Two Little Lemmas. Derek Ou, advised by Nicolas Templier

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1 Two Little Lemmas Dere Ou, advised by Nicolas Templier I: INTRODUCTION The n-body problem, first posed by Isaac Newton in his celebrated Philosophiae Naturalis Principia Mathematica in 1687, has become one of the great problems of celestial mechanics, and for general values of n remains unsolved [3]. We have seen the n-body problem many times during the course of this seminar, but for completeness, we will restate it here. Let our system have n particles. Let m i be the mass of the i th particle, r l be the distance between the th and l th particles, and i be the position vector of the i th particle. The n-body problem is represented by the following ODE: m = l m m l ( l ) r 3 l The left hand is euivalent to U, the partial derivative of the potential U with respect to. While the solution for the case of n = 2 was solved by Johann Bernoulli in 1710, the solution for even n = 3 eluded the efforts of great minds such as Euler, Lagrange, Cauchy and Poincaré for two centuries [3]. It remained unsolved until 1907, when a little-nown Finnish mathematician by the name of Karl Sundman published a series solution to the 3-body problem in the Acta Societatis Scientiarum Fennicae [2]. Even then, the solution did not become well-nown in mathematical circles until 1912, when Sundman published it again, after refinement, in a more illustrious journal, the Acta Mathematica [2]. Unfortunately this solution proved to be of no practical use due to the remarably slow rate at which it converges. Since describing motions for even small periods of time reuires summing up millions of terms, the solution is unable to usefully describe the dynamics of the 3-body system; this has led to its general decline into obscurity in the present day [3]. Nevertheless, Sundman s solution holds great mathematical interest, being the first of its ind; this led mathematicians lie Siegel to push for its study in the 1940s [2]. We discuss two lemmas Sundman used in his proof; even by themselves, they shed some light on the 3-body system. Both lemmas assume the components of angular momentum are not all 0. They are as follows: 1: The perimeter of the triangle formed by the 3 bodies stays above a positive constant. 2: The velocity of the body opposite the shortest side of our triangle stays below a finite constant. The arguments in this paper will follow in detail the proofs presented in [1]; we mae an attempt to reorganize and clarify this text, with reader comprehension as our primary goal. These steps were taen because of the observation that our text, while certainly elegantly written and probably a good tool for experts, may leave less sophisticated readers at times a bit perplexed. II: NOTATION Assumption: We assume that n = 3, since we are interested only in the system of 3 particles. Definition: Let L = (L x, L y, L z ) be the angular momentum vector. Define a constant η L 2 3. Definition: Define a constant τ to be the time corresponding to the initial or boundary value. Furthermore, let any uantity with the subscript τ denote the value of the uantity at t = τ. Assumption: Assume the center of mass of our system lies at the origin. This is valid since the euations of motion are invariant under the transformation by which we rigidly translate the center of mass, as we observed in Lecture 10. Define P 1, P 2, and P 3 to be our three particles. Definition: Let be the triangle formed by P 1, P 2, and P 3. Let be the position vector of P. Let r be the distance from O to P. When necessary, r (t) will specify r at time t. Let r j be the distance from P j to P. Finally, let σ be the perimeter of the triangle. 1

2 Definition: Define I m 2 = 3 m r 2, where m is the mass of P. Let M = m 1 +m 2 +m 3. 1 Definition: Recall the energy integral T = U + h, where h is the energy constant, T is the inetic energy, and U is the potential energy. Recall also the ineuality 2IT η from Lecture 10. Assumption: As stated before, assume that L x, L y, and L z are not all 0. Then η > 0. Definition: Define a constant A > 0 that satisfies the following ineualities: (i) I τ < A (ii) U τ < A (iii) h < A (iv) η 1 < A A can be found since (i) and (ii) have fixed values at t = τ and (iii) and (iv) are finite constants. In his proofs, Siegel zealously defined some 38 positive constants which can be expressed in terms of particle masses and A. Sundman explicitly defined his constants, but Siegel did not. In the name of precision, we explicitly define our constants in terms of A, particle masses, or other constants. III: SUNDMAN S FIRST LEMMA Lemma 1.1: σ stays above a positive constant for all time. Outline: First, we show that Lemma 1.1 holds if I remains above a positive constant defined in terms of A and m. Then we brea the problem into the cases h 0 and h < 0. The former will be (fairly) straightforward; to prove the latter, we explore the function F (x) = ηx 1/2 2hx 1/2 and prove two appropriately constructed ineualities. Without further ado, we proceed to the proof! Proposition 1.2: Iσ 2 is bounded from both above and below by positive values. Proof: It is elementary that the origin, the center of mass, is inside, since triangles are convex. Thus, the triangle j formed by the origin, P j, and P shares one edge (namely, that connecting P j with P ) with. Since j lies inside, σ j is strictly smaller than σ. Hence, the sum of the lengths of the other two edges of j is strictly smaller than the sums of the other two edges of. This gives us the following three ineualities: (i) r 1 + r 2 r 13 + r 23 (ii) r 1 + r 3 r 12 + r 23 (iii) r 2 + r 3 r 12 + r 13. Summing these three ineualities and dividing by 2, we see that r 1 + r 2 + r 3 σ. Define m max max(m 1, m 2, m 3 ), then I = m 1 r1 2 + m 2 r2 2 + m 3 r3 2 m max (r1 2 + r2 2 + r3) 2 m max (r 1 + r 2 + r 3 ) 2 m max σ 2. Thus, we have our upper bound: Iσ 2 m max. Since the sum of the lengths of any two sides of a triangle is greater than the length of the third side, the following are immediate: (iv) r 12 r 1 + r 2 (v) r 13 r 1 + r 3 (vi) r 23 r 2 + r 3. Summing these, we get σ 2(r 1 + r 2 + r 3 ). Suaring both sides and applying the Schwarz Ineuality yields σ 2 4(r 1 + r 2 + r 3 ) 2 = 4( 3 m 1/2 m 1/2 r ) (m 1/2 r ) 2 3 (m 1/2 ) 2 = 4I 3 m 1. =1 This gives a lower bound Iσ 2 (4(m 1 3 )) 1 > 0. Corollary 1.3: If I is bounded from below by a positive value, then Lemma 1.1 holds. Proof: Suppose that we can find some constant c 0 > 0 such that I > c 0 for all time. Since Iσ 2 m max, we see that σ 2 Im 1 max > c 0 m 1 max, and hence σ > (c 0 m 1 max) 1/2 > 0, as desired. Proposition 1.4: Define the function K(I) ( I 2 +4η)I 1/2 8hI 1/2. Then K and I are either both monotonically decreasing or both monotonically increasing as functions of time, t. Proof: Recall the following two euations, which were discussed and proven in Lecture 10. (i) 2Ï 1 = T + h = U + 2h (Lagrange s Formula) (ii) 2IT 1 4I 2 + η By (ii), T (2I) 1 ( 1 4I 2 +η); substituting this into (i) and multiplying by 2 yields Ï I 1 ( 1 4I 2 + η) + 2h. After rearrangement, we obtain (iii) Ï 1 4I 2 I 1 I 1 η 2h 0. Multiplying the left side of (iii) by 2II 1/2 gives (iv) 2Ï II 1/2 1 2I 3 I 3/2 2II 1/2 η 4II 1/2 h. It is simple to verify that this expression is the derivative of K with respect to I. Hence, K is the product of a non-negative uantity (the left side of (iii)) and 2II 1/2 ; 2I 1/2 is non-negative by definition, so K has the same sign as I everywhere, with the possible exception of at critical points of K but not I, corresponding to the case when 2I 1/2 (Ï 1 4I 2 I 1 I 1 η 2h) = 0. Fortunately, at these points, K is neither increasing nor decreasing and hence, I and K are either both monotonically increasing or decreasing, as desired. Remar 1.5: A simple but useful observation is that since U = m m l <l r l, where each term is a product of positive uantities, we have that < l, where, l {1, 2, 3}, m m l r lτ U τ < A. =1 =1 =1 2

3 Hence, r lτ > m m l A > 0. Let c 1 max( m1 A, m1m3 A, m3 A ); then, l {1, 2, 3}, r lτ > c 1. Remar 1.6: Another simple observation is that there exists a constant c 2 > 0 such that I τ > c 2. Recall the lower bound from Proposition 1.2: Iσ 2 (4(m 1 3 )) 1 > 0. Taing the value at t = τ, we obtain the expression I τ 1 4 (r 12τ + r 13τ + r 23τ ) 2 (m 1 3 ) 1 > 0. By Remar 1.5, I τ 9 4 c2 1(m 1 3 ) 1 > 9 5 c2 1(m 1 3 ) 1 c 2 > 0, as desired. Remar 1.7: η > 0, so by Proposition 1.4(ii), 2IT 1 4I 2 + η > 1 4I 2. Hence, 8IT > I 2 for all t, and in particular, 8I τ T τ I τ 2. But 8I τ T τ = 8I τ (U τ + h) < 8(A)(A + A) = 16A 2, hence I τ 2 < 16A 2. Now, our preliminary wor is done, and we will proceed to our two cases: h 0 and h < 0. Proposition 1.8: In the case that h 0, Lemma 1.1 holds. Proof: Since h 0 and U 0 by definition, Proposition 1.4(i) gives that 2Ï 1 = U + 2h > 0. Hence, the second derivative of I is positive, and I is a convex function (i.e. I has one critical point, which will be a global minimum). In the elementary case when I τ = 0, then the global minimum of I falls at t = τ, and we have by Remar 1.6 that I I τ > c 2 ; hence c 2 is the desired constant. Now suppose that I τ < 0. Then we may find an interval [τ, t 1 ] upon which I is monotonically decreasing. By Proposition 1.4, K is also monotonically decreasing on [τ, t 1 ]. Furthermore, since h 0, K + 8hI 1/2 is also decreasing monotonically on [τ, t 1 ]. Hence, in our interval, ( I 2 + 4η)I 1/2 = K + 8hI 1/2 K τ +8hIτ 1/2 = ( I τ 2 +4η)Iτ 1/2. II 2 is non-negative, so 4ηI 1/2 ( I τ 2 +4η)Iτ 1/2. Rearranging and multiplying by a clever form of 1, we get I 1/2 (4η)Iτ 1/2 ( I τ 2 +4η) 1 = Iτ 1/2 ( I τ 2 (4η) 1 +1) 1. Noting that all the terms on the right are positive, we suare everything while conserving the direction of the ineuality, and find that I I τ ( I τ 2 (4η) 1 + 1) 2. Now, applying Remars 1.6 and 1.7, we get I I τ ( I τ 2 (4η) 1 + 1) 2 > c 2 (4A 2 A + 1) 2 = c 2 (1 + 4A 3 ) 2 c 3. I is convex, so our lower bound for intervals of monotone decrease in fact wors globally! The case I τ > 0 is identical; we may even replace t with t to reestablish that I τ < 0, and proceed as above. (The only difference is whether the minimum happens before or after t = τ.) Statement: Now, we proceed to the case h < 0. Here, we lose the convexity condition, which made the h 0 case relatively easy. In fact, I can have an unspecified number (up to infinitely many!) of extrema. As before, it suffices to find a positive lower bound for I for all t τ. Proposition 1.9: Suppose that h < 0 and I 0 for all t τ, then Lemma 1.1 holds. Proof: I is monotonically increasing on the half-line t τ, so t τ, I I τ > c 2. Lemma 1.10: Let = 2h. If h < 0 and for some t > τ, I < 0, then in any interval [t0, t 1 ] in which I is monotonically decreasing, if I 0 is the value of I at t 0, then I I 0 (1 + η I η I0 2 ) 2. Proof: For some t > τ, I < 0, and hence we can find intervals [t0, t 1 ], where t 0 τ, in which I is monotonically decreasing. By Proposition 1.4, K is also monotonically decreasing in this interval, and hence (i) t (t 0, t 1 ), ( I 2 + 4η)(I 1/2 ) + 4I 1/2 ( I η)I 1/ I 1/2 0. Noting that > 0 and I 2 I 1/2 > 0, we find that (4ηI 1/2 ) ( I η)(I 1/2 0 ) + 4I 1/2 0. Rearranging and suaring (this is osher since every term is positive), we derive I I 0 (1 + I 0 η 1 + (4η) 1 I2 0 ) 2, as desired. In particular, this holds when we choose t 0 to be the smallest value, for each fixed t 1, such that I monotonically decreases on [t 0, t 1 ]. This will generally be our practice. Corollary 1.11: If t 0 = τ, then we have a lower bound for I on the interval [τ, t 1 ]. Proof: By Lemma 1.10, I I τ (1 + I τ η 1 + (4η) 1 I2 τ ) 2 in the interval. Using Remar 1.6 to bound I τ from below, A to bound I τ, η 1, and 2 = h from above, and Remar 1.7 to bound I τ from above, we see that I c 2 (1 + (2A)(A)(A) (A)A2 ) 2 > c 2 (1 + 3A 3 ) 2 c 4 > 0. Proposition 1.12: We note three properties of the function f(x) = ηx 1/2 + x 1/2, defined for positive values of x: (i) f( η2 x ) = f(x) (ii) f(x) has one critical point, a minimum at η 2 (iii)f(x) is monotonically decreasing in the interval (0, η ] and monotonically increasing in the interval [ η, ). Proof: (i) is simple to verify by substitution. To verify (ii), examine the first derivative f (x) = 1 2 ηx 3/2 + 2 x 1/2, which for positive x euals 0 only at η. We observe that this is a minimum by noting that at this critical point, f(x) has the finite value 2η 1/2 1/2, but f(x) approaches infinity for both very small and very large positive x. (iii) is a direct conseuence of (ii). 3

4 Proposition 1.13: If t 0 τ, we get the lower bound I η2 I 0 for t [t 2 0, t 1 ], where I 0 = I(t 0 ). Proof: If t 0 τ, then t 0 > τ (remember, we are only looing at the half-line to the right of τ). It must be that I has a local maximum at t 0. In this case, I0 = 0, and substituting this into Lemma 1.10(i), we get that t (t 0, t 1 ), ( I 2 + 4η)(I 1/2 ) + 4I 1/2 4ηI 1/ I 1/2 0. Noting that I 2 I 1/2 > 0, we subtract this from the left hand side without altering the ineuality, and we then divide across by 4 to arrive at the ineuality (i) ηi 1/2 + I 1/2 ηi 1/2 0 + I 1/2 0, which holds in [t 0, t 1 ] because Lemma 1.10(i) holds in that interval. Note that for t [t 0, t 1 ], we have I < I 0 since I 0 is a maximum in the interval. We note that since I > 0, f(i) is defined for all I; hence, (i) is euivalent to the condition that f(i) f(i 0 ). Hence, it cannot be that I 0 η since f defined at I 0 is greater than f defined at I < I 0, and f is monotonically decreasing on the interval (0, η ] by Proposition 1.12(iii). Thus, I 0 > η. By η 2 I 0 2 Proposition 1.12(ii), f( η2 I 0 ) = f(i 2 0 ). However, = ( η )( η I 1 0 ) < ( η ), as I 0 > η. Hence, we see that f(x) > f(i 0 ) when x < η2 I 0. Thus, I cannot be < η2 2 I 0 ; hence I η2 2 I 0 for t [t 2 0, t 1 ]. Remar 1.14: If I 0 1, then by Proposition 1.13, I η 2 > A 2 in the interval [t 2 0, t 1 ]. If I 0 > 2 and I 2 > (2A) 2 in [t 0, t 1 ], then we have an obvious lower bound for I in the interval. Otherwise, t 2 (t 0, t 1 ) such that I 2 I(t 2 ) = 2 by the intermediate value theorem. Fact 1.15: If t 2 (t 0, t 1 ) s.t. I 2 I(t 2 ) = 2, then I ( + η 1 + (4η) 1 I2 2 ) 2 in (t 0, t 1 ). Proof: Since I is monotonically decreasing on the interval [t 0, t 1 ], we have on [t 0, t 2 ] that I 2 > (2A) 2. Since in [t 2, t 1 ], I is monotonically decreasing, by Lemma 1.10, I 2 (1 + η η I0 2 ) 2 = (after distributing) ( + η η I 0 2 ) 2 = ( + η 1 + 4η I2 2 ) 2, as desired. Since I monotonically decreases in all of (t 0, t 1 ), this serves as a lower bound in the whole interval. Proposition 1.16: If we can find c 5 > 0 such that I 2 2 < c 5 1, then Lemma 1.1 holds. Proof: Suppose that we can find c 5 > 0 such that (i) I 2 2 < c 5 1. Then we may substitute (i) into our ineuality from Fact 1.15 to achieve I ( + η 1 + (4η) 1 I2 2 ) 2 > ( + η 1 + (4η) 1 c 5 1 ) 2 > (A + A + AA(4) 1 c 5 (2A) 1 ) 2 = (2A Ac 5) 2 c 6 > 0. Combining this with our results from Remar 1.14 we would have a satisfactory lower bound for I in any interval [t 0, t 1 ] in which I is monotonically decreasing and I has a maximum at t = t 0. We will go through several cases to show why this proves Lemma 1.1. Let t 0 be the time corresponding to the first maximum of I in the interval (τ, ). If I decreases in [t 0, ), then our lower bound c 6 wors for all time after t 0 since [t 0, ) is an interval of monotone decrease. Otherwise, I has a first minimum at some t = t 1 ; then there are multiple maxima. Assume the maxima do not accumulate in finite time; if they do, then there is an accumulation of zeros of I in finite time, and analytic continuation implies I = 0 everywhere; however, we dealt with this case in Proposition 1.9. Assuming the maxima do not accumulate, I > c 6 holds in any interval of monotone decrease, thus in any interval between successive maxima (Proof: If we have an interval t a < t b < t c where t a and t b are successive maxima and t c is a minimum, our bound wors on [t a, t b ] since I decreases monotonically there; since I increases monotonically in [t b, t c ], our bound wors on the whole interval). So c 6 is a lower bound for I for all t > t 0. Now, we examine the interval (τ, t 0 ). If there is no minimum in this interval, then I τ must be a minimum in [τ, t 0 ] since a continuous function has a minimum in a closed bounded set, and t 0 is a maximum, not a minimum. Hence, in [τ, t 0 ], I I τ > c 2. If this interval contains a minimum at some t 1 (there can be at most one, otherwise t 0 will not be the first maximum) then since t 0 is the first maximum, I decreases in [τ, t 1 ], and increases in [t 1, t 0 ]. We have the lower bound c 4 on [τ, t 1 ] from Corollary 1.11; since I increases monotonically in [t 1, t 0 ], c 4 is a suitable lower bound in all of [τ, t 0 ]. Taing the minimum of c 4 and c 6 gives a lower bound for t τ, as desired. Now, you might as the uestion: what if I has no maxima for any t > τ? We again have two cases. Suppose that I is not continually decreasing. If at any point t 1 it begins to increase, then I increases monotonically for t > t 1 or else a maximum will form. Hence, we need not worry about anything after t 1. But then [τ, t 1 ] is an interval of monotonic decrease, and Corollary 1.11 gives us a lower bound on c 4 on [τ, ). If I is continually decreasing, then the interval [τ, ) is an 4

5 interval of monotonic decrease and Corollary 1.11 again gives a lower bound I > c 4. Now taing C min(c 2, c 3, c 4, c 6 ), we have a constant C, defined only in terms of A, m, and c 5, which we will prove is also defined only in terms of A and m, such that I > C whenever the angular momentum constants are not all 0. Then, by Corollary 1.3, we see that σ > c (C m 1 M )1/2, and thus we have a positive lower bound c for σ. Hence, Lemma 1.1 is proven if we show that I 2 2 < c 5 1 for some c 5 > 0, when h < 0, I 2 = 2 and I 2 < 0. In order to see this, we prove an instrumental ineuality: Proposition 1.17: I 2M r 2 ṙ 2 < c 7 1/2 for some positive constant c 7. Proof: The center of mass does not move, so 0 = 3 =1 m (ẋ x 3 + ẏ y 3 + ż z 3 ). Subtracting this from 1 2I = m, we get (i) 1 2I = m 1 1 ( 1 3 ) + m 2 2 ( 2 3 ). We express 2 3 in terms of 2 and 1 3 as follows: m m m 3 3 = 0 implies that 0 = m m m 3 3 m m 1 3 m m 1 2 m m 3 2 = m 1 ( 1 3 ) + (m 1 + m 3 )( 3 2 ) + M 2. Isolating 2 3, we get that 2 3 = m1 ( 1 3 ) + M 2. Substituting this into (i), we get (ii) 1 2I = m m 1 1 ( 1 3 ) + m 2 2 ( 1 ( 1 3 ) + M 2 ) = m 1 ( Define v = max(v 1, v 2 ), where v is the velocity of P ; then j v for j = 1, 2, and hence by the Schwarz Ineuality m 1 ( )( 1 3 ) = m 1 ( m1+m )( 1 3 ) m1 r 13 ((m 1 + m 3 )v + m 2 v) = m1m vr )( 1 3 )+ M Hence, after multiplying (ii) by 2, maing an appropriate subtraction, and taing absolute values, we get (iii) I 2M 2 2 = m 1 ( )( 1 3 ) 2m1M vr 13. Since we assume h < 0, T = U + h < U. Recall U = m m l <l r l ; since r 13 is the shortest side, m r l r 13 < l. Hence, r 13 T r 13 U = r m l 13 <l r l < m 1 m 2 + m 1 m 3 + m 2 m 3 c 8. It follows that r 13 v 2 is also bounded, since r 13 T = r 13 (m 1 v1 2 + m 2 v2 2 + m 3 v3) 2 > r 13 (m 2 v2), 2 hence r 13 v2 2 < r 13 T m 1 2 < c 8 m 1 2 c 9. Hence r13v 2 2 r 13 c 9, and r 13 v (r 13 c 9 ) 1/2. We observe 0 2T = 2U, hence 2U, and hence 2 1 U 1, so using r 13 U < c 8, we write r 13 < c 8 U 1 < 2c 8 1. Hence, r 13 v (r 13 c 9 ) 1/2 < (2c 8 c 9 1 ) 1/2 = (2c 8 c 9 ) 1/2 1/2. Now, by definition, 2 2 = r2, 2 and simple differentiation yields that (iv) 2 2 = ṙ 2 r 2. Combining (iii) with our upper bound for vr 13 and substituting in (iv) gives us I 2M r 2 ṙ 2 < 2m 1M (2c 8 c 9 ) 1/2 1/2. Defining c 7 2m1M (2c 8 c 9 ) 1/2 completes our proof of our claim. Corollary 1.18: I2 2 < c 5 1 for some c 5 > 0 if r 2 (t 2 ) r 2 (t 2 ) < c 10 1/2 for some c 10 > 0 Proof: Recall that at t = t 2, I = 2 and I is decreasing monotonically so I < 0. Since I 2M r 2 ṙ 2 < c 7 1/2, an upper bound of the form c 10 1/2 > r 2 (t 2 )ṙ 2 (t 2 ), where c 10 > 0, would give us 0 < I 2 < (c 7 + c 10 ) 1/2, and conseuently I 2 2 < c 5 1, for c 5 (c 7 + c 10 ) 2 > 0. Plan and Definition: We must now prove that r 2 (t 2 ) r 2 (t 2 ) < c 10 1/2 for some c 10 > 0. We first mae some observations about the relationships between the sides of our triangle; we assume WLOG that the shortest side has endpoints P 1 and P 3, so has length r 13. Subseuently, we find a negative lower bound for r 2 ; together, these will let us find our upper bound for r 2 (t 2 ) r 2 (t 2 ). Proposition 1.19: We find upper and lower bounds for r 2 dependent only on m, r 12, and r 23. Proof: By a crude application of the the triangle ineuality, we get the expression r 2 < r 13 +r 23 2r 23 ; we can see this because we can extend the segment from P 2 to O until it hits the shortest side of the triangle at the point P 4 ; the triangle ineuality then ensures that r 2 < r 2 + r 4 r 14 + r 23 r 13 + r 23. By the same process, we achieve the ineuality r 2 < 2r 12. Now, we see a lower bound for r 2. Recall the center of mass integral m m m 3 3 = 0. Recalling M = m 1 + m 2 + m 3, a simple rearrangement gives us M 2 = m 1 ( 2 1 ) + m 3 ( 2 3 ). Taing norms and suaring both sides, we get M = m 1 ( 2 1 ) + m 3 ( 2 3 ) 2. It follows that M 2 r2 2 = m 2 1r m 2 3r m 1 m 3 ( 2 1 ) ( 2 3 ). Applying the dot product, we obtain 5

6 M 2 r2 2 = m 2 1r m 2 3r m 1 m 3 r 12 r 23 cos(θ), where θ is the angle at P 2. Since r 13 is the shortest side, θ is at most = π/3. Hence, the cosine of this angle is 1 2. Hence, M 2 r2 2 m 2 1( 2 1 ) 2 + m 2 3( 2 3 ) 2 + m 1 m 3 r 12 r 23 > 1 2 (m 1r 12 + m 3 r 23 ) 2. Taing the suare root of both sides tells us that (i) 2Mr 2 > 2 1/2 Mr 2 > m 1 r 12 + m 3 r 23. Hence, r2 2 > (2M) 1 m 1 r 12 + m 3 r 23 and r 2 > (2M) 1/2 (m 1 r 12 + m 3 r 23 ) 1/2, giving us our desired bound. Corollary 1.20: We can find positive upper and lower bounds for r 12 /r 2 and r 23 /r 2. Proof: We found the bounds r 2 < 2r 12 and r 2 < 2r 23 in Proposition Rearranging these, we get that r 12 /r 2 > 1/2 > 0 and r 23 /r 2 > 1/2 > 0, giving us lower bounds for both. To find upper bounds, we first deduce from the fact that r 13 is the shortest side of the triangle that r 12 /2 r 23 2r 12. By Proposition 1.19(i), we have that 2Mr 2 > 2 1/2 Mr 2 > m 1 r 12 + m 3 r 23. Hence, 2Mr 2 > m 1 r 12 +m 3 r 12 /2 and 2Mr 2 > m 3 r 23 +m 1 r 23 /2. From the first, we see that r 2 /r 12 > (2M) 1 (m 1 + m 3 /2). From the second, we see that r 2 /r 23 > (2M) 1 (m 3 + m 1 /2). Rearrangement gives the desired upper bounds r 12 /r 2 < 2M(m 1 + m 3 /2) 1 and r 23 /r 2 < 2M(m 3 + m 1 /2) 1. Proposition 1.21: We can find a negative lower bound for r 2 2 of the form c 11 r2 2. Proof: First, we will find a lower bound for r 2 ; note that the lower bound we find will be negative. To do this, we differentiate 2 2 = r2 2 a second time to get ( ) 2 = r 2 r 2 + ṙ2. 2 Using the euations of motion = v and m = U and the definition U = m m l ( r 3 l ), l which were covered in Lecture 4, we find that = 2 (m m 3 2 r ) and 2 r = v2; 2 substituting, we get (i) 2 (m m 3 2 r 3 3 )+v 2 12 r = r 2 r 2 +ṙ2. 2 Proposition 1.19 gave us a positive lower bound for r 2, so we may divide the euation ṙ2r 2 2 = 2 2 across by r 2 ; taing the suare, we see that ṙ2 2 = ( 2 2 r 2 ) 2 ; the Schwarz ineuality then gives us that (ii) ṙ2 2 ( 2)r ( 2) 2 = 2 2 = v2. 2 Taing = 1, 3, we note that (iii) 2 r 3 2 r 2 2 < 4r2 2, where we have used our upper bounds from Proposition 1.19, r 2 < 2r 23 and r 2 < 2r 12 to establish the final ineuality. Substituting (ii) into (i), we get that 2 (m m 3 2 r 3 3 )+v 2 12 r r 2 r 2 +v2 2 and thus 2 (m m 3 2 r 3 3 ) r 12 r r 2. Since we are finding a negative lower bound, we would want to find an upper bound for the absolute values of the terms in the parentheses on the right; thanfully, (iii) gives us this, and we have the ineuality 2 (m 1 4r2 2 + m 3 4r2 2 ) r 2r 2. Finally, we note that since (iv) 2 r 2, we may divide out the 2 terms on the left and the r 2 term on the right without changing the ineuality. Hence, (m 1 4r2 2 + m 3 4r2 2 ) r 2, hence r 2 4(m 1 + m 3 )r2 2 > 8(m 1 + m 3 )r2 2. Defining c 11 8(m 1 + m 3 ), we have that r 2 > c 11 r2 2 and have therefore proven our claim. Proposition 1.22: Finally, we can indeed find some c 10 > 0 such that r 2 (t 2 )ṙ 2 (t 2 ) < c 10 1/2. Proof: We only need to consider the case when ṙ 2 (t 2 ) < 0, since if ṙ 2 (t 2 ) > 0, then r 2 (t 2 )ṙ 2 (t 2 ) < 0 and any positive bound would wor. Now, if ṙ 2 (t 2 ) < 0, then we can find some interval whose upper boundary is t 2, which we call [t 1, t 2 ], such that in the interval, both ṙ 2 < 0 and r 13 is still the shortest side of the triangle. Multiplying the result of Proposition 1.21 by 2ṙ 2, and noting that this reverses the ineuality, we arrive at 2ṙ 2 r 2 2c 11 ṙ 2 r2 2, hence 2ṙ 2 r 2 + 2c 11 ṙ 2 r2 2 0; this holds in [t 1, t 2 ]. Now, we note that the expression on the left is the derivative of (i) ṙ2 2 2c 11 r2 1 with respect to t, which is easily verified by a simple application of the chain rule; since the derivative is nonpositive, (i) is monotone decreasing. Hence, we have that ṙ2(t 2 2 ) 2c 11 r2 1 (t 2) ṙ2(t) 2 2c 11 r2 1 (t) for t [t 1, t 2 ]. In particular, this is true when t = t 1. It follows, since 2c 11 r2 1 is positive in [t 1, t 2 ], that ṙ2(t 2 2 ) < ṙ2(t 2 1 ) + 2c 11 r2 1 (t 2). Multiplying across by r2(t 2 2 ), we find that ṙ2(t 2 2 )r2(t 2 2 ) < ṙ2(t 2 1 )r2(t 2 2 ) + 2c 11 r2(t 2 2 )r2 1 (t 2). Noting ṙ 2 0 in our interval, we observe that r 2 (t 2 ) r 2 (t 1 ); hence, we can bound the right side again to achieve (ii) ṙ2(t 2 2 )r2(t 2 2 ) < ṙ2(t 2 1 )r2(t 2 1 ) + 2c 11 r 2 (t 2 ). We would lie to find a bound for the right side of (ii). We introduce a bound for the second term as follows. Since I = m 1 r1+m 2 2 r2+m 2 3 r3, 2 we certainly have that I > m 2 r2, 2 and hence I 2 > m 2 r2(t 2 2 ), and hence r 2 (t 2 ) < (I 2 m 1 2 )1/2 ; recalling that I 2 = 2, we see that 2c 11 r 2 (t 2 ) < 2c 11 1 m 1/2 2. l 6

7 To find bounds for the first term of the right side of (ii), we consider the following cases: Case I: Suppose that t 0 as previously defined (a local maximum) is a possible value for t 1. Then I 1 = 0, and by Proposition 1.17, I 1 2M r 2 (t 1 )ṙ 2 (t 1 ) < c 7 1/2, hence 2M r 2 (t 1 )ṙ 2 (t 1 ) < c 7 1/2, hence ṙ 2 (t 1 )r 2 (t 1 ) < ( 2M ) 1 c 7 1/2 and (ṙ 2 (t 1 )r 2 (t 1 )) 2 < ( 2M ) 2 c 7 1. Case II: Now suppose that t 0 is not a valid value for t 1 ; then we pic the smallest possible value of t 1 so that t 0 < t 1 t 2 and at t 1, either ṙ 2 (t 1 ) = 0 or at t 1, r 13 stops being the shortest side of the triangle, since these are the only reasons why any value smaller than t 1 might be unsatisfactory. If ṙ 2 (t 1 ) = 0, then (ṙ 2 (t 1 )r 2 (t 1 )) 2 = 0 < ( 2M ) 2 c 7 1 so our bound still holds. Hence, in both Cases I and II, ṙ2(t 2 2 )r2(t 2 2 ) < ( 2M ) 2 c c 11 1 m 1/2 2 = (( 2M ) 2 c 7 + 2c 11 m 1/2 2 ) 1. Case III: The only remaining possibility is that at t 1, r 13 stops being the shortest side of the triangle. If this happens, then at t = t 1 we have two sides that are both the shortest, so both have length r 13 ; by the triangle ineuality, we now have that r 12 2r 13 and r 23 2r 13. Now, recall from Corollary 1.20 that we have a lower bound = 1/2 for r 2 /r 2, = 1, 3, and hence an upper bound 2 for r 2 /r 2. Since r 13 = r 2 for one of = 1, 3, we see that this upper bound also serves well for r 2 /r 13 at t = t 1. Recall from Proposition 1.17 our bound r 13 U < c 8 ; we can multiply by r 2 /r 13 to achieve, at t = t 1, the ineuality (iii) r 2 U < r2 r 13 c 8 < 2c 8. Now we can say that at t = t 1, (r 2 ṙ 2 ) 2 (by Proposition 1.21(ii)) r2v r22t 2 m 1 2 (since T = 1 2 m 1v m 2v m 3v3 2 > 1 2 m 2v2) 2 < 2r2Um (since in Proposition 1.17, we noted that h < 0 implies T < U) < 2r2U 2 2 U 1 m 1 2 < (by (iii)) 2(2c 8 ) 2 U 1 m 1 2. Now, we have (from Proposition 1.17) 2U, and hence 2 1 U 1, so we bound our ineuality by 4(2c 8 ) 2 1 m 1 2 = c 12 1 where c 12 4(2c 8 ) 2 m 1 2. Along with our earlier bound for the second term of the right side of (ii), we get that ṙ2(t 2 2 )r2(t 2 2 ) < (c c 11 m 1/2 2 ) 1. We have deduced an upper bound of the form c 13 1 for (r 2 (t 2 )ṙ 2 (t 2 )) 2 in all cases, where c 13 is whichever of (c c 11 m 1/2 2 ) or (( 2M ) 2 c 7 + 2c 11 m 1/2 2 ) is applicable; this gives us an upper bound for r 2 (t 2 )ṙ 2 (t 2 ) of the form c 10 1/2 in all cases, where c 10 c 1/2 13, and the claim is proven. Summary: This concludes our rather computationally involved proof for Sundman s first lemma. To recapitulate, we began by proving that our statement followed if I is bounded from below by a positive value; we then considered the cases h 0 and h < 0. In order to prove the latter case, we proved two ineualities that forced a deeper consideration of the behavior of I. Throughout this proof, a common tric was to, with a desired ineuality in mind, craft a function whose integral would yield the desired ineuality. This techniue was observed by the student during Lecture 10, and seems to be a simple but fundamental and very useful method of analyzing differential euations. We now proceed to the second lemma, which more or less follows from the first. IV: SUNDMAN S SECOND LEMMA Lemma 2.1: The velocity of the particle opposite the shortest side of the triangle remains below a finite constant for all time t. Outline: Before we begin, we will define some notation specific to this proof. We assume without loss of generality that the shortest side is that opposite P 2 ; this will have length r 13 r. We will denote by v the velocity of P 2. Any other notation remains unchanged. Let s outline the proof. We first prove the case when r 13 c/4, then we address the case when r 13 < c/4. In this latter case, we will investigate the behavior of the function r2(v 2 2 ṙ2); 2 this will involve an exploration of the angular momentum integrals, which will allow us to create a bound for v 2 ṙ2. 2 From there, we will see a bound for ṙ2, 2 which will allow us to bound v 2 and therefore complete our proof. Proposition 2.2: If for all time, r 1 4c, then Lemma 2.1 holds. Proof: By Lemma 1.1, σ = r 12 + r 13 + r 23 > c for some c > 0. Now, denote by r the shortest side of the triangle formed by the three particles. If for all time, r c/4, then we have that T = U + h < m m l <l r l + h < 4m m l <l c + h < 12m 2 max/c + A, where m max again denotes the greatest mass. This gives a bound on all velocities as T = m v, 2 hence, T > m v 2 for = 1, 2, 3, hence T m 1 > v 2, hence v < ((A + (12m 2 max)/c)m 1 )1/2, proving Lemma 2.1 for this case. We note that c/4 is arbitrary; it simply gives us a lower bound and could have been any number. 7

8 Fact 2.3: r2(v 2 2 ṙ2) 2 = (x 2 ẏ 2 y 2 ẋ 2 ) 2 + (y 2 ż 2 z 2 ẏ 2 ) 2 + (z 2 ẋ 2 x 2 ż 2 ) 2. Proof: We now treat the case when r < c/4. We consider the expression r2(v 2 2 ṙ2); 2 substituting the definition 2 2 = r2 2 and its derivative 2 2 = ṙ 2 r 2, as well as the definition 2 = v 2, all of which we used earlier, we obtain the expression ( 2 2 ) 2. If we expand this out as (x y2 2 + z2)(ẋ ẏ2 2 + ż2) 2 (x 2 ẋ 2 + y 2 ẏ 2 + z 2 ż 2 ) 2 and rearrange, we get the expression (x 2 ẏ 2 y 2 ẋ 2 ) 2 + (y 2 ż 2 z 2 ẏ 2 ) 2 + (z 2 ẋ 2 x 2 ż 2 ) 2, which resembles the angular momentum integrals. Fact 2.4: L z = m 1 ((x 1 x 3 )(ẏ 1 + ẏ 2 ) (y 1 y 3 )(ẋ 1 + ẋ 2 ))+ M (x 2 ẏ 2 y 2 ẋ 2 ). Proof: Consider the angular momentum integral L z = m 1 (x 1 ẏ 1 y 1 ẋ 1 ) + m 2 (x 2 ẏ 2 y 2 ẋ 2 ) + m 3 (x 3 ẏ 3 y 3 ẋ 3 ). We change it into m 1 ((x 1 x 3 )ẏ 1 (y 1 y 3 )ẋ 1 ))+m 2 ((x 2 x 3 )ẏ 2 (y 2 y 3 )ẋ 2 )) by noting that m 3 x 3 ẏ 3 = m 1 x 3 ẏ 1 m 2 x 3 ẏ 2 and m 3 y 3 ẋ 3 = m 1 y 3 ẋ 1 +m 2 y 3 ẋ 2 ; this is because in each case, the three terms sum to zero, because the center of mass is 0. In Proposition 1.17, we derived that 2 3 = m1 ( 1 3 )+ M 2. We can use this to eliminate the terms x 2 x 3 and y 2 y 3 ; this gives us the expression m 1 ((x 1 x 3 )(ẏ 1 + ẏ 2 ) (y 1 y 3 )(ẋ 1 + ẋ 2 ))+ M (x 2 ẏ 2 y 2 ẋ 2 ). Proposition 2.5: In the case that r < c/4, we can find c 19 > 0 such that 0 (v 2 ṙ2) 2 < c 19 r2 2. Proof: Consider our expression for L z from Fact 2.4. We bound the absolute value of the first term lie so: m 1 ((x 1 x 3 )(ẏ 1 + ẏ 2 ) (y 1 y 3 )(ẋ 1 + ẋ 2 )) m 1 (x 1 x 3 )(ẏ 1 + ẏ 2 ) + m 1 (y 1 y 3 )(ẋ 1 + ẋ 2 )) (Triangle Ineuality) m 1 (r)( (ẏ 1 + ẏ 2 ) + (ẋ 1 + ẋ 2 ) ) (since (x 1 x 3 ) r and (y 1 y 3 ) r) M m m 1 (r)( )( ẏ 1 + ẏ 2 + ẋ 1 + ẋ 1 ) (since 2 < M and 1 < M ) Mm1r ( x 1 + x 2 + y 1 + y 2 ) (Triangle Ineuality) Mm1r T 1/2 (2m 1/ m 1/2 2 ) (by definition of T ) = c 14 rt 1/2 where c 14 Mm1 (2m 1/ m 1/2 2 ). In turn, c 14 rt 1/2 c 14 r(u + h ) 1/2 c 14 ru 1/2 + c 14 r h 1/2 < c 14 ru 1/2 + c 14 ra 1/2. Recalling that ru < c 8 from the proof to Proposition 1.17, we conclude r 2 U < 1 4 c 8c, and so ru 1/2 < 1 2 (c 8c) 1/2, 1 and so the absolute value of the first term is bounded from above by c 14 2 (c 8c) 1/ c 14cA 1/2 c 15. In Lecture 10, it was proven that 2IT η; this shows that L z 2 6I τ T τ 6I τ (U τ + h ) < 12A 2. Hence, L z < 3 1/2 2A < 6A. This allows us to place a bound on (x 2 ẏ 2 y 2 ẋ 2 ), since L z = m 1 ((x 1 x 3 )(ẏ 1 + ẏ 2 ) (y 1 y 3 )(ẋ 1 + ẋ 2 )) + M (x 2 ẏ 2 y 2 ẋ 2 ) gives us m 2M (x 2 ẏ 2 y 2 ẋ 2 ) m 1 ((x 1 x 3 )(ẏ 1 + ẏ 2 ) (y 1 y 3 )(ẋ 1 + ẋ 2 )) + L z < 6A + c 15 and hence (x 2 ẏ 2 y 2 ẋ 2 ) < (6A+c 15 ) m1+m3 m c 2M 16. An identical process gives c 17 and c 18 such that (y 2 ż 2 z 2 ẏ 2 ) < c 17 and (z 2 ẋ 2 x 2 ż 2 ) < c 18. Hence, 0 (v 2 ṙ2) 2 < r2 2 (c c c 2 18) c 19 r2 2. Proposition 2.6: If r < c/4, we can find some c 20 > 0 such that 0 v 2 ṙ2 2 < c 20 c 19 r2 1. Proof: The conditions that r < c/4 and r 12 +r+r 23 c tell us that r 12 > c/4 and r 23 > c/4 by the triangle ineuality, and hence r12 1 < 4c 1 and r23 1 < 4c 1. Recall the upper bounds for r 12 /r 2 and r 23 /r 2 in Corollary 1.20; these are (2M)(m 1 + m 3 /2) 1 and (2M)(m 3 + m 1 /2) 1, respectively. Now, we bound r2 1 by min(4c 1 (2M)(m 1 + m 3 /2) 1, 4c 1 (2M)(m 3 + m 1 /2) 1 ) c 20. So we have (i) 0 v 2 ṙ2 2 < c 20 c 19 r2 1, as desired. Proposition 2.7: In turn, we can construct the bound r 2 < r 2 Proof: Now we recall from Proposition 1.21 the euation ṙ 2 2. Rearranging, we get 2 (m r 3 12 Proposition 2.6 and the triangle ineuality, that As in the proof to Proposition 1.21, 2 r 3 2 c 20 c 19 r2 1 r 2 (m 1 4r2 2 + m 3 4r2 2 ) + c 20c 19 r2 1 2 (4m 1 + 4m 3 + c 20 c 19 ) +m )+v 2 r = r 2 r (m r m ) + v 2 r ṙ2 2 = r 2 r 2, which gives us, when we apply 2 (m r m ) + c r c 19 r2 1 > r 2 r 2. < 4r 2 2, and hence we derive (m 14r m 3 4r 1 2 (m 1 4r m 3 4r 2 2 ) + c 20c 19 r 1 Noting r 2 > 0, this gives us the desired upper bound r 2 < r 2 2 (4m 1 + 4m 3 + c 20 c 19 ). 2 ) + 2 > r 2 r 2. 8

9 Proposition 2.8: In the case that r < c/4, Lemma 2.1 holds. Proof: As in our proofs to Lemma 1.1, it suffices to consider the cases when t τ. Suppose that at a particular time, ṙ 2 = 0. Then Proposition 2.6 and our upper bound c 20 for r2 1 gives us that 0 v 2 c 20 c 19 c 20, and hence we have the bound v < (c 20 c 1/2 19 )/2 and we are done. Now, consider a time t when ṙ 2 does not vanish; then we can find an interval t 1 < t < t 2 in which r < c/4 holds and in which ṙ 2 does not vanish. Since r < c/4 holds, the side corresponding to r will still be the shortest side of the triangle. Then we can see from Proposition 2.7 that (i) 2ṙ 2 r 2 < 2 ṙ 2 r2 2 (4m 1 + 4m 3 + c 20 c 19 ) in our interval. Since ṙ 2 0 in our interval, ṙ 2 must preserve its sign, and hence we see that (ii) ṙ2(t) ṙ 2 2(t 2 1 ) < 2(4m 1 +4m 3 +c 20 c 19 ) r2 1 (t) r 1 2 (t 1) ; this is because the change in ṙ 2 must be less than the change in 2(4m 1 +4m 3 +c 20 c 19 )(r 1 ), since by (i), the absolute value of the derivative of the former is less than the absolute value of the derivative of the latter. In the proof to Proposition 2.6, 0 r2 1 < c 20 ; this holds for r2 1 (t) and r 1 2 (t 1), so r2 1 (t) r 1 2 (t 1) < c 20. Thus, from (ii), we obtain (iii) ṙ2(t) 2 < ṙ2(t 2 1 ) + 2(4m 1 + 4m 3 + c 20 c 19 )c 20. Now taing t 1 to be the smallest value that satisfies the above conditions and the extra condition that t 1 τ. If t 1 = τ, then Proposition 2.6 gives us that ṙ2(t 2 1 ) vτ 2 ; in turn, we see that this is 2m 1 2 T τ 2m 1 2 U τ + h < 4m 1 2 A c 21. Otherwise, if t 1 > τ, then it must be that either ṙ 2 (t 1 ) = 0 or that r(t 1 ) = c/4. In the first case, we have that ṙ 2 (t 1 ) = 0 < c 21. In the second case, we bounce bac to the first case that we dealt with in the theorem, since at t 1, we see that again, T is bounded by U + h <l (4m m l )/c + h < (12m 2 max)/c + A c 22, and thus again, we have that ṙ2(t 2 1 ) v 2 (t 1 ) 2m 1 1 T 1 < 2 1 c 22 c 23. Hence, we see that Proposition 2.6 gives us 0 v 2 < c 20 c 19 r2 1 + ṙ2 2 (by (iii)) c 2 20c 19 + ṙ2(t 2 1 ) + 2(4m 1 + 4m 3 + c 20 c 19 )c 20 < c 2 20c 19 + c (4m 1 + 4m 3 + c 20 c 19 )c 20 c 25, where c 24 is whichever of c 21 or c 23 applies. Now, we have v < c 1/2 25, and this completes our proof. V: CONCLUDING REMARKS We have proven the two major lemmas that Sundman used to help prove his series solution for the 3-body problem. These lemmas have interesting implications: indeed, the first says that if we have three bodies, then if two are very close, then the third must be somewhat far away in order to satisfy the perimeter condition. The second tells us that furthermore, the speed at which this particle cannot move too uicly towards the other two. These lemmas illustrate a result of Sundman s solution: a system whose angular momentum components are not all 0 not only can have no triple collision (as we proved in Lecture 10), but is bounded away from such a collision. VI: REFERENCES [1] C.L. Siegel and J.K. Moser, Lectures on Celestial Mechanics, Springer-Verlag Berlin Heidelberg, [2] J. Barrow-Green, The dramatic episode of Sundman, Historia Mathematica 37(2) (2010), [3] F. Diacu, The solution of the n-body problem, The Mathematical Intelligencer 18 (1996), no. 3,