How to create a path defined by the gradient vector?

[rakuenso]

Suppose I have a function [math]f(x,y)[/math], along with a gradient at some arbitrary point [math](x_0, y_0)[/math]

 

The surface is smooth and [math]C^1[/math], there thus also exists some gradient vector:

 

[math]<\frac{df}{dx}(x_0, y_0) , \frac{df}{dy}(x_0, y_0)>[/math] I want to "follow" this vector to the next point [math](x_1, y_1)[/math], here there is another gradient vector [math]<\frac{df}{dx}(x_1, y_1) , \frac{df}{dy}(x_1, y_1)>[/math], I repeat this process for up to [math](x_n, y_n)[/math].

 

How do I write an expression for the curve? It seems like I would have to integrate along a vector somehow?

Edited May 19, 2009 by rakuenso

[ajb]

You need to solve equations like

 

[math] \frac{d x^{A}}{dt} = X^{A}(x(t))[/math]

 

in a given chart for the curves. You need some initial condition [math]x^{A}(0) = x^{A}_{0}[/math] and then by the existence and uniqueness theorem of ODEs you are guaranteed a solution (at lest locally).

 

It maybe possible that the integral curve is only defined for some subset of [math]\mathbb{R}[/math]. You will have to be careful that your parameter [math]t[/math] does not exceed the interval.

[rakuenso]

Are [math]x^A[/math] my variables? like [math]x=x^1,y=x^2[/math] etc. And what erxactly is [math]X^A[/math]?

 

How do you arbitrarily parametrize [math]x^A[/math] wrt to t? Or is this [math]x^A(t)[/math] exactly what I'm trying to solve?

 

I also have difficulties seeing where the integration along a "vector" part comes in.

 

For two dimensions, so far I've written the [math]n[/math] th point [math](x_n, y_n)[/math] as the following, over some interval tiny interval [math]t[/math] (in the case of the unit vector it is equal to the magnitude of the vector)":

 

[math] x_n = x_0 + \frac{1}{t} [\frac{\partial f}{\partial x}(x_0,y_0) + \frac{\partial f}{\partial x}(x_1,y_1) + ... + \frac{\partial f}{\partial x}(x_n,y_n) ] [/math]

 

[math] y_n = y_0 + \frac{1}{t} [\frac{\partial f}{\partial y}(x_0,y_0) + \frac{\partial f}{\partial y}(x_1,y_1) + ... + \frac{\partial f}{\partial y}(x_n,y_n) ] [/math]

 

note that [math](x_1,y_1), (x_2,y_2) .. etc [/math] can be found using the above definition

 

The path then would be comprised of the points [math] (x_0,y_0), (x_1,y_1) ... (x_n, y_n) [/math], which when taken over the [math]\displaystyle\lim_{n\to\infty}\displaystyle\lim_{t\to\infty}[/math] becomes continuous and never ending. Surely, given some function [math] f(x,y) [/math] this path can be traced out using numerical methods, but a method of finding an analytic solution seems much more difficult.

Edited May 19, 2009 by rakuenso

[ajb]

By curve you are looking for [math]x^{A}(t)[/math].

 

I think you need to do some reading on vector fields, integral curves and flows. All fundamental things in differential geometry.

[rakuenso]

eh? I've done calc II and we went over quite a bit of vector calculus. But all the stuff we did dealt with flows, and integrals of dot products (which is just a scalar), and stuff like stoke's theorem. here it seems like I'm integrating an actual vector as opposed to a dot product. So the problem is analagous to finding the integral curves for a vector field (which here is the gradient) using ODEs?

 

nm I think I get what you mean now, if my vector field is [math]< \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} >[/math], say [math]V=\frac{1}{|x-y|} [/math] then I would form the autonomous system, as it is independent of time:

 

[math] \frac{dx}{dt} = \frac{\partial V}{\partial x} = \frac{-1}{|x-y|(x-y)} [/math]

 

[math] \frac{dy}{dt} = \frac{\partial V}{\partial y} = \frac{1}{|x-y|(x-y)} [/math]

 

so I would have to look for an integral curve with components (x(t), y(t)) that obeys the above equations

Edited May 19, 2009 by rakuenso

[ajb]

Yes. You will need some initial condition, which I think you have stated in your original post. You can appeal to known theorems of ODEs to help you if needed.