Calculus of Variations MCQ Quiz - Objective Question with Answer for Calculus of Variations - Download Free PDF
Last updated on Jul 7, 2025
Latest Calculus of Variations MCQ Objective Questions
Calculus of Variations Question 1:
Define S : = {y ∈ C1[-1, 1] : y(-1) = -1, y(1) = 3}
Let φ be the extremal of the functional J : S → R given by
Define ||y||∞:=
Answer (Detailed Solution Below)
Calculus of Variations Question 1 Detailed Solution
Calculus of Variations Question 2:
For any b ∈ ℝ, let S(b) denote the set of all broken extremals with one corner of the variational problem
minimize J[y] =
subject to y(0) = 0, y(1) = b
Then which of the following statements are true?
Answer (Detailed Solution Below)
Calculus of Variations Question 2 Detailed Solution
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
Consider the roots of
Set derivative
Critical points are at
⇒
Also,
So, possible slopes are
Check for
Possible segments: try to sum contributions of constant slopes on [0, c] and [c, 1].
Each slope is constant, so
Consider slopes
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
∴
Now for
Repeat same method:
Total displacement:
⇒
⇒
⇒
This is feasible in [0, 1].
Now, the slopes
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.
Calculus of Variations Question 3:
Let S denote the set of all solutions of the Euler-Lagrange equation of the variational problem:
minimize J|y| =
subject to y(0) = 0, y(1) = 0,
Then the set
Answer (Detailed Solution Below)
Calculus of Variations Question 3 Detailed Solution
Calculus of Variations Question 4:
Let λ ∈ ℝ and K : [0, 1] × [0, 1] → ℝ be a function such that every solution of the boundary value problem
Satisfies the integral equation
Then
Answer (Detailed Solution Below)
Calculus of Variations Question 4 Detailed Solution
Calculus of Variations Question 5:
The extremal of the functional
Answer (Detailed Solution Below)
Calculus of Variations Question 5 Detailed Solution
Concept:
If
Explanation:
According to the question
⇒
⇒ x + 6y'' = 0
⇒
after integrating with respect to x, we get
again integrating with respect to x, we get
now,
and
Hence our extremum would be
Hence option (2) is correct
Top Calculus of Variations MCQ Objective Questions
Consider the variational problem (P)
J(y(x)) =
Which of the following statements is correct?
Answer (Detailed Solution Below)
Calculus of Variations Question 6 Detailed Solution
Download Solution PDFConcept:
Euler-Lagrange Equation: The extremal of the functional J[y] =
Explanation:
J(y(x)) =
If y > 0 then
f(x, y, y') = (y')2 − y2y' + xy
So using
-2yy' + x -
- 2yy' + x - 2y'' +2yy' = 0
y'' = x/2....(i)
If y < 0 then
f(x, y, y') = (y')2 + y2y' + xy
So using
2yy' + x -
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 =
Hence solution is
y =
Option (3) is correct
The extremal of the functional
Answer (Detailed Solution Below)
Calculus of Variations Question 7 Detailed Solution
Download Solution PDFConcept: If
Explanation:
According to the question
⇒
⇒ 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
Calculus of Variations Question 8:
Which of the following is an extremal of the functional
Answer (Detailed Solution Below)
Calculus of Variations Question 8 Detailed Solution
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
Calculation:
Given: F( x , y , y' ) = y'2 -2xy
Now by Euler's equation
-2x - 2y'' = 0
y'' = - x
Now integrating on both sides with respect to x, we get
y' =
Now again, integrating on both sides with respect to x, we get
y( x) =
where A and B are constant.
Using boundary conditions y(1) = 1 and y( -1) = -1
Now from equation 1, we get
y(1) =
1 =
A + B =
y( -1) =
-1 =
- A + B =
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) =
Option (3) is correct
Calculus of Variations Question 9:
The cardinality of the set of extremals of
subject to
y(0) = 1, y(1) = 6,
is
Answer (Detailed Solution Below)
Calculus of Variations Question 9 Detailed Solution
Concept:
A typical isoperimetric problem is to find an extremum of
I(y) =
where H = F + λH
Explanation:
Let H =
Then using
⇒ λ -
⇒ 2y'' = λ
⇒ y'' = λ/2
Integrating both sides we get
y' =
Again integrating
⇒ y =
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 =
Also,
⇒
⇒
⇒
⇒
⇒ 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
Calculus of Variations Question 10:
Let y = ϕ(x) be the extremizing function for the functional
Answer (Detailed Solution Below)
Calculus of Variations Question 10 Detailed Solution
Concept:
We know that if I(y) =
Explanation:
Here F =
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
Therefore we get ϕ(x) = √(x). Hence ϕ(1/4) = √(1/4) ⇒ ϕ(1/4) = 1/2
Option (1) is correct
Calculus of Variations Question 11:
The extremum of the functional I[y(x)] =
Answer (Detailed Solution Below)
Calculus of Variations Question 11 Detailed Solution
Concept:
The extremum of functional I[y(x)] =
(i) strong maxima if
(ii) weak maxima if
(iii) strong minima if
(iv) weak minima if
Explanation:
I[y(x)] =
So f = y'10
So
Hence the extremum of the functional attains weak minima.
Option (4) is correct
Calculus of Variations Question 12:
If J(y) =
Answer (Detailed Solution Below)
Calculus of Variations Question 12 Detailed Solution
Concept:
The extremal of the functional J(y) =
Explanation:
J(y) =
⇒ 2y' + 2y5 = 0
⇒ y' + y5 = 0
⇒
Integrating we get -
Now, y(0) = e2 ⇒ -1/4e8 = c
Hence
(3) is correct
Calculus of Variations Question 13:
The infimum of
Answer (Detailed Solution Below)
Calculus of Variations Question 13 Detailed Solution
Concept:
The extremal of the functional
If F is independent of t and u then u = at + b is the extremize function.
Explanation:
J(u) =
F(t, u, u') =
So u = at + b is the general solution
u(0) = 0 implies b = 0
Hence u(t) = at
So, 1 = a × 1 ⇒ a = 1
Solution is u(t) = t ⇒ u'(t) = 1
Therefore infimum value
=
So infimum value is 1
(3) is correct
Calculus of Variations Question 14:
The cardinality of the set of extremals of
subject to
y(0) = 2, y(1) = 5,
Answer (Detailed Solution Below)
Calculus of Variations Question 14 Detailed Solution
Concept:
A typical isoperimetric problem is to find an extremum of
I(y) =
where H = F + λH
Explanation:
Let H =
Then using
⇒ λ -
⇒ 2y'' = λ
⇒ y'' = λ/2
Integrating both sides we get
y' =
Again integrating
⇒ y =
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 =
Also,
⇒
⇒
⇒
⇒
⇒ 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
Calculus of Variations Question 15:
For any two continuous functions f, g : ℝ → ℝ, define
Which of the following is the value of f ⋆ g(t) when f(t) = exp(-t) and g(t) = sin(t)?
Answer (Detailed Solution Below)
Calculus of Variations Question 15 Detailed Solution
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