KindiakMath

The Jacobian

Let \mathbf f \equiv \mathbf f(\mathbf x): \mathbb R^n \to \mathbb R^m be Frechét-differentiable. In a previous post, we have shown that

\mathbf f' = \begin{bmatrix} \displaystyle \frac{\partial f_1}{\partial x_1} & \cdots & \displaystyle \frac{\partial f_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \displaystyle \frac{\partial f_m}{\partial x_1} & \cdots & \displaystyle \frac{\partial f_m}{\partial x_n} \end{bmatrix} =: \mathbf J_{\mathbf f} \equiv \displaystyle \mathbf J_{\mathbf f} (\mathbf x) \equiv \frac{\partial \mathbf f}{\partial \mathbf x} \equiv \frac{\partial(f_1,\dots,f_m)}{\partial(x_1,\dots,x_n)}.

We call \mathbf J_{\mathbf f} the Jacobian matrix of \mathbf f, and its determinant |\mathbf J_{\mathbf f}| the Jacobian of \mathbf f. Observe that in the special case n = 1, \mathbf J_f = \nabla^{\mathrm T}f.

Problem 1. Evaluate the Jacobian of the polar coordinates map \mathbf F : \mathbb R^2 \to \mathbb R^2 defined by

\displaystyle \begin{bmatrix} x \\ y \end{bmatrix} \equiv \mathbf F\left( \begin{bmatrix} r \\ \theta \end{bmatrix} \right) := \begin{bmatrix} r \cos(\theta) \\ r \sin(\theta) \end{bmatrix} .

(Click for Solution)

Solution. We first evaluate the Jacobian matrix \mathbf J_{\mathbf F}:

\mathbf J_{\mathbf F} = \begin{bmatrix} \displaystyle \frac{\partial}{\partial r}(r \cos(\theta)) & \displaystyle \frac{\partial}{\partial \theta}(r \cos(\theta)) \\ \displaystyle \frac{\partial}{\partial r}(r \sin(\theta)) & \displaystyle \frac{\partial}{\partial \theta}(r \sin(\theta)) \end{bmatrix} = \begin{bmatrix} \cos(\theta) & -r\sin(\theta) \\ \sin(\theta) & r \cos(\theta) \end{bmatrix}.

Taking the Jacobian determinant,

\begin{aligned} |\mathbf J_{\mathbf F}| &= \left| \begin{matrix} \cos(\theta) & -r\sin(\theta) \\ \sin(\theta) & r \cos(\theta) \end{matrix} \right| \\ &= r \cos^2(\theta) - r(-{\sin^2(\theta)} ) \\ &= r (\cos^2(\theta) + \sin^2(\theta)) = r. \end{aligned}

Problem 2. Evaluate the Jacobian of the spherical coordinates map \mathbf G : \mathbb R^3 \to \mathbb R^3 defined by

\displaystyle \begin{bmatrix} x \\ y \\ z \end{bmatrix} \equiv \mathbf G\left( \begin{bmatrix} \rho \\ \varphi \\ \theta \end{bmatrix} \right) := \begin{bmatrix} \rho \sin(\varphi) \cos(\theta) \\ \rho \sin(\varphi) \sin(\theta) \\ \rho \cos(\varphi) \end{bmatrix} .

Evaluate the determinant | \mathbf J_{\mathbf S} | of the Jacobian matrix of \mathbf S.

(Click for Solution)

Solution. We first evaluate the Jacobian matrix \mathbf J_{\mathbf G}:

\begin{aligned} \mathbf J_{\mathbf G} &= \begin{bmatrix} \displaystyle \frac{\partial }{\partial \rho} (\rho \sin(\varphi) \cos(\theta)) & \displaystyle \frac{\partial }{\partial \varphi} (\rho \sin(\varphi) \cos(\theta)) & \displaystyle \frac{\partial }{\partial \theta} (\rho \sin(\varphi) \cos(\theta)) \\ \displaystyle \frac{\partial }{\partial \rho}(\rho \sin(\varphi) \sin(\theta)) & \displaystyle \frac{\partial }{\partial \varphi}(\rho \sin(\varphi) \sin(\theta)) & \displaystyle \frac{\partial }{\partial \theta}(\rho \sin(\varphi) \sin(\theta)) \\ \displaystyle \frac{\partial }{\partial \rho}(\rho \cos(\varphi)) & \displaystyle \frac{\partial }{\partial \varphi}(\rho \cos(\varphi)) & \displaystyle \frac{\partial }{\partial \theta}(\rho \cos(\varphi)) \end{bmatrix} \\ &= \begin{bmatrix} \sin(\varphi) \cos(\theta) & \rho \cos(\varphi) \cos(\theta) & \displaystyle -\rho \sin(\varphi) \sin(\theta) \\ \displaystyle \sin(\varphi) \sin(\theta) & \rho \cos(\varphi) \sin(\theta) & \displaystyle \rho \sin(\varphi) \cos(\theta) \\ \displaystyle \cos(\varphi) & -\sin(\varphi) & 0 \end{bmatrix}. \end{aligned}

Taking the Jacobian determinant by applying cofactor expansion on the third row and using the multi-linearity of the determinant,

\begin{aligned} |\mathbf J_{\mathbf G}| &= \left| \begin{matrix} \sin(\varphi) \cos(\theta) & \rho \cos(\varphi) \cos(\theta) & \displaystyle -\rho \sin(\varphi) \sin(\theta) \\ \displaystyle \sin(\varphi) \sin(\theta) & \rho \cos(\varphi) \sin(\theta) & \displaystyle \rho \sin(\varphi) \cos(\theta) \\ \displaystyle \cos(\varphi) & -{\sin(\varphi)} & 0 \end{matrix} \right| \\ &= \cos(\varphi) \cdot \left| \begin{matrix} \rho \cos(\varphi) \cos(\theta) & -\rho \sin(\varphi) \sin(\theta) \\ \rho \cos(\varphi) \sin(\theta) & \rho \sin(\varphi) \cos(\theta) \end{matrix} \right| \\ &\phantom{==} - (-{\sin(\varphi)}) \cdot \left| \begin{matrix} \sin(\varphi) \cos(\theta) & -\rho \sin(\varphi) \sin(\theta) \\ \sin(\varphi) \sin(\theta) & \rho \sin(\varphi) \cos(\theta) \end{matrix} \right| + 0 \cdot | \cdots | \\ &= \cos(\varphi) \cdot \rho \cos(\varphi) \cdot \rho \sin(\varphi) \left| \begin{matrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{matrix} \right| \\ &\phantom{==} - (-{\sin(\varphi)}) \cdot \rho \sin^2 (\varphi) \cdot \left| \begin{matrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{matrix} \right| \\ &= \rho^2 \sin(\varphi) \cdot \underbrace{ (\cos^2(\varphi) + \sin^2(\varphi)) }_1 \cdot \underbrace{ \left| \begin{matrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{matrix} \right| }_1 = \rho^2 \sin(\varphi).\end{aligned}

Problem 3. Given the nonzero constant \alpha, evaluate the Jacobian of the map \mathbf H : \mathbb R^2 \to \mathbb R^2 defined by

\displaystyle \begin{bmatrix} x \\ y \end{bmatrix} \equiv \mathbf H\left( \begin{bmatrix} u \\ v \end{bmatrix} \right) := \begin{bmatrix} u+\alpha v \\ u-\alpha v \end{bmatrix}.

(Click for Solution)

Solution. We first evaluate the Jacobian matrix \mathbf J_{\mathbf F}:

\mathbf J_{\mathbf H} = \begin{bmatrix} \displaystyle \frac{\partial}{\partial u}(u+\alpha v) & \displaystyle \frac{\partial}{\partial v}(u+\alpha v) \\ \displaystyle \frac{\partial}{\partial u}(u-\alpha v) & \displaystyle \frac{\partial}{\partial v}(u-\alpha v) \end{bmatrix} = \begin{bmatrix} 1 &\alpha \\ 1 & -\alpha \end{bmatrix}.

Taking the Jacobian determinant,

|\mathbf J_{\mathbf H}| = \left| \begin{matrix} 1 &\alpha \\ 1 & -\alpha \end{matrix} \right| = -\alpha - \alpha = -2\alpha.

Remark 1. Observe that

\displaystyle \mathbf H \left( \begin{bmatrix} u \\ v \end{bmatrix} \right) = \begin{bmatrix} 1 &\alpha \\ 1 & -\alpha \end{bmatrix}\begin{bmatrix} u \\ v \end{bmatrix} = \mathbf J_{\mathbf H} \begin{bmatrix} u \\ v \end{bmatrix},

so that \mathbf H is a linear transformation with associated matrix \mathbf J_{\mathbf H}. In particular, \mathbf H is bijective if and only if \alpha \neq 0.

Problem 4. Suppose \mathbf F : \mathbb R^2 \to \mathbb R^2, \mathbf F(u,v) = (x,y) is bijective and Frechét-differentiable with Jacobian \mathbf J_{\mathbf F}. Given that g : \mathbb R^2 \to \mathbb R is continuous, the Jacobian transformation tells us that

\begin{aligned} \iint_D g(x,y)\, \mathrm dx\, \mathrm dy &= \iint_{\mathbf F^{-1}(D) } (g \circ \mathbf F)(u,v) \cdot \left| \mathbf J_{\mathbf F} \right| \, \mathrm du\, \mathrm dv. \end{aligned}

This result can be verified using standard tools in measure theory. Defining \mathbf F as per Problem 1, suppose

\mathbf F^{-1}(D) = [a,b] \times [\alpha,\beta].

Prove that for any continuously differentiable f,

\begin{aligned} \iint_D f'(x^2 + y^2)\, \mathrm dx\, \mathrm dy &= \frac 12 \cdot (\beta - \alpha) \cdot (f(b^2) - f(a^2)). \end{aligned}

(Click for Solution)

Solution. Using the details in Problem 1, define g(x,y) := x^2 + y^2. Then

\begin{aligned} g(x,y) &= g(\mathbf F(r,\theta)) = g(r \cos \theta, r\sin \theta) \\ &= (r \cos \theta)^2 + (r \sin \theta)^2 = r^2 \cos^2 \theta + r^2 \sin^2 \theta \\ &= r^2 (\cos^2 \theta + \sin^2 \theta) = r^2 \cdot 1 = r^2 \end{aligned}

so that

\begin{aligned} \iint_D f'(x^2 + y^2)\, \mathrm dx\, \mathrm dy &= \iint_D (f' \circ g)(x,y)\, \mathrm dx\, \mathrm dy \\ &= \iint_{\mathbf F^{-1}(D) } (f' \circ g)(\mathbf F(r,\theta)) \cdot |\mathbf J_{\mathbf F}|\, \mathrm dr\, \mathrm d\theta \\ &= \iint_{\mathbf F^{-1}(D) } f' (g(\mathbf F(r,\theta)) \cdot r\, \mathrm dr\, \mathrm d\theta \\ &= \iint_{[a,b]\times[\alpha,\beta] } f' (r^2) \cdot r\, \mathrm dr\, \mathrm d\theta \\ &= \int_\alpha^\beta \int_a^b f' (r^2) \cdot r\, \mathrm dr\, \mathrm d\theta \\ &= \int_\alpha^\beta 1\, \mathrm d\theta \cdot \int_a^b f' (r^2) \cdot r\, \mathrm dr \\ &= (\beta - \alpha) \cdot \frac 12 [f(r^2)]_a^b \\ &= \frac 12 \cdot (\beta - \alpha) \cdot (f(b^2) - f(a^2)). \end{aligned}

—Joel Kindiak, 16 Sept 25, 1326H

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