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. \(\blacksquare\)