Planar Transformations and Jacobians

Learning Objectives

  • Determine the image of a region under a given transformation of variables.
  • Compute the Jacobian of a given transformation.

Planar Transformations

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

On the left-hand side of this figure, there is a region G with point (u, v) given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with x = g(u, v) and y = h(u, v). On the right-hand side of this figure there is a region R with point (x, y) given in the Cartesian xy- plane.

Figure 1. The transformation of a region [latex]G[/latex] in the [latex]uv[/latex]-plane into a region [latex]R[/latex] in the [latex]xy[/latex]-plane.

definition


A transformation [latex]{T}{:} \ {G}{\rightarrow}{R}[/latex], defined as [latex]T(u, v)=(x, y)[/latex], is said to be a one-to-one transformation if no two points map to the same image point.

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

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

Example: determining how the transformation works

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

Example: finding the image under T

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

try it

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

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 [latex]g_u[/latex], [latex]g_v[/latex], [latex]h_u[/latex], and [latex]h_v[/latex] exist and are also continuous. A transformation that has this property is called a [latex]C^1[/latex] transformation (here [latex]C[/latex] denotes continuous). Let [latex]T(u, v)=(g(u, v), h(u, v))[/latex], where [latex]x=g(u, v)[/latex] and [latex]y=h(u, v)[/latex], be a one-to-one [latex]C^1[/latex] transformation. We want to see how it transforms a small rectangular region [latex]S[/latex], [latex]{\Delta}{u}[/latex] units by [latex]{\Delta}{v}[/latex] units, in the [latex]uv[/latex]-plane (see the following figure).

On the left-hand side of this figure, there is a region S with lower right corner point (u sub 0, v sub 0), height Delta v, and length Delta u given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with T. On the right-hand side of this figure there is a region R with point (x sub 0, y sub 0) given in the Cartesian x y-plane with sides r(u, v sub 0) along the bottom and r(u sub 0, v) along the left.

Figure 4. A small rectangle [latex]S[/latex] in the [latex]uv[/latex]-plane is transformed into a region [latex]R[/latex] in the [latex]xy[/latex]-plane.

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

[latex]\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}}}[/latex].

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

[latex]\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}}}[/latex].

Now, note that

[latex]\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}}[/latex].

Similarly,

[latex]\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}}[/latex].

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

[latex]\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}}}[/latex].

Since [latex]{\mid}{\bf{k}}{\mid} = {1}[/latex], we have [latex]{\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}[/latex].

definition


The Jacobian of the [latex]C^1[/latex] transformation [latex]T(u, v)=(g(u, v), h(u, v))[/latex] is denoted by [latex]J(u, v)[/latex] and is defined by the [latex]2\times 2[/latex] determinant

[latex]\large{J(u,v)=\Bigg|\frac{\partial(x,y)}{\partial(u,v)}\Bigg|=\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)}[/latex].

Using the definition, we have

[latex]\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}}.[/latex]

Note that the Jacobian is frequently denoted simply by

[latex]\large{{J}{({u},{v})} = {\frac{{\partial}{(x,y)}}{{\partial}{(u,v)}}}}.[/latex]

Note also that

[latex]\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}}[/latex]

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

Example: finding the jacobian

Find the Jacobian of the transformation given in Example “Determining How the Transformation Works”.

Example: finding the jacobian

Find the Jacobian of the transformation given in Example “Finding the Image under [latex]T[/latex]“.

try it

Find the Jacobian of the transformation given in the previous Try It: [latex]T(u, v)=(u+v, 3v)[/latex].

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

You can view the transcript for “CP 5.44” here (opens in new window).