$ explodes if $ t $ is close to $T$. This means that there exists $\delta>0$ such that $\|x(t)\|$ is not bounded on $[T-\delta,T[$. On the other hand, for $t\in [0,T]$, \begin{align*} \|x(t)\|&\le \int^t_0 \|\dot{x}(\sigma)\|d\sigma\cr & \le \int^t_0 \|\dot{x}(0)\| e^{-C \sigma}d\sigma\cr & \le \frac{\|\dot{x}(0)\|}{C}.\end{align*} This is a contradiction. Thus $T=+\infty$. Then \begin{align*} \|\dot{x}(t)\|\le \|\dot{x}(0)\| e^{-C t},\qquad\forall t\in [0,+\infty).\end{align*} This implies that the improper integral \begin{align*}\int^{+\infty}_0 \|\dot{x}(t)\|dt\end{align*} converges. Thus $x$ has a limit $\ell$ at $+\infty$. On the other hand, as $\dot{x}(t)\to 0$ as $t\to 0,$ then $f(\ell)=0$.

3) If we replace $ f $ by $ f-y $ which satisfies the same hypotheses, we see that $ f $ is a surjective from $\mathbb{R}^n$ to $\mathbb{R}^n$. According to the first hypothesis on $f,$ we deduce that $f$ is also injective. Then it realizes a bijection from $\mathbb{R}^n$ to $\mathbb{R}^n$ . By using Cauchy-Schwarz inequality we have \begin{align*} \|f(x)-f(y)\|\ge C \|x-y\|,\qquad \forall (x,y)\in \mathbb{R}^n\times \mathbb{R}^n.\end{align*} From this we deduce that \begin{align*} \|f^{-1}(x)-f^{-1}(y)\|\le \frac{1}{C} \|x-y\|,\qquad \forall (x,y)\in \mathbb{R}^n\times \mathbb{R}^n.\end{align*} This implies that $f^{-1}$ is continuous.

Differential calculus in Banach spaces is a very important part of mathematics. In fact, the treatment of partial differential equations strongly depends on this theory. Think about the heat equation. We mention that such equations use function spaces as a workspace. The latter is a Banach space, such as the Lebesgue space, the space of continuous and bounded functions. Hence the importance of differentiation in Banach spaces.

A Banach space $E$ is a vector space endowed with a norm $\|\cdot\|$ for which Cauchy sequences converge in $E$. On the other hand, we say that $E$ is complete with respect to the norm $\|\cdot\|$.

We shall denote by $\mathcal{L}(E)$ the space of linear continuous applications from $E$ into $E$, bounded operators. If $T\in \mathcal{L}(E) $, there exists $M\ge 1$ such that $$\|Tx\|\le M \|x\|,\quad \forall  x\in E.$$ As we will see in the sequel, the differentiability of an map is a linear continuous applications. Contrary to the one dimension case, where a derivative is just a real number.

Differentiation in the real number set

Let us start by defining the differentiation of functions of one variable. We say that a function $f:D\subset \mathbb{R}\to \mathbb{R}$ is differentiable at $c\in D$ if the following limit\begin{align*}\ell=\lim_{x\to c\,x\neq c}\frac{f(x)-f(c)}{x-c}\end{align*} exists. In this case, we set $\ell:f'(c)$; called the derivative of $f$ at the point $c$.

Also, we can write \begin{align*} f(c+h)=f(c)+f'(c)h+\varepsilon(h),\end{align*} where $ \varepsilon(h)/h \to 0$ as $h\to 0$.

We say that $f$ is differentiable on $D$ if $f$ is differentiable at all points of $D$. In this case, we define the differential function of $f$ by $f’:D\to \mathbb{R}$.

The important thing to remember is that the derivative at a point $ c $ is a real number, that is $f'(c)\in\mathbb{R}$.

Differential calculus in Banach spaces

In this part, we shall work with an infinite-dimensional Banach space $(E,\|\cdot\|)$. Let $f: E\to E$ be an application, not necessarily linear, and $x_0,h\in E$. We say that $f$ is differentiable at $x_0$ if there exists a linear continuous application $L\in \mathbb{L}(E)$ such that \begin{align*}f(x_0+h)=f(x_0)+Lh+\varepsilon(h)\end{align*}such that\begin{align*}\lim_{h\to 0}\frac{ \varepsilon(h)}{\|h\|}=0.\end{align*}

In this case we say that $f $ is differentiable at $x_0$ and $L$ the differential of $f$ at $x_0$, denoted by $L:=Df(x_0)$.

Contrary to $\mathbb{R}$, the differential of $f$ at a point $x_0$ is a linear continuous application from $E$ to $E$; that is $Df(x_0)\in\mathcal{L}(E)$.

Illustrate this definition: differentiability in the matrix spaces 

To that purpose, denote by $E=\mathcal{M}_{m,n}(\mathbb{R})$ the Euclidian spaces of matrices of order $m\times n$ endowed with the inner product\begin{align*}(A,B)\in E\times E\mapsto \langle A,B \rangle:={\rm Tr}(A^T B),\end{align*}where ${\rm Tr}$ is the trace. For a function $f:U\subset E\to \mathbb{R}$ defined on a open set $U$ of $E$ and differentiable in $X\in U,$ we denote by $\nabla f(X)$ the gradient of $f$ in $X$.

General matrices

Fix $A\in E$ and let $f_A$ the application\begin{align*}f_A: E\to \mathbb{R},\quad X\mapsto f_A(X)={\rm Tr}(A^T X).\end{align*}Let us show that $f_A$ is differentiable on $E$ and determine $\nabla f_A(X)$. In fact, we know that ${\rm Tr}(\cdot)$ is linear. Then $f_A$ is a linear application, which is continuous because it takes values in $\mathbb{R}$. Thus $f_A$ is differentiable on $E$ and the differential application of $f_A$ is $Df_A(X)=f_A$ for any $X\in E$. We then have\begin{align*}Df_A(X)H=f_A(X)={\rm Tr}(A^T H)=\langle A,H \rangle.\end{align*}Hence, we obtain\begin{align*}\nabla f_A(X)=A,\qquad \forall X\in E.\end{align*}

Differentiability of the space of invertible matrices 

Assume that $m=n$, then we will work with square matrices of order $n$. Let $A\in E:=\mathcal{M}_n(\mathbb{R})$. Denote by  $U$ the set of invertible matrices of $E$. Let $g:U\subset E\to \mathbb{R}$ defined by\begin{align*}g_A(X)={\rm Tr}(X^{-1} A).\end{align*}Now we prove that $g_A$ is differentiable on all $X\in U$ and calculate $\nabla g_A(X)$. We recall that the application $\varphi(X)=X^{-1}$ is differentiable on all $X\in U$. In addition, the differential $D\varphi(X)H=-X^{-1}HX^{-1}$ for all $X\in U$ and $H\in E$. We can write \begin{align*}g_A=\psi\circ \varphi,\end{align*}where $\varphi: U\to E$ and $\psi:E\to \mathbb{R}$ are such that $\varphi(X)=X^{-1}$ and $\psi(Y)={\rm Tr}(YA)$. The application $psi$ is differentiable on each $Y\in E$ with $D\psi(Y)K={\rm Tr}(KA)$ for $K\in E$. On the other hand, the application $\varphi$ is differentiable on each $X\in U$ and\begin{align*}D\varphi(X)H=-X^{-1}HX^{-1}.\end{align*}Thus $g_A=\psi\circ\varphi$ is differentiable on each $X\in U$ and\begin{align*}Dg_A(X)H&=D\psi(\varphi(X))(D\varphi(X)H)\cr &= {\rm Tr}(-X^{-1}HX^{-1} A)\cr &= -{\rm Tr}(-X^{-1}AX^{-1} H).\end{align*}This implies that \begin{align*}\nabla g_A(X)=-(X^{-1}AX^{-1})^T.\end{align*}

Using a canonical inner product 

Let $a\in\mathbb{R}^n,$ $U$ and $E$ as in the previous question.  We define an application $h_a:U\to \mathbb{R}$ by\begin{align*}h_a(X):=\langle X^{-1}a,a\rangle.\end{align*} Here $\langle\cdot,\cdot\rangle$ is the canonical inner product in $\mathbb{R}^n$. Prove that $h_a$ is differentiable on all $X\in U,$ and determine $\nabla h_a(X)$. In fact, we write $h_a=\xi\circ \varphi$ where \begin{align*}xi: E\to \mathbb{R},\quad \xi(Y)=\langle Ya,a\rangle.\end{align*}Remark that $\xi$ is linear and continuous. Then it is differentiable on $E$ and $D\xi(Y)=\xi$. Thus $h_a$ is differentiable in all $X\in U$ and\begin{align*}Dh_a(X)H&=\langle -X^{-1}HX^{-1}a,a\rangle\cr &= -a^T\left(X^{-1}AX^{-1}\right)a\cr &= -{\rm Tr}(X^{-1}aa^T X^{-1}H).\end{align*}Hence\begin{align*}\nabla h_a(X)=-(X^{-1}aa^T X^{-1})^T.\end{align*}

In this article, we learn how to solve the heat equation using the Fourier series. The heat equation belongs to the class of partial differential equations widely used in physics. It describes the diffusion of heat over a region of space. Of course, there are many approaches to studying this equation. Here we mainly use the simplest approach.

The heat equation model

The Fourier series was introduced by the French mathematician and politician Fourier to solve the heat equation. The latter is modeled as follows: let us consider a metal bar. Knowing, at the initial instant, the temperature at each point of the bar and, at all times; the temperature at both ends; can we determine; at any time and at any point; the temperature of the bar?

This problem has been modeled and the temperature $ u $ which is a function of time $ t $ and the point $ x $ is the solution of a partial differential equation called heat equation.

We can assume that the bar is the segment $ [0, L] $. Denote $ Q =(0, L)\times(0, + \infty) $, and $ \overline{Q} = [0, L] \times [0, + \infty) $. The problem to solve is as follows: find a function $ u $ such that \begin{align*}\tag{H1}u\in \mathcal{C}^0(\overline{Q}),\quad u\in \mathcal{C}^2(Q),\end{align*}\begin{align*}\tag{H2}\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}\quad\text{on}\quad Q,\end{align*} \begin{align*}\tag{H3} u(0,t)=u(L,t)=0,\qquad t\in [0,+\infty),\end{align*}\begin{align*}\tag{H4} u(x,0)=h(x),\quad x\in [0,L],\end{align*} where $h$ is a $C^1$ function on the closed interval $[0,L]$ satisfying $h(0)=h(L)=0,$ this is needed to give more regularity to the solution $u$.

The heat equation using the Fourier series approach

We write the solution of (H2) as $$ u (x, t) = f (x) g (t).$$ Then equation $(H2)$ is equivalent to $f(x)g'(t)=f”(x)g(t)$. Let us now find a solution $u$ correponding to the case $f(x)\neq 0$ pour any $x\in ]0,L[$, and $g(t)\neq 0$ for all $t>0$. We deduce that\begin{align*}\frac{f”(x)}{f(x)}=\frac{g'(t)}{g(t)},\quad \forall x\in (0,L),\;t>0.\end{align*}This is true only if the two sides of this equality are constants. This means that, there exists a constant $\lambda\in \mathbb{R}$ such that for any $xin (0,L)$ and $t>0,$ \begin{align*} f”(x)=\lambda f(x)\quad\text{and}\quad g'(t)=\lambda g(t).\end{align*}

In the case of $\lambda>,$ the solution of the first equation is given by \begin{align*} f(x)=a e^{\sqrt{ \lambda }x}+b e^{-\sqrt{ \lambda }x}.\end{align*} The condition $(H3)$ implies that $a+b=0$ and $ a e^{\sqrt{ \lambda }L}+b e^{-\sqrt{ \lambda }L} =0$ . Hence $a=b=0,$ and then $u(x,t)=0,$ this is a contradiction with the condition $(H4)$. On the other hand, if $\lambda=0,$ we have $f”(x)=0$. Then $f(x)=ax+b$. But, due to $(H3)$, we get $u=0,$ and this is a contacdition. The only case that remains is $\lambda <0$. We can wrire $\lambda=-\xi^2<0$. In this case, \begin{align*}f(x)=a\cos \xi x+b\sin \xi x,\quad g(t)=ce^{-\xi^2 t}.\end{align*}

Now using the condition $(H3)$, we deduce that $a=0$ and $\xi=\frac{n\pi}{L}$ for $n\in\mathbb{Z}$. We then have a family of solutions of the form\begin{align*} u_n(x,t)=b_n \sin\left( \frac{n\pi}{L}x\right) e^{- \frac{n^2\pi^2}{L^2}t }. \end{align*}

As the equation $(H2)$ is linear then any sum of $u_n$ is also a solution. We then introduce \begin{align*} u(x,t)=\sum_{n=1}^{+\infty} b_n \sin\left( \frac{n\pi}{L}x\right) e^{- \frac{n^2\pi^2}{L^2}t } .\end{align*}

Remark: Let us mention the isometry of the Fourier transform on the Hilbert space $L^2$ can help in solving the PDE. In fact, using the Fourier transform can help transform the heat equation as PDE into an ordinary differential equation. Then we can use elementary materials to solve the equation. On the other hand, one can use semigroup theory to solve the heat equation.

In this article, we state and prove Gronwall lemma and give some of its applications. In fact, we will use this lemma for the stability of the solution to the differential equations. In particular, the Lyapunov stability of nonlinear systems.

The proof of Gronwall’s lemma

Many proofs of known theorems in mathematics are based on Gronwall’s lemma. In fact, this lemma is used to prove the uniqueness of the solution of differential equations, as well as the stability of equilibrium points.

Lemma: Let $\varphi:[a,b]\to [0,+\infty)$ be a continuous function and assume that there exist two real constants $\alpha,\beta\ge 0$ such that for any $c\in [a,b]$, \begin{align*}\varphi(t)\le \alpha+\beta\left|\int^t_c \varphi(s)ds\right|,\qquad \forall t\in [a,b].\end{align*}Then \begin{align*}\varphi(t)\le \alpha e^{\beta |t-c|},\quad\forall t\in [a,b].\end{align*}

Proof: For simplicity, we suppose that $t\ge c$. We select\begin{align*}F(t)= \alpha e^{\beta |t-c|},\quad\forall t\in [a,b]. \end{align*}Then $F$ is a $C^1$ function and $F'(t)=\beta \varphi(t)\le \beta F(t)$. By multiplying $e^{-\beta t}$ on the both sides of this inequality, we obtain \begin{align*} e^{-\beta t} F'(t)- e^{-\beta t} \beta F(t)\le 0.\end{align*} Hence, \begin{align*} \frac{d}{ds} \left(e^{-s\beta}F(s)\right)\le 0.\end{align*} By integrating between $c$ and $t$, the result then follows.

Another version of the lemma: Let $\varphi,\psi:[a,b]\to [0,+\infty)$ be a continuous function and assume that there exist two real constants $\alpha,\beta\ge 0$ such that for any $c\in [a,b]$, \begin{align*}\varphi(t)\le \alpha+\beta\left|\int^t_c \psi(s)\varphi(s)ds\right|,\qquad \forall t\in [a,b].\end{align*}Then \begin{align*}\varphi(t)\le \alpha \exp\left(\beta\left|\int^t_c \psi(s)ds\right|\right),\quad\forall t\in [a,b].\end{align*}

Applications to the stability of the solution of differential equations

In this paragraph, we give an application of the Gronwall lemma to the stability of nonlinear differential equations. In fact, let $\lambda>0$ be a real number, and consider the following semilinear Cauchy problem \begin{align*}\tag{CP} \begin{cases}\dot{u}(t)=-\lambda u(t)+\arctan(t^2 u(t)^5) \frac{\sin{u(t)}}{1+2 u(t)^2},&t\ge 0,\cr u(0)=\frac{1}{2}.\end{cases}\end{align*} If we select \begin{align*}f(t,x)=\lambda x+\arctan(t^2 x^5)\frac{\sin(x)}{1+2x^2},\quad (t,x)\in \mathbb{R}^+\times \mathbb{R},\end{align*} then our equation has the following standard form \begin{align*}\tag{CP} \begin{cases}\dot{u}(t)=f(t,u(t)),& t\ge 0,\cr u(0)=\frac{1}{2}.\end{cases}\end{align*} Clearly $f$ is a $C^1$ function, then locally Lipschitz function. Thus by the Cauchy-Lipschitz theorem, the Cauchy problem (CP) admits a unique maximal solution $u:[0,T)\to \mathbb{R}$. This solution is global, i.e. defined on all $[0,+\infty)$ because $|f(t,x)|\le \lambda |x|+1$ for any $(t,x)\in\mathbb{R}^+\times \mathbb{R}$.

By computing the derivative $(e^{\lambda t}u(t))’$ and using the expression of $\dot{u}(t)$ in (CP), we can prove that \begin{align*}u(t)=\frac{1}{2}e^{-\lambda t}+\int^t_0 e^{-\lambda (t-s)}\arctan(t^2 u(s)^5) \frac{\sin{u(s)}}{1+2 u(s)^2}.\end{align*} Now using the facts that $|\arctan(y)|\le \frac{\pi}{2},$ $|\sin(y)|\le |y|$ and $\frac{1}{1+y^2}\le 1,$ for any $y\in\mathbb{R}$, we obtain \begin{align*} |u(t)|\le \frac{1}{2}e^{-\lambda t}+\frac{\pi}{2} e^{-\lambda t}\int^t_0 e^{\lambda s}|u(s)|ds,\quad t\ge 0.\end{align*} It follows that \begin{align*} e^{\lambda t}|u(t)|\le \frac{1}{2}+\frac{\pi}{2}\int^t_0 e^{\lambda s}|u(s)|ds,\quad t\ge 0.\end{align*} Now by using the Gronwall lemma we have \begin{align*} e^{\lambda t}|u(t)|\le \frac{1}{2}e^{\frac{\pi}{2}t}.\end{align*} Hence $$ |u(t)|\le \frac{1}{2}e^{-\left(\lambda-\frac{\pi}{2}\right)t},\quad \forall t\ge 0.$$ We deduce that the solution, of the equilibrium point $0$, is exponentially stable if $\lambda>\frac{\pi}{2}$.

The stability analysis of solutions of differential equations is one of the most important axes in ODE. Here w gives a concise course on the stability of equilibrium (critical) points of differential equations.

The stability analysis of solutions of differential equations

The existence of solutions is guaranteed by two fundamental theorems, Peano’s theorem which gives on the existence of solutions, and the Cauchy-Lipschitz theorem which gives the existence and the uniqueness of solutions. We also mention that several types of solutions are introduced for nonlinear Cauchy problems. In fact, the local solution; the maximal solution, and the global solution. For stability, the solution must be global.

Critical points of functions

A critical (or an equilibrium) point of a continuous function $F:\mathbb{R}^d\to \mathbb{R}^d$ is an element $x^\ast\in \mathbb{R}^d$ such that $F(x^\ast)=0$.

Let us consider the differential equation \begin{align*}\tag{Eq}\dot{u}(t)=F(u(t)),\quad t\ge t_0,\quad u(t_0)=x_0.\end{align*} As $F$ is continuous, by Peano’s theorem we know that there exists a maximal solution $u:[t_0,T)\to \mathbb{R}^d$.

If $x^\ast\in\mathbb{R}^d$ is a critical point of $F,$ then the function $r(t)=x^\ast$ for any $t\ge 0$ satisfies $\dot{r}(t)=F(r(t))$. The function $t\mapsto r(t)$ is called the reference or the stationary solution of the differential equation.

Having the stationary solution, one wonders if the other solutions to the differential equations are somehow close to this reference solution. This is the starting point of the theory of stability, which we study below.

Stability analysis of solutions to differential equations

The critical point $x^\ast$ is called exponentially stable if there exist constants $\delta,\omega,M\in (0,\infty),$ with $\delta$ small enough, such that if the initial state $u(t_0)=x_0$ is closed to the critical point $x^\ast$, i.e. $\|x_0-x^\ast\| \le \delta,$ then

• the solution $u(t)$ is well defined for any $t\ge t_0$. i.e. $u$ is a global solution, and
• $\|u(t)-x^\ast\|\le M e^{-\omega t}\|x_0-x^\ast\|$ for any $t\ge 0$.

The linear case: We assume that $F(x)=Ax,$ for $x\in\mathbb{R}^p,$ where $A$ is a square matrix of order $p$. As $F$ is linear, then $F(0)=0,$ so that $0$ is an equilibrium point for $F$. On the other hand, the solution of the differential equation $\dot{u}(t)=A u(t)$ is given by the following exponential of the matrix $A,$ \begin{align*}u(t)=e^{t A}u(0):=\sum_{n=0}^\infty \frac{t^n}{n!} A^n u(0),\qquad \forall t\ge 0.\end{align*}

If $A$ is a diagonal matrix with complex nombers $\lambda_1,\lambda_2,\cdots,\lambda_p$ at the diagonal, then it is clear that the matrix $e^{t A}$ is diagonal with $e^{\lambda_i}$ for $i=1,cdots,p$ at the diagonal. Thus \begin{align*}\|u(t)\|\le e^{(\sup_{1\le i\le p} {\rm Re}\lambda_i)t}.\end{align*}

In this case, the equilibrium point $0$ is exponentially stable if and only if the spectral bound $s(A):=\sup_{1\le i\le p} {\rm Re}\lambda_i < 0$. More generally, we have the following result

Lyapunov stability theorem: linear differential equation

Let $A$ be a square matrix with spectrum $\sigma(A)$, the set of eigenvalues of $A$. Then there exist constants $\omega >0$ a $M>0$ such that \begin{align*}\|e^{tA}\|\le M e^{-\omega t},\qquad \forall t\ge 0,\end{align*} if and only if the spectral bounded satisfies \begin{align*}s(A)=\sup\{ {\rm Re}\lambda: \lambda\in \sigma(A)\}.\end{align*}

Sketch of the proof: If $A$ is diagonalizable, then there exists an invertible matrix $P$ such that $A=P^{-1}D P,$ where $P$ is a diagonal matrix. Remark that for all $n\in \mathbb{N},$ $A^n=P^{-1}D^n P$. This implies that \begin{align*}e^{tA}=P^{-1} e^{tD} P,\quad \forall t\ge 0.\end{align*}Now, assume that $s(A) < 0. $ Using the previous result, for any $\omega\in (0,-s(A))$, we have \begin{align*}\left\| e^{t A}\right\|\le M e^{-\omega t},\quad\forall t\ge 0,\end{align*} where $M=\|P^{-1}\|\|P\|$. Conversely, by contraposition assume that $s(A)\ge 0,$ then there exists $\lambda\in\sigma(A)$ such that ${\rm Re}\lambda\ge 0$. Let $x\in \mathbb{R}^p$ the eigenvector associated with $\lambda,$ so that $Ax=\lambda x$. Then $A^n=\lambda^n x$. This implies that $e^{t A}x=e^{\lambda t}x$. Thus $\|e^{t A}x\|=e^{{\rm \lambda} t}\|x\|$ dont converge to zero as $t\to +\infty$. This ends the proof.

If $A$ is not diagonalizable, i.e. when $\sigma(A)=\{\lambda_1,\cdots,\lambda_r\}$ with $r < p$ and each $\lambda_i$ has a multiplicity $n_i$ such that $n_1+\cdots+n_r=p$. In this case, there exists a square invertible matrix $S$ such $A=S^{-1}JS$, where $J$ is a matrix of Jordan, $J$ is diagonalizable by block $J=diag(J_{\lambda_1},\cdots,J_{\lambda_r})$ with each block $J_{\lambda_i}$ is an upper triangular matrix with $\lambda_i$ repeated in the diagonal. We also have \begin{align*}e^{tA}=S^{-1} diag\left(e^{t J_{\lambda_1}},\cdots,e^{t J_{\lambda_r}}\right)S.\end{align*}

We propose a nice proof of Peano existence theorem. This theorem shows that the continuity of the vector field suffices for the existence of solutions to the ODE; ordinary nonlinear differential equations. We notice that this theorem does not guarantee the uniqueness of the solution.

Local solutions

Let $I$ be an interval of $\mathbb{R}$ and $\Omega$ an open set of $\mathbb{R}^d$ with $d\in\mathbb{N}^\ast$. Let $f:I\times \Omega\to \mathbb{R}^d$ be a continuous function. Let $(t_0,x_0)\in I\times\Omega$ and consider the Cauchy problem \begin{align*}(CP)\quad\begin{cases} \dot{u}(t)=f(t,u(t)),& t\in I,\cr u(t_0)=x_0.\end{cases}\end{align*}

A local solution to the Cauchy problem $(CP)$ is a couple $(J,\varphi)$, where $J$ is a subinterval of $I$ containing $t_0$, $\varphi:J\to \Omega$ is a function of class $C^1$ with $u(t_0)=x_0$ and satisfying the differential equation in $(CP)$.

The proof of Peano existence theorem

Peano’s theorem says that if $f$ is continuous then the Cauchy problem $(CP)$ admits a least a local solution $(J,u)$.

The idea of the proof of this theorem: Let $r>0$ such that the closed ball $\overline{B}(x_0,r)\subset \Omega$. By reducing $I$ to a compact, we can assume that $f$ is bounded by a constant $M>0$ on $I\times \overline{B}(x_0,r) $. We set \begin{align*}J:=I\cap \left[t_0-\frac{r}{M}, t_0+\frac{r}{M}\right].\end{align*}

Let $(f_k)_k$ be a sequence of functions of class $C^1$ from $\mathbb{R}\times \mathbb{R}^d$ to $\mathbb{R}^d$ be converging to $f$ on $J\times \overline{B}(x_0,r)$, bounded by $M$. Each $f_k$ is locally Lipschitz, we can then use a version of Cauchy-Lipschitz theorem which gives the existence of a solution $u_k:J\to \overline{B}(x_0,r)$ such that\begin{align*} \forall t\in J,\quad \dot{u}_k(t)=f_k(t,u_k(t)),\quad u_k(t_0)=x_0.\end{align*}

By taking the integral between $t_0$ and $t$, we obtain \begin{align*}\forall t\in J,\quad u_k(t)=x_0+\int^t_{t_0} f_k(s,u_k(s))ds.\end{align*}On the other hand, a simple argument shows that the sequence of function $(u_k)_k$ is equicontinuous and equi-bounded. Then by using Ascoli theorem, there exists a subsequence $(u_{n_k})_k$ of $(u_k)_k$ that converge on $J$ to an application $u$ continue on $J$ into $ \overline{B}(x_0,r) $. This is the local solution to the Cauchy problem $(CP)$.

The uniqueness of the solution is not assured by Peano’s theorem

The continuity of $f$ in the proof of the Peano theorem is not sufficient for the uniqueness of the solution.

The following example strengthens this remark: the Cauchy problem \begin{align*} x'(t)=\sqrt{x(t)},\quad x(0)=0.\end{align*}The problem admits two solution; the null solution and the following solution \begin{align*}t\mapsto \begin{cases}\frac{t^2}{4},& t\ge 0,\cr -\frac{t^2}{4},& t\le 0.\end{cases}\end{align*}

In this post, we study Fourier transform properties and give some applications. In fact, this transformation helps in converting partial differential equations to ODE. A simple method to solve the heat equation is the Fourier transform technique.

The Riemann-Lebesgue Lemma

We denote by $L^1(\mathbb{R})$ the Lebesgue space of measurable functions $f:\mathbb{R}\to \mathbb{R} $ such that\begin{align*} \|f\|_1^2:=\int_{\mathbb{R}} |f(x)|dx<\infty.\end{align*}

Also, we recall that $\|\cdot\|_1$ is a norm on $ L^1(\mathbb{R})$ and that $( L^1(\mathbb{R}), \|\cdot\|_1 ) $ is a Banach space, no reflexive. On the other hand, the space of continuously differentiable functions with compact support $\mathcal{C}_c^1(\mathbb{R})$ is dense in the space $ L^1(\mathbb{R}) $.

As for any $t,x\in\mathbb{R}$, we have $|e^{-itx}f(x)|=|f(x)|$, then for $f\in L^1(\mathbb{R})$, the function $(x\mapsto e^{-itx}f(x) )\in L^1(\mathbb{R}) $. We then have the following definition:

Definition: For $f\in L^1(\mathbb{R}) $ we define the Fourier transform of $f$ by \begin{align*}\hat{f}(t)=\int_{\mathbb{R}} e^{-itx}f(x) dx,\qquad t\in \mathbb{R}.\end{align*}

Clearly $f\mapsto\hat{f}$ is a linear application on $ L^1(\mathbb{R})$. We have the following Riemann-Lebesgue result.

Theorem: The Fourier transform is an application from $ L^1(\mathbb{R})$ to $\mathcal{C}_0(\mathbb{R})$, where \begin{align*} \mathcal{C}_0(\mathbb{R}) :=\{f\in \mathcal{C}(\mathbb{R}) : \lim_{|t|\to +\infty}f(t)=0\}.\end{align*}

Proof: Here we use the density of $\mathcal{C}_c^1(\mathbb{R})$ is dense in the space $ L^1(\mathbb{R}) $. Hence, in the first step, we will prove the result for $f\in \mathcal{C}_c^1(\mathbb{R}) $. In this case, and by using integration by part, we get for any $t\neq 0,$ \begin{align*}\hat{f}(t)=\frac{1}{it}\int_{\mathbb{R}}e^{-itx}f'(x)dx.\end{align*}This implies that\begin{align*} | \hat{f}(t) |\le \frac{1}{|t|}|f’|_1.\end{align*}

If $f\in L^1(\mathbb{R}) ,$ by density there exists a sequence $(f_n)_n\subset \mathcal{C}_c^1(\mathbb{R}) $ that converges to $f$ in $ L^1(\mathbb{R}) $. Thus, for any $\varepsilon>0,$ there exists $m\in\mathbb{N}$ such that $\|f_m-f\|_1\le \frac{\varepsilon}{2}$. Moreover, remark that\begin{align*} \hat{f}(t) = \hat{f_m}(t) + \hat{f}(t) – \hat{f_m}(t),\end{align*} so that \begin{align*} | \hat{f}(t) |&\le \frac{1}{|t|}\|f_m’\|_1+\|f-f_m\|_1\cr & \le \frac{1}{|t|}\|f_m’\|_1 + \frac{\varepsilon}{2}. \end{align*} This ends the proof.

Fourier Transform in the Schwartz space

We define the Schwartz space by\begin{align*}\mathcal{S}(\mathbb{R})=\{f\in \mathcal{C}^\infty(\mathbb{R}):\forall p,q\in\mathbb{N}&,\;\exists C_{p,q}>0:\cr & |x^pf^{(q)}(x)|\le C_{pq},\; \forall x\in\mathbb{R}\}.\end{align*} The first remark is that $ \mathcal{S}(\mathbb{R}) \subset L^1(\mathbb{R})$ continuously and densely. Hence we can prove all properties of the Fourier transform in $\mathcal{S}(\mathbb{R})$. These properties are translated to $L^1(\mathbb{R})$ by density.

As example we have $(x\mapsto e^{-x^2})\in \mathcal{S}(\mathbb{R}) $. On the other hand, we can easily prove that if $f\in \mathcal{S}(\mathbb{R}) ,$ then also $f’$ and $x\mapsto xf(x)$ are in $ \mathcal{S}(\mathbb{R}) $.

Theorem: If $f\in \mathcal{S}(\mathbb{R}) $ then the Fourier transform $\hat{f}\in \mathcal{S}(\mathbb{R}) $.