rakuenso Posted May 19, 2009 Share Posted May 19, 2009 (edited) 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 Link to comment Share on other sites More sharing options...

ajb Posted May 19, 2009 Share Posted May 19, 2009 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. Link to comment Share on other sites More sharing options...

rakuenso Posted May 19, 2009 Author Share Posted May 19, 2009 (edited) 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 Link to comment Share on other sites More sharing options...

ajb Posted May 19, 2009 Share Posted May 19, 2009 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. Link to comment Share on other sites More sharing options...

rakuenso Posted May 19, 2009 Author Share Posted May 19, 2009 (edited) 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 Link to comment Share on other sites More sharing options...

ajb Posted May 19, 2009 Share Posted May 19, 2009 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. Link to comment Share on other sites More sharing options...

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