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I'm just trying to get my head around calculus of variations so I gave myself a simple example to work with:
Suppose I want to get between two points (a and b) through a mountain range with height described by the scalar field h(x,y) whilst staying as low as possible. I want to know what shaped path I should take.
So I start by saying I want to minimise:
[itex]\int h(x,y) ds = \int h(x,y) \sqrt{1 + \dot{y^2}} [/itex]
Which I can solve using the Euler–Lagrange equations.
Noting that h and ds are not explicitly dependent on [itex]\dot{y}[/itex] and y respectively I think the E-L equation can reduce to:
[itex]\frac{\dot{y}}{1 + \dot{y^2}} \frac{dh(x,y)}{dx} = \frac{dh(x,y)}{dy} [/itex]
Assuming some form of h(x,y), my difficulty is going from here to my path through the mountain range, y(x). I can solve the following quadratic in [itex]\dot{y}[/itex] to get [itex]\dot{y}(x,y)[/itex] but I'm unsure about the following integral to get to y(x). At first I thought I could ignore the y in the h(x,y) when I'm integrating with respect to x but since I'm after the path y(x) I can hardly say they are independent.
I just wanted to check that I've had the right idea up to this point and that I can't use any arguments to justify ignoring the y in h(x,y) when I've got ∫f(h(x,y))dx, meaning I'm stuck with a horrible implicit integration by parts.
Thanks for any help :-) Sorry my post is a little convoluted.
Suppose I want to get between two points (a and b) through a mountain range with height described by the scalar field h(x,y) whilst staying as low as possible. I want to know what shaped path I should take.
So I start by saying I want to minimise:
[itex]\int h(x,y) ds = \int h(x,y) \sqrt{1 + \dot{y^2}} [/itex]
Which I can solve using the Euler–Lagrange equations.
Noting that h and ds are not explicitly dependent on [itex]\dot{y}[/itex] and y respectively I think the E-L equation can reduce to:
[itex]\frac{\dot{y}}{1 + \dot{y^2}} \frac{dh(x,y)}{dx} = \frac{dh(x,y)}{dy} [/itex]
Assuming some form of h(x,y), my difficulty is going from here to my path through the mountain range, y(x). I can solve the following quadratic in [itex]\dot{y}[/itex] to get [itex]\dot{y}(x,y)[/itex] but I'm unsure about the following integral to get to y(x). At first I thought I could ignore the y in the h(x,y) when I'm integrating with respect to x but since I'm after the path y(x) I can hardly say they are independent.
I just wanted to check that I've had the right idea up to this point and that I can't use any arguments to justify ignoring the y in h(x,y) when I've got ∫f(h(x,y))dx, meaning I'm stuck with a horrible implicit integration by parts.
Thanks for any help :-) Sorry my post is a little convoluted.