Mathematical Methods For Physicists Webber/Arfken Selected Solutions Ch. 8

Physics 451

Fall 2004 Homework Assignment #12 — Solutions

Textbook problems: Ch. 8: 8.2.2, 8.2.5, 8.2.6, 8.2.10, 8.2.16 Chapter 8 8.2.2 The Laplace transform of Bessel’s equation (n = 0) leads to (s2 + 1)f 0 (s) + sf (s) = 0 Solve for f (s) This equation is amenable to separation of variables Z

s df =− 2 ds f s +1

⇒ ⇒

Z df s =− ds 2 f s +1 ln f = − 12 ln(s2 + 1) + c

Exponentiating this and redefining the consant, we obtain f (x) = √

C s2 + 1

8.2.5 A boat, coasting through the water, experiences a resisting force proportional to v n , v being the instantaneous velocity of the boat. Newton’s second law leads to m

dv = −kv n dt

With v(t = 0) = v0 , x(t = 0) = 0, integrate to find v as a function of time and then the distance. This equation is separable k dv = − dt n v m For n 6= 1, this may be integrated to give Z

v

v0

dv 0 k =− 0n v m

Z

t

dt

0

⇒

0

⇒

  1 1 1 k − − n−1 = − t n−1 n−1 v m v0  −1/(n−1) (n − 1)k −(n−1) v(t) = v0 + t m

(1)

This may be integrated once more to obtain x as a function of t −1/(n−1) Z t Z t (n − 1)k 0 −(n−1) 0 0 v0 v(t )dt = + t dt0 x(t) = m 0 0 Although this may look somewhat scary, it is in fact trivial to integrate, as it is essentially t0 (plus a constant) to some fractional power. The only difficulty is bookkeeping the various constants. For n 6= 2, the result is  1−1/(n−1) t 1 m (n − 1)k 0 −(n−1) x(t) = v0 + t 1 − 1/(n − 1) (n − 1)k m 0 (2) " # (n−2)/(n−1) m (n − 1)k −(n−1) −(n−2) = v0 + t − v0 (n − 2)k m If desired, the position and velocity, (2) and (1) may be rewritten as " # (n−2)/(n−1) m (n − 1)kv0n−1 x(t) = 1+ t −1 m (n − 2)kv0n−2 −1/(n−1)  (n − 1)kv0n−1 v(t) = v0 1 + t m As a result, it is possible to eliminate t and obtain the velocity as a function of position  −1/(n−2) (n − 2)kv0n−2 x v = v0 1 + (3) m Note that we may define xk =

m (n − 2)kv0n−2

which represents a length scale related to the resisting force and initial velocity. In terms of xk , the velocity and position relation may be given as  −1/(n−2)  v n−2 v x x 0 = 1+ or =1+ v0 xk v xk Note that, in fact, it is possible to obtain (3) directly from Newton’s second law by rewriting it as dv k k = − v dt = − dx v n−1 m m and then integrating   Z v Z dv 0 k x 0 1 1 1 k =− dx ⇒ − − n−2 = − x 0n−1 n−2 m 0 n−2 v m v0 v0 v  v n−2 (n − 2)kv0n−2 0 ⇒ =1+ x v m

So far, what we have done does not apply to the special cases n = 1 or n = 2. For n = 1, we have   k v k dv = − dt ⇒ ln =− t ⇒ v(t) = v0 e−(k/m)t v m v0 m Integrating once more yields x(t) =

mv0 (1 − e−(k/m)t ) k

v kx =1− v0 mv0

⇒

which is in fact consistent with setting n = 1 in (3). For n = 2, we have k dv = − dt 2 v m

⇒

1 1 k − + =− t v v0 m

Integrating this for position yields   kv0 m ln 1 + t x(t) = k m

⇒

v(t) =

v0 1 + (kv0 /m)t

v  kx 0 = ln m v

⇒

8.2.6 In the first-order differential equation dy/dx = f (x, y) the function f (x, y) is a function of the ratio y/x: y dy =g dx x Show that the substitution of u = y/x leads to a separable equation in u and x. If we let u = y/x, this means that y = xu. So, by the product rule du dy =x +u dx dx The above differential equation now becomes x

du + u(x) = g(u) dx

⇒

du dx = g(u) − u x

which is separated in u and x. 8.2.10 A certain differential equation has the form f (x)dx + g(x)h(y)dy = 0 with none of the functions f (x), g(x), h(y) identically zero. Show that a necessary and sufficient condition for this equation to be exact is that g(x) = const.

The check for exactness is ∂ ∂ f (x) = (g(x)h(y)) ∂y ∂x or 0=

dg(x) h(y) dx

Since h(y) is not identically zero, we may divide out by h(y) (at least in any domain away from isolated zeros of h), leading to dg(x)/dx = 0, which indicates that g(x) must be constant. 8.2.16 Bernoulli’s equation dy + f (x)y = g(x)y n dx is nonlinear for n 6= 0 or 1. Show that the substitution u = y 1−n reduces Bernoulli’s equation to a linear equation. For n 6= 1, the substitution u = y 1−n is equivalent to y = u1/(1−n) . Thus 1 1 dy du du = u1/(1−n)−1 = un/(1−n) dx 1−n dx 1−n dx Bernoulli’s equation then becomes 1 du un/(1−n) + f (x)u1/(1−n) = g(x)un/(1−n) 1−n dx Multiplying by u−n/(1−n) gives 1 du + f (x)u = g(x) 1 − n dx or

du + (1 − n)f (x)u = (1 − n)g(x) dx