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Say you have two maps of hills with a single peak each which have contour lines which are concentric rings of various shapes. Superimpose these maps on a single piece of paper so the contour lines intersect. Now starting at the top of one hill, make a path down the hill following the gradient tangent to the contour line.

But also make sure that you are simultaneously going up the other hill tangent to its contour lines. In other words you are tracing a path where the contour lines of one map are tangent to the contour lines of the other.

You will draw a unique path from the top of the hill on the first map to the hill on the second map.

Given two height functions $\phi_1(x)$ and $\phi_2(y)$ how can one construct the equation for that curve relating the heights of both maps $f(\phi_1,\phi_2)=0$? Is there a general method? (exclude any pathological cases). Is there any books about this?

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There are a couple of options, but let's say a fairly straightforward approach is to calculate the two gradients $$\nabla \phi_1(x) = \left(\frac{\partial \phi_1}{\partial x_1}(x), \frac{\partial \phi_1}{\partial x_2}(x)\right) \, \text{ and } \, \nabla \phi_2(x) = \left(\frac{\partial \phi_2}{\partial x_1}(x), \frac{\partial \phi_2}{\partial x_2}(x)\right)$$ and look for all points $x = (x_1, x_2) \, \in \, \mathbb{R}^2$ for which the angle between the two gradients $\nabla \phi_1(x)$ and $\nabla \phi_2(x)$ is $\pi$.

The dot product formula at any arbitrary point $x \in \mathbb{R}^2$ gives
$$\big( \, \nabla \phi_1(x) \cdot \nabla \phi_2(x) \, \big) = |\nabla \phi_1(x)| \, |\nabla \phi_2(x)| \, \cos{(\theta(x))}$$ where $\theta(x)$ is the angle between the gradient vectors $\nabla \phi_1(x)$ and $\nabla \phi_2(x)$ while $|\nabla \phi_1(x)| = \sqrt{\big( \, \nabla \phi_1(x) \cdot \nabla \phi_1(x) \, \big)}$ and $|\nabla \phi_2(x)| = \sqrt{\big( \, \nabla \phi_2(x) \cdot \nabla \phi_2(x) \, \big)}$ are the lengths of the gradient vectors. For all points $x = (x_1, x_2) \, \in \, \mathbb{R}^2$ for which the angle $\theta(x)$ between the two gradients $\nabla \phi_1(x)$ and $\nabla \phi_2(x)$ is $\pi$ the value of $\cos{(\theta(x))} = -1$ so the condition becomes

$$\big( \, \nabla \phi_1(x) \cdot \nabla \phi_2(x) \, \big) = - |\nabla \phi_1(x)| \, |\nabla \phi_2(x)|$$ which is the same as the equation for a curve in $\mathbb{R}^2$

$$F(x_1, x_2) = \big( \, \nabla \phi_1(x) \cdot \nabla \phi_2(x) \, \big) + |\nabla \phi_1(x)| \, |\nabla \phi_2(x)| = 0$$

Another option would be to take the determinant equation

$$\Delta(x_1, x_2) = \begin{vmatrix} \, \frac{\partial \phi_1}{\partial x_1}(x_1, x_2) \, & \, \frac{\partial \phi_1}{\partial x_2}(x_1, x_2) \, \\ \, \frac{\partial \phi_2}{\partial x_1}(x_1, x_2) \, & \, \frac{\partial \phi_2}{\partial x_2}(x_1, x_2) \, \\ \end{vmatrix} = 0$$ and take only the solutions where the two gradients oppose each other (the other solutions will have the two gradients pointing in the same direction)