brentw (Brent Watts of Hickory, NC.
2000-11-21)
How do I solve the following differential equations?
1) For the first DE, what you have is clearly the differential of [x sin(xy) + y]
so, the solution is: [x sin(xy) + y] = constant.
2) The second DE has a singular solution (the straight line x=0).
The general solution is obtained by noticing that the differential of
[x4y3
- 3x3y2
+ x4y2]
is x2 times the given differential expression.
The solutions to the DE are therefore either curves of equation
[x4y3
- 3x3y2
+ x4y2]such algebraic pieces and segments of the singular solution x=0
(joined at the points where they are tangent).
3) The last DE is a no-brainer, but a fairly annoying one.
The general idea is to make a linear change of variables so that the variables
separate easily
(i.e., to obtain a differential which equate a differential form involving one variable
to a differential form involving only the other.
For example, you may use x = u+v and y = (u-v)Ö3, which means:
4u = x + yÖ3
and
4y = x - yÖ3
Your DE then becomes something like (please check):
[2(Ö3+3)v + Ö3-5] du = [2(Ö3-3)u + Ö3+5] dv
If I did manipulate things correctly (please double check),
this means that the solutions are the curves for which the following expression is constant:
[x - yÖ3 + 8/Ö3 - 6]3-Ö3 [x + yÖ3 - 8/Ö3+6]3+Ö3
(2023-02-12) Runge-Kutta numerical methods (1901)
For approxumate solutions to: dy = f (x,y) dx
The error of the Runge-Kutta method of order n (RKn) is proportional to hn+1, where h
is the step size. RK1 is also known as the (forward)
Euler method.
As a good compromise between accuracy an complexity, RK4 seems especially popular.
Matthew A. (Yahoo!
2002-07-21)
Singular Change of Variable
Solving the differential equation:
y dx = ( y ey - 2x ) dy
This example illustrates some of the issues which arise when
a given differential equation reduces to a tame equation
(here a simple quadrature) after a change of variable
which presents some singularities...
If y is not zero (a big "if")
we may replace x by the variable u = x y2 .
du = y (y dx + 2x dy)
Multiplying both sides of the original equation by y turns it into:
Any value of the constant of integration C can be
used to obtain a solution over any domain in which y
does not vanish.
However, for a well-behaved solution over a domain which crosses the line
y = 0, we must use
C = -2.
x =
[ (y2 -2y + 2) exp(y) - 2 ] / y2
It's only for this particular value of C that the
function x(y) is smooth about y = 0.
For small values of y, x(y) is indeed infinitely smooth:
x = y/3 + y2/4 + y3/10 +
y4/36 + y5/168 + ...
Now, what about y as a function of x,
over the largest possible domain?
Well, the above function x(y) achieves a single negative minimum:
x = -A = -0.169998...
when y = -1.45123...
For x > -A, the solution has two branches.
One of those (y < -1.45123...)
has an asymptote at x = 0 and doesn't extend
to positive values of x.
The other branch (y > -1.45123...)
extends from x = -A to
+¥
and it goes through x = 0,
y = 0 without the slightest glitch.
(2007-07-30) Falling vertically in a fluid
(e.g., against air resistance).
The derivative of the downward speed is: v' =
g - ( 2 u v + v2 ) / l
In a vacuum, the downward speed
v increases at a rate g
equal to the acceleration of gravity.
The resistive forces exerted by the medium
(e.g., air) include a viscous resistance
(Stokes drag, proportional to v ) and a pressure term
dubbed quadratic drag,
proportional to the square of v.
With the notations introduced above, 2u is the speed at
which the quadratic drag becomes as large as the viscous term.
Loosely speaking, 2u marks the middle of a transitional
regime where the flow is no longer laminar, although the quadratic
term of the turbulent regime is not yet dominant
(2u is often dubbed "critical speed", although there's nothing
critical about it). The symbol
l is a length which is inversely
proportional to the magnitude of the quadratic drag coefficient at high speeds.
Let's recast the above expression:
l dv/dt =
( lg + u2 ) -
( u + v ) 2
Introducing a speed w > u defined by the relation
w2 = lg + u2 , we obtain:
ldv
=
[ w2 - (u+v)2 ] dt
Therefore :
2w dt
=
2 w l dv
=
l dv
+
l dv
w2 -
( u + v ) 2
w + u + v
w - u - v
This is readily integrated from the beginning of the fall
(v = 0) at time t=0 :
2 w t / l =
Log ( [w+u+v] / [w-u-v] ) -
Log ( [w+u] / [w-u] )
Solving for v
(HINT: take the exponentials of both sides) we obtain:
v = ( w - u )
[ 1 -
2 w
]
(w-u) + (w+u) exp ( 2wt / l )
By integration, this yields the
height traveled since the beginning of the fall:
h = ( w - u ) t +
l Log
[
w + u
+
w - u
exp ( - 2wt / l )]
2 w
2 w
In practice, the exponential vanishes quickly and the
dropped object travels at a uniform
terminal velocity (w-u) according to the following equation:
h = (w-u) t - D
The distance D = l [ Log 2 - Log (1+u/w) ]
can be interpreted either as a
time delay or as a correction in height
if you time the fall of an object dropped from an altitude Z.
The actual duration of the fall will be the time it would have taken
to travel the distance Z+D at a uniform speed equal to the terminal velocity.
According to the above, the terminal velocity (w-u)
is the positive solution of the following quadratic equation:
v 2 + 2 u v
- l g
= 0
That solution is best expressed by the following
robust formula (which never entails
any loss of precision in floating-point arithmetic, regardless
of the relative magnitudes of the parameters involved).
Terminal Velocity v = w-u
v =
v0
where (v0) 2
= l g
exp (argsh u/v0 )
This formula unifies and interpolates the following extreme regimes:
u << v0 :
Pure pressure drag. Terminal velocity is
v = v0
u >> v0 :
Pure viscous resistance.
v = (v0 ) 3/ u 2
Actual Physics :
The thrill of a closed solution to the above differential equation
does not make that differential equation physically meaningful
to begin with.
In fact, the two resistive terms in the equation correspond to
two separate speed regimes.
It turns out that each term is small in the regime where the
other is valid. Therefore,
adding up the two terms for transitional speeds is
mathematically appealing.
Such a nice approximation may lack a firm physical basis,
but I happen to like it.
A resistive force proportional to speed occurs in the low-speed (viscous) regime
where there's absolutely no turbulence in the wake of the moving object
(a so-called "laminar flow"). In that case, a formula was established
by Sir George Stokes (1819-1903)
giving the resistive force F exerted on a sphere
of radius r moving at speed v in a fluid of
dynamic viscosity
h (Stokes' Law,
1851):
F = 6p h r v
On the other hand,
quadratic drag (proportional to v 2 )
occurs in the turbulent regime normally associated with motion through air :
F = ½ r
(C S) v 2
= ½ r
(C p r 2 ) v 2
In this,
S = p r 2 is the cross-section
of the object, C is a dimensionless
drag coefficient (about 0.47 for a smooth sphere)
whereas r is the fluid's density.
Taking the above at face value for a sphere of mass m and
radius r, we may reconcile both equations with our original notations:
l =
2m / rCpr2
and
u =
6h / rCr
For dry air at 20°C (1 atm)
r = 1.204 kg/m3
and
h = 1.808 10-5 Pa.s.
So, for a sphere (C=0.47) of radius r = 1 mm,
we obtain u = 0.1917 m/s.
A sphere of density d has a mass
m = d 4pr3/3. So
l is proportional to r:
l =
r (8d / 3rC)
With the above numbers and
d = 1000 kg/m3, we obtain
l = 4.7 r.
For steel ball bearings (d = 7850 kg/m3 )
we would have l = 36.9 r.
(2010-12-12) Jet Propulsion in Outer Space
Rockets are propelled by expelling exhaust gases
at constant speed u.
Let m and v be the mass and speed of a rocket (assumed to
move along a straight line). As an infinitesimal mass -dm of gases
is expelled at relative speed -u (or absolute speed v-u) the
conservation of momentum implies:
m v = (v-u) (-dm) + (m+dm) (v+dv)
Neglecting dm dv and rearranging, we obtain
- u dm = m dv or, equivalently, dv = - u dm /m
which is readily integrated:
A speed can also be construed as an impulse per unit of mass.
Thus, the exhaust speed u is also known in the industry as
the specific impulse of a rocket engine.
It depends mostly on the type of fuel used, assuming
the velocity of the ejecta has a negligible sideways component.
For example, the
blackpowder propellant of a model rocket
may deliver an impulse of 800 N.s for each kg burned.
That's a specific impulse of 800 m/s,
which does corresponds to the exhaust speed of the ejecta.
Users of deprecated technical
systems of units may still give specific impulses in seconds !
Indeed, if you don't bother to distinguish between a pound of force (lbf)
and a pound of mass (lb), then a pound-second per pound
looks like a number of seconds.
For example, a composite
propellant containing about 60% of ammonium nitrate, 20% of magnesium,
and 20% of binder may be advertised as having a
"specific impulse" of 230 seconds.
What this means, of course, is that the true specific impulse is
equal to this number of seconds multiplied into the
standard acceleration of gravity
(namely 9.80665 m/sý, or about 32.17405 ft/sý). "230 seconds" thus
corresponds to an actual specific impulse of about 2255.5 m/s or 7400 ft/s
(again, that's simply the speed of the ejecta).
With ammonium perchlorate, rocket motors have a slightly higher specific impulse
of about 2400 m/s, or 2400 N.s per kilogram of propellant, which is three times
what ordinary black powder delivers.
Liquid rocket fuel is better still, but plasma propulsion is best for
critical propulsion requirement in outer space.
The earliest demonstration of the practicality of such engines
was made by NASA in 1998 with the flight of the
"Deep Space 1" probe, whose electrostatic ion engine had an exhaust speed of
32000 m/s, and was thus over 10 times more efficient than the chemical
fuel of previous rocket engines.
However, the thrust of current ion engines,
is so small that they're only practical in outer space, where the
total impulse they deliver can quietly add up over time.
More traditional engines must still be used before a low orbit is reached,
because sheer thrust is essential in getting the thing off the ground
and preventing it from falling back before orbital speed is reached.
Also, residual air resistance in low orbit has to become small compared to
the minuscule thrust of an ion engine,
which may only be a fraction of a newton (N).
(2012-07-28) Riccatti Equation (Jacopo Riccati, 1720)
y' = y 2 + a (x) y + b (x)
This is the standard form of a differential equation where the derivative of the
unknown (y) is a quadratic function of that unknown.
Any such equation can indeed be reduced to the above form, at least piecewise,
by changing the variable (x) so that the coefficients of y' and y 2
are equal and can thus be factored out (on any interval of x where they don't vanish).
Introducing a new function z defined via y = -z'/z we have:
-z''/z + (z')2/z2
=
(-z'/z)2 + a (x) (-z'/z)
+ b (x)
Cancelling like terms and multiplying by (-z) this boils down to:
0 = z''
- a (x) z'
+ b (x) z
This is just a second-order linear (and homogeneous) differential equation.
Conversely, any homogeneous second-order linear differential equation can be transformed into
a Riccati equation by the reverse process. Those two types of differential equations
are thus, essentially, of the same difficulty.
Note that the middle term can be eliminated by looking for z in the form
of a product of two functions z = k(x) u(x) which yields:
0 =
( k'' u + 2 k' u' + k u'' )
-
a (x) ( k' u + k u' )
+ b (x) k u
Let's choose for k a solution of the equation 2 k' = a (x) k .
We obtain:
0 = k u'' +
[ k'' -
a (x) k' + b (x) k ] u
Using k' = ½ a k
and
k'' = ½ a' k +
¼ a 2 k this becomes:
(2019-06-30) Sturm-Liouville Theory (1834)
Second-order differential equations of the form (p y')' + (q+lr) y = 0
p, q and r depend on the variable x and
l is a parameter.
The function r (sometimes denoted w) is called
weighing function, weight or density.
The main result is that, for prescribed boundary conditions, this equation has
nonzero solutions only for discrete values of the parameter l,
called eigenvalues. Those solutions are called eigen functions.
The eigenfunctions ym and yn associated to two different
eigenvalues lm
and ln are orthogonal in the following sense:
(2019-07-23) Frobenius method (about z = 0).
Solving the differential equation
u'' + u' p(z)/z + u q(z)/z2 = 0
In this, the functions p and q are assumed to be
analytic about z = 0, with a nonzero
radius of convergence.
(The discussion could be generalized to the neighborhood of any point z0
besides zero.)
3 definitions of power-series (of the dummy-variable z) are relevant:
A Taylor series (convergent or not) entail nonnegative powers of z.
Taylor series form a ring (with convergent Taylor series as a subring).
A Laurent series is the product of zm
into a Taylor series of nonzero constant. The integer m (possibly negative)
is called the index of the zerie. The zero series is also considered a Laurent series
(of undefined index) so they form an ordinary field.
A Frobenius series, is the generalization where the index m may have
any real value (it need not be an integer). Frobenius series also form a field.
(2019-07-23) Fuchs' Theorem (1901)
On the validity of Frobenius method.
About either an ordinary point or a regular-singular point of a second-order
differential equation, the method of Frobenius
gives a number of solutions equal to the number of roots of the associated indicial equation.
