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$$c=\int_{x_1}^{x_2}f_{gr}(x)\;dx$$

The integral is a time-like curve between $x_1$ and $x_2$ and at imagine fgf(x1) is a lower left corner of the rectangle and fgf(x2) is the upper right corner and $x_2-x_1$ is the length of the base of the rectangle. The geodesic is the shortest length curve parametrized by this on the condition that the area = c. The solution isn't always a strait line but could be some curved function. How do I find the function ? I know this is a variational problem involving a length integral and an area integral.

Do I need to find the 2x2 metric tensor for the 2-d parameterization curve and solve a pair of differential equations like here ?

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Are you trying to solve $\inf \{ l(f) | \int f = c \}$ (where $l(f) = \int \sqrt{1+(f')^2}$)? –  copper.hat Jul 29 '12 at 5:50
    
yea...trying to find f –  mygeodesic Jul 29 '12 at 6:44
    
Are there other constraints, because the solution to that will be $f(x) = \frac{c}{x_2-x_1}$? –  copper.hat Jul 29 '12 at 6:46
    
this is a horizontal line which gives the areas and is a minimal length, but does not ajoin the endpoints. A simple case would be if the area parameter were (x2-x1)*f(x1)+(x2-x1)*(f(x2)-f(x1))/2 and then geodesic would be of the form y=mx+b –  mygeodesic Jul 29 '12 at 7:11

1 Answer 1

You can incorporate the area constraint with the Lagrange multiplier method; for example, the book of John Oprea Differential geometry and its applications has an accessible treatment of such variational problems.

However I recommend to look for a curve in parametric form $x=f(t), y=g(t)$ rather than in the graph form $y=f(x)$. The reason is as follows. The free part of the boundary (the curve you are looking for) wants to have constant curvature. Indeed, if a curve has different curvature $\kappa_1>\kappa_2$ at two points $p_1,p_2$, then we can push a little at the point $p_1$ to decrease the curvature, and push from the opposite side at $p_2$ to increase it. This perturbation will decrease the length without changing the area. A planar curve of constant curvature is a circular arc. So, the boundary wants to be a circular arc with given endpoints, bounding the given area. When such an arc is a graph $y=f(x)$, it is the solution. But in some cases (depending on your parameters) the extremal circular arc is not a graph, because it intersects some vertical line (either $x=x_1$ or $x=x_2$) twice. If you insist on the graph form of the solution, you create an obstacle problem, which makes the problem substantially more difficult.

On the geometric grounds (without doing the computations) I expect that when the obstacle comes into effect, there will not be a minimizer satisfying the given boundary conditions. Instead, a minimizing sequence will converge to a function which violates one or both boundary conditions, and whose graph is a circular arc tangent to the effective obstacle(s). For a simple example, take $x_1=-1, x_2=1$, prescribe the boundary conditions $f(-1)=f(1)=0$ and demand the area to be $10+\pi/2$. The minimizing functions will converge to $f(x)=5+\sqrt{1-x^2}$, which fails the boundary conditions. To recover the existence of minimizer, you can relax the problem by adding the length of vertical segments to the length of the proper graph: that is, the new variational problem is to minimize $|f(x_1)-y_1|+|f(x_2)-y_2|+\int_{x_1}^{x_2}\sqrt{1+f\,'(x)^2}\,dx$ subject to $\int_{x_1}^{x_2} f(x)\,dx=c$, and no boundary conditions. Still, an additional complication arises when the prescribed area $c$ is so small that $f$ wants to be negative.

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thanks for your answer. i spent all day looking for pdf files looking for examples similar to my problem but only found one here: ocw.mit.edu/courses/mathematics/… solving for the parameters(m,c,d) of the ellipse doesn't see to be possible. i imagine this problem is too difficult to be included in most texts –  mygeodesic Jul 30 '12 at 3:19
    
@mygeodesic It's not going to be an ellipse, because ellipses do not have constant curvature. –  user31373 Jul 30 '12 at 4:00

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