Two Little Lemmas
Derek 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 mi be the mass of the ith particle, rkl be the distance between the k th and lth particles, and qi
be the position vector of the ith particle. The n-body problem is represented by the following ODE:
mk q̈ =
X mk ml (ql − qk )
3
rkl
l6=k
The left hand is equivalent to Uqk , the partial derivative of the potential U with respect to qk .
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-known 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-known 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
remarkably slow rate at which it converges. Since describing motions for even small periods of time
requires 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 kind;
this led mathematicians like 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 make an attempt
to reorganize and clarify this text, with reader comprehension as our primary goal. These steps
were taken 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.
2
Definition: Let L = (Lx , Ly , Lz ) be the angular momentum vector. Define a constant η ≡ ||L||
3 .
Definition: Define a constant τ to be the time corresponding to the initial or boundary value.
Furthermore, let any quantity with the subscript τ denote the value of the quantity at t = τ .
Assumption: Assume the center of mass of our system lies at the origin. This is valid since the
equations of motion are invariant under the transformation by which we rigidly translate the center
of mass, as we observed in Lecture 10. Define P1 , P2 , and P3 to be our three particles.
Definition: Let ∆ be the triangle formed by P1 , P2 , and P3 . Let qk be the position vector of
Pk . Let rk be the distance from O to Pk . When necessary, rk (t) will specify rk at time t. Let rjk
be the distance from Pj to Pk . Finally, let σ be the perimeter of the triangle ∆.
1
Definition: Define I ≡
P
mq 2 =
q
3
P
1
mk rk2 , where mk is the mass of Pk . Let M = m1 +m2 +m3 .
Definition: Recall the energy integral T = U + h, where h is the energy constant, T is the
kinetic energy, and U is the potential energy. Recall also the inequality 2IT ≥ η from Lecture 10.
Assumption: As stated before, assume that |Lx |, |Ly |, and |Lz | are not all 0. Then η > 0.
Definition: Define a constant A > 0 that satisfies the following inequalities:
(i) Iτ < A
(ii) Uτ < 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 mk . Then we break 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 − 2hx1/2 and
prove two appropriately constructed inequalities. 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 ∆jk formed by the origin, Pj , and Pk shares one edge (namely, that connecting
Pj with Pk ) with ∆. Since ∆jk lies inside ∆, σjk is strictly smaller than σ. Hence, the sum of the
lengths of the other two edges of ∆jk is strictly smaller than the sums of the other two edges of ∆.
This gives us the following three inequalities: (i) r1 + r2 ≤ r13 + r23 (ii) r1 + r3 ≤ r12 + r23 (iii)
r2 + r3 ≤ r12 + r13 . Summing these three inequalities and dividing by 2, we see that r1 + r2 + r3 ≤ σ.
Define mmax ≡ max(m1 , m2 , m3 ), then I = m1 r12 + m2 r22 + m3 r32 ≤ mmax (r12 + r22 + r32 ) ≤
mmax (r1 + r2 + r3 )2 ≤ mmax σ 2 . Thus, we have our upper bound: Iσ −2 ≤ mmax .
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) r12 ≤ r1 + r2 (v) r13 ≤ r1 + r3 (vi) r23 ≤ r2 + r3 . Summing
these, we get σ ≤ 2(r1 + r2 + r3 ). Squaring both sides and applying the Schwarz Inequality yields
3
3
3
3
P
P
P
P
1/2 −1/2
1/2
−1/2
σ 2 ≤ 4(r1 + r2 + r3 )2 = 4(
mk mk rk )2 ≤ 4
(mk rk )2
(mk )2 = 4I
m−1
k .
k=1
k=1
k=1
k=1
−1
−1 −1
This gives a lower bound Iσ −2 ≥ (4(m−1
> 0.
1 + m2 + m3 ))
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 c0 > 0 such that I > c0 for all time. Since
−1 1/2
−1
> 0, as desired.
Iσ −2 ≤ mmax , we see that σ 2 ≥ Im−1
max > c0 mmax , and hence σ > (c0 mmax )
2
−1/2
1/2
˙
Proposition 1.4: Define the function K(I) ≡ (I + 4η)I
− 8hI . Then K and I are either
both monotonically decreasing or both monotonically increasing as functions of time, t.
Proof: Recall the following two equations, which were discussed and proven in Lecture 10.
(ii) 2IT ≥ 41 I˙2 + η
(i) 21 I¨ = T + h = U + 2h (Lagrange’s Formula)
−1 1 ˙2
By (ii), T ≥ (2I) ( 4 I + η); substituting this into (i) and multiplying by 2 yields I¨ ≥ I −1 ( 14 I˙2 +
η) + 2h. After rearrangement, we obtain (iii) I¨ − 14 I˙2 I −1 − I −1 η − 2h ≥ 0.
˙ −1/2 gives (iv) 2I¨II
˙ −1/2 − 1 I˙3 I −3/2 −2II
˙ −1/2 η −4II
˙ −1/2 h.
Multiplying the left side of (iii) by 2II
2
It is simple to verify that this expression is the derivative of K with respect to I. Hence, K̇ is the
˙ −1/2 ; 2I −1/2 is non-negative by
product of a non-negative quantity (the left side of (iii)) and 2II
definition, so K̇ 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 (I¨ − 14 I˙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.
P
Remark 1.5: A simple but useful observation is that since U = k<l mrkklml , where each term
ml
is a product of positive quantities, we have that ∀k < l, where k, l ∈ {1, 2, 3}, mrkklτ
≤ Uτ < A.
2
Hence, rklτ > mkAml > 0. Let c1 ≡ max( m1Am2 , m1Am3 , m2Am3 ); then ∀k, l ∈ {1, 2, 3}, rklτ > c1 .
Remark 1.6: Another simple observation is that there exists a constant c2 > 0 such that Iτ > c2 .
−1
−1 −1
Recall the lower bound from Proposition 1.2: Iσ −2 ≥ (4(m−1
> 0. Taking the
1 + m2 + m3 ))
−1
1
−1
+
m−1
> 0. By
value at t = τ , we obtain the expression Iτ ≥ 4 (r12τ + r13τ + r23τ )2 (m1 + m−1
2
3 )
−1
−1
−1 −1
−1
−1
−1 −1
9 2
9 2
Remark 1.5, Iτ ≥ 4 c1 (m1 + m2 + m3 ) > 5 c1 (m1 + m2 + m3 ) ≡ c2 > 0, as desired.
Remark 1.7: η > 0, so by Proposition 1.4(ii), 2IT ≥ 41 I˙2 + η > 41 I˙2 . Hence, 8IT > I˙2 for all t,
and in particular, 8Iτ Tτ ≥ I˙τ2 . But 8Iτ Tτ = 8Iτ (Uτ + h) < 8(A)(A + A) = 16A2 , hence I˙τ2 < 16A2 .
Now, our preliminary work 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 21 I¨ = 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 Remark 1.6 that I ≥ Iτ > c2 ; hence c2 is the desired constant.
Now suppose that I˙τ < 0. Then we may find an interval [τ, t1 ] upon which I is monotonically decreasing. By Proposition 1.4, K is also monotonically decreasing on [τ, t1 ]. Furthermore, since h ≥ 0,
K + 8hI 1/2 is also decreasing monotonically on [τ, t1 ]. Hence, in our interval, (I˙2 + 4η)I −1/2 = K +
−1/2
−1/2
1/2
. Rearrang. I I˙2 is non-negative, so 4ηI −1/2 ≤ (I˙τ2 +4η)Iτ
8hI 1/2 ≤ Kτ +8hIτ = (I˙τ2 +4η)Iτ
1/2 ˙2
1/2 ˙2
−1
1/2
ing and multiplying by a clever form of 1, we get I
≥ (4η)Iτ (Iτ + 4η) = Iτ (Iτ (4η)−1 + 1)−1 .
Noting that all the terms on the right are positive, we square everything while conserving the direction of the inequality, and find that I ≥ Iτ (I˙τ2 (4η)−1 + 1)−2 . Now, applying Remarks 1.6 and 1.7,
we get I ≥ Iτ (I˙τ2 (4η)−1 + 1)−2 > c2 (4A2 A + 1)−2 = c2 (1 + 4A3 )−2 ≡ c3 .
I is convex, so our lower bound for intervals of monotone decrease in fact works 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 ≥ Iτ > c2 .
Lemma 1.10: Let k = −2h. If h < 0 and for some t > τ , I˙ < 0, then in any interval [t0 , t1 ] in
1 ˙2 −2
which I is monotonically decreasing, if I0 is the value of I at t0 , then I ≥ I0 (1 + kη I0 + 4η
I0 ) .
˙
Proof: For some t > τ , I < 0, and hence we can find intervals [t0 , t1 ], where t0 ≥ τ , in which I
is monotonically decreasing. By Proposition 1.4, K is also monotonically decreasing in this interval,
−1/2
1/2
and hence (i) ∀ t ∈ (t0 , t1 ), (I˙2 + 4η)(I −1/2 ) + 4kI 1/2 ≤ (I˙02 + 4η)I0
+ 4kI0 . Noting that k > 0
−1/2
1/2
and I˙2 I −1/2 > 0, we find that (4ηI −1/2 ) ≤ (I˙02 + 4η)(I0
) + 4kI0 . Rearranging and squaring
(this is kosher since every term is positive), we derive I ≥ I0 (1 + kI0 η −1 + (4η)−1 I˙02 )−2 , as desired.
In particular, this holds when we choose t0 to be the smallest value, for each fixed t1 , such that
I monotonically decreases on [t0 , t1 ]. This will generally be our practice.
Corollary 1.11: If t0 = τ , then we have a lower bound for I on the interval [τ, t1 ].
Proof: By Lemma 1.10, I ≥ Iτ (1 + kIτ η −1 + (4η)−1 I˙τ2 )−2 in the interval. Using Remark 1.6 to
bound Iτ from below, A to bound Iτ , η −1 , and k2 = −h from above, and Remark 1.7 to bound I˙τ
from above, we see that I ≥ c2 (1 + (2A)(A)(A) + 14 (A)A2 )−2 > c2 (1 + 3A3 )−2 ≡ c4 > 0.
Proposition 1.12: We note three properties of the function f (x) = ηx−1/2 + kx1/2 , defined for
η2
η
positive values of x: (i) f ( xk
2 ) = f (x) (ii) f (x) has one critical point, a minimum at k (iii)f (x) is
η
monotonically decreasing in the interval (0, k ] and monotonically increasing in the interval [ kη , ∞).
Proof: (i) is simple to verify by substitution. To verify (ii), examine the first derivative f 0 (x) =
1
− 2 ηx−3/2 + k2 x−1/2 , which for positive x equals 0 only at kη . We observe that this is a minimum by
noting that at this critical point, f (x) has the finite value 2η 1/2 k 1/2 , but f (x) approaches infinity
for both very small and very large positive x. (iii) is a direct consequence of (ii).
3
2
Proposition 1.13: If t0 6= τ , we get the lower bound I ≥ I0ηk2 for t ∈ [t0 , t1 ], where I0 = I(t0 ).
Proof: If t0 6= τ , then t0 > τ (remember, we are only looking at the half-line to the right of
τ ). It must be that I has a local maximum at t0 . In this case, I˙0 = 0, and substituting this into
−1/2
1/2
Lemma 1.10(i), we get that ∀ t ∈ (t0 , t1 ), (I˙2 + 4η)(I −1/2 ) + 4kI 1/2 ≤ 4ηI0
+ 4kI0 . Noting that
2 −1/2
˙
I I
> 0, we subtract this from the left hand side without altering the inequality, and we then
−1/2
1/2
divide across by 4 to arrive at the inequality (i) ηI −1/2 + kI 1/2 ≤ ηI0
+ kI0 , which holds in
[t0 , t1 ] because Lemma 1.10(i) holds in that interval.
Note that for t ∈ [t0 , t1 ], we have I < I0 since I0 is a maximum in the interval. We note that
since I > 0, f (I) is defined for all I; hence, (i) is equivalent to the condition that f (I) ≤ f (I0 ).
Hence, it cannot be that I0 ≤ kη since f defined at I0 is greater than f defined at I < I0 , and
f is monotonically decreasing on the interval (0, kη ] by Proposition 1.12(iii). Thus, I0 > kη . By
2
Proposition 1.12(ii), f ( I0ηk2 ) = f (I0 ). However,
2
η2
I0 k2
= ( kη )( kη I0−1 ) < ( kη ), as I0 > kη . Hence, we see
2
2
that f (x) > f (I0 ) when x < I0ηk2 . Thus, I cannot be < I0ηk2 ; hence I ≥ I0ηk2 for t ∈ [t0 , t1 ].
Remark 1.14: If I0 ≤ k12 , then by Proposition 1.13, I ≥ η 2 > A−2 in the interval [t0 , t1 ].
If I0 > k −2 and I ≥ k −2 > (2A)−2 in [t0 , t1 ], then we have an obvious lower bound for I in the
interval. Otherwise, ∃ t2 ∈ (t0 , t1 ) such that I2 ≡ I(t2 ) = k −2 by the intermediate value theorem.
Fact 1.15: If ∃ t2 ∈ (t0 , t1 ) s.t. I2 ≡ I(t2 ) = k −2 , then I ≥ (k + η −1 + k(4η)−1 I˙22 )−2 in (t0 , t1 ).
Proof: Since I is monotonically decreasing on the interval [t0 , t1 ], we have on [t0 , t2 ] that
I ≥ k −2 > (2A)−2 . Since in [t2 , t1 ], I is monotonically decreasing, by Lemma 1.10, I ≥ k −2 (1 +
k −2
1 ˙2 −2
1
k ˙2 −2
+ 4η
I0 ) = (after distributing) (k + kη kk −2 + 4η
k I˙02 )−2 = (k + η −1 + 4η
I2 ) , as desired.
ηk
Since I monotonically decreases in all of (t0 , t1 ), this serves as a lower bound in the whole interval.
Proposition 1.16: If we can find c5 > 0 such that I˙22 < c5 k −1 , then Lemma 1.1 holds.
Proof: Suppose that we can find c5 > 0 such that (i) I˙22 < c5 k −1 . Then we may substitute
(i) into our inequality from Fact 1.15 to achieve I ≥ (k + η −1 + k(4η)−1 I˙22 )−2 > (k + η −1 +
k(4η)−1 c5 k −1 )−2 > (A + A + AA(4)−1 c5 (2A)−1 )−2 = (2A + 18 Ac5 )−2 ≡ c6 > 0.
Combining this with our results from Remark 1.14 we would have a satisfactory lower bound for
I in any interval [t0 , t1 ] in which I is monotonically decreasing and I has a maximum at t = t0 .
We will go through several cases to show why this proves Lemma 1.1.
Let t0 be the time corresponding to the first maximum of I in the interval (τ, ∞).
If I decreases in [t0 , ∞), then our lower bound c6 works for all time after t0 since [t0 , ∞) is an
interval of monotone decrease. Otherwise, I has a first minimum at some t = t1 ; 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 > c6
holds in any interval of monotone decrease, thus in any interval between successive maxima (Proof:
If we have an interval ta < tb < tc where ta and tb are successive maxima and tc is a minimum,
our bound works on [ta , tb ] since I decreases monotonically there; since I increases monotonically
in [tb , tc ], our bound works on the whole interval ). So c6 is a lower bound for I for all t > t0 .
Now, we examine the interval (τ, t0 ). If there is no minimum in this interval, then Iτ must be a
minimum in [τ, t0 ] since a continuous function has a minimum in a closed bounded set, and t0 is a
maximum, not a minimum. Hence, in [τ, t0 ], I ≥ Iτ > c2 . If this interval contains a minimum at
some t1 (there can be at most one, otherwise t0 will not be the first maximum) then since t0 is the
first maximum, I decreases in [τ, t1 ], and increases in [t1 , t0 ]. We have the lower bound c4 on [τ, t1 ]
from Corollary 1.11; since I increases monotonically in [t1 , t0 ], c4 is a suitable lower bound in all of
[τ, t0 ]. Taking the minimum of c4 and c6 gives a lower bound for t ≥ τ , as desired.
Now, you might ask the question: 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 t1 it begins to increase,
then I increases monotonically for t > t1 or else a maximum will form. Hence, we need not worry
about anything after t1 . But then [τ, t1 ] is an interval of monotonic decrease, and Corollary 1.11
gives us a lower bound on c4 on [τ, ∞). If I is continually decreasing, then the interval [τ, ∞) is an
4
interval of monotonic decrease and Corollary 1.11 again gives a lower bound I > c4 . Now taking
C 0 ≡ min(c2 , c3 , c4 , c6 ), we have a constant C 0 , defined only in terms of A, mk , and c5 , which we will
prove is also defined only in terms of A and mk , such that I > C 0 whenever the angular momentum
1/2
constants are not all 0. Then, by Corollary 1.3, we see that σ > c ≡ (C 0 m−1
, and thus we have
M )
2
˙
a positive lower bound c for σ. Hence, Lemma 1.1 is proven−if we show that I2 < c5 k −1 for some
c5 > 0, when h < 0, I2 = k −2 and I˙2 < 0. In order to see this, we prove an instrumental inequality:
2M
r2 ṙ2 | < c7 k −1/2 for some positive constant c7 .
Proposition 1.17: |I˙ − m2m
1 +m3
3
P
Proof: The center of mass does not move, so 0 =
mk (ẋk x3 + ẏk y3 + żk z3 ). Subtracting this
k=1
P
P
from 21 I˙ =
mq̇q, we get (i) 21 I˙ =
m1 q̇1 (q1 − q3 ) + m2 q̇2 (q2 − q3 ).
q
q
We express q2 − q3 in terms of q2 and q1 − q3 as follows: m1 q1 + m2 q2 + m3 q3 = 0 implies that
0 = m1 q1 + m2 q2 + m3 q3 − m1 q3 + m1 q3 − m1 q2 + m1 q2 − m3 q2 + m3 q2 = m1 (q1 − q3 ) + (m1 +
1
(q1 − q3 ) + m1M
m3 )(q3 − q2 ) + M q2 . Isolating q2 − q3 , we get that q2 − q3 = m1m+m
+m3 q2 .
3
P
1 ˙
1
Substituting this into (i), we get (ii) 2 I = m1 q̇1 (q1 − q3 ) + m2 q̇2 ( m1m+m
(q
−
q3 ) + m1M
1
+m3 q2 )
3
q
P
P
2
2M
= m1 (q̇1 + m1m+m
q̇2 )(q1 − q3 )+ mm1 +m
q̇2 q2 . Define v = max(v1 , v2 ), where vk is the velocity of
3
3
q
q
P
P
2
Pk ; then | q̇j | ≤ v for j = 1, 2, and hence by the Schwarz Inequality | m1 (q̇1 + m1m+m
q̇2 )(q1 − q3 )|
3
q
q
P m1 +m3
2
1
1M
= |m1 ( m1 +m3 q̇1 + m1m+m
q̇2 )(q1 − q3 )| ≤ m1m+m
r13 ((m1 + m3 )v + m2 v) = mm1 +m
vr13 .
3
3
3
q
Hence, after multiplying
making an appropriate subtraction, and taking absolute values,
P (ii) by 2,P
2M
2
1M
we get (iii) |I˙ − m2m
q̇
q
|
=
|
m1 (q̇1 + m1m+m
q̇2 )(q1 − q3 )| ≤ m2m
vr13 .
2
2
1 +m3
3
1 +m3
q
q
P
Since we assume h < 0, T = U + h < U . Recall U = k<l mrkklml ; since r13 is the shortest side,
P
rkl ≥ r13 ∀ k < l. Hence, r13 T ≤ r13 U = r13 k<l mrkklml < m1 m2 + m1 m3 + m2 m3 ≡ c8 .
It follows that r13 v 2 is also bounded, since r13 T = r13 (m1 v12 + m2 v22 + m3 v32 ) > r13 (m2 v22 ), hence
−1
2 2
1/2
r13 v22 < r13 T m−1
.
2 < c8 m2 ≡ c9 . Hence r13 v ≤ r13 c9 , and r13 v ≤ (r13 c9 )
−1
−1
We observe 0 ≤ 2T = 2U − k, hence 2U ≥ k, and hence 2k ≥ U , so using r13 U < c8 , we
−1
1/2 −1/2
write r13 < c8 U −1 < 2cP
. Hence, r13 v ≤ (r13 c9 )1/2 < (2c8 c9 k −1 )1/2 = (2cP
k
.
8k
8 c9 )
2
2
Now, by definition,
q2 = r2 , and simple differentiation yields that (iv) q̇2 q2 = ṙ2 r2 .
q
q
2M
Combining (iii) with our upper bound for vr13 and substituting in (iv) gives us |I˙ − m2m
r2 ṙ2 | <
1 +m3
2m1 M
2m1 M
1/2 −1/2
1/2
(2c
c
)
k
.
Defining
c
≡
(2c
c
)
completes
our
proof
of
our
claim.
8 9
7
8 9
m1 +m3
m1 +m3
Corollary 1.18: I˙22 < c5 k −1 for some c5 > 0 if −r2 (t2 )r˙2 (t2 ) < c10 k −1/2 for some c10 > 0
Proof: Recall that at t = t2 , I = k −2 and I is decreasing monotonically so I˙ < 0. Since
2M
˙
|I − m2m
r2 ṙ2 | < c7 k −1/2 , an upper bound of the form c10 k −1/2 > −r2 (t2 )ṙ2 (t2 ), where c10 > 0,
1 +m3
would give us 0 < |I˙2 | < (c7 + c10 )k −1/2 , and consequently I˙22 < c5 k −1 , for c5 ≡ (c7 + c10 )2 > 0.
Plan and Definition: We must now prove that −r2 (t2 )r˙2 (t2 ) < c10 k −1/2 for some c10 > 0. We
first make some observations about the relationships between the sides of our triangle; we assume
WLOG that the shortest side has endpoints P1 and P3 , so has length r13 . Subsequently, we find a
negative lower bound for r̈2 ; together, these will let us find our upper bound for −r2 (t2 )r˙2 (t2 ).
Proposition 1.19: We find upper and lower bounds for r2 dependent only on mk , r12 , and r23 .
Proof: By a crude application of the the triangle inequality, we get the expression r2 < r13 +r23 ≤
2r23 ; we can see this because we can extend the segment from P2 to O until it hits the shortest side
of the triangle at the point P4 ; the triangle inequality then ensures that r2 < r2 + r4 ≤ r14 + r23 ≤
r13 + r23 . By the same process, we achieve the inequality r2 < 2r12 .
Now, we seek a lower bound for r2 . Recall the center of mass integral m1 q1 + m2 q2 + m3 q3 = 0.
Recalling M = m1 + m2 + m3 , a simple rearrangement gives us M q2 = m1 (q2 − q1 ) + m3 (q2 − q3 ).
Taking norms and squaring both sides, we get |M 2 q22 | = |m1 (q2 − q1 ) + m3 (q2 − q3 )|2 . It follows
2
2
+ m23 r23
+ 2m1 m3 (q2 − q1 ) · (q2 − q3 ). Applying the dot product, we obtain
that M 2 r22 = m21 r12
5
2
2
M 2 r22 = m21 r12
+ m23 r23
+ 2m1 m3 r12 r23 cos(θ), where θ is the angle at P2 . Since r13 is the shortest
side, θ is at most = π/3. Hence, the cosine of this angle is ≥ 21 . Hence, M 2 r22 ≥ m21 (q2 −
q1 )2 + m23 (q2 − q3 )2 + m1 m3 r12 r23 > 12 (m1 r12 + m3 r23 )2 . Taking the square root of both sides
tells us that (i) 2M r2 > 21/2 M r2 > m1 r12 + m3 r23 . Hence, r22 > (2M )−1 m1 r12 + m3 r23 and
r2 > (2M )−1/2 (m1 r12 + m3 r23 )1/2 , giving us our desired bound.
Corollary 1.20: We can find positive upper and lower bounds for r12 /r2 and r23 /r2 .
Proof: We found the bounds r2 < 2r12 and r2 < 2r23 in Proposition 1.19. Rearranging
these, we get that r12 /r2 > 1/2 > 0 and r23 /r2 > 1/2 > 0, giving us lower bounds for both.
To find upper bounds, we first deduce from the fact that r13 is the shortest side of the triangle
that r12 /2 ≤ r23 ≤ 2r12 . By Proposition 1.19(i), we have that 2M r2 > 21/2 M r2 > m1 r12 + m3 r23 .
Hence, 2M r2 > m1 r12 +m3 r12 /2 and 2M r2 > m3 r23 +m1 r23 /2. From the first, we see that r2 /r12 >
(2M )−1 (m1 + m3 /2). From the second, we see that r2 /r23 > (2M )−1 (m3 + m1 /2). Rearrangement
gives the desired upper bounds r12 /r2 < 2M (m1 + m3 /2)−1 and r23 /r2 < 2M (m3 + m1 /2)−1 .
Proposition 1.21: We can find a negative lower bound for r̈22 of the form −c11 r2−2 .
Proof: First, we will find a lower
for r̈2 ; note that theP
lower bound we find will be
P bound
q22 = r22 a second time to get (q̈2 q2 + q̇22 ) = r̈2 r2 + ṙ22 .
negative. To do this, we differentiate
q
q
P mk ml
(ql − qk ),
Using the equations of motion q̇ = v and mq̈ = Uq and the definition Uqk =
r3
kl
l6=k
q1 −q2
q3 −q2
2
2
which were covered in Lecture 4, we find that q̈ = q2 (m1 r3 + m3 r3 ) and q̇2 = v2 ; substituting,
12
23
P
2
2
+m3 q3r−q
)+v22 = r̈2 r2 + ṙ22 . Proposition 1.19 gave us a positive lower bound
we get (i)
q2 (m1 q1r−q
3
3
12
23
q
P
for r2 , so we may divide the equation ṙ22 r2 =
q̇2 q2 across by r2 ; taking the square, we see that
q
P
P
P
P
ṙ22 = ( q̇2 rq22 )2 ; the Schwarz inequality then gives us that (ii) ṙ22 ≤ ( q̇22 )r2−2 ( q22 ) = q̇22 = v22 .
q
q
q
q
−3
−2
Taking k = 1, 3, we note that (iii)|qk −q2 |rk2
≤ rk2
< 4r22 , where we have used our upper bounds
from Proposition
(ii) into
23 and r2 < 2r12 to establish the final
P1.19, r2 < 2r
Pinequality. Substituting
2
2
2
2
(i), we get that q2 (m1 q1r−q
+m3 q3r−q
)+v22 ≤ r̈2 r2 +v22 and thus q2 (m1 q1r−q
+m3 q3r−q
) ≤ r̈2 r2 .
3
3
3
3
q
12
23
q
12
23
Since we are finding a negative lower bound, we would want to find an upper bound for the absolute
values of the P
terms in the parentheses on the right; thankfully, (iii) gives us this, and we have the
inequality − q2 (m1 4r2−2 + m3 4r2−2 ) ≤ r̈2 r2 . Finally, we note that since (iv) |q2 | ≤ r2 , we may
q
divide out the q2 terms on the left and the r2 term on the right without changing the inequality.
Hence, −(m1 4r2−2 + m3 4r2−2 ) ≤ r̈2 , hence r¨2 ≥ −4(m1 + m3 )r2−2 > −8(m1 + m3 )r2−2 . Defining
c11 ≡ 8(m1 + m3 ), we have that r¨2 > −c11 r2−2 and have therefore proven our claim.
Proposition 1.22: Finally, we can indeed find some c10 > 0 such that −r2 (t2 )ṙ2 (t2 ) < c10 k −1/2 .
Proof: We only need to consider the case when ṙ2 (t2 ) < 0, since if ṙ2 (t2 ) > 0, then −r2 (t2 )ṙ2 (t2 ) <
0 and any positive bound would work. Now, if ṙ2 (t2 ) < 0, then we can find some interval whose
upper boundary is t2 , which we call [t1 , t2 ], such that in the interval, both ṙ2 < 0 and r13 is still the
shortest side of the triangle. Multiplying the result of Proposition 1.21 by 2ṙ2 , and noting that this
reverses the inequality, we arrive at 2ṙ2 r̈2 ≤ −2c11 ṙ2 r2−2 , hence 2ṙ2 r̈2 + 2c11 ṙ2 r2−2 ≤ 0; this holds in
[t1 , t2 ]. Now, we note that the expression on the left is the derivative of (i) ṙ22 − 2c11 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 ṙ22 (t2 ) − 2c11 r2−1 (t2 ) ≤ ṙ22 (t) − 2c11 r2−1 (t)
for t ∈ [t1 , t2 ]. In particular, this is true when t = t1 . It follows, since 2c11 r2−1 is positive in
[t1 , t2 ], that ṙ22 (t2 ) < ṙ22 (t1 ) + 2c11 r2−1 (t2 ). Multiplying across by r22 (t2 ), we find that ṙ22 (t2 )r22 (t2 ) <
ṙ22 (t1 )r22 (t2 ) + 2c11 r22 (t2 )r2−1 (t2 ). Noting ṙ2 ≤ 0 in our interval, we observe that r2 (t2 ) ≤ r2 (t1 );
hence, we can bound the right side again to achieve (ii) ṙ22 (t2 )r22 (t2 ) < ṙ22 (t1 )r22 (t1 ) + 2c11 r2 (t2 ).
We would like to find a bound for the right side of (ii). We introduce a bound for the second term
as follows. Since I = m1 r12 +m2 r22 +m3 r32 , we certainly have that I > m2 r22 , and hence I2 > m2 r22 (t2 ),
−1/2
1/2
and hence r2 (t2 ) < (I2 m−1
; recalling that I2 = k −2 , we see that 2c11 r2 (t2 ) < 2c11 k −1 m2 .
2 )
6
To find bounds for the first term of the right side of (ii), we consider the following cases:
Case I: Suppose that t0 as previously defined (a local maximum) is a possible value for t1 . Then
2M
2M
r2 (t1 )ṙ2 (t1 )| < c7 k −1/2 , hence | m2m
r2 (t1 )ṙ2 (t1 )| <
I˙1 = 0, and by Proposition 1.17, |I˙1 − m2m
1 +m3
1 +m3
2m2 M −1
2m2 M −2
−1/2
−1/2
2
c7 k
, hence |ṙ2 (t1 )r2 (t1 )| < ( m1 +m3 ) c7 k
and (ṙ2 (t1 )r2 (t1 )) < ( m1 +m3 ) c7 k −1 .
Case II: Now suppose that t0 is not a valid value for t1 ; then we pick the smallest possible value
of t1 so that t0 < t1 ≤ t2 and at t1 , either ṙ2 (t1 ) = 0 or at t1 , r13 stops being the shortest side of
the triangle, since these are the only reasons why any value smaller than t1 might be unsatisfactory.
2M
)−2 c7 k −1 so our bound still holds. Hence, in both
If ṙ2 (t1 ) = 0, then (ṙ2 (t1 )r2 (t1 ))2 = 0 < ( m2m
1 +m3
−1/2
−1/2
2M
2M
Cases I and II, ṙ22 (t2 )r22 (t2 ) < ( m2m
)−2 c7 k −1 + 2c11 k −1 m2
= (( m2m
)−2 c7 + 2c11 m2 )k −1 .
1 +m3
1 +m3
Case III: The only remaining possibility is that at t1 , r13 stops being the shortest side of the
triangle. If this happens, then at t = t1 we have two sides that are both the shortest, so both have
length r13 ; by the triangle inequality, we now have that r12 ≤ 2r13 and r23 ≤ 2r13 . Now, recall from
Corollary 1.20 that we have a lower bound = 1/2 for rk2 /r2 , k = 1, 3, and hence an upper bound 2
for r2 /rk2 . Since r13 = rk2 for one of k = 1, 3, we see that this upper bound also serves well for r2 /r13
at t = t1 . Recall from Proposition 1.17 our bound r13 U < c8 ; we can multiply by r2 /r13 to achieve,
2
c8 < 2c8 . Now we can say that at t = t1 , (r2 ṙ2 )2 ≤ (by
at t = t1 , the inequality (iii) r2 U < rr13
−1
Proposition 1.21(ii)) r22 v22 ≤ r22 2T m2 (since T = 21 m1 v12 + 12 m2 v22 + 12 m3 v32 > 21 m2 v22 ) < 2r22 U m−1
2
(since in Proposition 1.17, we noted that h < 0 implies T < U ) < 2r22 U 2 U −1 m−1
< (by (iii))
2
−1
2(2c8 )2 U −1 m−1
≥ U −1 , so we
2 . Now, we have (from Proposition 1.17) 2U ≥ k, and hence 2k
2 −1 −1
−1
2 −1
bound our inequality by ≤ 4(2c8 ) k m2 = c12 k where c12 ≡ 4(2c8 ) m2 . Along with our earlier
−1/2
bound for the second term of the right side of (ii), we get that ṙ22 (t2 )r22 (t2 ) < (c12 + 2c11 m2 )k −1 .
−1
2
We have deduced an upper bound of the form c13 k for (r2 (t2 )ṙ2 (t2 )) in all cases, where c13 is
−1/2
−1/2
2M
)−2 c7 + 2c11 m2 ) is applicable; this gives us an upper
whichever of (c12 + 2c11 m2 ) or (( m2m
1 +m3
1/2
bound for −r2 (t2 )ṙ2 (t2 ) of the form c10 k −1/2 in all cases, where c10 ≡ c13 , 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 inequalities that forced a deeper consideration of the behavior of I.
Throughout this proof, a common trick was to, with a desired inequality in mind, craft a function
whose integral would yield the desired inequality. This technique was observed by the student during
Lecture 10, and seems to be a simple but fundamental and very useful method of analyzing differential
equations. 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 P2 ; this will have length r13 ≡ r. We will
denote by v the velocity of P2 . Any other notation remains unchanged. Let’s outline the proof.
We first prove the case when r13 ≥ c/4, then we address the case when r13 < c/4. In this latter
case, we will investigate the behavior of the function r22 (v 2 − ṙ22 ); this will involve an exploration of
the angular momentum integrals, which will allow us to create a bound for |v 2 − ṙ22 |. From there,
we will seek a bound for ṙ22 , which will allow us to bound v 2 and therefore complete our proof.
Proposition 2.2: If for all time, r ≥ 41 c, then Lemma 2.1 holds.
Proof: By Lemma 1.1, σ = r12 + r13 + r23 > c for some c > 0. Now, denote by r the shortest
side of the triangle
formed by the
P three particles. If for all time, r ≥ c/4, then we have that
P
T = U + h < k<l mrkklml + h < k<l 4mck ml + h < 12m2max /c + A, where mmax again denotes the
P
greatest mass. This gives a bound on all velocities as T = mq vq2 , hence, T > mk vk2 for k = 1, 2, 3,
q
−1 1/2
2
2
hence T m−1
, proving Lemma 2.1 for this case.
k > vk , hence vk < ((A + (12mmax )/c)mk )
We note that c/4 is arbitrary; it simply gives us a lower bound and could have been any number.
7
Fact 2.3: r22 (v 2 − ṙ22 ) = (x2 ẏ2 − y2 ẋ2 )2 + (y2 ż2 − z2 ẏ2 )2 + (z2 ẋ2 − x2 ż2 )2 .
2 2
2
Proof: WePnow treat the case when r < c/4.
P We consider the expression r2 (v −ṙ2 ); substituting
P 2
the definition
q22 = r22 and its derivative
q̇2 q2 = ṙ2 r2 , as well as the definition
q̇ = v 2 ,
q
q
q
P
P 2P 2
all of which we used earlier, we obtain the expression
q2 q̇2 − ( q̇2 q2 )2 . If we expand this
q
q
q
out as (x22 + y22 + z22 )(ẋ22 + ẏ22 + ż22 ) − (x2 ẋ2 + y2 ẏ2 + z2 ż2 )2 and rearrange, we get the expression
(x2 ẏ2 − y2 ẋ2 )2 + (y2 ż2 − z2 ẏ2 )2 + (z2 ẋ2 − x2 ż2 )2 , which resembles the angular momentum integrals.
2
2M
2
ẏ2 ) − (y1 − y3 )(ẋ1 + m1m+m
ẋ2 )) + mm1 +m
(x2 ẏ2 − y2 ẋ2 ).
Fact 2.4: |Lz | = m1 ((x1 − x3 )(ẏ1 + m1m+m
3
3
3
Proof: Consider the angular momentum integral |Lz | = m1 (x1 ẏ1 − y1 ẋ1 ) + m2 (x2 ẏ2 − y2 ẋ2 ) +
m3 (x3 ẏ3 −y3 ẋ3 ). We change it into m1 ((x1 −x3 )ẏ1 −(y1 −y3 )ẋ1 ))+m2 ((x2 −x3 )ẏ2 −(y2 −y3 )ẋ2 )) by
noting that m3 x3 ẏ3 = −m1 x3 ẏ1 − m2 x3 ẏ2 and −m3 y3 ẋ3 = m1 y3 ẋ1 + m2 y3 ẋ2 ; 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
1
(q1 − q3 ) + m1M
q2 − q3 = m1m+m
+m3 q2 . We can use this to eliminate the terms x2 − x3 and y2 − y3 ; this
3
2
2M
2
ẏ2 )−(y1 −y3 )(ẋ1 + m1m+m
ẋ2 ))+ mm1 +m
(x2 ẏ2 −y2 ẋ2 ).
gives us the expression m1 ((x1 −x3 )(ẏ1 + m1m+m
3
3
3
Proposition 2.5: In the case that r < c/4, we can find c19 > 0 such that 0 ≤ (v 2 − ṙ22 ) < c19 r2−2 .
Proof: Consider our expression for |Lz | from Fact 2.4. We bound the absolute value of the
2
2
first term like so: |m1 ((x1 − x3 )(ẏ1 + m1m+m
ẏ2 ) − (y1 − y3 )(ẋ1 + m1m+m
ẋ2 ))|
3
3
m2
2
≤ |m1 (x1 − x3 )(ẏ1 + m1 +m3 ẏ2 )| + |m1 (y1 − y3 )(ẋ1 + m1m+m
ẋ
))|
(Triangle Inequality)
2
3
m2
2
ẏ
)|
+
|(
ẋ
+
ẋ
)|)
(since
|(x
−
x
)|
≤
r
and |(y1 − y3 )| ≤ r)
≤ m1 (r)(|(ẏ1 + m1m+m
2
1
1
3
m1 +m3 2
3
m2
M
M
≤ m1 (r)( m1 +m3 )(|ẏ1 + ẏ2 | + |ẋ1 + ẋ1 |)
(since m1 +m3 < m1 +m3 and 1 < m1M
+m3 )
M m1 r
≤ m1 +m3 (|x˙1 | + |x˙2 | + |y˙1 | + |y˙2 |)
(Triangle Inequality)
−1/2
−1/2
M m1 r
1/2
(2m1
+ 2m2 )
m1 +m3 T
−1/2
−1/2
m1
= c14 rT 1/2 where c14 ≡ mM1 +m
(2m1
+ 2m2 ).
3
In turn, c14 rT 1/2 ≤ c14 r(U + |h|)1/2 ≤ c14 rU 1/2 + c14 r|h|1/2
that rU < c8 from the proof to Proposition 1.17, we conclude r2 U
≤
(by definition of T )
< c14 rU 1/2 + c14 rA1/2 . Recalling
< 14 c8 c, and so rU 1/2 < 12 (c8 c)1/2 ,
and so the absolute value of the first term is bounded from above by c14 21 (c8 c)1/2 + 14 c14 cA1/2 ≡ c15 .
In Lecture 10, it was proven that 2IT ≥ η; this shows that |Lz |2 ≤ 6Iτ Tτ ≤ 6Iτ (Uτ + |h|) <
12A2 . Hence, |Lz | < 31/2 2A < 6A. This allows us to place a bound on |(x2 ẏ2 − y2 ẋ2 )|, since
2
2
2M
|Lz | = m1 ((x1 − x3 )(ẏ1 + m1m+m
ẏ2 ) − (y1 − y3 )(ẋ1 + m1m+m
ẋ2 )) + mm1 +m
(x2 ẏ2 − y2 ẋ2 ) gives us
3
3
3
m2
m2
m2 M
| m1 +m3 (x2 ẏ2 − y2 ẋ2 )| ≤ |m1 ((x1 − x3 )(ẏ1 + m1 +m3 ẏ2 ) − (y1 − y3 )(ẋ1 + m1 +m3 ẋ2 ))| + |Lz | < 6A + c15
3
and hence |(x2 ẏ2 − y2 ẋ2 )| < (6A + c15 ) mm1 +m
≡ c16 . An identical process gives c17 and c18 such that
2M
|(y2 ż2 − z2 ẏ2 )| < c17 and |(z2 ẋ2 − x2 ż2 )| < c18 . Hence, 0 ≤ (v 2 − ṙ22 ) < r2−2 (c216 + c217 + c218 ) ≡ c19 r2−2 .
Proposition 2.6: If r < c/4, we can find some c20 > 0 such that 0 ≤ v 2 − ṙ22 < c20 c19 r2−1 .
Proof: The conditions that r < c/4 and r12 +r +r23 ≥ c tell us that r12 > c/4 and r23 > c/4 by
−1
−1
the triangle inequality, and hence r12
< 4c−1 and r23
< 4c−1 . Recall the upper bounds for r12 /r2
and r23 /r2 in Corollary 1.20; these are (2M )(m1 + m3 /2)−1 and (2M )(m3 + m1 /2)−1 , respectively.
Now, we bound r2−1 by min(4c−1 (2M )(m1 + m3 /2)−1 , 4c−1 (2M )(m3 + m1 /2)−1 ) ≡ c20 . So we have
(i) 0 ≤ v 2 − ṙ22 < c20 c19 r2−1 , as desired.
Proposition 2.7: In turn, we can construct the bound P
|r̈2 | < r2−2 (4m1 + 4m3 + c20 c19 )
2
2
+m3 q3r−q
)+v22 = r̈2 r2 +
Proof: Now we recall from Proposition 1.21 the equation q2 (m1 q1r−q
3
3
12
23
q
P
2
2
ṙ22 . Rearranging, we get
q2 (m1 q1r−q
+ m3 q3r−q
) + v22 − ṙ22 = r̈2 r2 , which gives us, when we apply
3
3
12
23
q
P
2
2
Proposition 2.6 and the triangle inequality, that | q2 (m1 q1r−q
+ m3 q3r−q
)| + c20 c19 r2−1 > |r̈2 r2 |.
3
3
q
12
23
−3
−2
As in the proof to Proposition 1.21, |qk − q2 |rk2
< 4r
we derive (m1 4r2−1 + m3 4r2−1 ) +
P2 , and hence
−1
−2
−2
−1
−2
c20 c19 r2 ≥ r2 (m1 4r2 + m3 4r2 ) + c20 c19 r2 ≥ | q2 ||(m1 4r2 + m3 4r2−2 )| + c20 c19 r2−1 > |r̈2 r2 |.
q
Noting r2 > 0, this gives us the desired upper bound |r̈2 | < r2−2 (4m1 + 4m3 + c20 c19 ).
8
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, ṙ2 = 0. Then Proposition 2.6 and our upper bound c20 for r2−1 gives us
1/2
that 0 ≤ v 2 ≤ c20 c19 c20 , and hence we have the bound v < (c20 c19 )/2 and we are done.
Now, consider a time t when ṙ2 does not vanish; then we can find an interval t1 < t < t2 in
which r < c/4 holds and in which ṙ2 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)|2ṙ2 r̈2 | < 2|ṙ2 |r2−2 (4m1 + 4m3 + c20 c19 ) in our interval. Since ṙ2 6= 0 in our interval, ṙ2 must
preserve its sign, and hence we see that (ii)|ṙ22 (t) − ṙ22 (t1 )| < 2(4m1 + 4m3 + c20 c19 )|r2−1 (t) − r2−1 (t1 )|;
this is because the change in ṙ2 must be less than the change in 2(4m1 + 4m3 + c20 c19 )(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 < c20 ; this holds for r2−1 (t) and r2−1 (t1 ), so
|r2−1 (t) − r2−1 (t1 )| < c20 . Thus, from (ii), we obtain (iii) ṙ22 (t) < ṙ22 (t1 ) + 2(4m1 + 4m3 + c20 c19 )c20 .
Now taking t1 to be the smallest value that satisfies the above conditions and the extra condition
that t1 ≥ τ . If t1 = τ , then Proposition 2.6 gives us that ṙ22 (t1 ) ≤ vτ2 ; in turn, we see that this is
−1
−1
≤ 2m−1
2 Tτ ≤ 2m2 |Uτ | + |h| < 4m2 A ≡ c21 . Otherwise, if t1 > τ , then it must be that either
ṙ2 (t1 ) = 0 or that r(t1 ) = c/4. In the first case, we have that ṙ2 (t1 ) = 0 < c21 . In the second case,
we bounce back to the firstPcase that we dealt with in the theorem, since at t1 , we see that again,
T is bounded by U + h ≤ k<l (4mk ml )/c + h < (12m2max )/c + A ≡ c22 , and thus again, we have
−1
that ṙ22 (t1 ) ≤ v 2 (t1 ) ≤ 2m−1
1 T1 < 2m2 c22 ≡ c23 .
Hence, we see that Proposition 2.6 gives us 0 ≤ v 2 < c20 c19 r2−1 + ṙ22 ≤ (by (iii)) c220 c19 + ṙ22 (t1 ) +
2(4m1 + 4m3 + c20 c19 )c20 < c220 c19 + c24 + 2(4m1 + 4m3 + c20 c19 )c20 ≡ c25 , where c24 is whichever
1/2
of c21 or c23 applies. Now, we have v < c25 , 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 quickly 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, 1995.
[2] J. Barrow-Green, The dramatic episode of Sundman, Historia Mathematica 37(2) (2010), 164–203.
[3] F. Diacu, The solution of the n-body problem, The Mathematical Intelligencer 18 (1996), no. 3, 66–70.
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