We offer exercises on differential calculus with detailed answers. In fact, our objectives are as follows: knowing how to calculate the derivative of a function according to a vector; calculating the differential of a function at a point; and showing that a function is continuously differentiable.

We recall that differential calculus is very important in differential equations theory.

Exercises on differential calculus

Exercise: Let $f:\mathbb{R}^2\to \mathbb{R}$ be the function defined by \begin{align*} f(x,y)=\begin{cases} \frac{x^2}{y},& y\neq 0,\cr 0,& y=0.\end{cases} \end{align*} Prove that $f$ admits directional derivatives at $(0,0)$ along all vectors $h\in\mathbb{R}^2\setminus\{(0,0)\}$ that we calculate.

Solution: Let $h=(h_1,h_2)\in\mathbb{R}^2\setminus\{(0,0)\}$. We consider the partial function \begin{align*} \varphi(t)&=f(th)=f(th_1,th_2)\cr &=\begin{cases} t f(h),& t\neq 0,\cr 0,&t=0.\end{cases} \end{align*} We remark that \begin{align*} \varphi(t)=t f(h),\qquad \forall t\in \mathbb{R}. \end{align*} This function is differential in $0$ and $\dot{\varphi}(t)=f(h)$. Hence the directional derivative at $(0,0)$ along $h$ is \begin{align*} D_hf(0,0)=f(h). \end{align*}

Exercise: Prove that the function $\mathbb{R}^2\to \mathbb{R}$ defined by \begin{align*} f(x,y)= \begin{cases} \frac{x^4+x^4}{x^2+y^2},& (x,y)\neq (0,0),\cr 0,& \text{if not.} \end{cases} \end{align*} is of class $\mathcal{C}^1$ on $\mathbb{R}^2$.

Solution: As we have a rational fraction involving $x^2+y^2$ we shall use polar coordinates to prove continuity at $(0,0)$.

Let us first prove that the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exist and equal to $0$ at $(0,0)$. In fact, the functions $\varphi:t\mapsto f\left((0,0)+t(1,0)\right)=t^2$ and $\psi:t\mapsto f\left((0,0)+t(0,1)\right)=t^2$ are differentiable at $0$ and $\varphi'(0)=\psi'(0)=0$. This proves the first claim.

Second, we prove continuity of partial derivatives of $f$ at $(0,0)$. Let $(x,y)\in\mathbb{R}^2\backslash\{(0,0)\}$. Then \begin{align*} \frac{\partial f}{\partial x}(x,y)=\frac{2x^5+4x^3y^2-2xy^4}{(x^2+y^2)^2}. \end{align*} For reals $r>0$ and $\theta\in\mathbb{R},$ we have \begin{align*} &\left|\frac{\partial f}{\partial x}(r\cos(\theta),r\sin(x))\right|\cr & = \left|\frac{r^5(2\cos^5(\theta)+4\cos^3(\theta)\sin^2(\theta)-2\cos(\theta)\sin^4(\theta))}{r^4}\right|\cr & \le 8r, \end{align*} which means that \begin{align*} \left|\frac{\partial f}{\partial x}(x,y)\right|\le 8\sqrt{x^2+y^2}. \end{align*} This implies that \begin{align*} \lim_{(x,y)\to (0,0)} \frac{\partial f}{\partial x}(x,y)=0. \end{align*} Hence $\frac{\partial f}{\partial x}$ is continuous at $(0,0)$. Remark that $x$ et $y$ play the same role, then we have also \begin{align*} \left|\frac{\partial f}{\partial y}(x,y)\right|\le 8\sqrt{x^2+y^2}. \end{align*} This implies that \begin{align*} \lim_{(x,y)\to (0,0)} \frac{\partial f}{\partial y}(x,y)=0. \end{align*} Hence $\frac{\partial f}{\partial y}$ is continuous at $(0,0)$. All these imply that $f$ is a class $\mathcal{C}^1$ on $\mathbb{R}^2$.

One of the classical examples in the exercises on differential calculus is the following:

Exercise: Let $E$ be an euclidian space endowed with the norm $\|x\|:=\sqrt{\langle x,x\rangle}$. Study the differentiability of the function $f(x)=\|x\|$.

Solution: We know that the function square root is differentiable at $0$, then we shall first prove that $f$ is not differentiable at $0_E$. In fact the function $t\mapsto f(th)=\|th\|=|t|\|h\|$ is not differentiable at $0$. This implies that $f$ does not admit a directional derivative along $h\in E\backslash\{0\}$ in $0$.

Assume that $x\in E\backslash\{0\}$. By using the fact that $\sqrt{1+u}=1+\frac{1}{2}u+o(u),$ as $u\to 0$, we obtain \begin{align*} \|x+h\|&=\sqrt{\langle x+h,x+h\rangle}\cr &=\sqrt{ \|x\|^2+2\langle x,h\rangle+\|h\|^2}\cr &= \|x\|\sqrt{1+\frac{2\langle x,h\rangle}{\|x\|^2}+\underset{\|h\|\to 0}{o}(\|h\|)}\cr &= \|x\|\left( 1+\frac{\langle x,h\rangle}{\|x\|^2}+\underset{\|h\|\to 0}{o}(\|h\|) \right)\cr &= \|x\|+\frac{\langle x,h\rangle}{\|x\|}+\underset{\|h\|\to 0}{o}(\|h\|) \end{align*} We define the application \begin{align*} Df(x):E\to \mathbb{R},\quad h\mapsto Df(x)h=\frac{\langle x,h\rangle}{\|x\|}. \end{align*} This application is linear and continuous since \begin{align*} |Df(x)h|\le \frac{|\langle x,h\rangle|}{\|x\|}\le \|h\|. \end{align*} We also have \begin{align*} f(x+h)=f(x)+Df(x)h+\underset{\|h\|\to 0}{o}(\|h\|). \end{align*} Thus $f$ is differentiable on $E\backslash\{0\}$ with differential application \begin{align*}Df(x):E\to \mathbb{R},\quad h\mapsto Df(x)h=\frac{\langle x,h\rangle}{\|x\|},\qquad x\neq 0. \end{align*}

You can also see details on the derivative functions.

The maximal solution to Cauchy’s problems is a solution that gives full information on the physical model. In fact, we obtain this solution by extending local solutions. In this post, we give some theorem that shows the existence of the uniqueness of the maximal solution.

What is the maximal solution to Cauchy’s problems?

Let $f: I\times \Omega\to \mathbb{R}^b$ be a continuous function, where $I$ is an interval of $\mathbb{R}$ and $Omega$ an open set of $\mathbb{R}^d$. Moreover, 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 maximal solution of the Cauchy problem (CP) that can not admit an extension to another solution.

Peano’s theorem gives the existence of the maximal solution under the continuity assumption of the vector field $f$ on $I\times \Omega$. On the other hand, if in addition, we assume that $f$ is locally Lipschitz with respect to its second variable, then the Cauchy problem admits a unique maximal solution. This result is called the Cauchy-Lipschitz theorem.

We recall that a function $f$ is locally Lipschitz with respect to its second variable, if for any $(t_0,x_0)\in I\times \Omega$ there exists a neighborhood $V_{t_0,x_0}$ of $(t_0,x_0) $ and a constant $C>0$ such that for any $(t,x)$ and $(s,y)$ in $V_{t_0,x_0}$ we have \begin{align*}\|f(t,x)-f(t,y)\|\le C \|x-y\|.\end{align*}

We mention that the maximal solution is necessarily defined on an open interval of the form $(\alpha,\beta)$.

The global solutions to Cauchy problems

A solution $u: J\to \Omega$ of the Cauchy problem (CP) is called global if $J=I$. It is then important to look for conditions that guarantee the existence of global solutions. In fact, there exists a nice theorem, called the explosion theorem, that gives these conditions.

Assume that $I=(a,b)$ with $a$ can possibly be $-\infty$, and $b$ can also be $+\infty$. In addition, we suppose that $\Omega=\mathbb{R}^d$. Let $u:(\alpha,\beta)\to\mathbb{R}^d$ be a maximal solution. Then we have the following cases:

• $\beta=b$, if not then $|u(t)|\to+\infty$ as $t\to \beta$. This means that if the solution is not left global then it is unbounded in a neighborhood of $\beta$.
• $\alpha=a$, if not $|u(t)|\to+\infty$ as $t\to \alpha$. This means that if the solution is not right global, then it is unbounded in a neighborhood of $\alpha$.

The delay equations are an important class of differential equations. In fact, these equations are studied by both mathematicians and engineers because most systems are affected by a delay.

Imagine a basketball game transmission from Italy to the United States. Sometimes a lag between the voice and the television images occurs. Indeed, the voice arrives before the image, so there is a delay for the image to coincide with the voice.

In the following, we will see that the translation operator plays a key role in the delay equation existence of solutions.

Functions of bounded variations

Let $X$ be a Banach space, and denote by $\mathcal{L}(X)$ the Banach algebra of all linear bounded operators from $X$ to $X,$ endowed with uniform topology.

A function $\mu:[-r,0]\to \mathcal{L}(X)$ is called a bounded variation functions if \begin{align*}|\mu|([-r,0])=\sup\left\{\sum_{i=1}^N |\mu(\tau_i)-\mu(\tau_{i-1})|, 0=\tau_0>\tau_1\cdots>\tau_N=-r\right\}\end{align*} is finite. We not $|\mu|$ is a Borel measure on $[-r,0]$.

For a continuous function $f:[-r,0]\to X,$ we define the following Riemann-Stieltjes integral \begin{align*}L f=\int^0_{-r}d\mu(\theta)f(\theta).\end{align*} We note that \begin{align*}\|Lf\|&\le \int^0_{-r}|f(\theta)|d|\mu|( \theta )\cr & \le \gamma \|f\|_\infty, \end{align*} where $\gamma:= |\mu|([-r,0]) $ and $\|f\|_\infty:=\sup_{\theta\in [-r,0]}|f(\theta)|$.

The heat delay equations

Here we consider a classical example of delay equations. We select $X=L^2([0,L])$ and define the operator \begin{align*} Au=u”,\quad D(A)=\{u\in W^{2,2}([0,L]):u(0)=u(L)=0\}.\end{align*} We recall from semigroup theory, Lumer-Phillips theorem, that $(A,D(A))$ generates a strongly continuous semigroup $(T(t))_{t\ge 0}$ on $X$. That is \begin{align*} & T(t)\in\mathcal{L}(X),\quad T(0)=Id,\cr & T(t+s)=T(t)T(s),\qquad \forall t,s\ge 0,\cr & \lim_{t\to 0}\|T(t)f-f\|=0,\qquad \forall f\in X.\end{align*} Moreover, $f\in D(A)$ if and only if the following limit \begin{align*}Af=\lim_{t\to 0} \frac{T(t)f-f}{t}\end{align*}exists.

Consider the delayed heat equation \begin{align*}\tag{Eq} \begin{cases}\dot{u}(t)=Au(t)+\displaystyle\int^0_{-r}d\mu(\theta)u(t+\theta),& t\ge 0,\cr u(0)=h,\cr u(t)=f(t),& -r\le t\le 0.\end{cases}\end{align*} Here $h\in X$ and $f\in L^2([-r,0],X)$.

We show you how to calculate integrals using elementary methods. In addition, we teach you to study the properties of functions defined by an integral.

It is very important to be able to calculate an integral easily because it interferes in the study of differential equations another important subject of mathematical analysis.

Exercises on how to calculate integrals

We propose several exercises with detailed solutions to teach you how to calculate integrals.

Exercise: Determine the value of the following integrals \begin{align*} I=\int^{\frac{\pi}{2}}_0 e^{2x}\cos(x)dx,\qquad J=\int^{\frac{\pi}{2}}_0 \frac{\sin(\theta)}{2+\cos(\theta)}d\theta. \end{align*}

Solution: To compute $I$ we shall use the integration by parts method. In fact, we can write \begin{align*} I&= \int^{\frac{\pi}{2}}_0 \left(\frac{e^{2x}}{2}\right)’\cos(x)dx\cr &= \left[ \frac{e^{2x}}{2} \cos(x)\right]^{\frac{\pi}{2}}_0- \int^{\frac{\pi}{2}}_0 \frac{e^{2x}}{2}\cos'(x)dx\cr & = \frac{1}{2}+ \frac{1}{2} \int^{\frac{\pi}{2}}_0 e^{2x} \sin(x)dx\cr & = \frac{1}{2}+ \frac{1}{2}\left( \left[ \frac{e^{2x}}{2} \sin(x)\right]^{\frac{\pi}{2}}_0-\int^{\frac{\pi}{2}}_0 \frac{e^{2x}}{2} \sin'(x)dx\right)\cr &= \frac{1}{2}+\frac{e^{\pi}}{4}- \frac{1}{4} \int^{\frac{\pi}{2}}_0 e^{2x}\cos(x)dx\cr &= \frac{1}{2}+\frac{e^{\pi}}{4}- \frac{1}{4} I. \end{align*} We deduce that \begin{align*} I+ \frac{1}{4} I= \frac{1}{2}+\frac{e^{\pi}}{4}. \end{align*} Finally, \begin{align*} I=\frac{2+e^\pi}{5}. \end{align*}

To calculate $J$ we will use the integration by parts technique. We put $t=\cos(\theta)$. We then have $dt=-\sin(\theta)d\theta$. Then \begin{align*} J&=\int^{\cos(\frac{pi}{2})}_{\cos(0)} \frac{-dt}{2+t}\cr &= -\int^0_1 \frac{dt}{2+t}\cr &= \int^1_0 \frac{dt}{2+t}\cr &= \left[\ln(2+t)\right]^1_0\cr & = \ln(3)-\ln(2)=\ln\left(\frac{3}{2}\right). \end{align*}

Exercise: Let consider the function \begin{align*} g(x)=\int^x_{\frac{1}{x}} \frac{\ln(t)}{t}dt. \end{align*}

• Determine the domain of definition $D_g$ of $g$.
• Prove that $g$ is differentiable on $D_g$ and compute $g'(x)$ for any $x\in D_g$.
• Deduce $g$ is the null function.

Solution: 1) We define the function \begin{align*} f(t)=\frac{\ln(t)}{t}. \end{align*} clearly, the function $f$ is only defined in $(0,+\infty)$. From the expression of $g$ we then conclude that $g(x)$ is well defined if and only if $x\in (0,+\infty)$. Hence the domain of definition of $g$ is $D_g=(0,+\infty)$.

2) Denote by $F$ the primitive of $f$, $F$ exists because the function $f$ is continuous on $(0,+\infty)$. We select \begin{align*} F(x)=\int^x_c f(t)dt \end{align*} for any constant $c>0$. The function $F$ is differentiable on $(0,+\infty)$ and $F'(x)=f(x)$ for all $x\in (0,+\infty)$. On the other hand, we can write \begin{align*} g(x)&=\int^x_c f(t)dt+\int^c_{\frac{1}{x}}f(t)dt\cr &= F(x)-F\left(\frac{1}{x}\right) \end{align*} for all $x\in (0,+\infty)$. Hence $g$ is differentiable on $(0,+\infty)$ as composition and sum of differentiable functions. Moreover, for all $x>0,$ \begin{align*} g'(x)&=F'(x)-\left(F\left(\frac{1}{x}\right)\right)’\cr &= f(x)- \left(\frac{1}{x}\right)’ F’\left(\frac{1}{x}\right)\cr &= \frac{\ln(x)}{x}+\frac{1}{x^2} f\left(\frac{1}{x}\right)\cr &= \frac{\ln(x)}{x}-\frac{\ln(x)}{x}=0. \end{align*}

3) An the derivative of $g$ on $(0,+\infty)$ is zero, then $g$ is the constant function on $(0,+\infty)$. But $g(1)=0$. Then $g(x)=0$ for any $x\in (0,+\infty)$.

We discuss some facts about first-order differential equations for beginners. Such equations are important because many problems in our real life can be modeled as a differential equation.

We assume that the reader is familiar with the concept of continuous function primitives.

First-order differential equations with constant coefficients

In algebra, we already studied algebraic equations where a variable is a number. Here we study equations where the variable is a function.

Let $a$ be a real number and $f:\mathbb{R}\to \mathbb{R}$ be a continuous function. We look for differential functions $u:\mathbb{R}\to \mathbb{R}$ such that\begin{align*}\dot{u}(x)=a u(x)+f(x).\end{align*} Here we denote $\dot{u}(x)=\frac{d}{dx}u(x)$, the derivative of the function $u$.

We recall that $\frac{d}{dx}e^{-a x}=ae^{-ax}.$ By multiplying the both sides of the above differential equation by $e^{-ax}$, we obtain $ e^{-ax} \dot{u}(x)-a e^{-ax} u(x)= e^{-ax} f(x) $. Also, we write\begin{align*}\frac{d}{dx}\left( e^{-ax} u(x) \right)= e^{-ax} f(x) .\end{align*}By taking the integral between $0$ and $x$ in the both sides of this equation, we get \begin{align*} e^{-ax} u(x) =u(0)+\int^x_0 e^{-as}f(s)ds.\end{align*} Now by multiplying the both sides of the avove equality br $e^{ax},$ we obtain \begin{align*} u(x)=e^{ax}u(0)+\int^x_0 e^{a(x-s)}f(s)ds.\end{align*}

Cauchy problem: A first-order differential is a Cauchy problem if it take the following form \begin{align*} \begin{cases} \dot{u}(t)=a u(t),& t\in\mathbb{R},\cr u(t_0)=x.\end{cases}\end{align*} Here $t_0$ is the initial time and $x$ is the initial state. The solution to this Cauchy problem is $u(t)=e^{t a}x$.

Equations with variable coefficients

In most cases the coefficients of differential equations are functions. These equations take the following form \begin{align*}a(x)\dot{u}(x)+b(x)u(x)=0,\end{align*}where $a(\cdot)$ and $b(\cdot)$ are continuous function such that $a(x)\neq 0$. This equation can be rewritten as\begin{align*} \frac{\dot{u}(x)}{u(x)}=-\frac{b(x)}{a(x)}.\end{align*} On the other hand, we recall that \begin{align*} \frac{d}{dx}\ln(|u(x)|)= \frac{\dot{u}(x)}{u(x)} .\end{align*} We then obtain \begin{align*} \ln(|u(x)|) =\int -\frac{b(x)}{a(x)} dx.\end{align*} Hence, we have\begin{align*} u(x)=e^{- \displaystyle \int \frac{b(x)}{a(x)} dx}.\end{align*}

Let consider the first example $(x-1)u’-2u=0$, here we have $a(x)=x-1$ and $b(x)=-2$. Then the solution is given by \begin{align*} u(x)= e^{\displaystyle \int \frac{2}{x-1} dx} =e^{ (2\ln(|x-1|+C)}= A e^{\ln((x-1)^2)}.\end{align*} Thus the solution of the differential equation is $u(x)=A(x-1)^2$, where $A$ is a real constante.

In this article, we state and prove the Cauchy-Lipschitz theorem for the existence and uniqueness of solutions to nonlinear ordinary differential equations. The key proof of this theorem is the Banach-Picard fixed point theorem. We give some applications of this theorem.

Local and maximal solutions to nonlinear Cauchy problems

Throughout this section, $I$ is an interval of $\mathbb{R}$, and $\Omega$ is an open set of $\mathbb{R}^d$. In addition, let $(t_0,x_0)\in \Omega$ et $f:I\times \Omega\to\mathbb{R}^n$ be a continuous function.

We look for additional conditions on $f$ so as the following Cauchy problem \begin{align*}\tag{Eq} u(t_0)=x_0,\quad \dot{u}(t)=f(t,u(t)),\quad t\in I,\end{align*}admits a “kind” of solutions.

By a solution to $({\rm Eq})$ we mean a couple $(J,u),$ where $J\subset I$ is an interval such that $t_0\in J,$ and $u:J\to \Omega$ is a $C^1$ function that satisfies $({\rm Eq})$.

On the other hand, we define an order in the set of all solutions to the Cauchy problem $({\rm Eq})$. In fact, we say that a solution $(J_2,u_2)$ extends another solution $(J_1,u_1),$ of (Eq) if $J_1\subset J_2$ and $u_2(t)=u_1(t)$ for any $t\in J_1$.

A maximum solution of $({\rm Eq})$ is a solution that does not admit an extension to another solution.

Remark: Niote that $(J,u)$ is a solution of $({\rm Eq})$ if and only if it satisfies the following integral equation \begin{align*}\tag{IE} u(t)=x_0+\int^t_{t_0}f(s,u(s))ds,\quad\forall t\in J.\end{align*}

A particular version of the Cauchy Lipschitz theorem

In this section, let $\alpha>0,\;r>0$ and $(t_0,x_0)\in I\times \Omega$ such that \begin{align*} \tag{H1}Q:=[t_0-\alpha,t_0+\alpha]\times \overline{B}(x_0,r)\subset I\times \Omega\end{align*}\begin{align*} \tag{H2} f:Q\to \mathbb{R}^d\; \text{is continuous, and }\;M=\sup_Q\|f\|<\infty.\end{align*}\begin{align*} \tag{H3} \exists C>0, \forall (t,x),(t,y)\in Q, \quad \|f(t,x)-f(t,y)\|\le C|x-y|.\end{align*}

Theorem: Under the condition $(H1)$ to $(H3)$, the Cauchy problem $({\rm Eq})$ admits a unique solution $(J,u)$ such that \begin{align*} J=[t_0-T,t_0+T]\quad\text{with}\quad T:=\min\left\{alpha,\frac{r}{M}\right\}.\end{align*}\begin{align*} (s,u(s))\in Q,\quad \forall s\in J.\end{align*}

Proof: We shall use Banach-Picard’s fixed point theorem. The latter said that if $E$ is a Banach space and $\Phi: E\to E$ is a contraction, that is there exists $\gamma\in (0,1)$ such that $\|\Phi(x)-\Phi(y)\|\le \gamma \|x-y\|$ for any $x,y\in E;$ then there exists a unique $u\in E$ such that $\Phi(u)=u$. In this case, we also have $\Phi^n(u)=u$ for any $n\in \mathbb{N}$, where $\Phi^n=\Phi\circ\Phi\circ\cdots\Phi$. Conversely, il there exists $m\in \mathbb{N}$ and a unique $u\in E$ such that $\Phi^m(u)=u$, then $u$ is a fixed point for $\Phi,$ that is $\Phi(u)=u$.

Now come back to the proof of the theorem. For $u\in E:=\mathcal{C}(J, \overline{B}(x_0,r) )$, we define for any $t\in J,$ \begin{align*}\left(\Phi(u)\right)(t)=x_0+\int^t_{t_0}f(s,u(s))ds.\end{align*}Then $\Phi:E\to E$. By recurrence we show that for any $t\in J$ and $n\in \mathbb{N}$ we have \begin{align*} \|\Phi^n(v)(t)- \Phi^n(w)(t) \|\le \frac{C^n}{n!}|t-t_0|^n \|u-v\|_\infty.\end{align*} For a large $n,$ we have \begin{align*}\gamma:= \frac{C^n}{n!} \left(\frac{r}{M}\right)^n < 1.\end{align*} Then $\Phi^n$ is a contraction. Thus there exists a unique $u\in E= \mathcal{C}(J, \overline{B}(x_0,r) )$ such that $\Phi(u)=u$. This means that $u:J\to \overline{B}(x_0,r) $ such that \begin{align*} u(t)=x_0+\int^t_{t_0}f(s,u(s))ds,\qquad \forall t\in J.\end{align*}This ends the proof.

You may also consult the concept of maximal solutions in detail.