Calculus of Variations solved problems

Calculus of Variations solved problems Pavel Pyrih June 4, 2012 ( public domain ) Acknowledgement. The following problems were solved using my own procedure in a program Maple V, release 5. All possible errors are my faults.

1

Solving the Euler equation

Theorem.(Euler) Suppose f (x, y, y 0 ) has continuous partial derivatives of the Rb second order on the interval [a, b]. If a functional F (y) = a f (x, y, y 0 )dx attains a weak relative extrema at y0 , then y0 is a solution of the following equation d  ∂f  ∂f − = 0. ∂y dx ∂y 0 It is called the Euler equation.

1

1.1

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ 12 x y(x) + ( y(x))2 dx ∂x a Hint: elementary Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = 12 x y(x) + ( y(x))2 ∂x ∂x

in the form f(x, y, z) = 12 x y + z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 12 x ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 2 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂2 f(x, y(x), y(x)) = 2 ( 2 y(x)) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂2 y(x)) = 0 ∂x2 We solve it using various tools and obtain the solution 12 x − 2 (

y(x) = x3 + C1 x + C2 Info. cubic polynomial Comment. no comment

2

1.2

Problem.

Using the Euler equation find the extremals for the following functional r Z b ∂ 3x + y(x) dx ∂x a Hint: elementary Solution. We denote auxiliary function f ∂ y(x)) = 3 x + f(x, y(x), ∂x

r

∂ y(x) ∂x

in the form √ f(x, y, z) = 3 x + z Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 0 ∂y ∂x and ∂ ∂ 1 1 f(x, y(x), y(x)) = q ∂z ∂x 2 ∂ ∂x y and 2

∂ ∂2 ∂ 1 ∂x2 y(x) f(x, y(x), y(x)) = − ∂ ∂x ∂z ∂x 4 ( ∂x y(x))(3/2) and finally obtain the Euler equation for our functional 2

∂ 1 ∂x2 y(x) =0 ∂ 4 ( ∂x y(x))(3/2) We solve it using various tools and obtain the solution

y(x) = C1 x + C2 Info. linear function Comment. no comment

3

1.3

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ x + y(x)2 + 3 ( y(x)) dx ∂x a Hint: elementary Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = x + y(x)2 + 3 ( y(x)) ∂x ∂x

in the form f(x, y, z) = x + y 2 + 3 z Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 2 y ∂y ∂x and ∂ ∂ f(x, y(x), y(x)) = 3 ∂z ∂x and ∂2 ∂ f(x, y(x), y(x)) = 0 ∂x ∂z ∂x and finally obtain the Euler equation for our functional 2 y(x) = 0 We solve it using various tools and obtain the solution y(x) = 0 Info. constant solution Comment. no comment

4

1.4

Problem.

Using the Euler equation find the extremals for the following functional Z b y(x)2 dx a

Hint: elementary Solution. We denote auxiliary function f f(x, y(x),

∂ y(x)) = y(x)2 ∂x

in the form f(x, y, z) = y 2 Substituting we x, y(x) and y (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives 0

∂ ∂ f(x, y(x), y(x)) = 2 y ∂y ∂x and ∂ ∂ f(x, y(x), y(x)) = 0 ∂z ∂x and ∂2 ∂ f(x, y(x), y(x)) = 0 ∂x ∂z ∂x and finally obtain the Euler equation for our functional 2 y(x) = 0 We solve it using various tools and obtain the solution y(x) = 0 Info. zero Comment. no comment

5

1.5

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ y(x)2 + x2 ( y(x)) dx ∂x a Hint: elementary Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = y(x)2 + x2 ( y(x)) ∂x ∂x

in the form f(x, y, z) = y 2 + x2 z Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 2 y ∂y ∂x and ∂ ∂ f(x, y(x), y(x)) = x2 ∂z ∂x and ∂2 ∂ f(x, y(x), y(x)) = 2 x ∂x ∂z ∂x and finally obtain the Euler equation for our functional 2 y(x) − 2 x = 0 We solve it using various tools and obtain the solution y(x) = x Info. linear Comment. no comment

6

1.6

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ y(x) + x ( y(x)) dx ∂x a Hint: elementary Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = y(x) + x ( y(x)) ∂x ∂x

in the form f(x, y, z) = y + x z Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 1 ∂y ∂x and ∂ ∂ f(x, y(x), y(x)) = x ∂z ∂x and ∂2 ∂ f(x, y(x), y(x)) = 1 ∂x ∂z ∂x and finally obtain the Euler equation for our functional 0=0 We solve it using various tools and obtain the solution y(x) = Y(x) Info. all functions Comment. no comment

7

1.7

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ y(x) dx ∂x a Hint: elementary Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = y(x) ∂x ∂x

in the form f(x, y, z) = z Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 0 ∂y ∂x and ∂ ∂ f(x, y(x), y(x)) = 1 ∂z ∂x and ∂2 ∂ f(x, y(x), y(x)) = 0 ∂x ∂z ∂x and finally obtain the Euler equation for our functional 0=0 We solve it using various tools and obtain the solution y(x) = Y(x) Info. all functions Comment. no comment

8

1.8

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ y(x)2 − ( y(x))2 dx ∂x a Hint: missing x Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = y(x)2 − ( y(x))2 ∂x ∂x

in the form f(x, y, z) = y 2 − z 2 Substituting we x, y(x) and y (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives 0

∂ ∂ f(x, y(x), y(x)) = 2 y ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = −2 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂2 f(x, y(x), y(x)) = −2 ( 2 y(x)) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂2 y(x)) = 0 ∂x2 We solve it using various tools and obtain the solution 2 y(x) + 2 (

y(x) = C1 cos(x) + C2 sin(x) Info. trig functions Comment. no comment

9

1.9

Problem.

Using the Euler equation find the extremals for the following functional Z br ∂ 1+( y(x))2 dx ∂x a Hint: missing x y Solution. We denote auxiliary function f ∂ y(x)) = f(x, y(x), ∂x

r 1+(

∂ y(x))2 ∂x

in the form p f(x, y, z) = 1 + z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 0 ∂y ∂x and ∂ y ∂ ∂ f(x, y(x), y(x)) = q ∂x ∂z ∂x ∂ 1 + ( ∂x y)2

and 2

2

∂ ∂ ( ∂ y(x))2 ( ∂x ∂ ∂2 2 y(x)) 2 y(x) q ∂x + f(x, y(x), y(x)) = − ∂x ∂ 2 (3/2) ∂x ∂z ∂x ∂ (1 + ( ∂x y(x)) ) 1 + ( ∂x y(x))2 and finally obtain the Euler equation for our functional 2

∂ ∂ ( ∂x y(x))2 ( ∂x 2 y(x))

(1 +

∂ ( ∂x

y(x))2 )(3/2)

−q

∂2 ∂x2

y(x)

∂ 1 + ( ∂x y(x))2 We solve it using various tools and obtain the solution

y(x) = C1 x + C2 Info. linear function Comment. no comment

10

=0

1.10

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ ∂ x( y(x)) + ( y(x))2 dx ∂x ∂x a Hint: missing y Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ ∂ y(x)) = x ( y(x)) + ( y(x))2 ∂x ∂x ∂x

in the form f(x, y, z) = x z + z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 0 ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = x + 2 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂2 f(x, y(x), y(x)) = 1 + 2 ( 2 y(x)) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂2 y(x)) = 0 ∂x2 We solve it using various tools and obtain the solution −1 − 2 (

1 y(x) = − x2 + C1 x + C2 4 Info. quadratic function Comment. no comment

11

1.11

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ (1 + x) ( y(x))2 dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = (1 + x) ( y(x))2 ∂x ∂x

in the form f(x, y, z) = (1 + x) z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 0 ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 2 (1 + x) ( y) ∂z ∂x ∂x and ∂2 ∂ ∂ ∂2 f(x, y(x), y(x)) = 2 ( y(x)) + 2 (1 + x) ( 2 y(x)) ∂x ∂z ∂x ∂x ∂x and finally obtain the Euler equation for our functional ∂2 ∂ y(x)) − 2 (1 + x) ( 2 y(x)) = 0 ∂x ∂x We solve it using various tools and obtain the solution −2 (

y(x) = C1 + C2 ln(1 + x) Info. not supplied Comment. no comment

12

1.12

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ 2 y(x) ex + y(x)2 + ( y(x))2 dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = 2 y(x) ex + y(x)2 + ( y(x))2 ∂x ∂x

in the form f(x, y, z) = 2 y ex + y 2 + z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 2 ex + 2 y ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 2 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂2 f(x, y(x), y(x)) = 2 ( 2 y(x)) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂2 y(x)) = 0 ∂x2 We solve it using various tools and obtain the solution 2 ex + 2 y(x) − 2 (

1 1 1 y(x) = ( cosh(x)2 + cosh(x) sinh(x) + x) sinh(x) 2 2 2 1 1 1 + (− cosh(x) sinh(x) + x − cosh(x)2 ) cosh(x) + C1 sinh(x) + C2 cosh(x) 2 2 2 Info. not supplied Comment. no comment

13

1.13

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ 2 y(x) + ( y(x))2 dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = 2 y(x) + ( y(x))2 ∂x ∂x

in the form f(x, y, z) = 2 y + z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 2 ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 2 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂2 f(x, y(x), y(x)) = 2 ( 2 y(x)) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂2 y(x)) = 0 ∂x2 We solve it using various tools and obtain the solution 2 − 2(

y(x) =

1 2 x + C1 x + C2 2

Info. not supplied Comment. no comment

14

1.14

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ 2( y(x))3 dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = 2 ( y(x))3 ∂x ∂x

in the form f(x, y, z) = 2 z 3 Substituting we x, y(x) and y (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives 0

∂ ∂ f(x, y(x), y(x)) = 0 ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 6 ( y)2 ∂z ∂x ∂x and ∂2 ∂ ∂ ∂2 f(x, y(x), y(x)) = 12 ( y(x)) ( 2 y(x)) ∂x ∂z ∂x ∂x ∂x and finally obtain the Euler equation for our functional ∂2 ∂ y(x)) ( 2 y(x)) = 0 ∂x ∂x We solve it using various tools and obtain the solution −12 (

y(x) = C1 Info. not supplied Comment. no comment

15

1.15

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ ∂ 4x( y(x)) + ( y(x))2 dx ∂x ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ ∂ y(x)) = 4 x ( y(x)) + ( y(x))2 ∂x ∂x ∂x

in the form f(x, y, z) = 4 x z + z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 0 ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 4 x + 2 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂2 f(x, y(x), y(x)) = 4 + 2 ( 2 y(x)) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂2 y(x)) = 0 ∂x2 We solve it using various tools and obtain the solution −4 − 2 (

y(x) = −x2 + C1 x + C2 Info. not supplied Comment. no comment

16

1.16

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ 7( y(x))3 dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = 7 ( y(x))3 ∂x ∂x

in the form f(x, y, z) = 7 z 3 Substituting we x, y(x) and y (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives 0

∂ ∂ f(x, y(x), y(x)) = 0 ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 21 ( y)2 ∂z ∂x ∂x and ∂2 ∂ ∂ ∂2 f(x, y(x), y(x)) = 42 ( y(x)) ( 2 y(x)) ∂x ∂z ∂x ∂x ∂x and finally obtain the Euler equation for our functional ∂2 ∂ y(x)) ( 2 y(x)) = 0 ∂x ∂x We solve it using various tools and obtain the solution −42 (

y(x) = C1 Info. not supplied Comment. no comment

17

1.17

Problem.

Using the Euler equation find the extremals for the following functional Z br ∂ y(x) (1 + ( y(x))2 ) dx ∂x a Hint: no hint Solution. We denote auxiliary function f ∂ y(x)) = f(x, y(x), ∂x

r y(x) (1 + (

∂ y(x))2 ) ∂x

in the form p f(x, y, z) = y (1 + z 2 ) Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ y)2 1 + ( ∂x ∂ ∂ 1 f(x, y(x), y(x)) = q ∂y ∂x 2 y (1 + ( ∂ y)2 ) ∂x

and ∂ y ( ∂x y) ∂ ∂ f(x, y(x), y(x)) = q ∂z ∂x ∂ y (1 + ( ∂x y)2 )

and ∂2 ∂ f(x, y(x), y(x)) = ∂x ∂z ∂x ∂ ∂ ∂ ∂2 ∂ y(x)) (( ∂x y(x)) (1 + ( ∂x y(x))2 ) + 2 y(x) ( ∂x y(x)) ( ∂x 1 y(x) ( ∂x 2 y(x))) − ∂ 2 (3/2) 2 (y(x) (1 + ( ∂x y(x)) )) 2

∂ ∂ ( ∂x y(x))2 y(x) ( ∂x 2 y(x)) +q +q ∂ ∂ y(x) (1 + ( ∂x y(x))2 ) y(x) (1 + ( ∂x y(x))2 ) and finally obtain the Euler equation for our functional 2

∂ ∂ ∂ ∂ ∂ y(x)) (( ∂x y(x)) (1 + ( ∂x y(x))2 ) + 2 y(x) ( ∂x y(x)) ( ∂x 1 1 1 y(x) ( ∂x 2 y(x))) p + ∂ 2 (3/2) 2 y(x) 2 (y(x) (1 + ( ∂x y(x)) )) 2

∂ ∂ ( ∂x y(x))2 y(x) ( ∂x 2 y(x)) −q −q =0 ∂ ∂ y(x) (1 + ( ∂x y(x) (1 + ( ∂x y(x))2 ) y(x))2 )

18

We solve it using various tools and obtain the solution y(x) =

1 4 + C1 2 x2 + 2 C1 2 x C2 + C2 2 C1 2 4 C1

Info. quadratic function Comment. no comment

19

1.18

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ x y(x) ( y(x)) dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = x y(x) ( y(x)) ∂x ∂x

in the form f(x, y, z) = x y z Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ ∂ f(x, y(x), y(x)) = x ( y) ∂y ∂x ∂x and ∂ ∂ f(x, y(x), y(x)) = x y ∂z ∂x and ∂2 ∂ ∂ f(x, y(x), y(x)) = y(x) + x ( y(x)) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂ y(x)) = 0 ∂x We solve it using various tools and obtain the solution −y(x) − x (

y(x) = 0 Info. not supplied Comment. no comment

20

1.19

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ x y(x) + 2 ( y(x))2 dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = x y(x) + 2 ( y(x))2 ∂x ∂x

in the form f(x, y, z) = x y + 2 z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = x ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 4 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂2 f(x, y(x), y(x)) = 4 ( 2 y(x)) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂2 y(x)) = 0 ∂x2 We solve it using various tools and obtain the solution x − 4(

y(x) =

1 3 x + C1 x + C2 24

Info. not supplied Comment. no comment

21

1.20

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ x y(x) + y(x)2 − 2 y(x)2 ( y(x)) dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = x y(x) + y(x)2 − 2 y(x)2 ( y(x)) ∂x ∂x

in the form f(x, y, z) = x y + y 2 − 2 y 2 z Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ ∂ f(x, y(x), y(x)) = x + 2 y − 4 y ( y) ∂y ∂x ∂x and ∂ ∂ f(x, y(x), y(x)) = −2 y 2 ∂z ∂x and ∂2 ∂ ∂ f(x, y(x), y(x)) = −4 y(x) ( y(x)) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂ y(x)) = 0 ∂x We solve it using various tools and obtain the solution x + 2 y(x) + 4 y(x) (

1 y(x) = − x 2 Info. no extremal Comment. no comment

22

1.21

Problem.

Using the Euler equation find the extremals for the following functional r Z b ∂ y(x) 1 + ( y(x))2 dx ∂x a Hint: no hint Solution. We denote auxiliary function f ∂ y(x)) = y(x) f(x, y(x), ∂x

r 1+(

∂ y(x))2 ∂x

in the form p f(x, y, z) = y 1 + z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives r ∂ ∂ ∂ f(x, y(x), y(x)) = 1 + ( y)2 ∂y ∂x ∂x and y ( ∂ y) ∂ ∂ f(x, y(x), y(x)) = q ∂x ∂z ∂x ∂ 1 + ( ∂x y)2 and ∂ ∂2 f(x, y(x), y(x)) = ∂x ∂z ∂x ∂ ∂2 ∂2 y(x) ( ∂x y(x))2 ( ∂x y(x) ( ∂x ( ∂ y(x))2 2 y(x)) 2 y(x)) q ∂x q − + ∂ 2 )(3/2) ∂ ∂ y(x)) (1 + ( ∂x 1 + ( ∂x y(x))2 1 + ( ∂x y(x))2 and finally obtain the Euler equation for our functional ∂ ( ∂x y(x))2

2

∂ ∂ y(x) ( ∂x y(x))2 ( ∂x 2 y(x))

2

∂ y(x) ( ∂x 2 y(x)) q 1− q + − =0 ∂ 2 (3/2) ∂ ∂ (1 + ( ∂x y(x)) ) 1 + ( ∂x y(x))2 1 + ( ∂x y(x))2 We solve it using various tools and obtain the solution √

1 1 + (e( C1 (x+ C2 )) )2 √ √ y(x) = 2 e( C1 (x+ C2 )) C1 Info. not supplied Comment. no comment 23

1.22

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ y(x) ( y(x)) dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = y(x) ( y(x)) ∂x ∂x

in the form f(x, y, z) = y z Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ ∂ f(x, y(x), y(x)) = y ∂y ∂x ∂x and ∂ ∂ f(x, y(x), y(x)) = y ∂z ∂x and ∂2 ∂ ∂ f(x, y(x), y(x)) = y(x) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂ y(x)) = 0 ∂x We solve it using various tools and obtain the solution −(

y(x) = Y(x) Info. not supplied Comment. no comment

24

1.23

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ y(x) ( y(x))2 dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = y(x) ( y(x))2 ∂x ∂x

in the form f(x, y, z) = y z 2 Substituting we x, y(x) and y (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives 0

∂ ∂ ∂ f(x, y(x), y(x)) = ( y)2 ∂y ∂x ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 2 y ( y) ∂z ∂x ∂x and ∂2 ∂ ∂ ∂2 f(x, y(x), y(x)) = 2 ( y(x))2 + 2 y(x) ( 2 y(x)) ∂x ∂z ∂x ∂x ∂x and finally obtain the Euler equation for our functional ∂2 ∂ y(x))2 − 2 y(x) ( 2 y(x)) = 0 ∂x ∂x We solve it using various tools and obtain the solution −2 (

y(x) = 0 Info. not supplied Comment. no comment

25

1.24

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ y(x)2 + 2 x y(x) ( y(x)) dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = y(x)2 + 2 x y(x) ( y(x)) ∂x ∂x

in the form f(x, y, z) = y 2 + 2 x y z Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ ∂ f(x, y(x), y(x)) = 2 y + 2 x ( y) ∂y ∂x ∂x and ∂ ∂ f(x, y(x), y(x)) = 2 x y ∂z ∂x and ∂2 ∂ ∂ f(x, y(x), y(x)) = 2 y(x) + 2 x ( y(x)) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂ y(x)) = 0 ∂x We solve it using various tools and obtain the solution −2 x (

y(x) = Y(x) Info. all functions Comment. no comment

26

1.25

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ ∂ y(x)2 + 4 y(x) ( y(x)) + 4 ( y(x))2 dx ∂x ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ ∂ y(x)) = y(x)2 + 4 y(x) ( y(x)) + 4 ( y(x))2 ∂x ∂x ∂x

in the form f(x, y, z) = y 2 + 4 y z + 4 z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ ∂ f(x, y(x), y(x)) = 2 y + 4 ( y) ∂y ∂x ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 4 y + 8 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂ ∂2 f(x, y(x), y(x)) = 4 ( y(x)) + 8 ( 2 y(x)) ∂x ∂z ∂x ∂x ∂x and finally obtain the Euler equation for our functional ∂2 ∂ y(x)) − 8 ( 2 y(x)) = 0 ∂x ∂x We solve it using various tools and obtain the solution 2 y(x) − 4 (

1 1 y(x) = C1 cosh( x) + C2 sinh( x) 2 2 Info. two exponentials Comment. no comment

27

1.26

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ ∂ y(x)2 + y(x) ( y(x)) + ( y(x))2 dx ∂x ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ ∂ y(x)) = y(x)2 + y(x) ( y(x)) + ( y(x))2 ∂x ∂x ∂x

in the form f(x, y, z) = y 2 + y z + z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ ∂ f(x, y(x), y(x)) = 2 y + ( y) ∂y ∂x ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = y + 2 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂ ∂2 f(x, y(x), y(x)) = ( y(x)) + 2 ( 2 y(x)) ∂x ∂z ∂x ∂x ∂x and finally obtain the Euler equation for our functional ∂2 ∂ y(x)) − 2 ( 2 y(x)) = 0 ∂x ∂x We solve it using various tools and obtain the solution 2 y(x) − (

y(x) = C1 sinh(x) + C2 cosh(x) Info. not supplied Comment. no comment

28

1.27

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ 2 y(x) ex + y(x)2 + ( y(x))2 dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = 2 y(x) ex + y(x)2 + ( y(x))2 ∂x ∂x

in the form f(x, y, z) = 2 y ex + y 2 + z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 2 ex + 2 y ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 2 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂2 f(x, y(x), y(x)) = 2 ( 2 y(x)) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂2 y(x)) = 0 ∂x2 We solve it using various tools and obtain the solution 2 ex + 2 y(x) − 2 (

1 1 1 y(x) = ( cosh(x)2 + cosh(x) sinh(x) + x) sinh(x) 2 2 2 1 1 1 + (− cosh(x) sinh(x) + x − cosh(x)2 ) cosh(x) + C1 sinh(x) + C2 cosh(x) 2 2 2 Info. not supplied Comment. no comment

29

1.28

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ y(x)2 + ( y(x))2 − 2 y(x) sin(x) dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = y(x)2 + ( y(x))2 − 2 y(x) sin(x) ∂x ∂x

in the form f(x, y, z) = y 2 + z 2 − 2 y sin(x) Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 2 y − 2 sin(x) ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 2 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂2 f(x, y(x), y(x)) = 2 ( 2 y(x)) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂2 y(x)) = 0 ∂x2 We solve it using various tools and obtain the solution 2 y(x) − 2 sin(x) − 2 (

1 1 1 1 y(x) = ( ex cos(x) − sin(x) ex + e(−x) cos(x) + e(−x) sin(x)) sinh(x) 4 4 4 4 1 x 1 1 (−x) 1 x + (− e cos(x) + sin(x) e + e cos(x) + e(−x) sin(x)) cosh(x) 4 4 4 4 + C1 sinh(x) + C2 cosh(x) Info. exponentials and sin Comment. no comment

30

1.29

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ ∂ y(x)2 − 4 y(x) ( y(x)) + 4 ( y(x))2 dx ∂x ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ ∂ y(x)) = y(x)2 − 4 y(x) ( y(x)) + 4 ( y(x))2 ∂x ∂x ∂x

in the form f(x, y, z) = y 2 − 4 y z + 4 z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ ∂ f(x, y(x), y(x)) = 2 y − 4 ( y) ∂y ∂x ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = −4 y + 8 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂ ∂2 f(x, y(x), y(x)) = −4 ( y(x)) + 8 ( 2 y(x)) ∂x ∂z ∂x ∂x ∂x and finally obtain the Euler equation for our functional ∂2 ∂ y(x)) − 8 ( 2 y(x)) = 0 ∂x ∂x We solve it using various tools and obtain the solution 2 y(x) + 4 (

1 1 y(x) = C1 cosh( x) + C2 sinh( x) 2 2 Info. not supplied Comment. no comment

31

1.30

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ ∂ 1 ∂ y(x) + y(x) ( y(x)) + ( y(x)) + ( y(x))2 dx ∂x ∂x 2 ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ 1 ∂ ∂ y(x)) = y(x) + y(x) ( y(x)) + ( y(x)) + ( y(x))2 ∂x ∂x ∂x 2 ∂x

in the form 1 2 z 2 0 Substituting we x, y(x) and y (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives f(x, y, z) = y + y z + z +

∂ ∂ ∂ f(x, y(x), y(x)) = 1 + ( y) ∂y ∂x ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = y + 1 + ( y) ∂z ∂x ∂x and ∂2 ∂ ∂ ∂2 f(x, y(x), y(x)) = ( y(x)) + ( 2 y(x)) ∂x ∂z ∂x ∂x ∂x and finally obtain the Euler equation for our functional ∂2 ∂ y(x)) − ( 2 y(x)) = 0 ∂x ∂x We solve it using various tools and obtain the solution 1−(

y(x) =

1 2 x + C1 x + C2 2

Info. quadratic function Comment. no comment

32

1.31

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ ∂ y(x) − y(x) ( y(x)) + x ( y(x))2 dx ∂x ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ ∂ y(x)) = y(x) − y(x) ( y(x)) + x ( y(x))2 ∂x ∂x ∂x

in the form f(x, y, z) = y − y z + x z 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ ∂ f(x, y(x), y(x)) = 1 − ( y) ∂y ∂x ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = −y + 2 x ( y) ∂z ∂x ∂x and ∂2 ∂ ∂ ∂2 f(x, y(x), y(x)) = ( y(x)) + 2 x ( 2 y(x)) ∂x ∂z ∂x ∂x ∂x and finally obtain the Euler equation for our functional ∂2 ∂ y(x)) − 2 x ( 2 y(x)) = 0 ∂x ∂x We solve it using various tools and obtain the solution 1−(

y(x) =

1 x + C1 + C2 ln(x) 2

Info. not supplied Comment. no comment

33

1.32

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ ∂ ( y(x)) (1 + x2 ( y(x))) dx ∂x ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ ∂ y(x)) = ( y(x)) (1 + x2 ( y(x))) ∂x ∂x ∂x

in the form f(x, y, z) = z (1 + x2 z) Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = 0 ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 1 + 2 x2 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂ ∂2 f(x, y(x), y(x)) = 4 x ( y(x)) + 2 x2 ( 2 y(x)) ∂x ∂z ∂x ∂x ∂x and finally obtain the Euler equation for our functional ∂2 ∂ y(x)) − 2 x2 ( 2 y(x)) = 0 ∂x ∂x We solve it using various tools and obtain the solution −4 x (

y(x) = C1 +

C2 x

Info. hyperbolic function Comment. no comment

34

1.33

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ ( ∂x y(x))2 dx x3 a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

( ∂ y(x))2 ∂ y(x)) = ∂x 3 ∂x x

in the form z2 x3 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives f(x, y, z) =

∂ ∂ f(x, y(x), y(x)) = 0 ∂y ∂x and ∂ y ∂ ∂ f(x, y(x), y(x)) = 2 ∂x3 ∂z ∂x x

and 2

∂ ∂ ∂2 2 y(x) f(x, y(x), y(x)) = 2 ∂x 3 −6 ∂x ∂z ∂x x and finally obtain the Euler equation for our functional ∂2 ∂x2

∂ y(x) ∂x y(x) + 6 =0 x3 x4 We solve it using various tools and obtain the solution

−2

y(x) = C1 + C2 x4 Info. not supplied Comment. no comment

35

∂ ∂x

y(x) x4

1.34

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ ∂ ( y(x))2 + 2 y(x) ( y(x)) − 16 y(x)2 dx ∂x ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ ∂ y(x)) = ( y(x))2 + 2 y(x) ( y(x)) − 16 y(x)2 ∂x ∂x ∂x

in the form f(x, y, z) = z 2 + 2 y z − 16 y 2 Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ ∂ f(x, y(x), y(x)) = 2 ( y) − 32 y ∂y ∂x ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 2 ( y) + 2 y ∂z ∂x ∂x and ∂2 ∂ ∂2 ∂ f(x, y(x), y(x)) = 2 ( 2 y(x)) + 2 ( y(x)) ∂x ∂z ∂x ∂x ∂x and finally obtain the Euler equation for our functional ∂ ∂2 y(x)) − 2 ( y(x)) = 0 ∂x2 ∂x We solve it using various tools and obtain the solution −32 y(x) − 2 (

y(x) = C1 cos(4 x) + C2 sin(4 x) Info. trig functions Comment. no comment

36

1.35

Problem.

Using the Euler equation find the extremals for the following functional Z b ∂ ( y(x))2 + cos(y(x)) dx ∂x a Hint: no hint Solution. We denote auxiliary function f f(x, y(x),

∂ ∂ y(x)) = ( y(x))2 + cos(y(x)) ∂x ∂x

in the form f(x, y, z) = z 2 + cos(y) Substituting we x, y(x) and y 0 (x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives ∂ ∂ f(x, y(x), y(x)) = −sin(y) ∂y ∂x and ∂ ∂ ∂ f(x, y(x), y(x)) = 2 ( y) ∂z ∂x ∂x and ∂2 ∂ ∂2 f(x, y(x), y(x)) = 2 ( 2 y(x)) ∂x ∂z ∂x ∂x and finally obtain the Euler equation for our functional ∂2 y(x)) = 0 ∂x2 We solve it using various tools and obtain the solution −sin(y(x)) − 2 (

y(x) = 0 Info. not supplied Comment. no comment

37