Module MA1132 (Frolov), Advanced Calculus
Tutorial Sheet 4
To be solved during the tutorial session Thursday/Friday, 11/12 February 2016
1. Consider the function
π
π
z = 2e 2 −y sin x − 3ex− 4 cos y
(a) Find
i)
∂z π π
( , ),
∂x 4 2
ii)
∂z π π
( , ).
∂y 4 2
Solution:
√
π
π
∂z
∂z π π
= 2e 2 −y cos x − 3ex− 4 cos y ⇒
( , ) = 2.
∂x
∂x 4 2
√
π
π
∂z
∂z π π
= −2e 2 −y sin x + 3ex− 4 sin y ⇒
( , ) = 3 − 2.
ii)
∂y
∂y 4 2
i)
π
π
(b) Find the slope of the surface z = 2e 2 −y sin x − 3ex− 4 cos y in the x-direction at the
point ( π2 , π6 );
Solution: The slope kx is equal to
√
∂z π π
3 3 π
kx =
( , )=−
e4 .
∂x 2 6
2
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.
(a) Find the equations of motion of a system of n particles moving in a Coulomb field
α
U (x1 , . . . , xn ) = ,
r
n
X
r=
xi e i .
i=1
Solution: We have
mi ẍi = −
∂U (x1 , . . . , xn )
∂ α
α ∂r
α xi
=−
= 2
= 2 .
∂xi
∂xi r
r ∂xi
r r
(b) Find the equations of motion of a system of n coupled harmonic oscillators
U (x1 , . . . , xn ) =
n−1
X
κ
i=1
1
2
(xi+1 − xi )2 ,
Solution: We have
n−1
n−1
X
∂ Xκ
∂U (x1 , . . . , xn )
2
=−
(xj+1 − xj ) = −κ
mi ẍi = −
(xj+1 − xj )(δi,j+1 − δij )
∂xi
∂xi j=1 2
j=1
= −κ
n−1
X
(xj+1 − xj )δi,j+1 − (xj+1 − xj )δij
j=1
= −κ
n−1
X
(xi − xi−1 )δi,j+1 − (xi+1 − xi )δij
j=1
= −κ(xi − xi−1 )
n−1
X
δi,j+1 + κ(xi+1 − xi )
j=1
n−1
X
δij
j=1
= −κ(xi − xi−1 ) + κ(xi − xi−1 )δi1 + κ(xi+1 − xi ) − κ(xi+1 − xi )δin
= −κ(2xi − xi−1 − xi+1 ) + κx1 δi1 + κxn δin ,
where δij is Kronecker’s delta δij = 1 if i = j, δij = 0 if i 6= j, and x0 = xn+1
Thus, more explicitly we get
m1 ẍ1 = −κ(x1 − x2 ) ,
mi ẍi = −κ(2xi − xi−1 − xi+1 ) ,
mn ẍn = −κ(xn − xn−1 ) .
if
i = 2, 3, . . . , n − 1 ,
(1)
= 0.
(2)
(c) Find the equations of motion of a system of n particles with pairwise interaction
n
X
U (x1 , . . . , xn ) =
V (xi − xj ) .
i,j=1,i6=j
Here V is an even function of a single variable, and we use the notation
n
X
aij ≡
n
n
X
X
aij =
j=1 i=1,i6=j
i,j=1,i6=j
n
n
X
X
aij .
i=1 j=1,j6=i
Solution: We have
∂U (x1 , . . . , xn )
∂
=−
mi ẍi = −
∂xi
∂xi
n
X
=−
V 0 (xi − xk ) +
k=1,k6=i
n
X
= −2
n
X
V (xj − xk ) = −
j,k=1,j6=k
n
X
n
X
V 0 (xj − xk )(δij − δik )
j,k=1,j6=k
V 0 (xj − xi )
j=1,j6=i
V 0 (xi − xj ) ,
j=1,j6=i
(3)
where δij is Kronecker’s delta δij = 1 if i = j, δij = 0 if i 6= j
2
3. The Taylor series is given by
f (~x) =
∞
X
k1 ,...,kn
∂1k1 · · · ∂nkn f (x~o ) k1
∆x1 · · · ∆xknn ,
k1 ! · · · kn !
=0
(4)
where we denote
f (xo1 , . . . , xon ) ≡ f (x~o ) ,
f (x1 , . . . , xn ) ≡ f (~x) ,
∂k f
∂xki
and ∂i0 f ≡ f ; ∂ik f ≡
xi − xoi ≡ ∆xi
(5)
is the k-th partial derivative of f with respect to xi .
The Taylor series can be equivalently written as
f (~x) =
∞
n
X
1 X
∂ q f (x~o )
∆xi1 · · · ∆xiq .
q!
∂x
·
·
·
∂x
i
i
q
1
q=0
i ,...,i =1
1
(6)
q
(a) Check the equality for functions of three variables by computing the Taylor series
expansion up to the third order.
Solution: We have (n = 3)
f (~x) =
∞
X
k1 ,k2 ,k3
3
X ∂f (x~o )
∂1k1 ∂2k2 ∂3k3 f (x~o ) k1 k2 k3
∆xi
∆x1 ∆x2 ∆x3 = f (x~o ) +
k
∂x
1 !k2 !k3 !
i
i=1
=0
3
3
X
1 X ∂ 2 f (x~o ) 2
∂ 2 f (x~o )
+
∆x
+
∆xi ∆xj
i
2 i=1 ∂x2i
∂xi ∂xj
1=i<j
+
3
3
3
1 X ∂ 3 f (x~o ) 3 1 X ∂ 3 f (x~o ) 2
1 X ∂ 3 f (x~o )
∆x
+
∆x
∆x
+
∆xi ∆x2j
j
i
i
3! i=1 ∂x3i
2 1=i<j ∂x2i ∂xj
2 1=i<j ∂xi ∂x2j
+
∂ 3 f (x~o )
∆x1 ∆x2 ∆x3 + O(∆x4 ) ,
∂x1 ∂x2 ∂x3
(7)
and
f (~x) =
∞
3
3
X
X
1 X
∂ k f (x~o )
∂f (x~o )
∆xi1 · · · ∆xik = f (x~o ) +
∆xi
k!
∂x
·
·
·
∂x
∂x
i
i
i
1
k
i ,...,i =1
i=1
k=0
1
k
3
X
3
∂ 2 f (x~o )
1
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 +
3
3
X
1 X ∂ 2 f (x~o ) 2
∂ 2 f (x~o )
∆x
+
∆xi ∆xj
i
2 i=1 ∂x2i
∂x
∂x
i
j
1=i<j
3
3
3
1 X ∂ 3 f (x~o )
1 X ∂ 3 f (x~o ) 3 1 X ∂ 3 f (x~o ) 2
+
∆xi +
∆xi ∆xj +
∆xi ∆x2j
3! i=1 ∂x3i
2 1=i<j ∂x2i ∂xj
2 1=i<j ∂xi ∂x2j
∂ 3 f (x~o )
+
∆x1 ∆x2 ∆x3 + O(∆x4 ) ,
∂x1 ∂x2 ∂x3
(8)
which proves the formula up to the third order.
3
4. Find the Taylor series expansion up to the quadratic order of the periodic Toda potential
n
X
U (x1 , . . . , xn ) =
eα(xi+1 −xi ) ,
xk+n ≡ xk ,
k ∈ Z,
i=1
about the point xi = a, i = 1, . . . , n.
Solution: We have
U (a, . . . , a) = n ,
∂U (a, . . . , a)
∂U (x1 , . . . , xn )
= αeα(xi −xi−1 ) − αeα(xi+1 −xi ) ⇒
= 0,
∂xi
∂xi
∂U (x1 , . . . , xn )
= α2 eα(xi −xi−1 ) (δij − δi−1,j ) − α2 eα(xi+1 −xi ) (δi+1,j − δij )
∂xi ∂xj
∂U (a, . . . , a)
= α2 (2δij − δi−1,j − δi+1,j ) ,
∂xi ∂xj
⇒
(9)
Thus, one gets
n
X
eα(xi+1 −xi ) = U0 + U1 + U2 ,
(10)
i=1
where
U0 = n ,
U2 =
=
U1 = 0 ,
(11)
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
(12)
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
4
ai ai+j−k
∀ j, k .
(13)