Calculus of Variations MCQ Quiz - Objective Question with Answer for Calculus of Variations - Download Free PDF

Concept:

Broken Extremals in Calculus of Variations:

• The functional is given by .
• The Euler-Lagrange equation is based on the derivative of the Lagrangian with respect to .
• Lagrangian derivative: .
• In broken extremals, momentum is constant on each smooth segment but can change at the corner point, provided corner conditions hold.
• The possible constant values of satisfy (constant).

Calculation:

Let .

Solve .

The possible constant solutions for solve .

Consider the roots of

Set derivative .

Critical points are at

⇒ .

Also, is a possible slope.

So, possible slopes are , , and 0.

Check for : starting at , ending at .

Possible segments: try to sum contributions of constant slopes on [0, c] and [c, 1].

Each slope is constant, so is linear on each segment.

Consider slopes and in some order.

Example broken extremal:

• On [0, t₁]: slope , contributes .
• On [t₁, 1]: slope , contributes .

Total displacement: .

Set equal to 2:

⇒

⇒

⇒ .

This is greater than 1, impossible.

So no broken extremal exists for .

∴ is empty.

Now for :

Repeat same method:

Total displacement:

⇒

⇒

⇒

This is feasible in [0, 1].

Now, the slopes and can be arranged in two ways: positive first or negative first.

Exactly two distinct broken extremals.

Correct statements:

• Option 1 is false (S(2) is empty).
• Option 2 is false (S(½) has two elements, not one).
• Option 3 is true (S(2) is empty).
• Option 4 is true (S(½) has exactly two elements).

∴ The correct answers are 3 and 4.

Concept: 

If  then its extremum is

Explanation:

  y(0) = 0, y(1) = 1,  

According to the question 

 = 0

⇒  = 0

⇒ x + 6y'' = 0

⇒ 

after integrating with respect to x, we get

again integrating with respect to x, we get

 where c and d are constants of integration

now, 

and 

Hence our extremum would be

Hence option (2) is correct

Concept:

Euler-Lagrange Equation: The extremal of the functional J[y] = f(x, y, y')dx satisfies  - () = 0

Explanation:

J(y(x)) = [(y')2 − y|y| y' + xy] dx, y(0) = 0, y(1) = 0

If y > 0 then

f(x, y, y') = (y')2 − y²y' + xy

So using   - () = 0 we get

-2yy' + x - (2y' - y²) = 0

- 2yy' + x - 2y'' +2yy' = 0

y'' = x/2....(i)

If y

f(x, y, y') = (y')2 + y2y' + xy

So using   - () = 0 we get

2yy' + x - (2y' + y2) = 0

 2yy' + x - 2y'' - 2yy' = 0

y'' = x/2....(ii)

Hence in both case we get

y'' = x/2

Integrating 

y' = 

Integrating again

y = 

Using  y(0) = 0, y(1) = 0 we get

c₂ = 0 and 0 =  + c₁ ⇒ c1 = - 

Hence solution is

y = 

Option (3) is correct

Concept: If  then its extremum is

Explanation:

  y(0)=0, y(1) = 1,  

According to the question 

 = 0

 = 0

⇒  = 0

⇒ x - 4y'' = 0

after integrating  with respect to x, we get

again integrating with respect to x, we get

where c and d are constants of integration

now, 

and 

Hence our extremum would be 

Hence option (4) is correct

Concept:

Euler Equation:

Let us examine the functional 

For an extreme value, the boundary points of the admissible curves being fixed  y( x₁ ) = y₁ and y( x₂ ) = y₂ 

then  - ( ) = 0 is called the Euler equation

Calculation:

Given: F( x , y , y' ) = y'² -2xy

Now by Euler's equation 

 - () = 0

-2x - 2y'' = 0 

y'' = - x 

Now integrating on both sides with respect to x, we get 

y' =  + A 

Now again, integrating on both sides with respect to x, we get 

y( x) = + Ax + B . . . . . . (1)

where A and B are constant.

Using boundary conditions y(1) = 1 and y( -1) = -1

Now from equation 1, we get 

y(1) = + A + B 

1 = + A +B 

A + B =  . . . . . . (2)

y( -1) =   - A + B

-1 =  - A + B

- A + B =  . . . . . . (3)

Adding equations 2 and 3, we get

B = 0 

Putting the value of B in equation (2), we get 

A = 

Now from equation (1), 

y(x)  =  +x

Option (3) is correct

Concept:

A typical isoperimetric problem is to find an extremum of 

I(y) =  subject to y(a) = y₁, y(b) = y₂, J(y) =  then the extrema satisfy the Euler-Lagrange's equation

 = 0

where H = F + λH

Explanation:

, y(0) = 1, y(1) = 6,

Let H = 

Then using

 = 0

⇒ λ -  = 0

⇒ 2y'' = λ

⇒ y'' = λ/2

Integrating both sides we get

y' = x + a

Again integrating

⇒ y =  + ax + b

y(0) = 1, y(1) = 6

⇒ b = 1

and

6 = λ/4 + a + 1

⇒ 5 = λ/4 + a

⇒ λ + 4a = 20.....(i)

So, we get y =  + ax + 1

Also, 

⇒  = 3

⇒  = 3

⇒  = 3

⇒  = 2

⇒ λ + 6a = 24....(ii)

Subtracting (i) from (ii) we get

2a = 4 ⇒ a = 2

Putting a = 2 in (i) we get

λ + 8 = 20 ⇒ λ = 12

Hence extremal is

y = 3x² + 2x + 1

Hence the functional has only one extremal.

Therefore the cardinality of the set of extremals is 1

(2) is correct

Concept:

We know that if I(y) =  F dx where F is function of y and y' only then F satisfy  = c ...(1)

Explanation:

Here F =  = y²y'²

So using (1)​we get

y2y'2 - y'2y2y' = c

⇒ - y2y'2= c 

⇒ yy' = c'

⇒ y dy = c' dx

Integrating both side

y² / 2 = c'x + d 

Substituting y(0) = 0 we get

d = 0

and substituting y(1) = 1 we get

1 / 2 = c'

Hence we get 

 ⇒ y² = x ⇒ y = √(x)

Therefore we get ϕ(x) = √(x). Hence ϕ(1/4) = √(1/4) ⇒ ϕ(1/4) = 1/2

Option (1) is correct   

Concept:

The extremum of functional I[y(x)] = , y(x₁) = y1, y(x2) = y1 attains 

(i) strong maxima if  

(ii) weak maxima if  

(iii) strong minima if  > 0 for all y'

(iv) weak minima if  > 0 for some y'

Explanation:

I[y(x)] = , y(0) = 0, y(a) = b

So f = y'¹⁰

So  > 0 if y' > 0

Hence the extremum of the functional attains weak minima.

Option (4) is correct

Concept:

The extremal of the functional J(y) = , y(a) = y₁, y(b) = arbitrary then satisfy Lagrange equation

 = 0 and  = 0

Explanation:

J(y) = ,  y(0) = e2 . Then

 = 0

⇒ 2y' + 2y⁵ = 0

⇒ y' + y⁵ = 0

⇒ 

Integrating we get -

Now, y(0) = e² ⇒ -1/4e⁸ = c

Hence  ⇒ 

(3) is correct

Concept:

The extremal of the functional , u(t1) = u1, u(t2) = u2, will satisfy Euler equation

 = 0

If F is independent of t and u then u = at + b is the extremize function.

Explanation:

J(u) = , u(0) = 0 and  

F(t, u, u') =  independent of t and u

So u = at + b is the general solution

u(0) = 0 implies b = 0

Hence u(t) = at

 i.e., at x = 1, u = 1

So, 1 = a × 1 ⇒ a = 1

Solution is u(t) = t ⇒ u'(t) = 1

Therefore infimum value

=  =  = [t]₀¹ = 1

So infimum value is 1

(3) is correct

Concept:

A typical isoperimetric problem is to find an extremum of 

I(y) =  subject to y(a) = y₁, y(b) = y₂, J(y) =  then the extrema satisfy the Euler-Lagrange's equation

 = 0

where H = F + λH

Explanation:

, y(0) = 2, y(1) = 5,

Let H = 

Then using

 = 0

⇒ λ -  = 0

⇒ 2y'' = λ

⇒ y'' = λ/2

Integrating both sides we get

y' = x + a

Again integrating

⇒ y =  + ax + b

y(0) = 2, y(1) = 5

⇒ b = 2

and

5 = λ/4 + a + 2

⇒ 3 = λ/4 + a

⇒ λ + 4a = 12.....(i)

So, we get y =  + ax + 2

Also, 

⇒  = 4

⇒  = 4

⇒  = 4

⇒  = 2

⇒ λ + 6a = 24....(ii)

Subtracting (i) from (ii) we get

2a = 12 ⇒ a = 6

Putting a = 6 in (i) we get

λ + 24 = 12 ⇒ λ = -12

Hence extremal is

y = -3x² + 6x + 1

Hence the functional has only one extremal.

Therefore the cardinality of the set of extremals is 1

(2) is correct

Explanation:

f(t) = exp(-t) = e^-t and g(t) = sin(t)

So, 

Let 

I = 

I = 

I = 0 + sin t - 

I = sin t - 

I = sin t - e^-t - cos t - I 

2I = sin t - e-t - cos t

I = 

Hence f⋆g(t) = 

Hence, option (1) is correct