In this article, we explore the extremal curves of the following integral: S = \int f(x,y) \sqrt{dx^2 + dy^2} = \int f(x,y) ds
This is essentially a geodesic problem. However, its form is relatively simple, and its physical meaning is more apparent. For example, if f(x,y) is a potential function, then this is a two-dimensional problem of finding the minimum potential energy; if f(x,y) is a friction function, then this is the problem of finding the path of minimum friction. Regardless of the interpretation, this problem has considerable practical value. Let us perform the variation:
\begin{aligned} \delta S &= \int \delta[f(x,y)\sqrt{dx^2+dy^2}] \\ &= \int \left[ ds \delta f(x,y) + f(x,y) \frac{\delta (dx^2+dy^2)}{2ds} \right] \\ &= \int \left[ ds \left( \frac{\partial f}{\partial x}\delta x + \frac{\partial f}{\partial y}\delta y \right) + f \frac{dx d(\delta x) + dy d(\delta y)}{ds} \right] \\ &= \int ds \left( \frac{\partial f}{\partial x}\delta x + \frac{\partial f}{\partial y}\delta y \right) + f \frac{dx}{ds} d(\delta x) + f \frac{dy}{ds} d(\delta y) \end{aligned}
Applying integration by parts to the d(\delta x) and d(\delta y) terms on the right side, we get: \delta S = \left[ f \frac{dx}{ds} (\delta x) + f \frac{dy}{ds} (\delta y) \right] + \int \left[ ds \left( \frac{\partial f}{\partial x}\delta x + \frac{\partial f}{\partial y}\delta y \right) - d\left( f \frac{dx}{ds} \right)(\delta x) - d\left( f \frac{dy}{ds} \right)(\delta y) \right]
According to the boundary conditions, we are selecting an appropriate curve from all curves passing through two fixed points, therefore f \frac{dx}{ds} (\delta x) + f \frac{dy}{ds} (\delta y) = 0, because at the boundaries we have \delta x = 0 and \delta y = 0. Let us rearrange the expression: \delta S = \int \left[ \frac{\partial f}{\partial x} ds - d\left( f \frac{dx}{ds} \right) \right] \delta x + \left[ \frac{\partial f}{\partial y} ds - d\left( f \frac{dy}{ds} \right) \right] \delta y
Since this equation must hold for any arbitrary \delta x and \delta y, we have: \begin{aligned} \frac{\partial f}{\partial x} ds - d\left( f \frac{dx}{ds} \right) = 0 \\ \frac{\partial f}{\partial y} ds - d\left( f \frac{dy}{ds} \right) = 0 \end{aligned}
Or: \begin{aligned} \frac{d}{ds} \left( f \frac{dx}{ds} \right) = \frac{\partial f}{\partial x} \\ \frac{d}{ds} \left( f \frac{dy}{ds} \right) = \frac{\partial f}{\partial y} \end{aligned}
These equations appear quite concise, where ds^2 = dx^2 + dy^2. However, this relationship does not necessarily need to be given in advance; it can be treated as one of the integrals of the above system of equations. This is because we have: \begin{aligned} \left( f \frac{dx}{ds} \right) \frac{d}{ds} \left( f \frac{dx}{ds} \right) = f \frac{\partial f}{\partial x} \frac{dx}{ds} \\ \left( f \frac{dy}{ds} \right) \frac{d}{ds} \left( f \frac{dy}{ds} \right) = f \frac{\partial f}{\partial y} \frac{dy}{ds} \end{aligned}
Adding the two equations yields: \begin{aligned} \left( f \frac{dx}{ds} \right) \frac{d}{ds} \left( f \frac{dx}{ds} \right) + \left( f \frac{dy}{ds} \right) \frac{d}{ds} \left( f \frac{dy}{ds} \right) = f \frac{\partial f}{\partial x} \frac{dx}{ds} + f \frac{\partial f}{\partial y} \frac{dy}{ds} \end{aligned}
Thus: \frac{d}{ds} \left[ f^2 \left( \frac{dx}{ds} \right)^2 + f^2 \left( \frac{dy}{ds} \right)^2 \right] = \frac{d}{ds} (f^2)
Which simplifies to: f^2 \left[ \left( \frac{dx}{ds} \right)^2 + \left( \frac{dy}{ds} \right)^2 - 1 \right] = C
The condition ds^2 = dx^2 + dy^2 corresponds to C = 0. This means that during the solution process, we can completely treat s as a parameter independent of x and y.
However, simple equations do not necessarily imply simple solutions. A clear fact is that the arc length of most curve equations does not have a simple expression. That is to say, even for very simple curves (such as the parabola y = x^2), the relationship between its arc length (circumference) and x, y themselves is very complex. This is because the square root appears in s = \int \sqrt{1 + (y')^2} dx; therefore, when using s as a parameter, it is difficult to obtain a concise analytical solution. Consequently, most of the solving work is performed by computers. For computers, these equations are not difficult. In fact, analytical solutions can only be obtained in very limited cases.
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