Conservative Vector Fields

Definition: Gradient Field

Let $\mathbf{v}:\mathcal{D}\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ be a real vector field.

We say that $\mathbf{v}$ is conservative if there exists a differentiable real scalar field $f:\mathcal{D}\subseteq {\mathbb{R}}^n\to \mathbb{R}$ whose gradient is $\mathbf{v}$.

$$\nabla f=\mathbf{v}$$

NOTE

Conservative vector fields are also known as gradient fields.

Gradient Theorem (Fundamental Theorem of Analysis for Line Integrals)

If $\mathbf{v}:\mathcal{D}\to {\mathbb{R}}^n$ is the gradient field of a real scalar field $f:\mathcal{D}\to \mathbb{R}$, then its line integral over any parametric curve $\gamma :[a;b]\to \mathcal{D}$ which is piecewise continuously differentiable on $(a;b)$ is given by

$${\int }_{\gamma }\mathbf{v}\cdot \mathrm{ds}=f(\gamma (b))-f(\gamma (a))$$

PROOF

By definition

$${\int }_{\gamma }\mathbf{v}\cdot \mathrm{ds}={\int }_a^b\mathbf{v}(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt}={\int }_a^b\nabla f(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt}$$

The chain rule for scalar fields tells us that the integrand is the derivative of $f(\gamma (t))$:

$$\frac{\mathrm{d}}{\mathrm{d}t}f(\gamma (t))=\nabla f(\gamma (t))\cdot {\gamma }^{\prime }(t)$$ $${\int }_a^b\nabla f(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt}={\int }_a^b\frac{\mathrm{d}}{\mathrm{d}t}f(\gamma (t))\mathrm{dt}$$

The fundamental theorem of real analysis then gives us

$${\int }_a^b\frac{\mathrm{d}}{\mathrm{d}t}f(\gamma (t))\mathrm{dt}=f(\gamma (t)){|}_a^b=f(\gamma (b))-f(\gamma (a))$$

Theorem: Path Independence of Line Integrals of Conservative Vector Fields

A continuous real vector field $\mathbf{v}:\mathcal{D}\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ is conservative if and only if its line integrals are equal over all injective, continuously differentiable parametric curves with the same endpoints.

$${\int }_{{\gamma }_1}\mathbf{v}\cdot \mathrm{ds}={\int }_{{\gamma }_2}\mathbf{v}\cdot \mathrm{ds}$$

PROOF

TODO

Theorem: Line Integrals of Conservative Vector Fields over Closed Curves

A continuous real vector field $\mathbf{v}:{\mathcal{D}}_{\mathbf{v}}\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ is conservative if and only if its line integral over every closed parametric curve $\gamma :{\mathcal{D}}_{\gamma }=[a;b]\subseteq \mathbb{R}\to {\mathcal{D}}_{\mathbf{v}}$ which is continuously differentiable on $(a;b)$ is zero.

$${\oint }_{\gamma }\mathbf{v}\cdot \mathrm{ds}=0$$

PROOF

We need to prove two things:

• (1) If $\mathbf{v}$ is conservative, then the line integral ${\oint }_{\gamma }\mathbf{v}\cdot \mathrm{ds}$ is zero for every closed parametric curve $\gamma :[a;b]\subseteq \mathbb{R}\to {\mathcal{D}}_{\mathbf{v}}$ which is continuously differentiable on $(a;b)$.
• (2) If the line integral ${\oint }_{\gamma }\mathbf{v}\cdot \mathrm{ds}$ is zero for every closed parametric curve $\gamma :[a;b]\subseteq \mathbb{R}\to {\mathcal{D}}_{\mathbf{v}}$ which is continuously differentiable on $(a;b)$, then $\mathbf{v}$ is conservative.

Proof of (1):

Since $\mathbf{v}$ is conservative, there exists a real scalar field $f:{\mathcal{D}}_{\mathbf{v}}\to \mathbb{R}$ such that $\mathbf{v}=\nabla f$. Substitute this into the definition of the line integral:

$${\oint }_{\gamma }\mathbf{v}\cdot \mathrm{ds}={\int }_a^b\mathbf{v}(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt}={\int }_a^b\nabla f(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt}$$

With the help of the chain rule, we notice that the integrand is the derivative of $f(\gamma (t))$:

$$(f\circ \gamma {)}^{\prime }(t)=\nabla f(\gamma (t))\cdot {\gamma }^{\prime }(t)$$

The integral thus becomes

$${\int }_a^b\nabla f(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt}={\int }_a^b(f\circ \gamma {)}^{\prime }(t)\mathrm{dt}$$

and using the fundamental theorem of analysis we obtain

$${\int }_a^b\nabla f(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt}={\int }_a^b(f\circ \gamma {)}^{\prime }(t)\mathrm{dt}=f(\gamma (b))-f(\gamma (a)).$$

Since $\gamma$ is closed, we know that $\gamma (b)=\gamma (a)$ and so $f(\gamma (b))-f(\gamma (a))=0$.

Proof of (2): TODO