Jacobi and the total time derivative

Motivation

A fairly common task, which quickly becomes second nature to physics students, is to compute the total derivative of a multi-dimensional time-dependent function. Here, we will only deal with “innocent” functions, that is, continuous functions that are totally differentiable everywhere in their domain.

Suppose we have the function $f: \R^4 \rightarrow \R$ given by

$$\begin{align*} f(\bm{x}(t), t) &= 2 \bigl(x_1(t)\bigr)^2 + \bigl(\ln(x_2(t))\bigr)^3 + 4 x_3(t) + \pi t,\\ &\quad x_1(t) = \sin(t), \quad x_2(t) = \cos(2t), \quad x_3(t) = \frac{1}{t} \end{align*}$$

which indirectly depends on $t$ via $x_1(t)$, $x_2(t)$ and $x_3(t)$, but also has a direct dependency on $t$ via the term $\pi t$. The total derivative is calculated by

$$\begin{align*} \dv{t} f\bigl(\bm{x}(t), t\bigr) &= \pdv{\bm{x}} f\bigl(\bm{x}(t), t\bigr) \cdot \dv{t} \bm{x}(t) + \pdv{t} f\bigl(\bm{x}(t), t\bigr)\\ &= \underbrace{\pdv{x_1} f\bigl(\bm{x}(t), t\bigr) \cdot \dv{t} x_1(t)}_{4 x_1(t) \cdot \dot{x}_1(t)} + \underbrace{\pdv{x_2} f\bigl(\bm{x}(t), t\bigr) \cdot \dv{t} x_2(t)}_{3 \bigl(\ln(x_2(t))\bigr)^2 \frac{1}{x_2(t)} \cdot \dot{x}_2(t)}\\ &\quad + \underbrace{\pdv{x_3} f\bigl(\bm{x}(t), t\bigr) \cdot \dv{t} x_3(t)}_{4 \dot{x}_3(t)} + \underbrace{\pdv{t} f\bigl(\bm{x}(t), t\bigr)}_{\pi}\\ &= 4 \sin(t) \cos(t) - 6\bigl(\ln(\cos(2t))\bigr)^2 \tan(2t) - \frac{4}{t^2} + \pi \end{align*}$$

One way to remember the formula is to form the total derivative with respect to all variables $x_1(t), x_2(t), x_3(t)$ and with respect to the independent variable $t$ itself. With $\dvv{f}{x_1} = \pdvv{f}{x_1} \cdot \dvv{x_1}{t}$ and likewise for $x_2$ and $x_3$, we then obtain the above formula.

However, where does this equation really come from? In Calculus II, the multivariable chain rule is introduced

$$\begin{align*} \boxed{J_{g\circ f}(\bm{x}) = J_g(f(\bm{x})) \cdot J_f(\bm{x})} \end{align*}$$

with $J_f$ being the Jacobian matrix of function $f$. We will give a recap on this matrix and the chain rule to then examine the relationship between the multivariable chain rule and the total time derivative of a function $f$.

Jacobian matrix

Definition

Recall that for a function $f: U \subopenin \R^n \rightarrow \R^m$ with $f$ being totally differentiable, we define the Jacobian matrix of $f$ at point $\bm{x}\in U$ as

$$\begin{align*} J_f(\bm{x}) &\coloneqq \Bigl((\partial_j f_i)(\bm{x})\Bigr) _{\substack{i=1,\dots, m\\j=1, \dots, n}}\\ &= \begin{pmatrix} (\partial_1 f_1)(\bm{x}) & (\partial_2 f_1)(\bm{x}) & \dots & (\partial_n f_1)(\bm{x})\\ \vdots & \vdots & \ddots & \vdots\\ (\partial_1 f_m)(\bm{x}) & (\partial_2 f_m)(\bm{x}) & \dots & (\partial_n f_m)(\bm{x}) \end{pmatrix} \in \Matrix_{m\times n}(\R) \end{align*}$$

where $\partial_j f_i$ is a shorthand for $\pdvv{f_i}{x_j}$. Common notations for the Jacobian matrix include ¹

$$\begin{align*} J_f(\bm{x}) = (Df)(\bm{x}) = \pdvv{(f_1, \dots, f_m)}{(x_1, \dots, x_n)}\,(\bm{x}) \end{align*}$$

Example

As an example, take the function $f: \R^3 \rightarrow \R^2$ given by $$\begin{align*} f(x_1, x_2, x_3) \coloneqq \bigl(x_1 + 42 x_2,\quad (x_1 + e^{x_3})^2 \bigr) \end{align*}$$

Its Jacobian matrix at $\bm{x} = (x_1, x_2, x_3)\in \R^3$ is

$$\begin{align*} J_f(\bm{x}) &= \begin{pmatrix} (\partial_1 f_1)(\bm{x}) & (\partial_2 f_1)(\bm{x}) & (\partial_3 f_1)(\bm{x})\\ (\partial_1 f_2)(\bm{x}) & (\partial_2 f_2)(\bm{x}) & (\partial_3 f_2)(\bm{x}) \end{pmatrix}\\ &= \begin{pmatrix} 1 & 42 & 0\\ 2 (x_1 + e^{x_3}) & 0 & 2 e^{x_3} (x_1 + e^{x_3}) \end{pmatrix} \in \Matrix_{2\times 3}(\R) \end{align*}$$

Multivariable chain rule

One variable

The well-known chain rule derived in Calculus I reads:

For functions $f: U\rightarrow V$ and $g: V \rightarrow W$, where $U,V,W\subseteq \R$, with $f$ being differentiable in $x\in U$ and $g$ being differentiable in $f(x)\in V$, the composite function $g(f(x))$ is differentiable and $\dv{x} \Bigl(g(f(x))\Bigr)$ is given by

$$\begin{align*} \dv{x} \Bigl(g(f(x))\Bigr) = g'(f(x)) \cdot f'(x) \end{align*}$$

Multiple variables

In multivariable calculus, the chain rule for one variable can be generalized to the following theorem (which we won’t prove here).

For $U\subopenin \R^n, V\subopenin \R^m$, let $f:U\rightarrow V$ be differentiable at point $\bm{x} \in U$. Let $g: V\rightarrow \R^k$ be differentiable at point $\bm{y} \coloneqq f(\bm{x}) \in V$. Visually, we are in the following situation:

The chain rule now states: $g\circ f: U\rightarrow \R^k$ is differentiable at $\bm{x}$ and

$$\boxed{J_{g\circ f}(\bm{x}) = J_g(f(\bm{x})) \cdot J_f(\bm{x}) \quad \in \Matrix_{k\times n}(\R)}$$

This is very similar to the case with one variable $x\in U\subseteq \R$ above, except now ($\bm{x}\in U\subseteq \R^n$) we make use of the Jacobian matrix and multiply matrices to account for the many variables $x_1, \dots, x_n$ that $\bm{x}$ consists of (thus multivariable chain rule).

Example

As an example, take the function $f: \R^3 \to \R^2$ from above: $$\begin{align*} f(x_1, x_2, x_3) \coloneqq \bigl(x_1 + 42 x_2,\quad (x_1 + e^{x_3})^2 \bigr) \end{align*}$$ and define $g: \R^2 \rightarrow \R,\, g(\bm{y}) \coloneqq y_1 \cdot y_2$, which puts us in this situation:

The Jacobian matrix for $g$ at $\bm{y} = (y_1, y_2) \in \R^2$ is: $$\begin{align*} J_g(\bm{y}) &= \begin{pmatrix} (\partial_1 g_1)(\bm{y}) & (\partial_2 g_1)(\bm{y}) \end{pmatrix} = \begin{pmatrix} y_2 & y_1 \end{pmatrix} \in \Matrix_{1\times 2}(\R) \end{align*}$$

and thus: $J_g(f(\bm{x})) = \begin{pmatrix} (x_1 + e^{x_3})^2 & x_1 + 42 x_2 \end{pmatrix}$.

With the chain rule, we obtain:

$$\begin{align*} &\underbrace{J_{g\circ f}(\bm{x})}_{\in \Matrix_{1\times 3}(\R)} = \underbrace{J_g(f(\bm{x}))}_{\in \Matrix_{1\times 2}(\R)} \cdot \underbrace{J_f(\bm{x})}_{\in \Matrix_{2\times 3}(\R)}\\ &= \begin{pmatrix} (x_1 + e^{x_3})^2 & x_1 + 42 x_2 \end{pmatrix} \cdot \begin{pmatrix} 1 & 42 & 0 \quad\\ 2 (x_1 + e^{x_3}) & 0 & 2 e^{x_3} (x_1 + e^{x_3}) \end{pmatrix}\\ &= \begin{pmatrix} (x_1 + e^{x_3})^2 + (x_1 + 42 x_2) \cdot 2(x_1 + e^{x_3})\phantom{-.}\\ 42 (x_1 + e^{x_3})^2\\ (x_1 + 42 x_2) \cdot 2 e^{x_3} (x_1 + e^{x_3})\\ \end{pmatrix}^T \end{align*}$$

Total derivative in physics

Putting everything together

Having recalled the Jacobian matrix as well as the multivariable chain rule, we can finally go back to our original function $f: \R^4 \rightarrow \R$ given by $$\begin{align*} f(\bm{x}(t), t) &= 2 \bigl(x_1(t)\bigr)^2 + \bigl(\ln(x_2(t))\bigr)^3 + 4 x_3(t) + \pi t,\\ &\quad x_1(t) = \sin(t), \quad x_2(t) = \cos(2t), \quad x_3(t) = \frac{1}{t} \end{align*}$$ The tricky part is to realize we can prefix $f$ by another function $\gamma$ which translates from the independent time variable $t$ to the $(n+1)$-dimensional input vector $\bigl(x_1(t), \dots, x_n(t), t\bigr)$ passed into $f$. Note that for $f$ in the introduction, we have $n=3$, yet we will leave $n$ generic here. The situation presents itself as follows in terms of a commutative diagram:

We define the function $$\begin{align*} \gamma:\, &I\rightarrow \R^{n+1},\\ &t\mapsto \gamma(t) \coloneqq \begin{pmatrix} x_1(t) & \cdots & x_n(t) & t \end{pmatrix} \end{align*}$$ where $I\subseteq \R$ is the interval of permitted time values. We often set $t \coloneqq 0$ for the start of a real measurement or thought experiment in physics and could therefore arbitrarily set $I \coloneqq [0, \infty)$. Furthermore, we demand every component of $\gamma$ to be continuous, i.e. $$\gamma_i: I \xrightarrow{\text{continuous}} \R \quad \forall i=1,\dots, n+1$$ With these properties, $\gamma$ is a curve and describes a trajectory. It just happens that as the last component of our $(n+1)$-dimensional space, $\gamma$ carries the time component itself, as we need to pass it into $f$ as well (as seen before, $f$ can directly depend on $t$, not just indirectly via $x_1(t)$ etc.).

The multivariable chain rule is now applicable to the composite function $f\circ \gamma$ (assuming differentiability of $f$ at the given point $\bm{x}(t)$ and differentiability of $\gamma$ at $t$). We obtain: $$\begin{align*} &\underbrace{J_{f\circ \gamma}(t)}_{\Matrix_{1\times 1}(\R) \cong \R} = \underbrace{J_f(\gamma(t))}_{\Matrix_{1\times (n+1)}(\R)} \cdot \underbrace{J_\gamma(t)}_{\Matrix_{(n+1) \times 1}(\R)}\\ &= \begin{pmatrix} \pdv{x_1} f(\gamma(t)) & \dots & \pdv{x_n} f(\gamma(t)) & \pdv{t} f (\gamma(t)) \end{pmatrix} \cdot \begin{pmatrix} \dv{t} x_1(t)\\ \vdots\\ \dv{t} x_n(t)\\ 1 \end{pmatrix}\\ &= \pdv{x_1} f(\gamma(t)) \cdot \dv{t} x_1(t) + \dots + \pdv{x_n} f(\gamma(t)) \cdot \dv{t} x_n(t) + \pdv{t} f(\gamma(t)) \cdot 1\\ &= \pdv{x_1} f(\underbrace{x_1(t), \dots, x_n(t)}_{\bm{x}(t)}, t) \cdot \dv{t} x_1(t) + \dots + \pdv{x_n} f(\bm{x}(t), t) \cdot \dv{t} x_n(t)\\ &\quad + \pdv{t} f(\bm{x}(t), t) \cdot 1\\ &= \pdv{\bm{x}} f(\bm{x}, t) \cdot \dv{t} \bm{x}(t) \:+\: \pdv{t} f(\bm{x}(t), t) \cdot 1\\ &\eqqcolon \dv{t} f(\bm{x}, t) \end{align*}$$

This shows how the Jacobian matrix, multivariable chain rule, and the total derivative are connected. We started with a function $f$ that had an explicit time dependency (in our case the term $\pi t$) and implicit time dependencies (since $x_1(t), \dots, x_n(t)$ are time-dependent). We then wanted to compute the total derivative of $f$ with respect to $t$, which is just asking for the time derivative of the composite function $f\circ \gamma$, where $\gamma$ is a curve passing in all the parameters to $f$, including time $t$ itself. This is in fact the definition of the total derivative of $f$ with respect to time $t$.

Note how the same formalism is applicable even when $f$ is not directly dependent on $t$. In this case, we still pass in $t$ as parameter to our function $f$, which is simply not using the variable at all. The term $\pdv{t} f(\bm{x}, t)$ will then evaluate to $0$ and we would get the same result as if we had just left out $t$ as last entry of $\gamma$ altogether. Therefore, our definition of $\gamma$ stays consistent.

More examples

Let’s take a look at some more examples. As is common in physics, we omit the parentheses when a function is only dependent on $t$, e.g. $x_1 = x_1(t)$ to not clutter the visual appearance. In addition, we also set $I \coloneqq \R$.

Example 1

One task might be to calculate the total time derivative of $$\begin{align*} &f\bigl(x_1, x_2, x_3, t\bigr) = 4 x_1^2 + 3 x_2^2 + \pi x_3 + 2t,\\ &x_1 = x_1(t) = \sin(t), \; x_2 = x_2(t) = \cos(t), \; x_3 = x_3(t) = \cosh(t) \end{align*}$$

With the chain rule and the common notation $\dv{t} x_1(t) = \dot{x}_1(t)$, we get: $$\begin{align*} &\dv{t} f\bigl(x_1, x_2, x_3, t\bigr)\\ &= \begin{pmatrix} \pdv{x_1} f(\gamma(t)) & \pdv{x_2} f(\gamma(t)) & \pdv{x_3} f(\gamma(t)) & \pdv{t} f(\gamma(t)) \end{pmatrix} \cdot \begin{pmatrix} \dot{x}_1\\ \dot{x}_2\\ \dot{x}_3\\ 1 \end{pmatrix}\\ &= \pdv{x_1} f(\gamma(t)) \cdot \dot{x}_1 + \pdv{x_2} f(\gamma(t)) \cdot \dot{x}_2 + \pdv{x_3} f(\gamma(t)) \cdot \dot{x}_3\\ &\quad + \pdv{t} f(\gamma(t)) \cdot 1\\ &= 8 x_1 \dot{x}_1 + 6x_2 \dot{x}_2 + \pi \dot{x}_3 + 2\\ &= 8 \sin(t) \cos(t) - 6 \cos(t) \sin(t) + \pi \sinh(t) + 2\\ &= \solution{2 \sin(t) \cos(t) + \pi \sinh(t) + 2}. \end{align*}$$

Example 2

For another physics problem, we might encounter the following function we want to calculate the total time derivative for: $$\begin{align*} f\bigl(x_1, \dot{x}_1, t\bigr) &= a \dot{x}_1^2 - b x_1 \dot{x}_1, \quad a,b\in \R, \, x_1=x_1(t), \, \dot{x}_1 = \dot{x}_1(t) \end{align*}$$

Again, employing the chain rule we obtain: $$\begin{align*} &\dv{t} f\bigl(x_1, \dot{x}_1, t\bigr) = \begin{pmatrix} \pdv{x_1} f(\gamma(t)) & \pdv{\dot{x}_1} f(\gamma(t)) & \pdv{t} f(\gamma(t)) \end{pmatrix} \cdot \begin{pmatrix} \dot{x}_1\\ \ddot{x}_1\\ 1 \end{pmatrix}\\ &= \pdv{x_1} f\bigl(x_1, \dot{x}_1, t\bigr) \cdot \dot{x}_1 + \pdv{\dot{x}_1} f\bigl(x_1, \dot{x}_1, t\bigr) \cdot \ddot{x}_1 + \pdv{t} f\bigl(x_1, \dot{x}_1, t\bigr) \cdot 1\\ &= -b \dot{x}_1 \dot{x}_1 + \bigl(2a\dot{x}_1 - b x_1\bigr) \ddot{x}_1 + 0\\ &= \solution{-b \dot{x}_1^2 + \bigl(2a\dot{x}_1 - b x_1\bigr) \ddot{x}_1} \end{align*}$$

Example 3

In another task, students were presented this function $$\begin{align*} f\bigl(x_1, \dot{x}_1, t\bigr) &= -a \cdot u(x_1) \cdot e^{-kt}, \quad a,k \in \R, \, x_1 = x_1(t),\, \dot{x}_1 = \dot{x}_1(t) \end{align*}$$

and were asked to calculate the total time derivative. However, $u(x_1(t))$ was never defined. As $f$ was not dependent on $u$ at all (see argument list of $f$), one had to assume $u$ would be replaced by some term containing $x_1$, e.g. $3 x_1 + 5$. Then, we can proceed as usual: $$\begin{align*} &\dv{t} f\bigl(x_1, \dot{x}_1, t\bigr)\\ &= \pdv{x_1} f\bigl(x_1, \dot{x}_1, t\bigr) \cdot \dot{x}_1 + \pdv{\dot{x}_1} f\bigl(x_1, \dot{x}_1, t\bigr) \cdot \ddot{x}_1 + \pdv{t} f\bigl(x_1, \dot{x}_1, t\bigr) \cdot 1\\ &= \solution{-a \cdot \dv{x}\Bigr\rvert_{x=x_1} u(x) \cdot e^{-kt} \cdot \dot{x}_1 + 0 + a k u(x_1) e^{-kt} \cdot 1} \end{align*}$$ where $\dv{x}\bigr\rvert_{x=x_1} u(x) = 3$ if $u(x) \coloneqq 3x + 5$ (as an example).

Thanks to Paul Obernolte for having reviewed this post.