Orthogonal families of curves

Branko Ćurgus


Flow of a vector field

Consider a vector field in ${\mathbb R}^2$: \begin{equation*} \vec{F}(x,y) = u(x,y) \vec{\imath} + v(x,y) \vec{\jmath} \end{equation*} A parameterized curve \begin{equation*} \vec{r}(t) = f(t) \vec{\imath} + g(t) \vec{\jmath} \end{equation*} is said to be a flow of the vector filed $\vec{F}$ if at each point of the curve its velocity vector equals the vector of the vector filed at that point. That is for all $t$ we have \begin{equation*} \vec{r}'(t) = f'(t) \vec{\imath} + g'(t) \vec{\jmath} = u\bigl(f(t),g(t)\bigr) \vec{\imath} + v\bigl(f(t),g(t)\bigr) \vec{\jmath} \end{equation*} In other words, to find a flow of the vector field $\vec{F}$ we need to solve the differential equations \begin{equation*} f'(t) = u\bigl(f(t),g(t)\bigr) \qquad \text{and} \qquad g'(t) = v\bigl(f(t),g(t)\bigr). \end{equation*} You are expected to know how to solve these differential equations for simple functions $u$ and $v.$ Since solutions of differential equations involve arbitrary constants, solving these differential equations leads to a family of curves. This family is called the flow of a give vector field.
Orthogonal vector fields - Orthogonal families of curves

When considering a single vector we have often considered a vector of the same magnitude which is orthogonal to the given vector. For example, in ${\mathbb R}^2$ a vector orthogonal to $2 \vec{\imath} + 1 \vec{\jmath}$ which has the same magnitude is $(-1) \vec{\imath} + 2 \vec{\jmath}.$ Similarly for a given vector field \begin{equation*} \vec{F}(x,y) = u(x,y) \vec{\imath} + v(x,y) \vec{\jmath} \end{equation*} we can consider the vector field which is orthogonal to it \begin{equation*} \vec{G}(x,y) = -v(x,y) \vec{\imath} + u(x,y) \vec{\jmath}. \end{equation*} Then, calculating the flow lines of these two vector fields will lead to two families of curves which are at each point orthogonal to each other.
Example 1

Consider the following constant vector field \begin{equation*} \vec{F}(x,y) = 2 \vec{\imath} + 1 \vec{\jmath} \end{equation*} Just looking at a plot of this vector field indicates that the flow lines of this vector field are parallel lines with the slope $1/2$. To get the explicit solutions for the flow lines we have to solve differential equations \begin{equation*} f'(t) = 2 \qquad \text{and} \qquad g'(t) = 1. \end{equation*} The solutions are \begin{equation*} x = f(t) = 2t +c_1 \qquad \text{and} \qquad y = g(t) = t + c_2. \end{equation*} Or, eliminating $t$ we have $y = (1/2) x + c$ where $c$ is an arbitrary constant. The orthogonal vector field to $\vec{F}$ is \begin{equation*} \vec{G}(x,y) = (-1) \vec{\imath} + 2 \vec{\jmath} \end{equation*} The flow of this vector field consists of parallel lines with slope $-2$. In this way we have obtained two families of mutually orthogonal lines At this point you are asking: What is a big deal? I hope next examples will show that.
Example 2

Consider the following vector field with the first component constant \begin{equation*} \vec{F}(x,y) = 1 \vec{\imath} + x \vec{\jmath} \end{equation*} Just looking at a plot of this vector field indicates that the flow lines of this vector field could be parabolas with vertices on the $y$-axis. To get the explicit solutions for the flow lines we have to find the parametric curve $x=f(t), y=g(t),$ by solving the differential equations \begin{equation*} f'(t) = 1 \qquad \text{and} \qquad g'(t) = f(t). \end{equation*} The solutions are \begin{equation*} x = f(t) = t +c_1 \qquad \text{and} \qquad y = g(t) = (1/2)t + c_1t + c_2. \end{equation*} Or, eliminating $t$ we have $y = (1/2) x^2 + c$ where $c$ is an arbitrary constant. The orthogonal vector field to $\vec{F}$ is \begin{equation*} \vec{G}(x,y) = (-x) \vec{\imath} + 1 \vec{\jmath} \end{equation*} Next we find the flow of this vector field: \begin{equation*} f'(t) = -f(t) \qquad \text{and} \qquad g'(t) = 1. \end{equation*} The solutions are \begin{equation*} x = f(t) = c_1 \exp(-t) \qquad \text{and} \qquad y = g(t) = t + c_2. \end{equation*} Or, eliminating $t$ we have $x = c \, \exp(-y)$ where $c$ is an arbitrary constant. In this way we have obtained two families of mutually orthogonal curves:
Example 3

Consider the rotational vector field \begin{equation*} \vec{F}(x,y) = (-y) \vec{\imath} + x \vec{\jmath} \end{equation*} Just looking at a plot of this vector field indicates that the flow lines of this vector field are concentric circles. To get the explicit solutions for the flow lines we have to find the parametric curve $x=f(t), y=g(t),$ by solving the differential equations \begin{equation*} f'(t) = -g(t) \qquad \text{and} \qquad g'(t) = f(t). \end{equation*} The solutions are \begin{equation*} x = f(t) = r \cos(t) \qquad \text{and} \qquad y = g(t) = r \sin(t) \end{equation*} where $r$ is an arbitrary positive constant. Or, eliminating $t$ we have $x^2 + y^2 = r$ where $r$ is an arbitrary positive constant. The orthogonal vector field to $\vec{F}$ is \begin{equation*} \vec{G}(x,y) = (-x) \vec{\imath} + (-y) \vec{\jmath} \end{equation*} Next we find the flow of this vector field: \begin{equation*} f'(t) = -f(t) \qquad \text{and} \qquad g'(t) = -g(t). \end{equation*} The solutions are \begin{equation*} x = f(t) = c_1 \exp(-t) \qquad \text{and} \qquad y = g(t) = c_2 \exp(-t). \end{equation*} Or, eliminating $t$ we have $y = c \, x$ where $c$ is an arbitrary constant. In this way we have obtained two families of mutually orthogonal curves:
Example 4

Consider the hyperbolic vector field \begin{equation*} \vec{F}(x,y) = y \vec{\imath} + x \vec{\jmath} \end{equation*} Just looking at a plot of this vector field indicates that the flow lines of this vector field are hyperbolas. To get the explicit solutions for the flow lines we have to find the parametric curve $x=f(t), y=g(t),$ by solving the differential equations \begin{equation*} f'(t) = g(t) \qquad \text{and} \qquad g'(t) = f(t). \end{equation*} The solutions are \begin{equation*} x = f(t) = r \cosh(t) \qquad \text{and} \qquad y = g(t) = r \sinh(t) \end{equation*} where $r$ is an arbitrary constant. Or, eliminating $t$ we have $x^2 - y^2 = c$ where $c$ is an arbitrary constant. The orthogonal vector field to $\vec{F}$ is \begin{equation*} \vec{G}(x,y) = (-x) \vec{\imath} + y \vec{\jmath} \end{equation*} Next we find the flow of this vector field: \begin{equation*} f'(t) = -f(t) \qquad \text{and} \qquad g'(t) = g(t). \end{equation*} The solutions are \begin{equation*} x = f(t) = c_1 \exp(-t) \qquad \text{and} \qquad y = g(t) = c_2 \exp(t). \end{equation*} Or, eliminating $t$ we have $xy = c$ where $c$ is an arbitrary constant. In this way we have obtained two families of mutually orthogonal curves:
Example 5

Consider the elliptic vector field \begin{equation*} \vec{F}(x,y) = (-y) \vec{\imath} + (2x) \vec{\jmath} \end{equation*} Just looking at a plot of this vector field indicates that the flow lines of this vector field are concentric ellipses. To get the explicit solutions for the flow lines we have to find the parametric curve $x=f(t), y=g(t),$ by solving the differential equations \begin{equation*} f'(t) = -g(t) \qquad \text{and} \qquad g'(t) = 2f(t). \end{equation*} Solutions are \begin{equation*} x = f(t) = r \cos\bigl(\sqrt{2}t\bigr) \qquad \text{and} \qquad y = g(t) = r \sqrt{2} \sin\bigl(\sqrt{2}t\bigr) \end{equation*} where $r$ is an arbitrary positive constant. Or, eliminating $t$ we have $x^2 + \frac{y^2}{2} = r$ where $r$ is an arbitrary positive constant. The orthogonal vector field to $\vec{F}$ is \begin{equation*} \vec{G}(x,y) = (-2x) \vec{\imath} + (-y) \vec{\jmath} \end{equation*} Next we find the flow of this vector field: \begin{equation*} f'(t) = -2f(t) \qquad \text{and} \qquad g'(t) = -g(t). \end{equation*} The solutions are \begin{equation*} x = f(t) = c_1 \exp(-2t) \qquad \text{and} \qquad y = g(t) = c_2 \exp(-t). \end{equation*} Or, eliminating $t$ we have $x = c y^2$ where $c$ is an arbitrary constant. In this way we have obtained two families of mutually orthogonal curves: Here we encountered the family of ellipses with the ratio of vertical and horizontal axis being $\sqrt{2}$. There is nothing exceptional about this family. So, the above picture suggests an animation of a collection of families of ellipses with the ratios of vertical and horizontal axis being between $1/2$ and $2$, each family accompanied by its orthogonal family of curves.

Place the cursor over the image to start the animation.


Problems