A Simple Second-Order Nonlinear System Addendum

A previous presentation considered the pair of nonlinear differential equations

$F^{″} = F G G^{″} = F^{2}$

In the process of developing that presentation, a number of interesting features of this system were identified but not applied. The purpose of this presentation is to document these features for future possible usefulness. Some parts of the previous presentation are repeated to give context.

Multiplying each equation by the function on its left-hand side, it is simple to see that

$F F^{″} = G G^{″}$

Using this equivalence, it is straightforward to show that

$\frac{d^{2}}{d x^{2}} (F^{2} - G^{2}) = 2 ({F^{'}}^{2} - {G^{'}}^{2})$

An exact constant for this system is found by adding a pair of equivalences and integrating both sides:

$\begin{array}{c} 2 F^{″} F^{'} = 2 F^{'} F G G^{″} G^{'} = F^{2} G^{'} \\ 2 F^{″} F^{'} + G^{″} G^{'} = 2 F^{'} F G + F^{2} G^{'} \\ {F^{'}}^{2} + \frac{1}{2} {G^{'}}^{2} = F^{2} G + \frac{1}{2} c \\ 2 {F^{'}}^{2} + {G^{'}}^{2} - 2 F^{2} G = c \end{array}$

The third term on the left-hand side can be written two different ways:

$2 F^{2} G = {\begin{cases} 2 F F^{″} = \frac{d^{2}}{d x^{2}} F^{2} - 2 {F^{'}}^{2} \\ 2 G G^{″} = \frac{d^{2}}{d x^{2}} G^{2} - 2 {G^{'}}^{2} \end{cases}$

With these three results, one can write the following compact equivalences

$\begin{array}{l} {F^{'}}^{2} = \frac{1}{2} \frac{d^{2}}{d x^{2}} G^{2} - \frac{3}{2} {G^{'}}^{2} + \frac{c}{2} \\ {G^{'}}^{2} = \frac{d^{2}}{d x^{2}} F^{2} - 4 {F^{'}}^{2} + c \\ \frac{d^{2}}{d x^{2}} F^{2} = 2 \frac{d^{2}}{d x^{2}} G^{2} - 5 {G^{'}}^{2} + c \\ \frac{d^{2}}{d x^{2}} G^{2} = 3 \frac{d^{2}}{d x^{2}} F^{2} - 10 {F^{'}}^{2} + 2 c \end{array}$

where each quantity is given in terms of two corresponding quantities of the other function.

It is a simple matter to write a provocative fourth-order differential equation for the function F:

$\frac{d^{2}}{d x^{2}} \frac{F^{″}}{F} = G^{″} = F^{2} \to \frac{1}{F} \frac{d^{2}}{d x^{2}} \frac{F^{″}}{F} = F$

The corresponding equation for the function G contains square roots of its second derivative, and it thus not as provocative.

Defining the sum and difference of the two functions with indicative symbols

$\begin{array}{l} P = F + G \\ M = F - G \end{array} \begin{array}{l} F = \frac{1}{2} (P + M) \\ G = \frac{1}{2} (P - M) \end{array}$

the differential system becomes

$P^{″} = F P M^{″} = - F M$

These equations are reminiscent of those for trigonometric functions with variable frequency.

Multiplying each left-hand side by the other function and adding one has

$M P^{″} + P M^{″} = 0 \to \frac{P^{″}}{P} + \frac{M^{″}}{M} = 0$

Using this expression, it is straightforward to show that

$\frac{d^{2}}{d x^{2}} P M = 2 P^{'} M^{'}$

which is immediately consistent with an equivalence above.

Including the definition of F, the rewritten system becomes

$\begin{array}{c} P^{″} = \frac{P^{2} + P M}{2} M^{″} = - \frac{M^{2} + P M}{2} \\ P^{″} - \frac{1}{2} P^{2} = \frac{1}{2} P M M^{″} + \frac{1}{2} M^{2} = - \frac{1}{2} P M \\ P^{″} + M^{″} = \frac{1}{2} (P^{2} - M^{2}) \end{array}$

The rewritten differential system shows a distinct emphasis on the function F. Noting that

$\frac{d^{2}}{d x^{2}} (F \pm G) = F^{″} \pm F^{2}$

one can immediately write the equations

$F^{″} + F^{2} - F P = 0 F^{″} - F^{2} + F M = 0$

which are reminiscent of Riccati equations. Given that

$F = \frac{P^{″}}{P} = - \frac{M^{″}}{M}$

one can write a variety of equations relating the two sum and difference functions:

$\frac{P^{″}}{P} - \frac{M^{″}}{M} = 2 F = P + M$

$\frac{P^{″}}{P} - P = \frac{M^{″}}{M} + M$

$2 \frac{P^{″}}{P} = P + M = - 2 \frac{M^{″}}{M}$

$\begin{array}{l} P^{″} - M^{″} = 2 F^{2} = 2 (\frac{P^{″}}{P})^{2} = 2 (\frac{M^{″}}{M})^{2} \\ = (\frac{P^{″}}{P})^{2} + (\frac{M^{″}}{M})^{2} = - 2 \frac{P^{″}}{P} \frac{M^{″}}{M} \end{array}$

$P^{″} = M^{″} + 2 (\frac{M^{″}}{M})^{2} M^{″} = P^{″} - 2 (\frac{P^{″}}{P})^{2}$

$P^{″} - (\frac{P^{″}}{P})^{2} = M^{″} + (\frac{M^{″}}{M})^{2}$

The sum and difference functions have fourth-order differential equations with clear symmetry. Using an equivalence above, first write

$P = - 2 \frac{M^{″}}{M} - M M = 2 \frac{P^{″}}{P} - P$

then insert each quantity into the other and simplify. This gives

$\frac{d^{2}}{d x^{2}} (\frac{P^{″}}{P} - P) + (\frac{P^{″}}{P})^{2} = 0 \frac{d^{2}}{d x^{2}} (\frac{M^{″}}{M} + M) + (\frac{M^{″}}{M})^{2} = 0$

which differ in one sign. Part of the motivation for introducing these functions is to attempt to use this symmetry for an analytic solution.

One odd thing about these two equations is the two single functions inside the parentheses, since the system appears to favor second derivatives of a function dived by itself. The function F is already given above as two such ratios. One then has

$G = \frac{F^{″}}{F} = \frac{(P^{″} / P)^{″}}{P^{″} / P} = \frac{(M^{″} / M)^{″}}{M^{″} / M}$

Multiplying the two equations of the system as initially written leads to

$\frac{F^{″}}{F} \frac{G^{″}}{G} = F^{2} = \frac{d^{2}}{d x^{2}} \frac{F^{″}}{F} \to \frac{G^{″}}{G} = \frac{(F^{″} / F)^{″}}{F^{″} / F}$

Surely such nested behavior must imply something about the solution. It would be most useful if one could write equations in each of these ratios only, but that does not appear possible.

One way to approach such ratios is by writing exponentiated forms of each function:

$F = f_{0} exp \int_{0}^{x} d x f (x) \to \frac{F^{″}}{F} = f^{'} + f^{2}$

This can lead to equations that look a somewhat simpler, such as

$\frac{P^{″}}{P} + \frac{M^{″}}{M} = 0 \to \begin{array}{c} p^{'} + p^{2} + m^{'} + m^{2} = 0 \\ \frac{d (p + m)}{p^{2} + m^{2}} = - d x \end{array}$

but it is not clear if this gets any closer to a full solution.