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Regarding BC's question?
"We all" know the soln. to this one: D^2 + aD + bX = 0 A more
realistic in some cases is: D^2 + a(D * D) + bX = 0 ** What's
the soln. for the oscillatory case?
bc been modeling it w/ leapfrog and would like the analytic in
addition to the numerical.
** More complete, but unnecessary and likely very difficult:
D^2 + aD + b(D*D) + c*signum(D)+ dX = 0
The DE:
D^2 + a*(D * D) + b*X = 0
can be generated by the "Lagrangian"
L = exp(2*a*X)*((1/2)*(D * D) - (b/(2*a))*(X - 1/(2*a))).
The "Hamiltonian" corresponding to this "Lagrangian" is conserved.
This gives us a first integral for the dynamics. The constant value
E of this "Hamiltonian" (in terms of X & D) is given by:
E = exp(2*a*X)*((1/2)*(D * D) + (b/(2*a))*(X - 1/(2*a))).
This last equation is first order in D. It can be solved formally
in terms of quadratures. But I don't know of a closed-form formula
for the X(t) function. But a formal solution for the inverse function
t(X) is given by:
t = +/- Integral[dX/sqrt(2*E*exp(-2*a*X) + (b/a)*(1/(2*a) - X))] .
BTW, the DE BC asked about is fully reversible w.r.t. the
independent variable t. But I suspect that he may have had in mind
a closely related *irreversible* DE showing damped oscillations.
That suspected DE is given by:
D^2 + a*(D * |D|) + b*X = 0 or equivalently
D^2 + a*signum(D)*(D * D) + b*X = 0
and it exhibits a quadratic velocity damping in both motional
directions (assuming, of course, that parameter a is positive).
However the DE BC asked about (and I partially solved above) is only
quadratically velocity-damped in one motional direction (i.e. the
positive D direction), and it is quadratically velocity-enhanced in
the other (negative D) direction. But it is fairly straight forward
to solve the irreversibly dissipative problem in terms of the
solution to the reversible dissipative/enhanced one by treating the
motion in each direction separately and piecewise splicing
alternately the half-cycle solutions for normal X values to the
half-cycle solutions for a formally negated X-value taken from the
solutions to the reversible problem, and appropriately matching the
boundary/initial conditions at each half-cycle boundary.
David Bowman
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