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Calculus variation of Solve Problem
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- Stanford University
Calculus variation of Solve Problem. Engineering mathematics problem Solving.
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Calculus of Variationssolved 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
∂f d ∂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
a ∂x
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
12 x − 2 ( y(x)) = 0
∂x2
We solve it using various tools and obtain the solution
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
Z b r
∂
3x + y(x) dx
a ∂x
Hint:
elementary
Solution.
We denote auxiliary function f
r
∂ ∂
f(x, y(x), y(x)) = 3 x + y(x)
∂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
a ∂x
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
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(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
a ∂x
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
a ∂x
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
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