Let's say we have the multivariable function $f(x,y)=-x^2-xy-y^2$. Then $\nabla f=\frac{\partial f}{\partial x} \hat{i}+\frac{\partial f}{\partial y}\hat{j}=(-2x-y)\hat{i}+(-x-2y)\hat{j}$.
$\nabla f$ gives the direction vector where the function's slope is highest.
Here, I've plotted the function on Wolfram Mathematica:
What I want to do is as follows:
Given an object to follow the vector with maximum slope and move along the surface $f$, I want to obtain an equation for the curve traced by that object (which I've sketched in $\color{red}{\text{red}}$), given some initial point. Let's say we start at the point $(x,y)=(10,10)$.
Below is my full attempt to do this:
I've calculated the point at which the maximum value of $f$ is by letting each partial derivative equal to zero, and solving the system. I have done this to verify whether the traced line I find will stop at the origin $(0,0,0)$.
$$\frac{\partial f}{\partial x}=-2x-y=0$$ $$\frac{\partial f}{\partial y}=-x-2y=0$$
Which gives me the point $(0,0)$, and thus the maximum value is $f=0$.
To solve for the curved line, I've thought of parametrizing the curve with respect to some variable $t$.
Therefore, I've made the following system of ODE's (From $\nabla f$):
$$\frac{dx}{dt}=-2x-y$$ $$\frac{dy}{dt}=-x-2y$$
With initial conditions $x(0)=10$ and $y(0)=10$.
From this, I've solved it using matrices and finding eigenvalues and eigenvectors:
$$\mathbf{Y'}=\begin{pmatrix} -2 & -1 \\ -1 & -2 \end{pmatrix} \mathbf{Y}$$
Where $\mathbf{Y}=\begin{pmatrix} x \\ y \end{pmatrix}$.
Using $\det(A-\lambda I)=0$, I got:
$$\begin{vmatrix} -2-\lambda & -1 \\ -1 & -2-\lambda \end{vmatrix}=(-2-\lambda)^2-1=\lambda^2+4\lambda+3=0$$
Solving this quadratic gives me the eigenvalues $\lambda_1=-3$ and $\lambda_2=-1$.
The corresponding eigenvectors are thus:
$$\vec{v_1}=\begin{pmatrix} 1 \\ 1 \end{pmatrix} \qquad \vec{v_2}=\begin{pmatrix} -1 \\ 1 \end{pmatrix}$$
Therefore, I've obtained the following general solution:
$$\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}=c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix}e^{-3t}+c_2 \begin{pmatrix} -1 \\ 1 \end{pmatrix}e^{-t}$$
When substituting initial conditions for $(x(0),y(0))=(10,10)$ I obtained the system:
$$\begin{cases} 10=c_1-c_2 \\ 10=c_1+c_2 \end{cases}$$
Which gave me $c_1=10$ and $c_2=0$.
Therefore, the specific solution for this case is:
$$x(t)=10e^{-3t}$$ $$y(t)=10e^{-3t}$$
I figured that I could substitute the solutions into $f(x,y)$ to obtain an expression for $f(t)$.
$$f(t)=-(10e^{-3t})^2-(10e^{-3t} \cdot 10e^{-3t})-(10e^{-3t})^2=-300e^{-6t}$$
Therefore, I figured that these parametrized curves would give me the red line. It seems correct to me, since as $t \to \infty$, $x=y=f=0$, which is what was predicted.
If I reverse the parametrization, I get the following curve (with restricted domain, of course):
$$x^2=y^2=-\frac{30000}{z}$$
However, I have several questions since I've made up this method myself (Even though it probably already exists):
1) Is my attempt correct? I am wondering since I am unsure whether the system of equations is actually correct to begin with.
2) Will the red line always be a geodesic (shortest route) along the surface given any initial conditions? If not, is it for the initial condition $(10,10)$?
3) Is there a simpler method to do this? Is there a more general equation to do this? I am aware of the geodesic equation (below), however I am unsure how to solve it, or if it is relevant at all.
$$\frac{d^2 \gamma^\lambda}{dt^2}+\Gamma^{\lambda}_{\mu \nu} \frac{d\gamma^{\mu}}{dt} \frac{d\gamma^{\nu}}{dt}=0$$
Please let me know if there are any problems with the questions. Any thoughts on the problem will be appreciated. Thanks in advance.