Non-Intersecting Graphs (Single Order) — Jemoka Knowledge Base

In this project, we aim to derive situations for the existence of a differential equation for when a family of functions do not intersect. We were able to derive a full solution for the result in linear equations, and we offer an exploration of a partial solution for non-linear cases. Function Families Fundamentally, function families are functions parameterized by some C, which has the shape:

\begin{equation} y(x, \dots, c) = f(x, \dots)+c \end{equation}

Through this result, we can figure a statement for “intersection.” If two functions intersect, their difference will be 0; if there is a non-trivial solution (that c_1\neq c_2 — that, they are not the same function—still makes y_{C_1} = y_{C_2}), the function family interact. We can test this by subtracting two arbitrary members from the desired family. If it results that c_1-c_2=0 \implies c_1=c_2, we can say that the family does not intersect: that there are no non-trivial solutions to the function having no difference. Single-Order Linear Differential Equations Here, we prove the fact that single-order linear differential equations do not produce solutions that intersect. We have the following single-order linear differential equation:

\begin{equation} \dv{y}{x} + P(x) = Q(x) \end{equation}

If, as desired, our function has a analytical solution (without an integral), we will make both terms differentiable.

\begin{equation} \dv{y}{x} + P’(x) = Q’(x) \end{equation}

Recall the general solution of this expression:

\begin{align} y &= e^{-\int P’(x)\dd{x}} \int e^{\int P’(x)\dd{x}} Q’(x)\dd{x} \\ &= e^{-(P(x)+C_1)} \int e^{P(x)+C_1} Q’(x)\dd{x} \end{align}

Of course, we can separate the constants e^{C_1} out.

\begin{align} y &= e^{-(P(x)+C_1)} \int e^{P(x)+C_1} Q’(x)\dd{x} \\ &= e^{-P(x)} \int e^{P(x)} Q’(x)\dd{x} \end{align}

Now, it is the case that, for the most part, e^{P(x)}Q’(x) may not be integral-differentiable. Applying the fundamental theorem, we still have that as the integral function, with some “differentiated” term which we will call a(x): below

\begin{align} y &= e^{-P(x)}(a(x) +C) \\ &= e^{-P(x)}a(x) +Ce^{-P(x)} \end{align}

Excellent. Now, let’s do the subtraction test devised above; if we have that C_1-C_2=0 given y_1-y_2=0, then we can ensure that the function family do not intersect.

\begin{align} y_1 - y_2 =0 &= (e^{-P(x)}a(x) +C_{1}e^{-P(x)})-(e^{-P(x)}a(x) +C_{2}e^{-P(x)}) \\ &= C_{1}e^{-P(x)}-C_{2}e^{-P(x)} \\ &= (C_{1}-C_{2})e^{-P(x)} \end{align}

We now have that:

\begin{equation} 0 = (C_1+C_2)e^{-P(x)} \end{equation}

Notably, the codomain of e^{x} is (0, \infty). Having never reached 0, we have that 0=C_1-C_2, as desired. \blacksquare