Dopo le ultime simulazioni di un semplice modellino lineare della particella, eccoci qui alle prese con l'equazione completa della dinamica della particella. Si suppongono numeri di Reynolds relativamente bassi e numeri di Knudsen prossimi allo 0. Ricordo che il numero di Reynolds non è altro che il rapporto tra le forze di inerzia e le forze viscose. Ovviamente la scelta di usare approssimazioni per bassi numeri di Reynolds è perfettamente giustificata, dal momento che la massa delle particelle prese in esame è estremamente ridotta. I numeri di Knudsen (il rapporto tra il ''mean free path'' ed il raggio equivalente) esprimono sostanzialmente quanto sia valida l'approssimazione continua del mezzo rispetto the size of the particle, which in our case despite the (small, tiny, tiny, microscopic nanoscale) is still significantly larger than the molecules that make up the air. The slip correction factor so it should be approximately equal to 1, but an excess of zeal and sadistic taste for the hurt to include the CPU and make it even depends on the height to which it has perticella. All particles are considered in the subsequent simulations are also non-Strakeriane. I remind the uninitiated that a particle is a particle Strakeriana of which are unknown (and can not be measured in any way with any instrument) the size, which does not obey any law of physics known or not, the dynamics of which is not expressible by any formalism past, present or future composition of which varied at will.
Here is the ugly beast of equation that we'll have to simulate (in these moments I thank Turing and von Neumann for his invention of the PC):
m_ {p} \\ frac {dv_ {z}} {dt} =
\\ frac {6 \\ pi \\ mu (z) R_ {eq} \\ chi} {C_ {c} (z)} \\ cdot (v_-u_ {z} {z})
+ \\ frac {9} {4} \\ pi \\ rho (z) \\ left (\\ frac {R_ {eq} \\ chi} {C_ {c} (z)} \\ right) ^ 2 (u_-v_ {z} {z}) u_{z}}{dt}-\frac{d v_{z}}{dt}\right)
+6R_{eq}^{2}\sqrt{\pi\mu(z) \rho(z)}\int\limits_{0}^{t}
\frac{\displaystyle \left.\left(\frac{d}{dt}(u_{z}-v_{z})\right)\right
is nothing more than the Stokes force acting on the particle moving in the fluid.
\\ frac {9} {4} \\ pi \\ rho (z) \\ left (\\ frac {R_ {eq} \\ chi} {C_ {c} (z)} \\ right) ^ 2 (u_ {z} -v_ {z}) accelerate.
\\ frac {V_ {p}} {2} \\ rho (z) \\ left (\\ frac {z} u_ {d} {dt} - \\ frac {d v_ {z}} {dt} \\ right)
expresses the effect of the potential flow around the particle under consideration, and serves to take into account the inertia of the fluid that surrounds the particle.
+6
R_ {eq} ^ {2} \\ sqrt {\\ pi \\ mu (z) \\ rho (z)} \\ int \\ limits_ {0} ^ {t}
\\ frac {\\ displaystyle \\ left. \\ Left (\\ frac {d} {dt} (u_-v_ {z} {z}) \\ right) \\ right density of the medium. In practice is the term that takes into account the vorticity of the fluid that is created due to the motion of the particle.
m_ {p} g
It 's the good old gravity (as said earlier, only particles are considered non-Strakeriane).
Once we've pitted all terms and all constants, we can not do any thing to solve the equation, since there is a singularity that lurks in the integral. Let me explain the term with the root in the denominator of the integral is discontinuous at t = τ Since there is over what is practically a constant, that we are in reckon we have an infinite time as integrating (constant / 0). No fear though, with a daring pass that math is to add and remove the same to kill a certain amount singularity ugly, and just create one that is numerically much more quiet (0 / 0):
\\ int \\ limits_ {0} ^ {t} \\ frac {\\ displaystyle \\ left. \\ left (\\ frac {d} {dt} (u_-v_ {z} {z}) \\ right) \\ right + \\ Int \\ limits_ {0} ^ {t} \\ frac {\\ displaystyle \\ left. \\ Left (\\ frac {d} {dt} (u_-v_ {z} {z}) \\ right) \\ right \\ Right \\ Left (\\ frac {d} {dt} (u_-v_ {z} {z}) \\ right) \\ cdot \\ int \\ limits_ {0} ^ {t} \\ frac {\\ displaystyle 1} {\\ sqrt {t - \\ tau}} d \\ tau =
\\ left (\\ frac {du_ {z}} {dt} - \\ frac {dv_ {z}} {dt} \\ right) \\ cdot 2 \\ sqrt {t}
Replacing the whole equation we get:
m_ {p} \\ frac {dv_ {z}} {dt} =
\\ frac {6 \\ pi \\ mu (z) R_ {eq} \\ chi} {C_ {c } (z)} \\ cdot (v_-u_ {z} {z})
+ \\ frac {9} {4} \\ pi \\ rho (z) \\ left (\\ frac {R_ {eq} \\ chi} {C_ {c} (z)} \\ right) ^ 2 (u_-v_ {z} {z})
+ \\ frac {V_ {p}} {2} \\ rho (z) \\ left (\\ frac {z} u_ {d} {dt} - \\ frac {d v_ {z}} {dt} \\ right)
+6 R_ {eq} ^ {2} \\ sqrt {\\ pi \\ mu (z) \\ rho (z)} \\ left (\\ int \\ limits_ left {0} ^ {t} \\ left [\\. \\ left (\\ frac {d} {dt} (u_ {z}-v_ {z}) \\ right) \\ right 2\sqrt{t} \right)-m_{p}g
Ora, dividiamo tutto per m
p
tanto per ricondurci a qualcosa di più o meno leggibile.
In linea di massima abbiamo una cosa del tipo:
\frac{dv_{z}}{dt}=
l(z)(u_{z}-v_{z})
+g(z)(u_{z}-v_{z})
with: \\ begin {array} {l}
f (z) = 6 \\ frac {R_ {eq} ^ {2} \\ sqrt {\\ pi \\ mu (z) \\ rho (z)} } {m_ {p}} \\ \\
\\ \\
\\ Psi \\ left (\\ tau, t, v_ {z} (\\ tau), \\ frac {dv_ {z}} {dt} (\\ tau) \\ right)
= \\ left [\\ left. \\ left (\\ frac {d} {dt} (u_-v_ {z} {z}) \\ right) \\ right R_ {eq} \\ chi} {C_ {c} (z) m_ {p}} \\ \\
\\ \\
g (z) = \\ frac {9} {4} m_ {p} \\ pi \\ rho (z) \\ left (\\ frac {R_ {eq} \\ chi} {C_ {c} (z)} \\ right) ^ 2
\\ end {array} But since we are not going to be fighting with integral-algebraic equations -differentials, with a daring pass math we get:
(1 + h (z) +2 \\ sqrt {t} f (z)) \\ frac {dv_ {z}} {dt} =
l (z) ( u_-v_ {z} {z})
+ g (z) (u_-v_ {z} {z}) d \\ tau-g
that with an equally bold step math becomes:
\\ frac {dv_ {z}} {dt} =
M (z, t) (u_-v_ {z} {z})
+ G (z, t) (u_-v_ {z} {z})
\\
H(z,t)=\frac{3 h(z)+2\sqrt{t}f(z)}{1+h(z)+2\sqrt{t}f(z)}\\
\\
F(z,t)=\frac{f(z)}{1+h(z)+2\sqrt{t}f(z)}\\
\\
P(z,t)=\frac{g}{1+h(z)+2\sqrt{t}f(z)}
\end{array}
Ora però è il momento di iniziare a pensare alla risoluzione del problema vero e proprio. In questo caso si procederà con una tecnica di risoluzione approssimata analitica, invece che puramente numerica dal momento che per gestire la memoria introdotta dal termine integrale nell'equazione sarebbe stato necessario scrivere codice da zero ed implementare a mano metodi numerici ad hoc for the equation. So at least this evening to explain the simulation will prefer an approach that is not the classic variation on the theme of the various Runge-Kutta and not exactly see every day. The approach in question is precisely the "Variational Iteration Method". But we see a little 'how it works: build since the last report that we have obtained a variational formulation of the problem, and a functional correction. Basically at this point the aim is to try to derive a formula from the equation originally iterative, allowing for further steps to arrive at the solution. At each step the solution will be improved, and as you progress, you move to a low point: the exact solution. It is also hoped that the method in question is focused and steady. Much easier to write than to explain. Now rewrite the equation so as to match it all to 0.
\\ frac {dv_ {z}} {dt} -
M (z, t) (u_-v_ {z} {z})
-G (z, t) (u_-v_ {z} {z }) = 0
But now we have to solve for the position along the z axis, so you replace v with z ':''
z - V (t, z, z')-F (z, t) \\ int \\ limits_ {0} ^ {t} \\ Psi (\\ tau, t, z ', z'') d \\ tau = 0
At this point, however, since it is not strictly necessary to bring us back to a system of first order for now we leave things as they are without adding time ausiliarie.Una equations written in the form so general and basic equation, we write now the functional correction that will work to find the approximate solution:
\\ xi_ {n +1 } (t) = \\ xi_ {n} (t) + \\ int \\ limits_ {0} ^ {t} \\ lambda (s) \\ left [\\ xi_ {n}''(s) - V (s, \\ xi_ {n}, \\ xi_ {n} ') -
F (\\ xi_ {n}, s) \\ int \\ limits_ {0} ^ {s} \\ Psi (\\ tau, s \\ tilde {\\ xi} _ {n} ', \\ tilde {\\ xi} _ {n}'') d \\ tau \\ right] ds
where λ i (s) are the generalized Lagrange multipliers (for now known). Who did the exams (or read about) the optimal control or optimization and dynamic programming knows what I'm talking about. These will be chosen so''good''compared to the variations of ξ
n, ie n δξ
.
Now if we write the change in the functional correction is obtained trivially that:
\\ delta \\ xi_ {n +1} (t) = \\ delta \\ xi_ {n} (t) + \\ delta \\ int \\ limits_ {0 } ^ {t} \\ lambda (s) \\ left [\\ xi_ {n}''(s) - V(s,\xi_{n},\xi_{n}')-
F(\xi_{n},s)\int\limits_{0}^{s}\Psi(\tau,s,\tilde{\xi}_{n}',\tilde{\xi}_{n}'')d\tau \right]ds
Ora dal momento che stiamo considerando variazioni limitate, il termine che contiene Ψ(.) sparisce e della funzione di cui sopra, e resta soltanto:
\delta\xi_{n+1}(t)=\delta\xi_{n}(t)+\delta\int\limits_{0}^{t}\lambda(s)\left[\xi_{n}''(s) - V(s,\xi_{n},\xi_{n}')\right]ds
Ora spezziamo l'integrale in due pezzi:
\delta\xi_{n+1}(t)=\delta\xi_{n}(t)+
\\ delta \\ int \\ limits_ {0} ^ {t} \\ lambda (s) \\ xi_ {n}''(s) ds-
\\ delta \\ int \\ limits_ {0} ^ {t} \\ lambda (s) V (s, \\ xi_ {n}, \\ xi_ {n} ') ds
now developed with integration by parts the first integral term twice:
\\ delta \\ xi_ {n +1} ( t) = \\ delta \\ xi_ {n} (t) +
\\ left. \\ lambda (s) \\ right
rewriting a bit 'first terms, highlighting δξ
n
get:
\\ delta \\ xi_ {n +1} (t) = \\ left (1 - \\ left. \\ Lambda' (s) \\ right minimum, and then impose the change anything). The thing is negligible considering the integral term with the function V, and then simply to impose a set of three conditions:
\\ left \\ {
\\ begin {array} {l}
\\ left (1 - \\ left. \\ Lambda '(s) \\ right - V (s, \\ xi_ {n}, \\ xi_ {n} ') - F (\\ xi_ {n}, s) \\ int \\ limits_ {0} ^ {s} \\ Psi (\\ tau, s \\ xi_ {n} ', \\ xi_ {n}'') d \\ tau \\ right] ds
smucinare Theoretically you could go in full-term with V (.), but you quickly realize that makes no sense because it has no linear terms in ξ
n and the coefficients are mostly implicit functions, so in any case it would leave out the terms which go to affect the setting of conditions.
Each iterative self-respecting, however, needs something to start with. Our method is no exception. As it happens we already have a little ' data to formulate an approximation of the initial choice of the approximate solution. If we remember the old simulations with the linear model, and the explicit solution of the differential equation associated with it, we soon realize that the game is almost done: the speed is stable and tends to that of the wind very quickly, the location is practically defined by a linear law.
is therefore practically we have:
\\ xi_ {0} (t) = z_ {0} + v_ {z} (0) t
At this point, however, if you want you can use a little trick from As is readily apparent that the function
\\ xi_ {0} (t) = Z_ {0} + u_ {t} z
in the case of constant wind speed is that which minimizes the time it reaches equilibrium immediately and subsequent iterations do not in any way change the result.
After several tests trying to preserve more the equation, however, concludes that so complicated an equation can not be handled by a convenient analytical method, especially because of the implicit function present.
For the numerical solution will proceed in the subsequent article by the method of placement.
