(The handout "Asymptotics of the Airy Equation" in dvi, postscript, or pdf formats.)
The spring may weaken, the dashpot material may harden:
y" + p(x)y' + q(x) = r(x) (*)
is still linear but without constant coefficients we lose the notion of a characteristic polynomial and we lose exponential solutions to the homogeneous equation. Linearity still tells us that we have a solution yp, that the homogeneous equation has a pair of linearly independent solutions y1, y2, and that for any such yp, y1, y2, the general solution has the form y = yp + ay1 + by2.
Most such equations do not have solutions which are elementary functions. A typical example is the Airy equation y" + xy = 0, which arises in optics and in the statics of a column. There is a handout on the long-term behavior of this equation, which is up to a change of variable an example of a "Bessel equation." One class of equations which does admit elementary solutions are the "Euler-Cauchy" equations
x^2 y" + pxy' + qy = r(x). (EC)
We'll study the homogeneous case r = 0; PS4.5 contains some hints about the inhomogeneous case. Differentiation decreases exponents by one, and multiplying by x restores the exponent. This may give the hint to try a solution of the form y = x^r for a constant r. When we make this substitution (EC) (with r = 0) becomes
r(r-1) + pr + q = 0, or r^2 + (p-1)r + q = 0.
This is what we need for x^r to be a solution. Three examples:
1. x^2 y" -2xy' + 2y = 0: r^2 - 3r + 2 = 0: r = 1, 2; so the general solution is ax + bx^2. Note that constants are not solutions.
2. x^2 y" + xy' + y = 0: r^2 + 1 = 0: r = i, -i: so x^i, x^{-i} are solutions. This is scary. Restrict to x>0, and write x = e^{ln x}; then x^i = e^{i ln x} = cos(ln x) + i sin(ln x), not so scary. [I didn't point out that this is not quite complete; x = e^{ln x + 2 pi k i} also, for any integer k; so just as the halfth power is 2-valued, the i-th power is infinitely valued: the given value can be multiplied by any integer power of e^{-2 pi}. But this doesn't change the fact that e^{i ln x} and e^{-i ln x} are independent solutions.]
3. x^2 y" - xy' + y = 0: r^2 - 2r + 1 = 0: r = 1, 1: repeated root. Now we get only one power-solution, y1 = x. (This is a solution.) To find a second solution we use a general technique:
REDUCTION OF ORDER: If you have one solution y1 to (*), look for a second of the form y = y1 u, where u is some unknown function. Find an ODE for u by substituting y1 u into (*):
Here we find xu" + u' = 0. This leads to a second general idea: this equation is REDUCIBLE: it's missing one of the ingredients u, u', u". You can regard it as a first order equation for v = u': xv' + v = 0. This is first order linear nonconstant coefficient. It's homogeneous, so separable. One would not use the integrating factor method on it, except that the LHS is obviously the derivative of a product as it stands: (xv)' = 0; xv = c. We are looking some solution - the general solution will be a linear combination of y1 and y2 in the end - so take c = 1: v = 1/x; u = ln|x| (again choosing a constant of integration); so finally y2 = x ln|x|.
Reduction of order uses one solution as a crowbar to split a second order equation into a sequence of two first order equations. It's equivalent to the Abel's equation method from homework problem 12.
Finally, a NONLINEAR example: A projectile is fired straight up with initial velocity v0 on the surface of the earth, a distance x0 from the center x=0; will it escape from the earth's gravity? F = ma reads
m x-dot-dot = -GMm/x^2
where M is the mass of the planet, m is the mass of the projectile, and G is a universal constant. This is a nonlinear second order equation, but it's reducible: it's missing x-dot. Physicists are smart, they don't try to solve explicitly for x as a function of time. Rather, regard v = x-dot as a function of x. Then by the chain rule, x-dot-dot = dv/dt = (dx/dt)(dv/dx) = vv', so the equation is
mvv' = -GMm/x^2.
This is a first order linear equation, now linear, but with nonconstant coefficients. It's separable, indeed, separated, and integrating both sides
mv^2/2 = GMm/x + c.
The constant of integration c is the ``total energy'' of the system; it's the sum of the kenetic energy and the potential energy. Since it's constant,
mv^2/2 - GMm/x = mv0^2/2 - GMm/x0.
This is an implicit solution to the ODE. Escape occurs if there is residual velocity when x goes to infinity: that is, provided
v0^2/2 is greater than GM/x0.
Now imagine a star imploding. As its radius x0 shrinks, the escape velocity increases. When it reaches the speed of light, c, even light won't escape. This occurs when
x0 = 2GM/c^2.
This is the radius of the event horizon of a black hole. This calculation should obviously be done using relativity, but I believe that when you do this you get the same answer. For the earth, x0 is a centimeter. Note that as x0 increases, M increases not as x0^3, as you might expect, but only as x0. Thus the bigger the black hole, the less dense it is. Other things being equal, dense implies hot. Tiny black holes are very hot. If they are small enough they are so hot they explode. Many were produced at the big bang but they all exploded right away. At the other extreme, you can compute the radius of a black hole whose mass is the mass of the universe. You get approximately the radius of the universe: there is some reason to say we are actually living inside the event horizon of a black hole.