home | index | units | counting | geometry | algebra | trigonometry | calculus | functions
analysis | sets & logic | number theory | recreational | misc | nomenclature & history | physics

Final Answers
© 2000-2023   Gérard P. Michon, Ph.D.

Differential Equations

 Alan M. Turing
Science is a differential equation.
Religion is a boundary condition.

 Alan Turing  (1912-1954)
 Michon
 
 border
 border
 border

See also:

Related Links (Outside this Site)

SOS Math - Differential Equations
Ordinary Differential Equations & Partial DE  by  Eric Weisstein
 
Wikipedia :   Differential Equations   |   Ordinary Differential Equations   |   Cauchy-Lipschitz theorem
Dirichlet boundary conditions   |   Neumann boundary conditions   |   Robin boundary conditions

A difficult differential equation (3:09)  by  Edward Teller  (Web of Stories).
 
Tourist's Guide to Differential Equations  by  Grant Sanderson  (3Blue1Brown):
Differential equations, studying the unsolvable (27:15)  DE1  (2019-03-31).
What is a partial differential equation? (17:38)  DE2  (2019-04-21).
Solving the heat equation (14:12)  DE3  2019-06-16).

 
border
border

Differential Equations  (DE)


brentw (Brent Watts of Hickory, NC. 2000-11-21)
How do I solve the following differential equations?
  1. [sin (xy) + xy cos (xy)] dx + [1 + x2 cos (xy)] dy = 0
  2. [4xy3 - 9y2 + 4xy2] dx + [3x2y2 -6xy + 2x2y] dy = 0
  3. (3x - y - 5) dx + (x - y + 1) dy = 0

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.

 Come back later, we're
 still working on this one...

Carl Runge (1956-1927) is pronounced "Rungue" as in "lingual"   |   Martin Kutta (1867-1944)
Runge-Kutta methods
 
Introduction to Runge-Kutta Methods (5:56)  by  Jacob Bishop  (2013-09-20).


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 y.

du   =   y (y dx + 2x dy)

Multiplying both sides of the original equation by  y  turns it into:

du  =   y2 exp(y)  dy

Integration by parts then gives  u,  so we obtain the following expression:

x y2   =   (y2 - 2y + 2)  exp(y)  +  C

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:

l dv    =     [ w2 - (u+v)2 ]  dt Therefore :
2 w dt   =     2 w l dv     =     l dv    +    l dv
Vinculum Vinculum Vinculum
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   ]
Vinculum
(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 ) ]
Vinculum Vinculum
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
Vinculum
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):  Stokes

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. 

Poiseuille's Law   |   Stokes' Law   |   Air Resistance   |   Terminal Velocity of a Skydiver


(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:

The Rocket Equation   (Moore, 1810.  Tsiolkovsky, 1903)
v   =   v0  +  u × Log  (  m0
Vinculum
m 
 ) or m0
Vinculum
m 
  =   exp  (  v - v0
Vinculum
u
 )

Rocket Engines :

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).

The rocket equation   |   Konstantin Tsiolkovsky (1857-1935).
 
The Rocket Equation (19:46)  by astronaut  Chris Hadfield  & mathematician  Matt Parker  (2020-09-04).


(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:

0   =   u''  +  ¼ [ 2 a'  -  a 2  +  4 b ]  u

This is a one-dimensional  Schrödinger equation  in the function  u.

Riccati equations  by  Mohamed Amine Khamsi  |  Riccati equation (Wikipedia)  |  Jacopo Riccati (1676-1754)


(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 functionweight  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:

 Come back later, we're
 still working on this one...

Sturm-Liouville theory   |   Rayleigh ratio   |   Spectrum
Charles-François Sturm (1803-1855)   |   Joseph Liouville (1809-1882)
 
Introduction  by  Ryan C. Daileda  (2012-04-10).
 
Sturm-Liouville Theorem and Proof (9:18)  by  Khan  (Faculty of Khan, 2016-08-12).
 
Henry Margenau's physics class (5:02)  by  Murray Gell-Mann  (Web of Stories, 1998).


(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:

  • Taylor series  (convergent or not)  entail nonnegative powers of z.  Taylor series form a  ring  (with convergent Taylor series as a subring).
  • 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.
  • 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.

 Come back later, we're
 still working on this one...

Example   |   Frobenius method   |   Georg Frobenius (1849-1917)


(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.

 Come back later, we're
 still working on this one...

Regular singular point   |   Fuchs' Theorem   |   Lazarus Fuchs (1833-1902)

border
border
visits since July 28, 2007
 (c) Copyright 2000-2023, Gerard P. Michon, Ph.D.