\(\def\pd{\partial}\def\Real{\mathbb{R}}

\)

- Let \(\Omega \subset \mathbb{R}^n\) be some Euclidean domain, and consider

a curve \( \gamma:\underbrace{[0,1]}_{I} \to \Omega \) given by

$$\gamma(t) = \begin{pmatrix} \gamma_1(t), & \ldots &, \gamma_n(t) \end{pmatrix}. $$Given a real-valued multivariate function \(f:\Omega\to \mathbb{R}\) we’ll try to reconstruct the Newton-Leibnitz formula.

Suppose that we have a partition

\(0=t_0 < t_1 <\ldots < t_K=1\), then

$$ f(\gamma(1)) – f(\gamma(0)) = \sum_{k=0}^K \big(f(\gamma(t_{k+1})-f(\gamma(t_{k})\big) $$

If \(f\) is continuously differentiable, then using the first Taylor approximation we get

$$ f(\gamma(t_{k+1})) – f(\gamma(t_k)) = \Delta_k\cdot \frac{d}{dt} f(\gamma(t))\Big|_{t=t_k} +

\text{o}(\Delta_k) ~~~~~~(**)$$

here \(\Delta_k = t_{k+1}-t_k\).

The composed derivative is given by the so-called total differential formula we know from calculus:

$$ \frac{d}{dt} f(\gamma(t)) = \sum_{l=1}^n \frac{\partial f}{\partial x_l} \frac{\partial \gamma_l}{\partial t} $$Of course, this is a special case of the composition of two differentiable functions

\( F:R^k\to R^l,~~~ G:R^l \to R^m\). The Jacobian matrix of the composed function is

the product of their respective Jacobians:

$$ J_{F\circ G} = J_F\cdot J_G. $$ - If we have the row vector

$$ \frac{\pd f}{\pd x}(\gamma):=\begin{pmatrix} \frac{\partial f(\gamma)}{\partial x_1},&\ldots&,\frac{\partial f(\gamma)}{\partial x_n} \end{pmatrix} $$

and the differential approximation \((**)\):

$$ f(\gamma(1))-f(\gamma(0)) \approx \sum_{k} \left(\frac{\pd f}{\pd x}\cdot \dot \gamma\right)\Big|_{t_k} \Delta_k$$

then in the limit we obtain

$$ f(\gamma(1))-f(\gamma(0)) = \int_{0}^1 \frac{df}{dt}\cdot \dot \gamma dt

= \int_{0}^1 df\cdot \dot \gamma $$

This integral holds for any (differentiable) re-parametrization of \(\gamma\) that

maintains the same trajectory. **Example:**let \(f(x,y) = x^2-y^2 \), and let \(\gamma(t) = (2t,t)\).

Clearly, \(f(\gamma(1)) -f(\gamma(0)) = 2^2-1 = 3 \). We will verify this by integration.

The differential of \(f\) is:

$$ df = 2x dx + 2y dy $$

and the tangent is:

$$ \dot \gamma(t) = 2 \partial_x + \partial_y $$

(where we use the fancy notation \(\partial_x,\partial_y\) for the standard basis in \(\Real^2\), and \(dx,dy\) for the standard dual basis.Hence

$$ \int_0^1 (df \cdot \dot \gamma) =

\int_0^1 \begin{pmatrix} 2x & 2y \end{pmatrix}

\begin{pmatrix} 2 \\1 \end{pmatrix} =

= \int_0^1 (4x(t)-2y(t)) dt = \int_0^1 8t-2t dt = 6 \int_0^1 t dt = 3 $$**Differential 1-forms**If we drop the requirements that the components of the covector (linear functional)

coupled with \(\dot\gamma\) are partial derivatives of a function, we arrive at the*notion of 1-form*

$$

\omega = \sum a_k(x) dx_k,

$$

with coefficients depending on \(x\). We can integrate a 1-form along a curve \(\gamma\) using exactly the same definition as above:

\[

\int_\gamma \omega:=\int_0^1 \sum a_k(\gamma(t))\frac{\partial \gamma_l}{\partial t}dt.

\]It is easy to check that this integral of 1-form along the curve does not change if we reparameterize the curve (that is if we represent \(t=t(s)\) as a function of another variable \(s\in I\)).**Differential 1-forms:**given the path integral

$$ \int_\gamma df = \int_\gamma \sum a_k(x) dx_k $$

a*differential 1-form*(at point \(x\)) is the expression

$$ \omega = \sum a_k(x) dx_k. $$

The components \(a_k(x)\) are continuous (or smooth) functions, \(dx_k\) are basis vectors

of linear functionals on tangent vectors. This is an arbitrary form, not necessarily

originated in a differential of a function.**Integration**over the 1-form \(\omega\) along a path \(\gamma\) is defined as

$$ \int_\gamma \omega = \int_0^1 \sum_k a_k(\gamma(t)) \frac{\partial \gamma_k}{\partial t} dt $$**Example:**let \(\omega = xdx + ydy\), and let \(\gamma(t):=(x_* t, y_* t) \).

Then

$$ \int_\gamma \omega

= (x_*t,y_*t)\begin{pmatrix} x_* \\ y_* \end{pmatrix} dt

= (x_*^2+y_*^2) \int_0^1 t dt = \frac{1}{2}(x_*^2+y_*^2) $$For another path \(\gamma’\) (not to be confused with derivative prime), defined by

$$ \gamma'(t) = \begin{cases} (2x_*t,0) & t\leq \frac{1}{2} \\ (2x_*,2y_*(t-\frac{1}{2})) & t\geq \frac{1}{2}

\end{cases} $$

the curve would have

$$ \int_\gamma \omega = \int_0^{0.5} …. + \int_{0.5}^1 … $$

but the end result will not change whatsoever.**Example:**integrate the form \(\omega = y dx – xdy \), along the curve

$$ \gamma(t) = (x_*t, y_*t) $$

$$ \int_\gamma \omega = \int (y,-x) \begin{pmatrix} x_* \\ y_* \end{pmatrix} dt = 0 $$

Now, do the same along the arc

$$ \gamma_a(t) = (\cos(\phi t),~~\sin (\phi t)) $$

we get

$$ \int_{\gamma_a} (\sin(\phi t), ~\cos(\phi t) \begin{pmatrix} -\sin(\phi t) \\ \cos(\phi t) \end{pmatrix} dt

= \phi \int_0^1 (-\sin^2(\phi t) – \cos^2 (\phi t)) dt = -\phi $$

This is an example of a form whose path integral depends on the integration path.**Conserving/exact forms**

consider a differential form, \( \sum a_k(x) dx_k, \) and its integral along some curve.If that integral depends only on the end points of the curve, we call the

the form*conservative*or*exact*.

Incidentally, a form \(\omega\) is a exact iff \(\omega = df\) for some \(f\).

We can explicitly construct the function \(f\) by

$$ f(x):=\int_\gamma \omega $$- A necessary condition for a form \(\omega = a_1(x) dx_1 + a_2(x) dx_2 \)

to be exact is that

$$ \frac{\partial a_1}{\partial x_2} = \frac{\partial a_2}{\partial x_1}. $$

This condition is not sufficient, for example the form

$$ \omega = \frac{y dx }{x^2+y^2} – \frac{x dy}{x^2+y^2} $$

is not exact, but satisfies the necessary condition nonetheless, expect for at \(x=y=0\) –

which is a singular point (we will learn more about that in the chapter about complex analysis). **Exercise**: Integrate \(dx/y\) over the circle (oriented counterclockwise) \(\{x^2+y^2=R^2\}\).

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