Mathematical Physics - Homework 5

Ying-Lun Kao(R06943183)
Graduate Institute of Electronics Engineering,
National Taiwan University

December 11, 2018

1. (a) Suppose we want to extremize a functional J with k unknown func-

tions y1 , y2 , · · · , yk :
Z x1
J= F (x, y1 , y10 , y2 , y20 , · · · , yk , yk0 )dx.
x0

Derive the Euler-Lagrange equations for this problem.

Solution. Let y ≡ (y1 , y2 , · · · , yk ) and the corresponding

y 0 = (y10 , y20 , · · · , yk0 ). We could reformulate the function J as
Z x1
J= F (x, y, y 0 )dx.
x0

Besides, we define the η vector as η = (η1 , η2 , · · · , ηk ). Using Taylor’s

expansion, we have
k k
X ∂F X ∂F 0
F (x, y + αη, y 0 + αη) = F (x, y, y 0 ) + αηi + 0 αηi + · · ·
i=1
∂yi i=1
∂yi

Therefore, the rate of change of functional J due to αη would be

∂J 1
≡ lim [J(x, y + αη, y 0 + αη) − J(x, y, y 0 )]
∂α α→0 α
Z x1 Xk k
!
∂F X ∂F 0
= lim ηi + η + · · · dx
α→0 x
0 i=1
∂yi i=1
∂yi0 i

k k
!
Z x1 X ∂F X ∂F 0
= ηi + 0 ηi dx
x0 i=1
∂yi i=1
∂yi

k
! k Z x1 "Xk  #
Z x1 
X ∂F X ∂F x1 d ∂F
= ηi dx + ηi  − ηi dx
x0 i=1
∂yi i=1
∂yi0 x0 x0 i=1
dx ∂yi0

k Z x1    k
X ∂F d ∂F X ∂F x1
= − ηi dx + ηi 
i=1 x0 ∂yi dx ∂yi0 i=1
∂yi0 x0

1
 For the stationary condition, it’s required that
 
∂F d ∂F
= ηi ,
∂yi dx ∂yi0

with the natural boundary conditions

∂F  ∂F 
= 0; = 0,
∂yi0 x=x0 ∂yi0 x=x1
 

for all i ∈ (1, 2, · · · , k). 

(b) Suppose we want to extremize a functional J with an unknown func-

tion y(x):
Z x1 
dy d2 y dk y

J= F x, y, , 2 , · · · , k dx.
x0 dx dx dx

Derive the Euler-Lagrange equation for this problem.

Solution.
dk y dk η
 
dy dη
F x, y + αη, + α ,··· , k + α k
dx dx dx dx
∂F ∂F ∂F
= F (x, y, y 0 , · · · , y (k) ) + αη + 0 αη 0 + · · · + (k) αη (k) + · · ·
∂y ∂y ∂y

2
Therefore,
∂J 1
≡ lim [J(x, y (i) + αη (i) ) − J(x, y (i) )]
∂α α→0 α
Z x1  
∂F ∂F ∂F
= η + 0 η 0 + · · · + (k) η (k) dx
x0 ∂y ∂y ∂y
Z x1  Z x1   
∂F ∂F x1 d ∂F
= ηdx + η − ηdx
x0 ∂y ∂y 0 x0 x0 dx ∂y 0
 Z x1    Z x1 " X k
#
∂F 0 x1 d ∂F 0 ∂F (n)
+ η − η dx + η dx
∂y 00 x0 x0 dx ∂y 00 x0 n=3
∂y (n)
Z x1      x1
∂F d ∂F ∂F ∂F
= − ηdx + η + 00 η 0

∂y dx ∂y 0 ∂y 0 ∂y

x0 
x0
" k
#
Z x1   Z x1
d ∂F X ∂F (n)
− η 0 dx + η dx
x0 dx ∂y 00 x0 n=3
∂y (n)

Z x1      x1
∂F d ∂F ∂F ∂F
= − ηdx + η + 00 η 0

∂y dx ∂y 0 ∂y 0 ∂y

x0 
x0
" k
#
x1 x1
d2
   Z   Z
d ∂F x1 ∂F X ∂F (n)
− η + ηdx + η dx
dx ∂y 00 x0 x0 dx2 ∂y 00 x0 n=3
∂y (n)
x1 
d2
Z    
∂F d ∂F ∂F
= − ηdx +
x0 ∂y dx ∂y 00 ∂y 0 dx2
    x1 Z x1 " X
k
#
∂F d ∂F ∂F 0  ∂F (n)
+ − η + 00 η  + η dx
∂y 0 dx ∂y 00 ∂y  x0 n=3
∂y (n)
x0

Obviously, if we use integration by parts to rearrange those remaing

integral terms, we obtain
Z x1 
d2 d(k)

∂J d
= Fy − Fy0 + 2 Fy00 + − · · · + (−1)k (k) Fy(k) ηdx
∂α x0 dx dx dx
     
∂F d ∂F ∂F d ∂F
+ − η+ − η0
∂y 0 dx ∂y 00 ∂y 00 dx ∂y (3)
    x1
∂F d ∂F 00 ∂F (k−1) 
+ − η + · · · + (k) η
∂y (3) dx ∂y (4) ∂y


x0

For the stationary condition, it’s required that

d d2 d(k)
Fy − Fy0 + 2 Fy00 + − · · · + (−1)k (k) Fy(k) = 0,
dx dx dx

3
 which is the Euler-Lagrange equation for this problem. Furthermore,
if the boundary values are fixed, then the boundary condition of η(x)
would be

η(x0 ) = η(x1 ) = η 0 (x0 ) = η 0 (x1 ) = · · · = η (k−1) (x0 ) = η (k−1) (x1 ) = 0.

2. Find the function y(x) which makes the following functional

Z 1
I= (3y 02 − 2y 0 y + 4y 0 + y)dx − 2y(0) − 3y(1)
0

stationary, (a) when y(0) = 4 and y(1)=7; or (b) when the values y(0)
and y(1) are not specified.

Solution.

(a) Let

F (x, y, y 0 ) ≡ 3y 02 − 2y 0 y + 4y 0 + y.

∂I 1
= lim [I(x, y + αη, y 0 + αη 0 ) − I(x, y, y 0 )]
∂α α→0 α
Z 1  
∂F d ∂F ∂F 1
= − ηdx + ηi  − 2η(0) − 3η(1)
0 ∂y dx ∂y 0 ∂y 0 0
Z 1 
d
= −2y 0 + 1 − (6y 0 − 2y + 4) ηdx
0 dx
   
∂F ∂F
− 2 + 0 (0) η(0) − 3 − 0 (1) η(1)
∂y ∂y
Z 1
= (1 − 6y 00 )ηdx − [2 + 6y 0 (0) − 2y(0) + 4]η(0)
0

− [3 − 6y 0 (1) + 2y(1) − 4]η(1)

Z 1
= (1 − 6y 00 )η(x)dx − [6y 0 (0) − 2y(0) + 6]η(0) + [6y 0 (1) − 2y(1) + 1]η(1)
0
Z 1
= (1 − 6y 00 )η(x)dx + [−6y 0 (0) + 2]η(0) + [6y 0 (1) − 13]η(1)
0

Therefore, we need 6y 00 = 1 and η(0) = η(1) = 0 to ensure that the

functional I is stationary when the boundary values are fixed.
1 2
y(x) = x + C1 x + C2
12
Since y(0) = 4 and y(1) = 7, we have
1 2 35
y(x) = x + x + 4.
12 12

4
(b) Since y(0) and y(1) are not fixed, η(0) and η(1) is arbitrary. There-
fore, we need the following constraints.

1 − 6y 00 = 0
6y 0 (0) − 2y(0) + 6 = 0

6y 0 (1) − 2y(1) + 1 = 0

That is,
1 2
y(x) = x + C1 x + C2
12
3y 0 (0) − y(0) + 3 = 3C1 − C2 + 3 = 0
   
0 1 1
6y (1) − 2y(1) + 1 = 6 + C1 − 2 + C1 + C2 + 1 = 0
6 12

Therefore,
1 2 25 13
y(x) = x − x− .
12 12 4