Conservative Vector Fields#

Definition: Conservative Vector Field

Let $U \subseteq \mathbb{R}^n$ be open.

A real vector field $f: U \to \mathbb{R}^n$ is conservative if there exists a totally differentiable real scalar field $F: U \to \mathbb{R}$ whose gradient is $f$:

\[f(\boldsymbol{x}) = \nabla F(\boldsymbol{x}) \qquad \boldsymbol{x} \in U\]

We also say that $f$ is a gradient field or a potential field.

Definition: Potential Function

Let $U \subseteq \mathbb{R}^n$ be open and let $f: U \to \mathbb{R}^n$ be a conservative real vector field.

A potential function of $f$ is any real scalar field $F: U \to \mathbb{R}$ whose negative gradient is $f$:

\[f(\boldsymbol{x}) = \nabla F(\boldsymbol{x}) \qquad \boldsymbol{x} \in U\]

Note

In pure mathematical contexts, potential functions are almost universally defined in the above way. However, in physics and applied mathematics, a minus sign is commonly used: $-\nabla F$.

Algorithm: Finding Potential Function

Let $U \subseteq \mathbb{R}^n$ be open and let $f: U \to \mathbb{R}^n$ be a conservative real vector field with component functions $f_1, \dotsc, f_n$. We want to find a potential function $F: U \to \mathbb{R}$:

\[f(\boldsymbol{x}) = \nabla F(\boldsymbol{x}) \qquad \boldsymbol{x} \in U\]

Set $\partial_1 F = f_1$ and antidifferentiate w.r.t. the first variable:

$$F(x_1, \dotsc, x_n) = \int f_1 (x_1, \dotsc, x_n) \,\mathrm{d}x_1 + C_1(x_2, \dotsc, x_n)$$

For $i \in \{2, \dotsc, n\}$:

Differentiate the expression for $F$ obtained from the previous step w.r.t. $x_i$ and set it equal to $f_i$.
Solve for $\partial_i C_{i-1}$ and antidifferentiate w.r.t. $x_i$ to find

$$C_{i-1}(x_i, \dotsc, x_n) = \int \partial_i C_{i-1}(x_i, \dotsc, x_n) \,\mathrm{d}x_i + C_i(x_{i+1}, \dotsc, x_n)$$

Example

Consider the real vector field $f: \mathbb{R}^3 \to \mathbb{R}^3$ defined as follows:

\[f(x, y, z) = \begin{bmatrix}\mathrm{e}^x y + 1 \\ \mathrm{e}^x + z \\ y\end{bmatrix}\]

It is conservative because its curl is zero and $U$ is simply connected:

\[\operatorname{curl} f(x, y, z) = \begin{bmatrix} \partial_y f_3(x,y,z) - \partial_z f_2(x,y,z) \\ \partial_z f_1(x,y,z) - \partial_x f_3(x,y,z) \\ \partial_x f_2(x,y,z) - \partial_y f_1(x,y,z)\end{bmatrix} = \begin{bmatrix} 1 - 1 \\ 0 - 0 \\ \mathrm{e}^x - \mathrm{e}^x\end{bmatrix} = \boldsymbol{0}\]

We set $\partial_1 F = f_1$ and antidifferentiate w.r.t. $x$:

\[\begin{aligned}F(x, y, z) & = \int f_1(x, y, z) \, \mathrm{d}x \\ & = \int \mathrm{e}^x y + 1 \, \mathrm{d}x \\ & = \mathrm{e}^x y + x + C_1(y, z)\end{aligned}\]

We set $\partial_2 F = f_2$:

\[\partial_{y} F(x,y,z) = \partial_y (\mathrm{e}^x y + x + C_1(y, z)) = \mathrm{e}^x + \partial_y C_1(y, z) = \mathrm{e}^x + z\]

From this, we get

\[\partial_y C_1(y,z) = z\]

and by antidifferentiating w.r.t. $y$, we get:

\[C_1(y,z) = yz + C_2(z)\]

Therefore:

\[F(x,y,z) = \mathrm{e}^x y + x + yz + C_2(z)\]

We set $\partial_3 F = f_3$:

\[\partial_{z} F(x,y,z) = \partial_z (\mathrm{e}^x y + x + yz + C_2(z)) = y + \partial_z C_2(z) = y\]

From this, we get

\[\partial_z C_2(z) = 0\]

and by antidifferentiating w.r.t. $z$, we get:

\[C_2(z) = C_3 \in \mathbb{R}\]

Therefore:

\[F(x,y,z) = \mathrm{e}^x y + x + yz + C_3, \qquad C_3 \in \mathbb{R}\]

Warning

We must ensure that $f$ is conservative before starting the algorithm. Otherwise, the outcome can be unpredictable: we might reach a contradiction and no longer be able to proceed or we might get a result but it won't be a potential function.

The Gradient Theorem (Fundamental Theorem of Calculus for Line Integrals I)

Let $U \subseteq \mathbb{R}^n$ be open, let $f: U \to \mathbb{R}^n$ be a real vector field and let $\gamma: [a,b] \subseteq \mathbb{R} \to \mathbb{R}^n$ be a continuous parametric curve with $\gamma([a,b]) \subseteq U$ which is differentiable on $(a,b)$.

If $f$ is conservative and its line integral over $\gamma$ exists, then this integral is given by

\[\int_{\gamma} f \cdot \mathrm{d}\boldsymbol{r} = F(\gamma(b)) - F(\gamma(a)),\]

where $F$ is any potential function of $f$ (meaning $\nabla F = f$).

Proof

TODO

Fundamental Theorem of Calculus for Line Integrals II

Let $U \subseteq \mathbb{R}^n$ be open and path-connected, let $f: U \to \mathbb{R}^n$ be a real vector field and let $\boldsymbol{p} \in U$.

If $f$ is continuous and conservative, then the real scalar field $F: U \to \mathbb{R}$ defined by the line integral

\[F(\boldsymbol{x}) = \int_{\gamma_{\boldsymbol{x}}} f \cdot \mathrm{d}\boldsymbol{r},\]

where $\gamma_{\boldsymbol{x}}: [a,b] \to U$ is any piecewise continuously differentiable parametric curve with $\gamma_{\boldsymbol{x}}(a) = \boldsymbol{p}$ and $\gamma_{\boldsymbol{x}}(b) = \boldsymbol{x}$, is a potential function of $f$ (meaning $\nabla F = f$).

Example

Let $U = \mathbb{R}^2$ and consider the real vector field $f: U \to \mathbb{R}^2$ defined as follows:

\[f(x, y) = \begin{bmatrix}y + 1 \\ x + 4y\end{bmatrix}\]

We see that $f$ is continuously differentiable on $U$ with the following Jacobian matrix:

\[J_f(x,y) = \begin{bmatrix}0 & 1 \\ 1 & 4\end{bmatrix}\]

Since $U$ is simply connected and since $J_f(x,y)$ is symmetric, we know that $f$ is conservative.

Let $\boldsymbol{p} = \begin{bmatrix} 0 & 0 \end{bmatrix}^{\mathsf{T}}$. Define the real scalar field $F: U \to \mathbb{R}$ via the line integral

\[F(x,y) = \int_{\gamma_{(x,y)}} f \cdot \mathrm{d}\boldsymbol{r},\]

where $\gamma_{(x,y)}: [0,1] \to \mathbb{R}^2$ is the parametric curve defined as follows:

\[\gamma_{(x,y)}(t) = \begin{bmatrix}t x \\ t y\end{bmatrix}\]

We see that $\gamma_{(x,y)}(0) = \begin{bmatrix} 0 & 0\end{bmatrix}^{\mathsf{T}} = \boldsymbol{p}$ and $\gamma_{(x,y)}(1) = \begin{bmatrix}x & y\end{bmatrix}^{\mathsf{T}}$. Since $\gamma$ is also continuously differentiable, we know that $F(x,y)$ is a potential function of $f$. We can also find it explicitly:

\[\begin{aligned}F(x,y) & = \int_{\gamma_{(x,y)}} f \cdot \mathrm{d}\boldsymbol{r} \\ & = \int_0^1 f(\gamma_{(x,y)}(t)) \cdot \gamma_{(x,y)}'(t) \,\mathrm{d}t \\ & = \int_0^1 \begin{bmatrix}ty + 1 \\ tx + 4ty\end{bmatrix} \cdot \begin{bmatrix}x \\ y\end{bmatrix} \,\mathrm{d}t \\ & = \int_0^1 (ty + 1)x + (tx + 4ty)y \,\mathrm{d}t \\ & = \int_0^1 tyx + x + txy + 4ty^2 \,\mathrm{d}t \\ & = \int_0^1 t(2yx + 4y^2) + x \,\mathrm{d}t \\ & = \left.\left( \frac{t^2}{2}(2xy + 4y^2) + tx \right)\right\vert_0^1 \\ & = \frac{1}{2}(2xy + 4y^2) + x \\ & = xy + 2y^2 + x\end{aligned}\]

To verify, let's find the gradient of $F$:

\[\nabla F(x,y) = \begin{bmatrix}y + 1 \\ x + 4y\end{bmatrix} = f(x,y)\]

Proof

TODO

Theorem: Path Independence of Line Integrals of Conservative Vector Fields

Let $U \subseteq \mathbb{R}^n$ be open and path-connected.

A continuous real vector field $f: U \subseteq \mathbb{R}^n \to \mathbb{R}^n$ is conservative if and only if the line integrals of $f$ over all piecewise continuously differentiable parametric curves $\gamma: [a,b] \subset \mathbb{R} \to U$ and $\varphi: [c,d] \subset \mathbb{R} \to U$ with $\gamma(a) = \varphi(c)$ and $\gamma(b) = \varphi(d)$ are equal:

\[\int_{\gamma} f \cdot \mathrm{d}\boldsymbol{r} = \int_{\varphi} f \cdot \mathrm{d}\boldsymbol{r}\]

Proof

TODO

Theorem: Line Integrals of Conservative Vector Fields over Closed Curves

Let $U \subseteq \mathbb{R}^n$ be open and path-connected.

A continuous real vector field $f: U \subseteq \mathbb{R}^n \to \mathbb{R}^n$ is conservative if and only if the line integrals of $f$ over all piecewise continuously differentiable closed parametric curves $\gamma: [a,b] \subset \mathbb{R} \to U$ are zero:

\[\oint_{\gamma} f \cdot \mathrm{d}\boldsymbol{r} = 0\]

Proof

TODO

Theorem: Gradient Field via Jacobian Matrix

Let $U \subseteq \mathbb{R}^n$ be open and simply connected.

A continuously differentiable real vector field $f: U \to \mathbb{R}^n$ is conservative if and only if its Jacobian matrix is symmetric everywhere:

\[J_f(\boldsymbol{x}) = J_f(\boldsymbol{x})^{\mathsf{T}} \qquad \forall \boldsymbol{x} \in U\]

Example

Let $U = \mathbb{R}^2 \setminus \{\boldsymbol{0}\}$ and consider the real vector field $f: U \to \mathbb{R}^2$ defined as follows:

\[f(x, y) = \frac{1}{x^2 + y^2}\begin{bmatrix} -y \\ x \end{bmatrix}\]

It is totally differentiable on $U$ with the following Jacobian matrix:

\[J_f(x, y) = \frac{1}{(x^2 + y^2)^2}\begin{bmatrix}2xy & y^2 - x^2 \\ y^2 - x^2 & -2xy\end{bmatrix}\]

It is obvious that $J_f(x,y)$ is symmetric for all $(x,y) \in U$. However, $f$ is not conservative.

Consider the closed parametric curve $\gamma: [0, 2\uppi] \to \mathbb{R}^2$ defined as follows:

\[\gamma(t) = \begin{bmatrix} \cos t \\ \sin t \end{bmatrix}\]

The line integral of $f$ over $\gamma$ is the following:

\[\begin{aligned}\oint_{\gamma} f \cdot \mathrm{d}\boldsymbol{r} & = \int_{0}^{2\uppi} f(\gamma(t)) \cdot \gamma'(t) \,\mathrm{d}t \\ & = \int_{0}^{2\uppi} \begin{bmatrix} -\sin t \\ \cos t \end{bmatrix} \cdot \begin{bmatrix} -\sin t \\ \cos t \end{bmatrix} \,\mathrm{d}t \\ & = \int_{0}^{2\uppi} (\sin^2 t + \cos^2 t) \,\mathrm{d}t \\ & = \int_{0}^{2\uppi} 1 \,\mathrm{d}t \\ & = 2\uppi\end{aligned}\]

Since this line integral is not zero, $f$ cannot be conservative.

The reason that the Jacobian matrix misled us is that $U$ is not simply connected and so the theorem cannot be applied.

Example

Let $U = \mathbb{R}^2 \setminus \{\boldsymbol{0}\}$ and consider the real vector field $f: U \to \mathbb{R}^2$ defined as follows:

\[f(x, y) = \frac{1}{x^2 + y^2}\begin{bmatrix} -y \\ x \end{bmatrix}\]

It is totally differentiable on $U$ with the following Jacobian matrix:

\[J_f(x, y) = \frac{1}{(x^2 + y^2)^2}\begin{bmatrix}2xy & y^2 - x^2 \\ y^2 - x^2 & -2xy\end{bmatrix}\]

It is obvious that $J_f(x,y)$ is symmetric for all $(x,y) \in U$.

Consider the closed parametric curve $\gamma: [0, 2\uppi] \to \mathbb{R}^2$ defined as follows:

\[\gamma(t) = \begin{bmatrix} 3 + \cos t \\ 4 + \sin t \end{bmatrix}\]

While $U$ itself is not simply connected, $\gamma([0, 2\uppi])$ is contained in the simply connected open ball $B_2(\begin{bmatrix}3, 4\end{bmatrix}^{\mathsf{T}})$. Therefore, the restriction of $f$ on $B_2(\begin{bmatrix}3, 4\end{bmatrix}^{\mathsf{T}})$ is conservative and we have the line integral of $f$ over $\gamma$ must be zero.

\[\oint_{\gamma} f \cdot \mathrm{d}\boldsymbol{r} = 0\]

Proof

TODO

Theorem: Gradient Field via Curl

Let $U \subseteq \mathbb{R}^3$ be open and simply connected.

A continuously differentiable real vector field $f: U \to \mathbb{R}^3$ is conservative if and only if its curl is zero everywhere:

\[\operatorname{curl} f(\boldsymbol{x}) = \boldsymbol{0} \qquad \forall \boldsymbol{x} \in U\]

Proof

TODO