The integral is independent from the road that unites A and B,there is a f...

[evinda]

Hey! ;)
I am looking at the proof of this theorem:
"Let $F$ be a vector field with components $M,N$ and $P$,$F=Mi + Nj+Pk$,$M,N$ and $P$ continuous at a region $D$.Then a necessary and sufficient condition,so that the integral $\int_A^BFdR$ is independent from the road that unites $A$ and $B$ at $D$ is that there is a differentiable function $f$,so that $F=\nabla{f}=if_x+jf_y+kf_z$ at $D$.
At this case,the value of the integral is given by:
$\int_A^BFdR=f(B)-f(A)$ "

For the one direction,the proof is the following:
We suppose that there is a differentiable function $f$,such that: $F=\nabla {f}=if_x+jf_y+kf_z$.
We see that $\frac{df}{dt}=f_x\frac{dx}{dt}+f_y\frac{dy}{dt}+f_z\frac{dz}{dt} \Rightarrow \frac{df}{dt}= \nabla {f} \frac{dR}{dt}$,where $R(t)=(x(t),y(t),z(t))$.
Since, $F=\nabla {f} \Rightarrow FdR=\nabla {f} dR \frac{dt}{dt}=df \Rightarrow \int_A^BFdR=f(B)-f(A)$

How can show the other direction??

 

[ThePerfectHacker]

Here is a suggestion. Pick a point $p$ in the region. Define the following function $f$ on your region. Given a point $x$, choose any path from $p$ to $x$, and define $f(x)$ to be the integral along that path of your vector field. Notice that $f(x)$ is well-defined because by assumption the vector field is path-independent. Then argue that $\nabla f$ recovers your vector-field.

 

[chisigma]

Hey! ;)
I am looking at the proof of this theorem:
"Let $F$ be a vector field with components $M,N$ and $P$,$F=Mi + Nj+Pk$,$M,N$ and $P$ continuous at a region $D$.Then a necessary and sufficient condition,so that the integral $\int_A^BFdR$ is independent from the road that unites $A$ and $B$ at $D$ is that there is a differentiable function $f$,so that $F=\nabla{f}=if_x+jf_y+kf_z$ at $D$.
At this case,the value of the integral is given by:
$\int_A^BFdR=f(B)-f(A)$ "

For the one direction,the proof is the following:
We suppose that there is a differentiable function $f$,such that: $F=\nabla {f}=if_x+jf_y+kf_z$.
We see that $\frac{df}{dt}=f_x\frac{dx}{dt}+f_y\frac{dy}{dt}+f_z\frac{dz}{dt} \Rightarrow \frac{df}{dt}= \nabla {f} \frac{dR}{dt}$,where $R(t)=(x(t),y(t),z(t))$.
Since, $F=\nabla {f} \Rightarrow FdR=\nabla {f} dR \frac{dt}{dt}=df \Rightarrow \int_A^BFdR=f(B)-f(A)$

How can show the other direction??

For semplicity let's suppose to have a function f (x,y) of two variables and define...

$\displaystyle M = \frac{\partial {f}}{\partial {x}}, N = \frac{\partial {f}}{\partial {y}}\ (1)$

If in a region D with contour C the following condition is verified... $\displaystyle \frac{\partial{M}}{\partial{y}} = \frac{\partial{N}}{\partial{x}}\ (2)$

... then is...

$\displaystyle \int_{C} (M\ d x + N\ d y) = 0\ (3)$

... so that the integral from A to B is independent to the path connecting A to B. All that is consequence of the Green's Theorem that extablishes that... $\displaystyle \int_{C} (M\ d x + N\ d y) = \int \int_{D} (\frac{\partial {N}}{\partial {x}} - \frac{\partial{M}}{\partial{y}})\ dA\ (4)$In the case of three [or more...] variables the details are a little more complex... Kind regards $\chi$ $\sigma$

 

[I like Serena]

For semplicity let's suppose to have a function f (x,y) of two variables and define...

$\displaystyle M = \frac{\partial {f}}{\partial {x}}, N = \frac{\partial {f}}{\partial {y}}\ (1)$

Hold on. This is not the other direction is it?
For the other direction this is what needs to be proven...

 

[chisigma]

For semplicity let's suppose to have a function f (x,y) of two variables and define...

$\displaystyle M = \frac{\partial {f}}{\partial {x}}, N = \frac{\partial {f}}{\partial {y}}\ (1)$

If in a region D with contour C the following condition is verified... $\displaystyle \frac{\partial{M}}{\partial{y}} = \frac{\partial{N}}{\partial{x}}\ (2)$

... then is...

$\displaystyle \int_{C} (M\ d x + N\ d y) = 0\ (3)$

... so that the integral from A to B is independent to the path connecting A to B. All that is consequence of the Green's Theorem that extablishes that... $\displaystyle \int_{C} (M\ d x + N\ d y) = \int \int_{D} (\frac{\partial {N}}{\partial {x}} - \frac{\partial{M}}{\partial{y}})\ dA\ (4)$In the case of three [or more...] variables the details are a little more complex...

I apologize for the fact that my answer has been incomplete and may be it produced some misundestanding. Remaining in the case of a two variables function, what I intended to say is that, if we start from an f(x,y) which is twice differentiable in a region D with a contour C, then, setting... $\displaystyle F = \nabla f = i\ f_{x} + j\ f_{y} = i\ M + j\ N\ (1)$

... the condition for which the integral $\displaystyle \int_{A}^{B} F\ dR$ doesn't depend from the path connecting A to B is that the following condition, consequence of the Green's Theorem, holds... $\displaystyle \frac{\partial {M}}{\partial {y}} = \frac {\partial {N}}{\partial {x}}\ (2)$

But (2) is verified only if $\displaystyle f_{x y} = f_{y x}$ and that is true only if the Schwartz Theorem Hypotheses are respected and in not true for all twice differentiable f(x,y). For more details see... http://www2.math.technion.ac.il/~mcwikel/h2m/SchwarzFxyFyx.pdfKind regards $\chi$ $\sigma$

 

[I like Serena]

If the integral is independent of the way then we can define
$$\phi(x,y,z) = \int_{(x_0,y_0, z_0)}^{(x,y,z)} \mathbf F \cdot d\mathbf R$$
where the integral is along any curve from a fixed point $(x_0,y_0, z_0)$ to $(x,y,z)$.
This is well-defined since it does not depend on which path we integrate along.
In particular we can pick a curve going from $(x_0,y_0, z_0)$ to $(a,y, z)$ and then along a straight line segment from $(a,y, z)$ to $(x,y,z)$:
$$\phi(x,y,z)
= \int_{(x_0,y_0, z_0)}^{(a,y,z)} \mathbf F \cdot d\mathbf R + \int_{(a,y, z)}^{(x,y,z)} \mathbf F \cdot d\mathbf R
= \int_{(x_0,y_0, z_0)}^{(a,y,z)} \mathbf F \cdot d\mathbf R + \int_a^x M(t,y,z) dt$$

The first integral is independent of x, so
$$\frac{\partial \phi}{\partial x}(x,y,z) = \frac{\partial}{\partial x}\int_a^x M(t,y,z) dt = M(x,y,z)$$

Similarly we can proof that \(\displaystyle \frac{\partial \phi}{\partial y}(x,y,z)=N\) and \(\displaystyle \frac{\partial \phi}{\partial z}(x,y,z)=P\).

In other words, if $\int_A^B \mathbf F \cdot d\mathbf R$ is independent of the path taken, then $\phi$ is the required differentiable function $f$.$\qquad \blacksquare$