Change of Variables: Learn It 2

Jacobians

We mentioned earlier that transformation functions must have continuous first partial derivatives—meaning [latex]g_u[/latex], [latex]g_v[/latex], [latex]h_u[/latex], and [latex]h_v[/latex] exist and are continuous. A transformation with this property is called a [latex]C^1[/latex] transformation (where [latex]C[/latex] denotes continuous).

Let [latex]T(u,v) = (g(u,v), h(u,v))[/latex] be a one-to-one [latex]C^1[/latex] transformation, where [latex]x = g(u,v)[/latex] and [latex]y = h(u,v)[/latex]. We want to understand how this transformation affects a small rectangular region [latex]S[/latex] in the [latex]uv[/latex]-plane, measuring [latex]\Delta u[/latex] units by [latex]\Delta v[/latex] units (see Figure 4).

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 can write the position vector of the image of point [latex](u, v)[/latex] as:

[latex]\mathbf{r}(u,v) = g(u,v)\mathbf{i} + h(u,v)\mathbf{j}[/latex]

Suppose [latex](u_0, v_0)[/latex] is the lower left corner of rectangle [latex]S[/latex], which maps to [latex](x_0, y_0) = T(u_0, v_0)[/latex].

The line [latex]v = v_0[/latex] in the [latex]uv[/latex]-plane maps to an image curve with vector function [latex]\mathbf{r}(u, v_0)[/latex]. The tangent vector at [latex](x_0, y_0)[/latex] to this image curve is:

[latex]\mathbf{r}_u = g_u(u_0, v_0)\mathbf{i} + h_u(u_0, v_0)\mathbf{j} = \frac{\partial x}{\partial u}\mathbf{i} + \frac{\partial y}{\partial u}\mathbf{j}[/latex]

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

[latex]\mathbf{r}_v = g_v(u_0, v_0)\mathbf{i} + h_v(u_0, v_0)\mathbf{j} = \frac{\partial x}{\partial v}\mathbf{i} + \frac{\partial y}{\partial v}\mathbf{j}[/latex]

Tangent Vectors and Derivatives

For a vector function [latex]\mathbf{r}(u)[/latex], the derivative [latex]\mathbf{r}_u[/latex] gives the tangent vector:

[latex]{{\bf{r}}_{u}} = {\displaystyle\lim_{{\Delta}{u}{\rightarrow}{0}}}{\dfrac{{\bf{r}}{({u_0} + {{\Delta}{u}},{v_0}) - {\bf{r}}{({u_0},{v_0})}}}{{\Delta}{u}}}[/latex]

This means [latex]\mathbf{r}(u_0 + \Delta u) - \mathbf{r}(u_0) \approx \Delta u \mathbf{r}_u[/latex] for small [latex]\Delta u[/latex].

From the definition of derivative, we have:

[latex]{{\bf{r}}_{u}} = {\displaystyle\lim_{{\Delta}{u}{\rightarrow}{0}}}{\dfrac{{\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]{{\bf{r}}_{v}} = {\displaystyle\lim_{{\Delta}{v}{\rightarrow}{0}}}{\dfrac{{\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].

Finding the Area Element

The small rectangle [latex]S[/latex] in the [latex]uv[/latex]-plane transforms into a curved region [latex]R[/latex] in the [latex]xy[/latex]-plane. We can estimate the area [latex]\Delta A[/latex] of image [latex]R[/latex] by finding the area of the parallelogram formed by the sides [latex]\Delta u \mathbf{r}_u[/latex] and [latex]\Delta v \mathbf{r}_v[/latex].

Area of a Parallelogram Using Cross Product

For vectors [latex]\mathbf{a}[/latex] and [latex]\mathbf{b}[/latex] in space, the area of the parallelogram they span is [latex]|\mathbf{a} \times \mathbf{b}|[/latex].

To compute the cross product of two-dimensional vectors, add a [latex]\mathbf{k}[/latex] component of zero to make them three-dimensional.

Using the cross product (by adding the [latex]\mathbf{k}[/latex] component as [latex]0[/latex]), the area [latex]\Delta A[/latex] of image [latex]R[/latex] is approximately:

[latex]|\Delta u \mathbf{r}_u \times \Delta v \mathbf{r}_v| = |\mathbf{r}_u \times \mathbf{r}_v| \Delta u \Delta v[/latex]

In determinant form, the cross product is:

[latex]\mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u} & 0 \\ \dfrac{\partial x}{\partial v} & \dfrac{\partial y}{\partial v} & 0 \end{vmatrix} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u} \\ \dfrac{\partial x}{\partial v} & \dfrac{\partial y}{\partial v} \end{vmatrix} \mathbf{k} = \left( \dfrac{\partial x}{\partial u} \dfrac{\partial y}{\partial v} - \dfrac{\partial x}{\partial v} \dfrac{\partial y}{\partial u} \right) \mathbf{k}[/latex]

Since [latex]|\mathbf{k}| = 1[/latex], we have:

[latex]\Delta A \approx |\mathbf{r}_u \times \mathbf{r}_v| \Delta u \Delta v = \left| \dfrac{\partial x}{\partial u} \dfrac{\partial y}{\partial v} - \dfrac{\partial x}{\partial v} \dfrac{\partial y}{\partial u} \right| \Delta u \Delta v[/latex]

This determinant expression is so important that it gets its own name: the Jacobian.

The Jacobian Determinant

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]J(u,v) = \left| \dfrac{\partial(x,y)}{\partial(u,v)} \right| = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u} \\ \dfrac{\partial x}{\partial v} & \dfrac{\partial y}{\partial v} \end{vmatrix} = \left( \dfrac{\partial x}{\partial u} \dfrac{\partial y}{\partial v} - \dfrac{\partial x}{\partial v} \dfrac{\partial y}{\partial u} \right)[/latex]

the Jacobian

For transformation [latex]T(u,v) = (x,y)[/latex] where [latex]x = g(u,v)[/latex] and [latex]y = h(u,v)[/latex]:

[latex]J(u,v) = \dfrac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u} \\ \dfrac{\partial x}{\partial v} & \dfrac{\partial y}{\partial v} \end{vmatrix}[/latex]

The Jacobian tells us how the transformation scales areas: [latex]\Delta A \approx |J(u,v)| \Delta u \Delta v[/latex]

Using this definition, we can write:

[latex]\Delta A \approx J(u,v) \Delta u \Delta v = \left| \dfrac{\partial(x,y)}{\partial(u,v)} \right| \Delta u \Delta v[/latex]

Note that the Jacobian is frequently written simply as:

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

Two Ways to Arrange the Jacobian

The determinant can be computed in two equivalent ways:

[latex]\begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u} \\ \dfrac{\partial x}{\partial v} & \dfrac{\partial y}{\partial v} \end{vmatrix} = \left( \dfrac{\partial x}{\partial u} \dfrac{\partial y}{\partial v} - \dfrac{\partial x}{\partial v} \dfrac{\partial y}{\partial u} \right) = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix}[/latex]

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

Consider the transformation [latex]T(r,\theta) = (x,y)[/latex] where [latex]x = r\cos\theta[/latex] and [latex]y = r\sin\theta[/latex]. Find the Jacobian of this transformation.

Consider the transformation [latex]T(u,v) = (x,y)[/latex] where [latex]x = u^2 + v^2[/latex] and [latex]y = uv[/latex]. Find the Jacobian of this transformation.