Let
\[\mathbf{F}\]
defined in a polygon ally connected open set \[B \subset \mathbb{R}^n\]
.If the line integral
\[\int_C \mathbf{F} \cdot d \mathbf{r}\]
is independent of the path in \[B\]
from \[x_0\]
to \[x\]
then the function \[f(x) = \int_{\mathbf{x_0}}^{\mathbf{x}} \mathbf{F} \cdot d \mathbf{x}\]
is continuously differentiable and throughout \[B\]
the relationship \[\mathbf{F}= grad \: f\]
is satisfied.Proof
Since
\[x \in B \subset \mathbb{R}^n\]
and \[B\]
us open there exists an open ball \[B(x, \delta ) \subset B\]
of radius \[\delta\]
with centre \[x\]
For all real numbers
\[ | \alpha | < \delta\]
and unit vector \[\mathbf{u}\]
, \[x+ \alpha \mathbf{u} \in B\]
The line integral
\[\int_C \mathbf{F} \cdot d \mathbf{x}\]
is independent of the path so we can choose any piecewise smooth curve in \[B\]
from \[x_0\]
to \[x\]
and extend it to \[\mathbf{x} + \alpha \mathbf{u}\]
by a linear segment.\[\begin{equation} \begin{aligned} f(\mathbf{x} + \alpha \mathbf{u} )- f(\mathbf{x}) &= \int^{\mathbf{x} + \alpha \mathbf{u}}_{\mathbf{x_0}} \mathbf{F} \cdot d \mathbf{x} - \int^{\mathbf{x}}_{\mathbf{x_0}} \mathbf{F} \cdot d \mathbf{x} \\ &= \int^{\mathbf{x} + \alpha \mathbf{u}}_{\mathbf{x}} \mathbf{F} \cdot d \mathbf{x} \\ &= \int^{\alpha}_{0} \mathbf{F}( \mathbf{x}+ \beta \mathbf{u} ) \cdot \mathbf{u} d \beta \end{aligned} \end{equation}\]
If
\[\mathbf{u}= \mathbf{e_k}\]
is a base vector in \[\mathbb{R}^n\]
then\[\begin{equation} \begin{aligned} \frac{\partial f}{\partial x_k} (\mathbf{x}) &= lim_{\alpha \rightarrow 0} \frac{f(\mathbf{x} + \alpha \mathbf{u}) - f (\mathbf{x})}{\alpha} \\ &= lim_{\alpha \rightarrow 0} \frac{1}{\alpha} \int^{\alpha}_0 \mathbf{F} ( \mathbf{x} + \beta \mathbf{e_k} ) \cdot \mathbf{e_k} d \beta \\ &= \frac{d}{d \alpha} \int^{\alpha}_0 \mathbf{F} ( \mathbf{x} + \beta \mathbf{e_k} ) \cdot \mathbf{e_k} d \beta |_{\alpha =0} \\ &= \mathbf{F}( \mathbf{x}) \cdot \mathbf{e_k }) \\ &= F_k ( \mathbf{x}) \end{aligned} \end{equation}\]
Since
\[\mathbf{F}\]
is differentiable, the partial derivatives are continuous,
Hence \[\mathbf{F} = grad \: f \]