Module MA1132 (Frolov), Advanced Calculus
Homework Sheet 4
Each set of homework questions is worth 100 marks
Due: at the beginning of the tutorial session Thursday/Friday, 18/19 February 2016
Name:
1. Consider the function
π
π
z = 3ey− 4 cos x − 2e 2 −x sin y
(a) Find
iii)
∂ 2z π π
( , ),
∂x∂y 2 4
iv)
∂ 2z π π
( , ).
∂y∂x 2 4
Solution:
√
π
π
∂ 2z π π
∂ z
∂ ∂z
∂
−x
y− π4
−x
y− π4
2
2
cos x − 2e
cos y = −3e
sin x+2e
cos y ⇒
iii)
=
=
3e
( , ) = −3+ 2 .
∂x∂y
∂x ∂y
∂x
∂x∂y 2 4
2
iv)
√
π
π
π
π
∂ 2z
∂ ∂z
∂
∂ 2z π π
=
=
−3ey− 4 sin x + 2e 2 −x sin y = −3ey− 4 sin x+2e 2 −x cos y ⇒
( , ) = −3+ 2 .
∂y∂x
∂y ∂x
∂y
∂x∂y 2 4
π
π
(b) Find the slope of the surface z = 3ey− 4 cos x − 2e 2 −x sin y in the y-direction at the
point ( π3 , π6 ).
Solution: The slope ky is equal to
ky =
π
π
∂z π π
3 π √ π
( , ) = 3ey− 4 cos x − 2e 2 −x cos y|x= π3 ,y= π6 = e− 12 − 3e 6 ≈ −1.76936 .
∂y 3 6
2
π
π
(c) Show that the function z = 3ey− 4 cos x − 2e 2 −x sin y satisfies Laplace’s equation
∂ 2z ∂ 2z
+
= 0.
∂x2 ∂y 2
Solution: To this end we compute the following derivatives
π
π
π
π
∂ 2z
∂ ∂z
∂
=
=
−3ey− 4 sin x + 2e 2 −x sin y = −3ey− 4 cos x − 2e 2 −x sin y ,
2
∂x
∂x ∂x
∂x
π
π
∂ 2z
∂ ∂z
∂
y− π4
−x
y− π4
−x
2
2
=
=
3e
cos
x
−
2e
cos
y
=
3e
cos
x
+
2e
sin y .
∂y 2
∂y ∂y
∂y
The sum of these two expressions is obviously 0.
2. The equations of motion of a system of n particles are given by
mi ẍi = −
∂U (x1 , . . . , xn )
,
∂xi
ẍi =
d 2 xi
,
dt2
i = 1, 2, . . . , n ,
where mi is the mass and xi is the coordinate of the i-th particle, and U (x1 , . . . , xn ) is
the potential energy of the system.
1
(a) Consider a system of n particles moving in a central field
U (x1 , . . . , xn ) = V (r) ,
n
X
r=
xi e i ,
i=1
where V is a smooth function of a single variable.
i.
ii.
iii.
iv.
v.
Find the equations of motion of the first particle (x1 ).
Find the equations of motion of the second particle (x2 ).
Find the equations of motion of the last particle (xn ).
Find the equations of motion of the i-th particle (xi ) for 1 < i < n.
Write the equations of motion of the i-th particle (xi ) for 1 ≤ i ≤ n by using
the Kronecker delta δij .
Solution: We have for all i’th
mi ẍi = −
∂
xi
∂U (x1 , . . . , xn )
=−
V (r) = −V 0 (r) ,
∂xi
∂xi
r
i = 1, . . . , n .
(b) Consider a system of n particles with the Toda potential
U (x1 , . . . , xn ) =
n−1
X
eα(xi+1 −xi ) .
i=1
i.
ii.
iii.
iv.
v.
Find the equations of motion of the first particle (x1 ).
Find the equations of motion of the second particle (x2 ).
Find the equations of motion of the last particle (xn ).
Find the equations of motion of the i-th particle (xi ) for 1 < i < n.
Write the equations of motion of the i-th particle (xi ) for 1 ≤ i ≤ n by using
the Kronecker delta δij .
Solution:
i.
ii.
n−1
∂U (x1 , . . . , xn )
∂ X α(xj+1 −xj )
m1 ẍ1 = −
=−
e
∂x1
∂x1 j=1
!
n−1
X
∂
=−
eα(x2 −x1 ) +
eα(xj+1 −xj ) = αeα(x2 −x1 ) ,
∂x1
j=2
(1)
n−1
∂ X α(xj+1 −xj )
∂U (x1 , . . . , xn )
m2 ẍ2 = −
=−
e
∂x2
∂x2 j=1
∂
=−
∂x2
= αe
eα(x2 −x1 ) + eα(x3 −x2 ) +
α(x3 −x2 )
n−1
X
j=3
− αe
2
α(x2 −x1 )
,
!
eα(xj+1 −xj )
(2)
=
iii.
iv.
n−1
∂U (x1 , . . . , xn )
∂ X α(xj+1 −xj )
mn ẍn = −
=−
e
∂xn
∂xn j=1
!
n−2
X
∂
α(xn −xn−1 )
α(xj+1 −xj )
e
+
=−
e
= −αeα(xn −xn−1 ) ,
∂xn
j=1
(3)
n−1
∂ X α(xj+1 −xj )
∂U (x1 , . . . , xn )
=−
e
=
mi ẍi = −
∂xi
∂xi j=1
∂
−
∂xi
= −αe
e
α(xi −xi−1 )
+e
α(xi+1 −xi )
n−1
X
+
!
e
α(xj+1 −xj )
=
(4)
j=1,j6=i−1,i
α(xi −xi−1 )
v.
mi ẍi = −
+ αe
α(xi+1 −xi )
,
n−1
∂ X α(xj+1 −xj )
∂U (x1 , . . . , xn )
=−
e
∂xi
∂xi j=1
= −α
n−1
X
eα(xj+1 −xj ) δi,j+1 − eα(xj+1 −xj ) δij
(5)
j=1
= −α
n−1
X
eα(xj+1 −xj ) δi,j+1 + α
j=1
n−1
X
eα(xj+1 −xj ) δij .
j=1
Since δi,j+1 = 1 if j = i − 1 and is 0 for all the other values of j, the first sum
gives
n−1
X
−α
eα(xj+1 −xj ) δi,j+1 = −αeα(xi −xi−1 ) + αeαx1 δi1 ,
(6)
j=1
where the last term αeαx1 δi1 appears because the first sum is 0 for i = 1.
Since δij = 1 if j = i and is 0 for all the other values of j, the second sum gives
α
n−1
X
eα(xj+1 −xj ) δij = αeα(xi+1 −xi ) − αe−αxn δi,n ,
(7)
j=1
where the last term −αe−αxn δi,n appears because the second sum is 0 for i = n.
Summing up the two contributions one gets
mi ẍi = −αeα(xi −xi−1 ) + αeα(xi+1 −xi ) + αeαx1 δi1 − αe−αxn δi,n ,
(8)
where x0 = 0 , xn+1 = 0.
(c) Find the equations of motion of a system of n particles with the rational CalogeroMoser potential
n
X
α
.
U (x1 , . . . , xn ) =
(xi − xj )2
i,j=1,i6=j
3
Solution: We have
∂
∂U (x1 , . . . , xn )
=−
mi ẍi = −
∂xi
∂xi
n
X
n
X
α
2α
=
(δij − δik )
(xj − xk )2 j,k=1,j6=k (xj − xk )3
j,k=1,j6=k
n
X
n
n
X
X
2α
2α
4α
=
−
=
.
3
3
(xi − xk )
(xj − xi )
(xi − xj )3
k=1,k6=i
j=1,j6=i
j=1,j6=i
(9)
3. Compute the differential df of
1
1
1
f (x1 , x2 , . . . , xn ) = ( + x1 )α1 ( + x2 )α2 · · · ( + xn )αn ,
2
2
2
and find its local linear approximation at ( 12 , 12 , . . . , 21 ).
Solution: The differential of the function is defined by
n
X
∂f
df =
dxi .
∂xi
i=1
Computing partial derivatives, one gets
∂ 1
∂f
1
1
=
( +x1 )α1 ( +x2 )α2 · · · ( +xn )αn =
∂xi
∂xi 2
2
2
1
2
αi 1
1
1
( +x1 )α1 ( +x2 )α2 · · · ( +xn )αn =
2
2
+ xi 2
1
2
The differential of the function is given by the formula
df = f
n
X
1
i=1 2
n
X
αi
1
dxi = f
αi d ln( + xi ) .
2
+ xi
i=1
The local linear approximation of the function at (1, . . . , 1) is given by the formula
n
X ∂f (1, . . . , 1)
1
1
1
L(x1 , x2 , . . . , xn ) = f ( , . . . , ) +
(xi − ) .
2
2
∂xi
2
i=1
We obviously have f ( 12 , . . . , 12 ) = 1, and
∂f ( 12 ,..., 12 )
∂xi
L(x1 , x2 , . . . , xn ) = 1 +
= αi . Thus
n
X
i=1
1
αi (xi − ) .
2
4. The Taylor series is given by
f (~x) =
∞
X
k1 ,...,kn
∂1k1 · · · ∂nkn f (x~o ) k1
∆x1 · · · ∆xknn ,
k1 ! · · · kn !
=0
(10)
where we denote
f (x1 , . . . , xn ) ≡ f (~x) ,
f (xo1 , . . . , xon ) ≡ f (x~o ) ,
4
xi − xoi ≡ ∆xi
(11)
αi
f.
+ xi
and ∂i0 f ≡ f ; ∂ik f ≡
∂k f
∂xki
is the k-th partial derivative of f with respect to xi .
The Taylor series can be equivalently written as
∞
n
X
1 X
∂ q f (x~o )
f (~x) =
∆xi1 · · · ∆xiq .
q! i ,...,i =1 ∂xi1 · · · ∂xiq
q=0
1
(12)
q
Check the equality by computing the Taylor series expansion up to the third order.
Solution: We have
∞
X
f (~x) =
+
n
X ∂f (x~o )
∂1k1 · · · ∂nkn f (x~o ) k1
∆x1 · · · ∆xknn = f (x~o ) +
∆xi
k
∂x
1 ! · · · kn !
i
i=1
=0
k1 ,...,kn
n
2
X
n
X
∂ f (x~o ) 2
∂ 2 f (x~o )
∆x
+
∆xi ∆xj
i
∂x2i
∂x
∂x
i
j
1=i<j
1
2
1
+
3!
i=1
n
X
i=1
n
X
n
n
1 X ∂ 3 f (x~o )
∂ 3 f (x~o ) 3 1 X ∂ 3 f (x~o ) 2
∆xi +
∆xi ∆xj +
∆xi ∆x2j
3
2
2
∂xi
2 1=i<j ∂xi ∂xj
2 1=i<j ∂xi ∂xj
(13)
∂ 3 f (x~o )
∆xi ∆xj ∆xk + O(∆x4 ) ,
+
∂xi ∂xj ∂xk
1=i<j<k
and
∞
n
n
X
X
∂ k f (x~o )
1 X
∂f (x~o )
o
f (~x) =
∆xi1 · · · ∆xik = f (x~ ) +
∆xi
k!
∂x
·
·
·
∂x
∂x
i
i
i
1
k
i ,...,i =1
i=1
k=0
1
k
n
n
1 X ∂ 2 f (x~o )
1 X ∂ 3 f (x~o )
+
∆xi ∆xj +
∆xi ∆xj ∆xk + O(∆x4 )
2 i,j=1 ∂xi ∂xj
3! i,j,k=1 ∂xi ∂xj ∂xj
= f (x~o ) +
n
X
∂f (x~o )
i=1
+
+
∂xi
∆xi +
n
n
X
1 X ∂ 2 f (x~o ) 2
∂ 2 f (x~o )
∆xi ∆xj
∆x
+
i
2 i=1 ∂x2i
∂x
∂x
i
j
1=i<j
n
n
n
1 X ∂ 3 f (x~o )
1 X ∂ 3 f (x~o ) 3 1 X ∂ 3 f (x~o ) 2
+
∆x
∆x
∆x
+
∆xi ∆x2j
j
i
i
3! i=1 ∂x3i
2 1=i<j ∂x2i ∂xj
2 1=i<j ∂xi ∂x2j
n
X
∂ 3 f (x~o )
∆xi ∆xj ∆xk + O(∆x4 ) ,
∂x
∂x
∂x
i
j
k
1=i<j<k
which proves the formula up to the third order.
5. Consider the “Higgs” potential
λ2
κ2
U (x1 , . . . , xn ) = − r2 + r4 ,
2
4
n
X
r=
xi e i ,
i=1
(a) Plot the potential for n = 1, and for κ = λ = 2.
Solution: The curve is shown below
5
κ > 0, λ > 0 .
(14)
0.5
-1.5
-1.0
0.5
-0.5
1.0
1.5
-0.5
-1.0
(b) Plot the potential for n = 2, and for κ = λ = 2.
Solution: The surface is shown below
It is a surface of revolution obtained by revolving the graph of the curve above about
the vertical axis.
(c) Find the Taylor series expansion of the “Higgs” potential about the point xo1 = λκ ,
xoi = 0, i = 2, . . . , n up to the fourth order in yi ≡ xi − xoi . Use Mathematica to
check your answer.
6
Solution: We have
κ
κ4
U ( , 0, . . . , 0) = − 2 ,
λ
4λ
∂U ( λκ , 0, . . . , 0)
∂U (x1 , . . . , xn )
2
2 2
= −κ xi + λ r xi ⇒
= 0,
∂xi
∂xi
∂ 2 U (x1 , . . . , xn )
= (−κ2 + λ2 r2 )δij + 2λ2 xi xj ⇒
∂xi ∂xj
2
∂ U ( λκ , 0, . . . , 0)
∂ 2 U ( λκ , 0, . . . , 0)
2
=
2κ
,
= 0 if i 6= 1, j 6= 1 ,
∂x21
∂xi ∂xj
∂ 3 U (x1 , . . . , xn )
= 2λ2 xk δij + 2λ2 xi δjk + 2λ2 xj δik ⇒
∂xi ∂xj ∂xk
3
∂ 3 U ( λκ , 0, . . . , 0)
∂ 3 U ( λκ , 0, . . . , 0)
∂ U ( λκ , 0, . . . , 0)
=
6κλ
,
=
0
,
= 2κλ ,
∂x31
∂x21 ∂xj
∂x1 ∂xj ∂xj
∂ 3 U ( λκ , 0, . . . , 0)
= 0 , i, j, k = 2, . . . , n ,
∂xi ∂xj ∂xk
(15)
j = 2, . . . , n
(16)
∂ 4 U (x1 , . . . , xn )
∂xi ∂xj ∂xk ∂xl
∂ 4 U ( λκ , 0, . . . , 0)
∂xi ∂xj ∂xk ∂xl
∂ 4 U ( λκ , 0, . . . , 0)
∂x4i
∂ 4 U ( λκ , 0, . . . , 0)
∂x3i ∂xl
∂ 4 U ( λκ , 0, . . . , 0)
∂x2i ∂x2k
∂ 4 U ( λκ , 0, . . . , 0)
∂x2i ∂xk ∂xl
∂ 4 U ( λκ , 0, . . . , 0)
∂xi ∂xj ∂xk ∂xl
= 2λ2 δkl δij + 2λ2 δil δjk + 2λ2 δjl δik
⇒
= 2λ2 (δkl δij + δil δjk + δjl δik ) ,
= 6λ2 ,
= 0,
= 2λ2
l 6= i ,
(17)
i 6= k ,
= 0 i 6= k, k 6= l ,
= 0 i 6= j, k, l .
Thus, one gets
U (x1 , . . . , xn ) = −
where ~y 2 ≡
Pn
k=1
n
X
κ4
1
2 2
3
+
κ
y
+
κλy
+
κλy
yk2 + λ2 (~y 2 )2 ,
1
1
1
2
4λ
4
k=2
yk2 .
6. Find the Taylor series expansion of the periodic Toda potential
U (x1 , . . . , xn ) =
n
X
eα(xi+1 −xi ) ,
i=1
7
xk+n ≡ xk ,
k ∈ Z,
(18)
about the point xi = a, i = 1, . . . , n up to the fourth order in xi − a. Use Mathematica
to check your answer.
Solution: We have
U (a, . . . , a) = n ,
∂U (x1 , . . . , xn )
∂U (a, . . . , a)
= αeα(xi −xi−1 ) − αeα(xi+1 −xi ) ⇒
= 0,
∂xi
∂xi
∂ 2 U (x1 , . . . , xn )
= α2 eα(xi −xi−1 ) (δij − δi−1,j ) − α2 eα(xi+1 −xi ) (δi+1,j − δij )
∂xi ∂xj
∂ 2 U (a, . . . , a)
= α2 (2δij − δi−1,j − δi+1,j ) ,
∂xi ∂xj
⇒
(19)
∂ 3 U (x1 , . . . , xn )
= α3 eα(xi −xi−1 ) (δij − δi−1,j )(δik − δi−1,k )
∂xi ∂xj ∂xk
− α3 eα(xi+1 −xi ) (δi+1,j − δij )(δi+1,k − δik )
⇒
(20)
3
∂ U (a, . . . , a)
= α3 (δij − δi−1,j )(δik − δi−1,k ) − (δi+1,j − δij )(δi+1,k − δik ) ,
∂xi ∂xj ∂xk
∂ 4 U (x1 , . . . , xn )
= α4 eα(xi −xi−1 ) (δij − δi−1,j )(δik − δi−1,k )(δil − δi−1,l )
∂xi ∂xj ∂xk ∂xl
− α4 eα(xi+1 −xi ) (δi+1,j − δij )(δi+1,k − δik )(δi+1,l − δil )
⇒
∂ 4 U (a, . . . , a)
= α4 (δij − δi−1,j )(δik − δi−1,k )(δil − δi−1,l ) − (δi+1,j − δij )(δi+1,k − δik )(δi+1,l − δil ) .
∂xi ∂xj ∂xk ∂xl
(21)
Thus, one gets
n
X
eα(xi+1 −xi ) = U0 + U1 + U2 + U3 + U4 ,
(22)
i=1
where
U0 = n ,
U2 =
=
U1 = 0 ,
(23)
n
α2 X
(2δij − δi−1,j − δi+1,j )(xi − a)(xj − a)
2 i,j=1
n
α2 X
2(xi − a)2 − (xi − a)(xi−1 − a) − (xi − a)(xi+1 − a)
2 i=1
n
α2 X
=
2x2i − xi xi−1 − xi xi+1 − 4axi + 2axi + axi−1 + axi+1
2 i=1
(24)
n
n
α2 X
α2 X 2
2
=
xi + xi+1 − 2xi xi+1 =
(xi+1 − xi )2 ,
2 i=1
2 i=1
where we used the periodicity condition xn+k ≡ xn which implies in particular that
n
X
i=1
ai+k =
n
X
i=1
ai ,
n
X
ai+k ai+j =
i=1
n
X
i=1
8
ai ai+j−k
∀ j, k .
(25)
Similarly one gets
U3 =
=
n
α3 X
(δij − δi−1,j )(δik − δi−1,k ) − (δi+1,j − δij )(δi+1,k − δik ) (xi − a)(xj − a)(xk − a)
3! i,j,k=1
n
α3 X
(−2(xi − a)2 (xi−1 − a) + (xi − a)(xi−1 − a)2 + 2(xi − a)2 (xi+1 − a) − (xi − a)(xi+1 − a)2 )
3! i=1
n
α3 X
(−2(xi+1 − a)2 (xi − a) + (xi+1 − a)(xi − a)2 + 2(xi − a)2 (xi+1 − a) − (xi − a)(xi+1 − a)2 )
=
3! i=1
n
n
α3 X
α3 X
2
2
=
(−3(xi+1 − a) (xi − a) + 3(xi+1 − a)(xi − a) ) =
((xi+1 − a)(xi − a)(xi − xi+1 )
3! i=1
2 i=1
=
n
n
n
α3 X
α3 X
α3 X
(xi+1 xi − a(xi + xi+1 ))(xi − xi+1 ) =
(3xi+1 x2i − 3x2i+1 xi ) =
(xi+1 − xi )3 ,
2 i=1
6 i=1
6 i=1
(26)
and
U4 =
n
α4 X
(δij − δi−1,j )(δik − δi−1,k )(δil − δi−1,l )
4! i,j,k,l=1
− (δi+1,j − δij )(δi+1,k − δik )(δi+1,l − δil ) (xi − a)(xj − a)(xk − a)(xl − a)
n
α4 X =
(xi − a)4 − 3(xi − a)3 (xi−1 − a) + 3(xi − a)2 (xi−1 − a)2 − (xi − a)(xi−1 − a)3
4! i=1
+ (xi+1 − a)4 − 3(xi − a)3 (xi+1 − a) + 3(xi − a)2 (xi+1 − a)2 − (xi − a)(xi+1 − a)3
n
α4 X (xi+1 − a)4 − 4(xi+1 − a)3 (xi − a) + 6(xi+1 − a)2 (xi − a)2 − 4(xi+1 − a)(xi − a)3 + (xi − a)4
=
4! i=1
n
α4 X
=
(xi+1 − xi )4 .
4! i=1
(27)
Thus,
n
X
n
n
n
α2 X
α3 X
α4 X
(xi+1 − xi )2 +
(xi+1 − xi )3 +
(xi+1 − xi )4 .
2
3
4
i=1
i=1
i=1
i=1
(28)
A much simpler way to get the result is to use the fact that the Taylor expansion is an
expansion in powers of xi − a, and that
eα(xi+1 −xi ) = n +
n
X
i=1
e
α(xi+1 −xi )
=
n
X
α (xi+1 −a)−(xi −a)
e
.
(29)
i=1
Thus the Taylor expansion about xi = a is equivalent to the Taylor expansion in powers
of α, and the result (28) follows immediately.
9