Planar Transformations

A planar transformation $T$ is a function that transforms a region $G$ in one plane into a region $R$ in another plane by a change of variables. Both $G$ and $R$ are subsets of $R^{2}$. For example, Figure 1 shows a region $G$ in the $uv$-plane transformed into a region $R$ in the $xy$-plane by the change of variables $x=g(u, v)$ and $y=h(u, v)$, or sometimes we write $x=x(u, v)$ and $y=y(u, v)$. We shall typically assume that each of these functions has continuous first partial derivatives, which means $g_u$, $g_v$, $h_u$, and $h_v$ exist and are also continuous. The need for this requirement will become clear soon.

Figure 1. The transformation of a region $G$ in the $uv$-plane into a region $R$ in the $xy$-plane.

To show that $T$ is a one-to-one transformation, we assume ${T}{({{u}_{1}},{{v}_{1}})} = {T}{({{u}_{2}},{{v}_{2}})}$ and show that as a consequence we obtain ${({{u}_{1}},{{v}_{1}})} = {({{u}_{2}},{{v}_{2}})}$. If the transformation $T$ is one-to-one in the domain $G$, then the inverse ${T}^{-1}$ exists with the domain $R$ such that ${T}^{-1}{\circ}{T}$ and ${T}{\circ}{T}^{-1}$ are identity functions.

Figure 1 shows the mapping ${T}{({u},{v})} = {({x},{y})}$ where $x$ and $y$ are related to $u$ and $v$ by the equations $x=g(u, v)$ and $y=h(u, v)$. The region $G$ is the domain of $T$ and the region $R$ is the range of $T$, also known as the image of $G$ under the transformation $T$.

Example: determining how the transformation works

Suppose a transformation $T$ is defined as ${T}{({r},{\theta})} = {({x},{y})}$ where ${x} = {r}{\cos}{\theta}, {y} = {r}{\sin}{\theta}$. Find the image of the polar rectangle ${G} = {\left \{{({r},{\theta})}{\mid}{0} \leq {r} \leq {{1},{0}} \leq {\theta} \leq {{\pi}/{2}} \right \}}$ in the ${r}{\theta}$-plane to a region $R$ in the $xy$-plane. Show that $T$ is a one-to-one transformation in $G$ and find ${{T}^{-1}}{({x},{y})}$.

Show Solution

Since $r$ varies from 0 to 1 in the ${r}{\theta}$-plane, we have a circular disc of radius 0 to 1 in the $xy$-plane. Because ${\theta}$ varies from 0 to ${\pi}/{2}$ in the ${r}{\theta}$-plane, we end up getting a quarter circle of radius 1 in the first quadrant of the $xy$-plane (Figure 2). Hence $R$ is a quarter circle bounded by $x^{2}+y^{2}=1$ in the first quadrant.

Figure 2. A rectangle in the $r\theta$-plane is mapped into a quarter circle in the $xy$-plane.

In order to show that $T$ is a one-to-one transformation, assume ${T}{({r_1},{{\theta}_1})} = {T}{({r_2},{{\theta}_2})}$ and show as a consequence that ${({r_1},{{\theta}_1})} = {({r_2},{{\theta}_2})}$. In this case, we have

$\begin{aligned} {T}{({r_1},{{\theta}_1})} & = {T}{({r_2},{{\theta}_2})} \\ {({x_1},{y_1})} & = {({x_2},{y_2})} \\ {({r_1}{\cos}{{\theta}_{1}},{r_1}{\sin}{{\theta}_{1}})} & = {({r_2}{\cos}{{\theta}_{2}},{r_2}{\sin}{{\theta}_{2}})} \\ {r_1}{\cos}{{\theta}_{1}} & = {r_2}{\cos}{{\theta}_{2}} \\ {r_1}{\sin}{{\theta}_{1}} & = {r_2}{\sin}{{\theta}_{2}} \end{aligned}$

Dividing, we obtain

$\begin{aligned} {\frac{{r_1}{\cos}{{\theta}_{1}}}{{r_1}{\sin}{{\theta}_{1}}}} & = {{\frac{{r_2}{\cos}{{\theta}_{2}}}{{r_2}{\sin}{{\theta}_{2}}}}} \\ {\frac{{\cos}{{\theta}_{1}}}{{\sin}{{\theta}_{1}}}} & = {\frac{{\cos}{{\theta}_{2}}}{{\sin}{{\theta}_{2}}}} \\ {\tan}{{\theta}_{1}} & = {\tan}{{\theta}_{2}} \\ {{\theta}_{1}} & = {{\theta}_{2}} \end{aligned}$

since the tangent function is one-one function in the interval ${0} \leq {\theta} \leq {{\pi}/{2}}$. Also, since ${0} < {r} \leq {1}$, we have ${{r}_{1}} = {{r}_{2}}, {{\theta}_{1}} = {{\theta}_{2}}$. Therefore, ${({{r}_{1}}, {{\theta}_{1}})} = {({{r}_{2}}, {{\theta}_{2}})}$ and $T$ is a one-to-one transformation from $G$ into $R$.

To find ${{T}^{-1}}{({x},{y})}$ solve for ${r},{\theta}$ in terms of $x, y$. We already know that ${{r}^{2}} = {{x}^{2}} + {{y}^{2}}$ and ${\tan}{\theta} = {\frac{y}{x}}$. Thus ${{T}^{-1}}{({x},{y})} = {({r},{\theta})}$ is defined as ${r} = {\sqrt{{x^{2}}+{y^{2}}}}$ and ${\theta} = {{\tan}^{-1}}{({\frac{y}{x}})}$.

Example: finding the image under T

Let the transformation $T$ be defined by $T(u, v)=(x, y)$ where $x=u^{2}+v^{2}$ and $y=uv$. Find the image of the triangle in the $uv$-plane with vertices $(0, 0)$, $(0, 1)$, and $(1, 1)$.

Show Solution

The triangle and its image are shown in Figure 3. To understand how the sides of the triangle transform, call the side that joins $(0, 0)$ and $(0, 1)$ side $A$, the side that joins $(0, 0)$ and $(1, 1)$ side $B$, and the side that joins $(1, 1)$ and $(0, 1)$ side $C$.

Figure 3. A triangular region in the $uv$-plane is transformed into an image in the $xy$-plane.

For the side $A$: $u=0$, ${0} \leq {v} \leq {1}$ transforms to $x=-v^{2}$, $y=0$ so this is the side ${A}^{\prime}$ that joins $(-1, 0)$ and $(0, 0)$.

For the side $B$: $u=v$, ${0} \leq {u} \leq {1}$ transforms to $x=0$, $y=u^{2}$ so this is the side ${B}^{\prime}$ that joins $(0, 0)$ and $(0, 1)$.

For the side $C$: ${0} \leq {u} \leq {1}$, $v=1$ transforms to $x=u^{2}-1$, $y=u$ (hence $x=y^{2}-1$) so this is the side ${C}{\prime}$ that makes the upper half of the parabolic arc joining $(-1, 0)$ and $(0, 1)$.

All the points in the entire region of the triangle in the $uv$-plane are mapped inside the parabolic region in the $xy$-plane.

try it

Let a transformation $T$  be defined as $T(u, v)=(x, y)$ where $x=u+v$, $y=3v$. Find the image of the rectangle ${G} = {\left \{ {({u,v})}: {0} \leq {u} \leq {1,0} \leq {v} \leq {2}\right \}}$ from the $uv$-plane after the transformation into a region $R$ in the $xy$-plane. Show that $T$ is a one-to-one transformation and find ${{T}^{-1}}{({x},{y})}$.

Show Solution

$T^{-1}(x,y)=(u,v)$ where $u=\frac{3x-y}3$ and $v=\frac{y}3$.

Jacobians

Recall that we mentioned near the beginning of this section that each of the component functions must have continuous first partial derivatives, which means that $g_u$, $g_v$, $h_u$, and $h_v$ exist and are also continuous. A transformation that has this property is called a $C^1$ transformation (here $C$ denotes continuous). Let $T(u, v)=(g(u, v), h(u, v))$, where $x=g(u, v)$ and $y=h(u, v)$, be a one-to-one $C^1$ transformation. We want to see how it transforms a small rectangular region $S$, ${\Delta}{u}$ units by ${\Delta}{v}$ units, in the $uv$-plane (see the following figure).

Figure 4. A small rectangle $S$ in the $uv$-plane is transformed into a region $R$ in the $xy$-plane.

Since $x=g(u, v)$ and $y=h(u, v)$, we have the position vector ${\bf{r}}{({u},{v})} = {g}{({u},{v})}{\bf{i}} + {h}{({u},{v})}{\bf{j}}$ of the image of the point $(u, v)$. Suppose that $(u_0, v_0)$ is the coordinate of the point at the lower left corner that mapped to $(x_0, y_0)=T(u_0, v_0)$. The line $v=v_0$ maps to the image curve with vector function $r(u, v_0)$, and the tangent vector at $(x_0, y_0)$ to the image curve is

$\large{{{\bf{r}}_{u}} = {{g}_{u}}{({u_0},{v_0})}{\bf{i}} + {h_{u}}{({u_0},{v_0})}{\bf{j}} = {\frac{{\partial}{x}}{{\partial}{u}}}{\bf{i}} + {\frac{{\partial}{y}}{{\partial}{u}}}{\bf{j}}}$.

Similarly, the line $u=u_0$ maps to the image curve with vector function ${\bf{r}}(u_0, v)$, and the tangent vector at $(x_0, y_0)$ to the image curve is

$\large{{{\bf{r}}_{v}} = {{g}_{v}}{({u_0},{v_0})}{\bf{i}} + {h_{v}}{({u_0},{v_0})}{\bf{j}} = {\frac{{\partial}{x}}{{\partial}{v}}}{\bf{i}} + {\frac{{\partial}{y}}{{\partial}{v}}}{\bf{j}}}$.

Now, note that

$\large{{{\bf{r}}_{u}} = {\displaystyle\lim_{{\Delta}{u}{\rightarrow}{0}}}{\frac{{\bf{r}}{({u_0} + {{\Delta}{u}},{v_0}) - {\bf{r}}{({u_0},{v_0})}}}{{\Delta}{u}}} \ {\text{so}} \ {{\bf{r}}{({u_0} + {{\Delta}{u}},{v_0}) - {\bf{r}}{({u_0},{v_0})}}}{\approx}{\Delta}{u}{\bf{r}}_{u}}$.

Similarly,

$\large{{{\bf{r}}_{v}} = {\displaystyle\lim_{{\Delta}{v}{\rightarrow}{0}}}{\frac{{\bf{r}}{({u_0},{v_0} + {{\Delta}{v}}) - {\bf{r}}{({u_0},{v_0})}}}{{\Delta}{v}}} \ {\text{so}} \ {\bf{r}}{({u_0},{v_0} + {{\Delta}{v}}) - {\bf{r}}{({u_0},{v_0})}}{\approx}{\Delta}{v}{\bf{r}}_{v}}$.

This allows us to estimate the area ${\Delta}{A}$ of the image $R$ by finding the area of the parallelogram formed by the sides ${\Delta}{v}{{\bf{r}}_{v}}$ and ${\Delta}{u}{{\bf{r}}_{u}}$. By using the cross product of these two vectors by adding the ${\bf{k}}$th component as 0, the area ${\Delta}{A}$ of the image $R$ (refer to The Cross Product) is approximately ${\mid}{\Delta}{u}{{\bf{r}}_{u}} \ {\times} \ {\Delta}{v}{{\bf{r}}_{v}}{\mid} = {\mid}{{\bf{r}}_{u}} \ {\times} \ {{\bf{r}}_{v}}{\mid} \ {\Delta}{u}{\Delta}{v}$. In determinant form, the cross product is

$\large{{\bf{r}}_u\times{\bf{r}}_v=\begin{vmatrix}{\bf{i}}&{\bf{j}}&{\bf{k}} \\ \frac{\partial{x}}{\partial{u}}&\frac{\partial{y}}{\partial{u}}&0 \\ \frac{\partial{x}}{\partial{v}}&\frac{\partial{y}}{\partial{v}}&0\end{vmatrix}=\begin{vmatrix}\frac{\partial{x}}{\partial{u}}&\frac{\partial{y}}{\partial{u}} \\ \frac{\partial{x}}{\partial{v}}&\frac{\partial{y}}{\partial{v}}\end{vmatrix}{\bf{k}}=\left(\frac{\partial{x}}{\partial{u}}\frac{\partial{y}}{\partial{v}}-\frac{\partial{x}}{\partial{v}}\frac{\partial{y}}{\partial{u}}\right){\bf{k}}}$.

Since ${\mid}{\bf{k}}{\mid} = {1}$, we have ${\Delta}{A} \ {\approx} \ {\mid}{{\bf{r}}_{u}} \ {\times} \ {{\bf{r}}_{v}}{\mid} \ {\Delta}{u}{\Delta}{v} = {\left ({\frac{{\partial}{x}}{{\partial}{u}}} \ {\frac{{\partial}{y}}{{\partial}{v}}} - {\frac{{\partial}{x}}{{\partial}{v}}} \ {\frac{{\partial}{y}}{{\partial}{u}}} \right )}{\Delta}{u}{\Delta}{v}$.

Using the definition, we have

$\large{{\Delta}{A}{\approx}{J}{({u},{v})}{\Delta}{u}{\Delta}{v} = {\bigg\vert}{\frac{{\partial}{(x,y)}}{{\partial}{(u,v)}}}{\bigg\vert}{\Delta}{u}{\Delta}{v}}.$

Note that the Jacobian is frequently denoted simply by

$\large{{J}{({u},{v})} = {\frac{{\partial}{(x,y)}}{{\partial}{(u,v)}}}}.$

Note also that

$\large{\begin{vmatrix}\frac{\partial{x}}{\partial{u}}&\frac{\partial{y}}{\partial{u}} \\ \frac{\partial{x}}{\partial{v}}&\frac{\partial{y}}{\partial{v}}\end{vmatrix}=\left(\frac{\partial{x}}{\partial{u}}\frac{\partial{y}}{\partial{v}}-\frac{\partial{x}}{\partial{v}}\frac{\partial{y}}{\partial{u}}\right)=\begin{vmatrix}\frac{\partial{x}}{\partial{u}}&\frac{\partial{x}}{\partial{v}} \\ \frac{\partial{y}}{\partial{u}}&\frac{\partial{y}}{\partial{v}}\end{vmatrix}}$

Hence the notation ${J}{({u},{v})} = {\frac{{\partial}{(x,y)}}{{\partial}{(u,v)}}}$ suggests that we can write the Jacobian determinant with partials of $x$ in the first row and partials of $y$ in the second row.

Watch the following video to see the worked solution to the above Try It