Master of Science in Mathematics

Lecture Notes. Accessed on: 2019-10-21 18:10:46

## Introduction to Vector Fields

**Definition**. A vector field in $\mathbb{R}^n$ is a function $\mathbf{V}$ that assigns a vector from each point in its domain. A vector field with domain $D$ in $\mathbb{R}^n$ has the form \begin{equation} {V}(x_1, \ldots, x_n) = \langle u_1 (x_1, \ldots,x_n), \ldots, u_n (x_1, \ldots, x_n) \rangle \end{equation} where the scalar functions $u_1,$,\ldots,$u_n$ are called the components of $\mathbf{V}.$

For example a vector field in $\mathbb{R}^2$ has the form \begin{equation} {V}(x,y)=u(x,y) {i}+v(x,y) {j}=\langle u,v \rangle \end{equation}

and in $\mathbb{R}^3$ has the form \begin{equation}{V}(x,y,z) = u(x,y,z) {i}+v(x,y,z) {j}+w(x,y,z) {k}=\langle u,v,w\rangle. \end{equation}

Common examples of vector fields include force fields, velocity fields, gravitational fields, magnetic fields, and electric fields. Vector fields can be used to quantify the amount of work done by a variable force acting on a moving body. Measuring the amount of force (fluid flow, electric charge, etc.) can sometimes be achieved by computing an integral of a vector field with respect to an orientable curve or surface.

## Gradient Fields

**Definition**. Let $f$ be a differentiable function. The vector field obtained by applying the del operator to $f$ is called the gradient field of $f.$

**Example**. Find the gradient field of the function $f(x,y)=\sin x+e^{x y}.$

**Solution**. Since $$ \nabla f(x,y)=\langle \cos x+y e^{x y},x e^{x y}\rangle $$ the gradient field of $f$ is \begin{equation} {V}(x,y)= \left(\cos x+y e^{x y} \right) {i}+ x e^{x y} {j}. \end{equation}

## Conservative Vector Fields

**Definition**. A vector field ${V}$ is said to be conservative in a region $D$ if ${V}=\nabla f$ for some scalar function $f$ in $D.$ The function $f$ is called a scalar potential of ${V}$ in $D.$

**Example**. Determine whether the vector field is conservative and if so, find a scalar potential function \begin{equation} {V}(x,y,z)=y^2{i}+\left(2x y+e^{3z}\right){j}+\left(3y e^{3z}\right){k} \end{equation}

**Solution**. If there is such a function $f$ then $f_x(x,y,z)=y^2$, $f_y(x,y,z)=2 x y+e^{3z}$, and $f_z(x,y,z)=3y e^{3z}.$ Integrating $f_x$ with respect to $x,$ $f(x,y,z)=x y^2+g(y,z).$ Then differentiating $f$ with respect to $y,$ we have $f_y(x,y,z)=2x y+g_y(y,z)$ and this yields $g_y(y,z)=e^{3z}.$ Thus $g(y,z)=y e^{3z}+h(z)$ and we have $$ f(x,y,z)=x y^2+y e^{3z}+h(z). $$ Finally, differentiating $f$ with respect to $z$ and comparing, we obtain $h'(z)=0$ and therefore, $h(z)=K,$ a constant. The desired scalar function is $$

f(x,y,z)=x y^2+y e^{3z}+K $$ with ${V}=\nabla f.$

**Definition**. A region $D$ in the plane is called connected (one piece) if it has the property: (i) any two points in the region can be connected by a piecewise smooth curve lying entirely within $D;$ and a simply connected region (no holes) is a connected region $D$ that has the property: (ii) every closed curve in $D$ encloses only points that are in $D.$

**Theorem**. (** Conservative in Space**) Suppose that the vector field ${V}$ and $\mathop{curl}{V}$ are both continuous in the simply connected region $D$ of $\mathbb{R}^3.$ Then ${V}$ is conservative in $D$ if and only if $\mathop{curl}{V}={0}.$

**Theorem**. (** Conservative in the Plane**) Consider the vector field $$ {V}(x,y)=u(x,y){i}+v(x,y){j} $$ where $u$ and $v$ have continuous first partials in the open simply connected region $D$ in the plane. Then ${V}(x,y)$ is conservative in $D$ if and only if \begin{equation} \frac{\partial u}{\partial y}=\frac{\partial v}{\partial x} \end{equation} on $D.$

**Proof**. Note that a vector field ${V}=u(x,y){i}+v(x,y){j}$ in $\mathbb{R}^2$ can be regarded as the vector field ${U} = u(x,y,0){i}+v(x,y,0){j}+0{k}$ in $\mathbb{R}^3.$ Since \begin{equation} \text{curl} \, {U}= \left| \begin{array}{ccc} {i} & {j} & {k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{ \partial }{\partial z} \\ u(x,y,0) & v(x,y,0) & 0 \end{array} \right| =0\, {i}+0\, {j}+\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right){k} \end{equation} we have $\text{curl} \, {U}=0$ if and only if $\frac{\partial v}{\partial x}=\frac{\partial u}{\partial y}.$

**Example**. Determine whether the vector field is conservative and if so, find a scalar potential function ${V}(x,y)=2 x y{i}+x y^3 {j}.$

**Solution**. Since $u(x,y)=2 x y$, $v(x,y)= x y ^3$, and $\frac{\partial u}{\partial x}=2x\neq y^3=\frac{\partial v}{\partial y},$ we see that ${V}$ is not a conservative vector field.

**Example**. Show that the vector field \begin{equation} {V}(x,y,z)=

\left(\frac{y}{1+x^2}+\tan ^{-1}z\right) {i} +\left(\tan ^{-1}x\right) {j}+\left(\frac{x}{1+z^2}\right) {k} \end{equation} is conservative and find a scalar potential function.

**Solution**. Since $\text{curl}{V}=0,$ it follows ${V}$ is conservative. Now we set out to find the scalar potential function $f.$ Since \begin{equation}\frac{\partial f}{\partial x}=\frac{y}{1+x^2}+\tan ^{-1}z\end{equation} we set \begin{equation}f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z+c(y,z).\end{equation} Since \begin{equation}\frac{\partial f}{\partial y}=\tan ^{-1}z=\frac{\partial }{\partial y}\left(y \tan ^{-1}x+x \tan ^{-1} z+c\right)=\tan ^{-1}x+\frac{\partial c}{\partial y},\end{equation} we find $\frac{\partial c}{\partial y}=0$ and $c=c_1(z)$ and so we set \begin{equation}f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z+c_1(z).\end{equation} Since \begin{equation}\frac{\partial f}{\partial z}=\frac{x}{1+z^2}=\frac{\partial }{\partial z}\left[y \tan ^{-1}x+x \tan ^{-1}z+c_1(z)\right] = \frac{x}{1+z^2} + c’_1(z),\end{equation} we then find $c_1 ‘(z)=0,$ $ c_1=0$ and so we obtain \begin{equation} f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z \end{equation} as desired.

## Exercises on Vector Fields

**Exercise**. If ${F}(x,y)=u(x,y){i}+v(x,y){j}$ where $u$ and $v$ are differentiable functions, show that $\text{curl}{F}=0$ if and only if $\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}.$

**Exercise**. Show that $\text{div}(f\nabla g)=f \text{div} \nabla g+\nabla f \cdot \nabla g.$

**Exercise**. Show that the curl of the gradient of a function is always $0.$

**Exercise**. Show that the divergence of the curl of a vector field is $0.$

**Exercise**. Let ${F}=.$ Either find a vector field ${G}$ such that ${F}=\text{curl}{G}$ or show that no such ${G}$ exists.

**Exercise**. Find the divergence and the curl for the following vector field.

$(1) \quad {F}(x,y)=x^2 {i}+ x y {j}+ z^3 {k}.$

$(2) \quad {F}(x,y,z)= z \, {i}-{j}+2y\, {k}.$

$(3) \quad {F}(x,y,z)= x y z\, {i}+y\, {j}+ x \, {k}.$

$(4) \quad {F}(x,y,z)= e^{-x}\sin y \, {i}+ e^{-x}\cos y\, {j}+{k}.$

$(5) \quad {F}(x,y)= x \, {i}+y \, {j}.$

$(6) \quad {F}(x,y)= x^2 \, {i}-y^2 \, {j}.$

$(7) \quad {F}(x,y,z)= x^2\, {i}+ y^2 \, {j}+ z^2 \, {k}.$

$(8) \quad {F}(x,y,z)= 2x z\, {i}+ y z^2\, {j}-\, {k}.$

$(9) \quad {F}(x,y,z)= z^2e^{-x}\, {i}+ y^3\ln z\, {j}+ x e^{-y} \, {k}.$

**Exercise**. Find the divergence of ${F}$ given that ${F}=\nabla f$ where $f(x,y,z)=x y^3z^2.$

**Exercise**. If ${F}(x,y,z)=x y \, {i}+y z \, {j}+z^2\, {k}$ and ${G}(x,y,z)=x \, {i}+y \, {j}-z \, {k}$ find $\text{curl}({F}\times {G}).$

**Exercise**. Determine whether or not the following vector fields are conservative.

$(1) \quad {F}(x,y)=y^2\, {i}+2 x y\, {j}$

$(2) \quad {F}(x,y)=2x y^3\, {i}+3y^2x^2\, {j}$

$(3) \quad {F}(x,y)=x e^{x y}\sin y \, {i}+ \left(e^{x y}\cos x y+y\right) {j} $

$(4) \quad {F}(x,y)=\left(-y+e^x\sin y\right){i}+\left((x+2)e^x\cos y\right){j}$

$(5) \quad {F}(x,y)=\left(y-x^2\right){i}+\left(2x+y^2\right){j}$

$(6) \quad {F}(x,y)= e^{2x}\sin y \, {i}+ e^{2x}\cos y {j} $