18.034 Problem Set #1Due by Friday, February 13, 2009, by NOON. 1. (a) Verify that y = xa solves the differential equation x2 y = 2y if the constant a satises the equation a2 a 2 = 0. Thus get the two solutions x2 and x1 . Note that the rst is valid on the whole interval < x < but the second on < x < 0 or 0 < x < only. This behavior is typical for a broad class of linear homogeneous equations known as equations of Euler type. For this class the substitution y = xa always lead to an algebraic equation for a. (b) The equation x2 y = 2y admits a solution y = xa , where a is a nonzero constant. What are the possible values of a? 2. Suppose a function y = f (x) satises the differential equation dy = 4y sin 2x dx and the initial condition y() = e. The purpose of this exercise is to nd y(/6). (a) Separate variables and integrate to obtain ln y = c 2 cos 2x, y > 0. By use of the initial condition show that c = 3, and then get y(/6). (b) The initial condition y() = e means that x = corresponds to y = e. Likewise, x = /6 corresponds to y = a, where a = y(/6). Integrating between corresponding limits gives a /6 dy = 4 sin 2x dx. e y Evaluate the denite integrals and solve the resulting equation for a. (c) If xdy + 3ydx = 0 and y() = e you cant nd y(). Why not? 3. Birkhoff-Rota, pp. 5, #3. 4. (a) Show that y1 (x) = 0 and y2 (x) = x3/2 are solutions for x 0 of the differential equation y = (3/2)y 1/3 . (b) Discuss that all nonnegative solutions of the differential equation in part (a) starting at (0, 0) lie between two solutions y1 and y2 and that the solution 0 for x < c, y(x) = 3/2 (x c) for x c lls out the funnel between them. 5. Birkhoff-Rota, pp. 11, #7. (Typo. k is n.) 6. (The Bernoulli equation.) It is a differential equation of the form y + p(x)y = q(x)y n with n = 1. (a) Show that it becomes linear by the change of variable u = y 1n . (Hint. Begin by dividing both sides of the equation by y n .) (b) Solve the Bernoulli equation y + y = xy 3 using the method in part (a).

This trick was found by Leibniz in 1696.

1

18.034 Solutions to Problemset 1

Spring 2009

1. (b) a = 1 or a = 4 2. (a) y(/6) = e2 (b) Same as part (a) (c) Because x = 0.dx x

is not integrable on any interval containing the point

3. (a) y is increasing because y = y 2 + 1 > 0. The formula is obtained by the separation of variables. (b) y(x) = tan(x c) is dened on the interval (c /2, c + /2). 4. (b) Let c = sup{x : y(x) = 0}. If c = +, then y = y1 . If c < +, then y(x) = 0 for x c and y(x) = (x c)3/2 for x > c by the separation of variables and uniqueness. 5. Let q(x) = an xn + an1 xn1 + . . . + a0 , an = 0 an The general solution y(x) = c/x+ n+1 xn + an1 xn1 +. . .+a0 , (x = 0) n where c is constant admits the only polynomial solution when c = 0.

6. (a) (b)

1 1n u + p(x)u = y = u1/2 where

q(x). u 2u = 2x for y = 0.

1

18.034 Problem Set #2Due by Friday, February 20, 2009, by NOON. Notation. = d/dx. 1. Let f (x) =

x/|x| k

for x = 0, for x = 0,

where k is a constant. Show that no matter how the constant k is chosen, the differential equation y = f (x) has no solution on an interval containing the origin. 2. Suppose that f be a continuous bounded function for the entire real axis. If f is continuous, then show that the nonzero solution of the initial value problem of y = yf (y) with y(0) = y0 = 0 exists for all x. (You may need to assume the uniqueness theorem. ) 3. Brikhoff-Rota, pp. 20, #9. 4. (The Ricatti equation) It is the differential equation of the form y = a(x) + b(x)y + c(x)y 2 . In general the Ricatti equation is not solvable by elementary means. However, (a) show that if y1 (x) is a solution then the general solution is y = y1 + u, where u is the general solution of a certain Bernoulli equation (cf. pset #1). (b) Solve the Ricatti equation y = 1 x2 + y 2 by the above method. Ly = y + y. We are going to nd the rest solution of the differential equation Ly = 3 sin 2x + 3 + 4ex . That is the solution with u(0) = u (0) = 0. (a) Find the general solution of Ly = 0. (b) Solve Ly = 3 sin 2x, Ly = 3, and Ly = 4ex by use of appropriate trial solutions. (c) Determine the constants in y(x) = c1 cos x + c2 sin x sin 2x + 3 + 2ex to nd the solution. 6. (Eulers equi-dimensional equation) It is a differential equation of the form x2 y +pxy +qy = 0, where p, q are constants. (a) Show that the setting x = et changes the differential equation into an equation with constant coefcients. (b) Use this to nd the general solution to x2 y + xy + y = 0. (c) For which values of p, the general solutions of x2 y + pxy + 2y = 0 are dened for the entire real axis (, )? 5. Let

This was shown by Liouville in 1841.

1

18.034 Solutions to Problemset 2

Spring 2009

1. y(x) =

x + c1 x < 0 for some c1 and c2 . x + c2 x>0

y is continuous at x = 0 c1 = c2 = 0. x + c1 x < 0 But y(x) = is not dierentiable at x = 0. x + c2 x>0 2. As long as the solution is dened, ness).y y

= f (y) (y is never zero by unique

(log |y|) |f (y)| M by assumption. Then |y(x)| |y0 |eM x (1)

For a > 0, consider the rectangle {(x, y) : |x| a, |y| |y0 |eM a }. If the solution does not exist on x (a, a), then |y(a)| = |y0 |eM a . This contradicts (??). 3. Let = a + ib, = c + id. In terms of polar coordinate functions, y = Im() cos() Re(( + i)ei ) Re(ei )bc a+d .

So y changes signs at slopes b/a, and

4. (a) u solves the DE y + (b(x) + 2c(x)y1 (x))y + c(x)y 2 = 0.1 (b) y1 (x) = x, u(x) = x+c .

5. (a) c1 sin x + c2 cos x (b) sin 2x, 3, 2ex (c) b sin x sin 2x + 3 + 2ex 1

6. (a) u + (p 1)u + qu = 0, = (b) sin log |x|, cos log |x| (c) No solutions

d dt

2

18.034 Problem Set #3Due by Friday, March 6, 2009, by NOON .

1. This problem pertains to the differential equation y + 2 y = sin 0 t, where = 0 and 0 is close to but different from . sin 0 t (a) Verify that y1 (t) = 2 2 is a particular solution. 0 (b) As 0 show that one of the initial conditions y1 (0) or y1 (0) becomes innite. sin 0 t sin t (c) Check that y2 (t) = is the particular solution for which the initial conditions 2 2 0 remain nite as 0 .

(d) By lHospitals rule show that the limit as 0 of y2 (t) gives a particular solution of y + 2 y = sin t. 2. Let f (x) and g(x) be two solutions of the differential equation y = F (x, y) in a domain where F satises the condition: y1 < y2 Show that implies F (x, y2 ) F (x, y1 ) L(y2 y1 ). if x > a.

|f (x) g(x)| eL(xa) |f (a) g(a)|

3. Very that (sin x)/x, x satisfy the following equations, respectively, and thus obtain the second solution. (a) xy + 2y + xy = 0 (x > 0), (b) (2x 1)y 4xy + 4y = 0 (2x > 1). 4. (a) Birkhoff-Rota, pp. 57, #4. (Typo. I(x) = q p2 /4 p /2.) (b) Birkhoff-Rota, pp. 57, #7(a). (Use part (a) instead of #6 as is suggested in the text.) (c) Birkhoff-Rota, pp. 57, #7(b). 5. Let (cosh x)y + (cos x)y = (1 + x2 )y for a < x < b and let y(a) = y(b) = 1. Show that 0 < y(x) < 1 for a < x < b. 6. (a) Birkhoff-Rota, pp. 75, #3, (b) Birkhoff-Rota, pp. 75, #4.

It is called a one-sided Lipschitz condition.1

18.034 Solutions to Problemset 3

Spring 2009

1. (b) y1 (0) = (c) y2 (0) =

1 +0 0 1 +0

+ as 0 . 21 .

(d) lim y2 (t) = 0

1 t cos t. 2

2. Birkho-Rota, pp. 28, Theorem 5. 3. (a) c1 cos x + c2 sin x x x (b) c1 x + c2 e2x 4. (c) 1 n(n + 1) + (n + y2 )2 . Compare the solution with 2 )2 (1 x 1 x2 cos(n + y2 )x.

5. Suppose u(x) > 1 at some point a < x < b. Then u takes a positive maximum > 1 at a < c < b. Observe that u(c) > 1, u (c) = 0 and u (c) 0. At c, the dierential equation reduces to (cosh c)u (c) = (1 + e2 )u(c) 0 >0 But this is a contradiction. The case u(x) < 0 at some point is completely analogous. 6. (a) c1 ex + c2 ex + c3 eix + c4 eix or c1 ex + c2 ex + c3 cos x + c4 sin x(1+i)x/ 2 + c e(1i)x/ 2 + c e(1+i)x/ 2 + c e(1i)x/ 2 or (b) c1 e 2 4 3 x/ 2 (c cos x/ 2+c sin x/ 2)+ex/ 2 (c cos x/ 2+c sin x/ 2) e 1 2 3 4

1

18.034 Problem Set #4Due by Friday, March 13, 2009, by NOON .

1. (a) Show that if u1 is a solution of the third-order variable-coefcient linear differential equation u + p1 (x)u + p2 (x)u + p3 (x)u = 0, then the substitution u(x) = u1 (x)v(x) leads to the second-order differential equation for v : u1 v + (3u + p1 u1 )v + (3u + 2p1 u + p2 u1 )v = 0. 1 1 1 (b) Verify that u1 (x) = ex is a solution of the differential equation (2x)u +(2x3)u xu +u = 0. Use the method in part (a) to nd the general solution of the differential equation. 2. Find a particular solution of the differential equation x2 u + (1 )xu + u = x2 f (x) (a) when = and (b) when = . 3. Birkhoff-Rota, pp. 62, #2. 4. (a) Find annihilators of xm ex , xm sin x, and ex cos x. (b) Find the general solution of (D2 + 1)(D 3)2 u = 12e3x (10x + 1). 5. Consider the nth order linear homogeneous differential equation u(n) + a1 u(n1) + + an1 u + an u = 0 with real constant coefcients. (a) Show that if the differential equation is asymptotically stable then a1 , a2 , , an > 0. (b) Show that the converse to part (a) is true if all roots of its associated characteristic polynomial are real. (Hint. Assume it is false, and seek for a contradiction.) (c) Show by a counter example that the converse to part (a) is in general false. 6. (a) Show that the function y 2 is not Lipschitzian on < y < . Discuss how this failure of the Lipschitz condition is reected in the behavior of the initial value problem y = y2, y(0) = y0 > 0.

(b) Show that the function y 2/3 is not Lipschitzian in any strip |y| < h containing the origin. Discuss how this failure of the Lipschitz condition is reected in the behavior of the initial value problem y = y 2/3 , y(0) = 0.

1

18.034 Solutions to Problemset 4

Spring 2009

1. (a) u = u v + u1 v , u = u v + 2u v + u1 v , 1 1 1 u = u v + 3u v + 3u v + u1 v . 1 1 1 (b) The equation for v reduces to (2 x)v + (3 x)v = 0, so that v = c1 xex + c2 x + c3 . Hence, u = c1 x + c2 xex + c3 ex . 2. (a) {x , x } is a basis of solution of the homogeneous equation. 1 1 1 u(x) = x f (x)x dx x f (x)x dx (b) {x , x log x} is a basis of the homogenous equation. 1 u(x) = x log x f (x)x dx x f (x)x1 log x dx 3. {cos kx, sin kx} is a basis of the homogenous equation. W (cos kx, sin kx) = 1 k. So the Greens function is G(x, t) = k (cos kx sin ktcos kt sin kx) = 1 k sin k(x t). 4. (a) (D )m+1 , (D2 + 2 )m+1 , D2 2D + (2 + 2 ), respectively. (b) {cos x, sin x, e3x , xe3x } is a basis of solutions of the homogenous equation. An annihilator for e3x (10x + 1) is (D 3)2 . So, design a particular solution as c1 cos x + c2 sin x + (c3 + c4 x + c5 x2 + c6 x3 )e3x A straightforward calculation shows that up (x) = x2 (2x 3)e3x . Hence, the general solution is u = c1 cos x + c2 sin x + c3 e3x + c4 xe3x + x2 (2x 3)e3x 1

5. (a) p() is factorized in the real eld into + a and 2 + p + q, (a, p, q R). Since the equation is asymptotically stable, a, p, q are all positive. This implies that all coecients of the dierential equations are positive. (b) Suppose it is not asymptotically stable. That means a 0 is a root of the characteristic polynomial. p(a) = an + a1 an1 + . . . + an an > 0 which leads to a contradiction. (c) A counterexample is u + u + u + u = 0. 6. (a) |y 2 0| = |y| as y . So, not Lipshitzian. The solution |y 0| 2 y0 y = y of is y = as t y10 . y(0) = y0 > 0 1 ty0 |y 2/3 0| = |y 1/3 | as y 0. So, not Lipschitzian. The |y 0| y = y 2/3 initial value problem has two solutions, y1 (t) = 0 y(0) = 0 and y2 (t) = (t/3)3 .

(b)

2

18.034 Problem Set #5(modied on March 30, 2009) Due by Friday, April 3, 2009, by NOON.

1. (a) Let fn (t), n = 1, 2, . . . be continuous functions on an interval [a, b] and {fn (t)} converge uniformly to f (t) on [a, b]. Show that b b lim fn (t)dt = f (t)dt.n a a

(b) Construct {fn (t)} on [0, 1] such that the above equality does not hold true. 2. For the initial value problem x = f (t, x) with x(t0 ) = x0 , where f is continuous and Lips chitzian in the rectangle |t t0 | T and |x x0 | K with the Lipschitzian constant L, suppose the exact solution x and the Picard iterates xn all exist over one and the same interval of t. Show that on such an interval T n+1 LT |x(t) xn (t)| M Ln e , (n + 1)! where |f (t, x)| M in |t t0 | T and |x x0 | K. 3. Let f be a real-valued continuous function in the rectangle |t t0 | T and |x x0 | K. Consider the initial value problem (1) x = f (t, x), x(t0 ) = x0 , x (t0 ) = x1 . (a) Show that is a solution of (1) if and only if is a solution of the integral equation t (2) x(t) = x0 + (t t0 )x1 + (t s)f (s, x)ds.t0

(b) Let {xn } be a successive approximation for (2). That is, x0 (t) = x0 and t xn (t) = x0 + (t t0 )x1 + (t s)f (s, xn1 )ds n = 1, 2, . . . .t0

If f (t, x) is continuous and Lipshitzian with respect to x in the rectangle |t t0 | T and |x x0 | K, show that {xn } converges on the interval |t t0 | min(T, K/B) to the solution of (1), where B = |x1 | + M T /2 and |f (t, x)| M in |t t0 | T and |x x0 | K. x2 4. (a) Show that L(1/ t)(s) = /s by using the well-known formula e dx = . 2 0 (b) Use part (a) to show that L( t)(s) = 3/2 for s > 0. 2s 5. (a) Show that L(et )(s) does not exist for any interval of the form s > a. (b) For what values of k, will L(1/tk ) exist? 6. Find the functions whose Laplace transforms are the following functions: 5s 6 2 9s + 3 (a) 2 + , (b) 2 . s +4 s 9s + 6s + 19A mathematician is one to whom that is as obvious as that twice two makes four is to you. Liouville was a mathe matician. Lord Kelvin1 2

18.034 Solutions to Problemset 5

Spring 2009

1. (a) Since fn f uniformly on [a, b], for > 0 given, there exists N Z+ such that |fn (t) f (t)| < /(b a) for t [a, b] whenever n N . For n N , b b b fn (t) dt f (t) dt |fn (t) f (t)| dt <

a a a

(b) fn (t) is given by

n

01

1/2n 1/n1 for all n, 2

1

fn (t) dt = 0 1

0

f (t) dt = 0. 2. In the course of the local existence theorem, 1 |xk (t) xk1 (t)|

M Lk1

|t t0 |k for k = 1, 2, . . . k! | (t) xn (t)| = x (xk (t) xk1 (t)) k=n+1 |xk (t) xk1 (t)| = =k=n+1 M (LT )k L k! k=n+1 M (LT )n+1 (LT )k L (n+1)! k! k=0 n T n+1 eLT M L (n+1)!

3. (a) () Let F (t) = f (t, (t)) and solve x = F (t) when x(t0 ) = x0 , x (t0 ) = x1 . t d t f ( Use ) f (s, t) ds = f (t, t) + (s, t) ds. dt t0 t0 t (b) Repeat the proof of the local existence theorem by showing t (t s)f (s, xn1 (s) ds (1) |xn (t) x0 | = |x1 | |t t0 | + t0 t |x1 | |t t0 | + M |t s| ds = |x1 | |t t0 | + M |t t0 |2 2 B | t0 | t t (2) |xn (t) xn1 (t)| (t s)| (s, xn1 (s)) f (s, xn2 (s))| ds f t0 t L (t s)|xn1 xn2 | ds |t t0 |2n (2n)! where L is the Lipshitz constant. 1 st 1 x2 s2x 2 4. (a) L = e dt = e dx, where x = st. x s t t 0 0 2 x2 = e dx = /s s 0 est est 1 st dt by parts. (b) L[ t] = e t dt = t s 0 s 2 t 0 0 1 1 = 0 + est dt = /2s3/2 2s 0 t M Ln1 2

t0 t0

5. (a)

t2 L[e ]

=

0

et

2 st

dt is indenite for every real value of s, no

matter how large, since t2 st > 0 for t > s and so t2 st t2 st 0 e dt > e dt > e dt = 0 s s

1 1 (b) L k = est k dt, (s > 0). The trouble here is when t = 0. t t 0 dt st 1 st 1 and therefore Near t = 0, e e dt = k tk 0 0 t t1k k= 1 . Therefore, L 1 exists for k < 1. 1 k tk 0 log t|0 k = 1 6. (a) 5 cos 2t 3 sin 2t + 2. (b) et/3 cos 2t.

3

18.034 Problem Set #6(modied on April 8, 2009) Due by Friday, April 10, 2009, by NOON.

1. (Laplace transform of tr ). The Gamma function is dened by the integral (r + 1) = et tr dt.0

(a) Show that the improper integral converges for all r > 1. (b) Show that (r + 1) = r(r) for r > 0. Show that (1) = 1, (1/2) = (c) For r > 1 show that L[tr ] = (r + 1)/sr+1 , s > 0 2. (a) Find the solution of the initial value problem y + 2 y = h(t) sin t h(t c) sin t, where c > 0 is a constant and 2 = 1. (b) Show that y(0) = y (0) = y (0) = 0. Show that y and y are continuous at t = c. (c) Show that y (c+) y (c) = sin c, which is 0 if and only if c = n for n an integer. This behavior is explained by that the function h(t c) sin t is continuous at c if and only if sin c = 0. 3. Find the Laplace transform of a full rectied wave f (t) = | sin t|. 4. Find the solution of the initial value problem y + y = A(t c), where A, a, b, c are constant and c > 0. (b) Show that y(t) = 0 for t c if and only if y(c) = 0 and A = a2 + b2 , a sin c b cos c; A = a2 + b2 , y(0) = a, y (0) = b, y(0) = y (0) = 0, .

a sin c b cos c.

To interpret, these amplitudes A and locations of impulse c cancel the oscillation. 5. (The Volterra integral equation). Consider the integral equation t 1 y(t) + (t s)y(s)ds = sin 2t. 4 0 (a) Show that the above integral equation is equivalent to the initial value problem y + y = sin 2t, y(0) = 0, 1 y (0) = . 2

(b) Solve the integral equation by using the Laplace transform.

It describes a direct current.

1

6. Consider the Bessel equation of order zero ty + y + ty = 0. Note that t = 0 is a singular point and thus solutions may become unbounded as t 0. Never theless, let us try to determine whether there are any solutions that remain nite at t = 0 and have nite derivatives there. (a) Show that Y (s) = L[y](s) satises (1 + s2 )Y (s) + sY (s) = 0. (b) Using the binomial series for (1 + s2 )1/2 for s > 1 show that y(t) = c (1)n t2n n=0

22n (n!)2

,

which is referred to as the Bessel function of the rst kind of order zero. Show that y(0) = 1 and y has nite derivatives for all orders at t = 0.

2

Problem set 6, Solution keys 1

1. (a)0

e t dt =0

t r

e t dt +1

t r

et tr dt := (I) + (II).

For r > 1. 1 (I) tr dt < +. (II) (b) (r + 1) =0 0

en nr < +

n=1

et tr dt = et tr | + o = r(r) (1/2) =

0

ret tr1 dt (Integration by parts)

2 dt = 2eu du = 0 0 0 1 L(r + 1) (c) L(tr ) = et tr dt = eu (u/s)r (1/s)du = ( r+1 ) eu (u)r du = s sr+1 0 0 0

(1) =

e dt = 1,

t

e t

t 1/2

2. (This is a long one.) (a) L[h(t c) sin t] = = e = e0 sc sc

est sin tdt L[sin(t + c)]

L[sin t cos c + cos t sin c] cos c + s sin c = esc s2 + 1 Taking the transform, L(y) = = So, y(t) = (b) (c) 1 1 cos c + s sin c ( esc ) s2 + w 2 s2 + 1 s2 + 1 1 1 1 ( )(1 esc (cos c + s sin c)) 2 2 1 s2 + 1 w s + w2 cos w(t c) sin c))

1 (sin t sin wt h(t c)(sin t sin w(tc) cos c w w w2 1 1 sin wc 1 (c) = y(c) = w2 1 (sin c w ), y (cos c cos wc). w2 1 (c+) y (c) = 1 ( sin c + w 2 sin c). y w2 1

1

3. Let f0 (t) = sin t Lf =Lf0 1es

0 t < .1 1+es . s2 +1 1es

Lf0 =

1 (1 s2 +1

+ es )

=

Alternatively, compute term by term in f (t) = sin t + 2

n=1

h(t n) sin(t n)

4. (a) Taking the transform, s2 Ly (sa + b) + Ly = Aesc. So, y(t) = Ah(t c) sin(t c) + a cos t + b sin t. (b) The reason why y(c) = 0 is necessary is: at a point y(c) = 0, the impulse can control the derivative and reduce y (c) to be zero. Then y(t) = 0 for t > c by uniqueness. But, at a point y(c) = 0, no choice of the impulse will make y (c) = 0. In general in y + ay + by = f (t), a, b constants, the eect of f (t) f (t) + (t) is the same as the eect of y (0) y (0) + c. t d t f 5. (a) Uses f (s, t)ds = f (t, t) + (s, t)dt. dt t0 t0 t (b) Taking the transform, Ly + So, y(t) =1 6 1 Ly s2 s = 1 s21 , and Ly = 1 (s2 +1)(s2 +4) . 2 +4 22

sin t + 1 sin 2t 3

d 6. (a) Uses L[y ] = ds (s2 Y (s)) = s2 Y (s) 2sY (s).

(b) (This is again long.)1 Y s = 2 , so Y (s) = c(1 + s2 ) 2 s +1 Y 1 (1)n (2n)! 2 ) 1 , Y (s) = c 2 s2n = c 2 Using the binomial series for (1+s . s n 22n (n!)2 s2n+1

n=0

n=0

(2n)! , y(t) = c s2n+1 22n (n!)2 n=0 0 k=2n+1 Clearly, y(0) = 0. y (k) (0) = (1)n (2n)! k=2n 22n (n!)2 Since L[t2n ] =

(1)n t2n

2

18.034 Problem Set #7(modied on April 13, 2009) Due by Friday, April 17, 2009, by NOON.

1. In many applications the input f and the output y are related by L2 y = L1 f , where L1 and L2 are linear differential operators. (a) If Pj is the characteristic polynomial of Lj , j = 1, 2, show that P2 (s)Y (s) = P1 (s)F (s) + P0 (s), where P0 (s)is a polynomial depending on the initial conditions of both f and y. (b) The null initial condition gives the transfer function W (s) = P1 (s)/P2 (s). If W (s) = Lw, express y by the convolution theorem. (c) We can have a function w as in part (b) only if the degree of P1 is less than that of P2 . Why? What happens to the formula of part (b) if P1 = P2 ? 2. One method of determining the ratio e/m of charge to mass for an electron leads to the system mx + Hey = eE, my Hex = 0. The initial values are x(0) = x (0) = y(0) = y (0) = 0 and m, e, H, E are positive constants. (a) Solve by the Laplace transform and show that the path of a particle in the coordinates (x, y) are given by x(t) = E(H)1 (1 cos t), y(t) = E(H)1 (t sin t), where = He/m. This is a cycloid generated by a circle of radius E(H)1 . (b) How can one determine the system-parameter e/m from knowledge of the path and the values E and H set by the experimenter? 3. Consider the system of differential equations Y = AY , where y11 (t) y12 (t) a11 (t) a12 (t) Y = , A = . y21 (t) y22 (t) a21 (t) a22 (t) (a) If U (t) is a fundamental solution and C is a nonsingular matrix, show that U C is also a funda mental solution. (b) If V (t) is a fundamental solution, show that U (t) = V (t)V (t0 )1 satises U (t0 ) = I and is also a fundamental solution. (c) Show that| | = D1 + D2 , where Y a11 y11 + a12 y21 a11 y12 + a12 y22 = a11 |Y | D1 = y21 y22 and D2 = a22 |Y |. Deduce the Liouville theorem. 4. (a) Show that if (y1 (t), y2 (t)) is a solution of t2 y1 = 2ty1 + 4y2 , ty2 = 2ty1 + 5y2

then both y1 and y2 are solutions of the Euler equation (t2 + 2t) = 8 + 5(t + 2).1

2t1 4t2 (b) Find a fundamental matrix for 2 5t1 1 2t 2 (c) For A = 2 2 obtain a fundamental matrix. t t + 2t 2 5. Solve the initial value problem d y1 1 4 y1 = , 2 1 y2 dt y2 y1 (0) 0 = y2 (0) 3

(a) by using the eigenvalues and (b) by using the Laplace transform. (c) For which values of a and b, the solution of the initial value problem d y1 1 4 y1 y1 (0) a = , = 2 1 y2 y2 (0) b dt y2 has the behavior limt (y1 (t), y2 (t))T = (0, 0)? 6. For which vector (a, b, c), is the solution of the initial value problem y 1 0 2 y1 y1 (0) a d 1 y2 (0) = b y2 = 0 1 0 y2 , dt y 1 1 1 y y (0) c3 3 3

periodic in t?

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Problem set 7, Solution keys

1. (a) (b) Y (s) =P1 (s) P2 (s) F (s)

= W (s)F (s).

y =wf

(c) Because W (s) 0 as s . If P1 = P2 then W (s) = 1 and w(t) = (t). So, y = f . 2. (a) (b) Once you have a cycloid:

E T = 2 W H is calculated by mesuring the distance between two cusps.

3. (a) (U C) = U C = (AU )C = A(U C) and |U C| = |U ||C| = 0 (b) V (t0 ) is a non-singular matrix. (c) Let Y = (y1 y2 ). |Y | = |y1 y2 | + |y1 y2 | = |Ay1 y2 | + |y1 Ay2 | = D1 + D2 = (a11 + a22 )|Y |. 4. (a) (b) By setting (t) = tm , (t)2 t

Liouviues theorem.

2 t2 t t3

.

1 (t) (c) By setting =e , 2 2 t t(t) t2 e t 2 1 3t 3t 5. (a) c1 e + c2 e . 1 1 a 2 (b) =c , c is an arbitrary const. b 1 3

a 6. 0 , a and c are arbitrary. c

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18.034 Problem Set #8(modied on April 27, 2009) Due by Friday, May 1, 2009, by NOON.

1. An n n complex matrix A is called Hermitian if A = A , where A is the conjugate transpose of A. That is, aij = aji for all 1 i, j n. When A is real A = AT and the terms Hermitian and symmetric mean the same thing. If = (u1 , . . . , un ) and = (v1 , . . . , vn ) are column vectors in u v n , then = u v + + u v and u2 = uu . R uv 1 1 n n If A = A show that all eienvalues of A are real. Furthermore, if A = A then eigenvectors u = 0. and corresponding to different eigenvalues and are orthogonal. That is, v uv 2. If A and B are n n matrices, compute eAt eBt eBt eAt . t0 t2 3. (a) Let A be a 3 3 matrix with eigenvalues 1 , 2 , 3 . Show that the nonzero columns of (A 2 I)(A 3 I) are eigenvectors for 1 . lim (b) A 3 3 matrix A has characteristic polynomial p() = (2 1). Find eAt . 4. (a) For x = 6x + y, y = 4x + 3y show that the origin is an unstable node. (b) If y = mx is a trajectory, show that m = 1 or m = 4. (c) Sketch the trajectories in the (x, y)-plane. 5. Repeat Problem 4 for x = 3x + 2y, y = 3x + 4y. 6. Consider the differential equation u + p(t)u + q(t)u = 0, where p(t), q(t) are continuous func tions on some interval of t. (a) Let u(t) = r(t) sin (t), u (t) = r(t) cos (t). Show that d/dt = cos2 + p(t) cos sin + q(t) sin2 , (1/r)dr/dt = p(t) cos2 + (1 q(t)) cos sin . (b) Using part (a) discuss that if q(t) > p2 (t)/4 then solutions are oscillatory and if q(t) < 0 then solutions are nonoscillatory.

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Problem set 8, Solution keys

1. (a) Let Au = u. Then u Au = u u = uu . On the other hand, u A = u A = (Au) = u . So, u Au = u u. (b) Let Au = u, Av = v, u = 0, v = 0, = Au = v u, u Av = u v. Using the argument above, v And v Au = v u. So v v = 0 2. eAt = I + tA +t2 A2 2

is real.

+ t3 R1 (t),2

eBt = I + tB +2 +B 2 2 +B 2

t2 B 2 2

+ t3 R2 (t)

So, eAt eBt = I + t(A + B) + t2 ( A

+ AB) + t3 R3 (t), + BA) + t3 R4 (t),

eBt eAt = I + t(A + B) + t2 ( A

2

Here R1 (t), , R4 (t) are continuous matrix-valued functions. Therfore, lim eAt eBt eBt eAt = AB BA (= [A, B]) t t2

3. (a) p() = ( 1 )( 2 )( 3 ) By Cayley-Hamilton, (A 1 I)(A 2 I)(A 3 I) = 0 (b) By Cayley-Hamilton, A(A2 I) = 0 eAt = I + tA + So, A2k+1 = A and A2k = A2 .

t2 A2 t3 A3 + + 2! 3! t3 t5 t2 t4 = I + A(t + + + ) + A2 ( + + ) 3! 5! 2! 4! 1 1 2 t = I + A2 + A(et et ) + A (e + et ). 2 2

4. (a)P () = 2 9 14 ... unstable node. (b) m =4+3m 6+m ,

m2 + 3m 4 = 0 So, m = 1 or m = 4.

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(c) 5. (a) Saddle. (b) m =1 2

or 3.

(c) 6. Birkho-Rota pp.50-51.

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18.034 Problem Set #9(modied on May 6, 2009) Due by Friday, May 8, 2009, by NOON.

1. (a) If V (x, y) is continuously differentiable, then each solution curve of the plane autonomous system V V x = (x, y), y = (x, y) y x lies on some level curve V (x, y) = constant. (b) Show that the solution curves of the autonomous system x = x(2y 3 x3 ), 2. Consider the plane autonomous system x = f (x, y), y = g(x, y), where f, g are continuously differentiable. (a) If limt (x(t), y(t)) = (x1 , y1 ), show that (x1 , y1 ) is a critical point. (b) Near a critical point (x0 , y0 ) the system can be written as x x0 a b x x0 = , y y0 c d y y0 where a, b, c, d are functions of (x, y) satisfying a b fx (x0 , y0 ) fy (x0 , y0 ) lim = . gx (x0 , y0 ) gy (x0 , y0 ) (x,y)(x0 ,y0 ) c d 3. Find the critical points of the competitive system x = x(3 2x y), y = y(3 x 2y) and determine the stability and behavior near the critical points. 4. Repeat Problem 3 for the predator-prey equation with self-limiting x = x(1 x + y), y = y(3 x y). 5. (a) Show that x = f (x) has an asymptotically stable critical point 0 if and only if 0 < |x| < implies xf (x) < 0 for some > 0. (b) Show that x = f (x) has an asymptotically stable critical point 0 if f (0) = 0 and f (0) < 0. 6. (a) Let x sin(1/x) f (x) = 0 x = 0, x = 0. y = y(2x3 y 3 ) are the curves x3 + y 3 3cxy = 0, where c is an arbitrary constant. Sketch typical solution curves.

Show that the critical point 0 of x + f (x) = 0 is neutrally stable, but not asymptotically stable. (b) Show that the critical point (0, 0) of the plane autonomous system x = y x3 , is stable, although its linearization is unstable.1

y = x3

Problem set 9, Solution keys

1. Birkho-Rota pp. 135-136, Theorem 1 and Example 3. Folia of Descartes are in gure 5.2. 2. (a) Suppose not. This means at least one of the inequalities f > 0, f < 0, g > 0, g < 0 holds at (x1 , y1 ). Without loss of generality, let f (x1 , y1 ) = 2 < 0. (other cases are similar). By continuity, f (x(t), y(t)) < for t > T for some large T > 0. Hence, x (t) for t > T , and x(t) t + for some for t > T . Then x(t) as t , which contradicts that x(t) x1 . (b) Without loss of generality, let x0 = y0 = 0. Let F (t) = f (xt, yt) in a disk r < for some > 0. Hence r = x2 + y 2 . By the Mean Value Theorem, F (1) F (0) = F ( ) for some 0 < < 1. And, F ( ) = fx (x, y )x + fy (x, y )y. Since fx and fy are continuous, |fx (x, y )fx (0, 0)| < (r), |fy (x, y )fy (0, 0)| < (r) and (r) 0 as r 0. Therefore a fx (0, 0), b fy (0, 0), as r 0. A similar argument applies to g. 3. (0, 0) unstable singular node ( 3 , 0), (0, 3 ) saddles. 2 2 (1,1) stable nodes.

4. (0, 0), (0, 3) saddles. (1,2) stable focus.

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5. (a) Birkho-Rota pp. 153. Theorem 5. (b) Let E(x) = x2 a Lyapunov function. E(0) = 0 and E(x) > 0 for x = 0. And E (x) = 2xf (x) < 0 for x = 0 and x near 0 since f (0) < 0. Therfore 0 is asymptotically stable.

1 6. (a) In addition to x = 0, x = n , n = 1, 2, 3, ... are critical points. Let o < |x(0)| 1 so 1 1 that (n+1) < |x(0)| n for some n large. 1 1 Then (n+1) < |x(t)| < n for all t since a non-stationary solution cannot pass through a critical point. Therefore |x(t)| < n+1 |x(0)| 2|x(0)|, and 0 is stable. n 1 1 Take xn (0) = n , n = 1, 2, 3... . Then xn (t) = n for all t. So ,0 is not asymptotically stable. x 0 1 x (b) The linear system is = y 0 0 y The solutions are x(t) = a + bt, y(t) = b. Therefore (0, 0) is unstable. For the nonlinear system, let E(x, y) = x4 + 2y 2 . Then, E(0, 0) = 0, E(x, y) > 0 for (x, y) = (0, 0). (x, y) = 4x3 (y x3 ) + 4y(x3 ) = 4x6 0. So, (0, 0) is stable. E

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