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 \( J[y] = \int_0^1 \left( (y')^4 - 3(y')^2 \right) dx \).
• The Euler-Lagrange equation is based on the derivative of the Lagrangian \( L(y') \) with respect to \( y' \).
• Lagrangian derivative: \( \frac{\partial L}{\partial y'} = 4(y')^3 - 6y' \).
• In broken extremals, momentum \( p(x) \) is constant on each smooth segment but can change at the corner point, provided corner conditions hold.
• The possible constant values of \( y' \) satisfy \( 4(y')^3 - 6y' = c \) (constant).

Calculation:

Let \( p = y' \).

Solve \( \frac{d}{dx}(4p^3 - 6p) = 0 \).

The possible constant solutions for \( p \) solve \( 4p^3 - 6p = k \).

Consider the roots of \( 4p^3 - 6p \):

Set derivative \( \frac{dL}{dp} = 12p^2 - 6 \).

Critical points are at \( p^2 = \frac{1}{2} \)

⇒ \( p = \pm \frac{1}{\sqrt{2}} \).

Also, \( p = 0 \) is a possible slope.

So, possible slopes are \( \frac{1}{\sqrt{2}} \), \( -\frac{1}{\sqrt{2}} \), and 0.

Check for \( S(2) \): starting at \( y(0) = 0 \), ending at \( y(1) = 2 \).

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

Each slope is constant, so \( y \) is linear on each segment.

Consider slopes \( \frac{1}{\sqrt{2}} \) and \( -\frac{1}{\sqrt{2}} \) in some order.

Example broken extremal:

• On [0, t₁]: slope \( \frac{1}{\sqrt{2}} \), contributes \( \frac{t_1}{\sqrt{2}} \).
• On [t₁, 1]: slope \( -\frac{1}{\sqrt{2}} \), contributes \( -\frac{1 - t_1}{\sqrt{2}} \).

Total displacement: \( \frac{t_1}{\sqrt{2}} - \frac{1 - t_1}{\sqrt{2}} = \frac{2t_1 - 1}{\sqrt{2}} \).

Set equal to 2:

\( \frac{2t_1 - 1}{\sqrt{2}} = 2 \)

⇒ \( 2t_1 - 1 = 2\sqrt{2} \)

⇒ \( 2t_1 = 1 + 2\sqrt{2} \)

⇒ \( t_1 = \frac{1 + 2\sqrt{2}}{2} \).

This is greater than 1, impossible.

So no broken extremal exists for \( b = 2 \).

∴ \( S(2) \) is empty.

Now for \( S \left( \frac{1}{2} \right) \):

Repeat same method:

Total displacement: \( \frac{2t_1 - 1}{\sqrt{2}} = \frac{1}{2} \)

⇒ \( 2t_1 - 1 = \frac{\sqrt{2}}{2} \)

⇒ \( 2t_1 = 1 + \frac{\sqrt{2}}{2} \)

⇒ \( t_1 = \frac{1}{2} + \frac{\sqrt{2}}{4} \)

This is feasible in [0, 1].

Now, the slopes \( \frac{1}{\sqrt{2}} \) and \( -\frac{1}{\sqrt{2}} \) 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 \(J(y)=\int_0^1[F(x,y,y')] d x,\) then its extremum is

\(\dfrac{\partial F}{\partial y}-\dfrac{d}{dx}(\dfrac{\partial F}{\partial y'})=0\)

Explanation:

\(J(y)=\int_0^1\left[3\left(y^{\prime}\right)^2-x y\right] d x,\)  y(0) = 0, y(1) = 1,  \(y\in C^2[0,1]\)

According to the question 

\(F(x,y,y')=3(y')^2-xy\)

\(\dfrac{\partial F}{\partial y}-\dfrac{d}{dx}(\dfrac{\partial F}{\partial y'})\) = 0

⇒ \(- x-\dfrac{d}{dx}(6y')\) = 0

⇒ x + 6y'' = 0

⇒ \(y''=-\dfrac{x}{6}\)

after integrating with respect to x, we get

\(y'=-\dfrac{x^2}{12}+c\)

again integrating with respect to x, we get

\(y(x)=-\dfrac{x^3}{36}+cx+d\) where c and d are constants of integration

now, \(y(0)=d=0\)

and \(y(1)=-\dfrac{1}{36}+c=1⇒c=\dfrac{37}{36}\)

Hence our extremum would be

 \(y(x)=-\dfrac{x^3}{36}+{37\over 36}x\)

Hence option (2) is correct

Concept:

Euler-Lagrange Equation: The extremal of the functional J[y] = \(\int_a^b\)f(x, y, y')dx satisfies \(\frac{\partial f}{\partial y}\) - \(\frac{d}{dx}\)(\(\frac{\partial f}{\partial y'}\)) = 0

Explanation:

J(y(x)) = \(\rm\displaystyle\int_0^1\)[(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  \(\frac{\partial f}{\partial y}\) - \(\frac{d}{dx}\)(\(\frac{\partial f}{\partial y'}\)) = 0 we get

-2yy' + x - \(\frac{d}{dx}\)(2y' - y²) = 0

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

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

If y < 0 then

f(x, y, y') = (y')2 + y2y' + xy

So using  \(\frac{\partial f}{\partial y}\) - \(\frac{d}{dx}\)(\(\frac{\partial f}{\partial y'}\)) = 0 we get

2yy' + x - \(\frac{d}{dx}\)(2y' + y2) = 0

 2yy' + x - 2y'' - 2yy' = 0

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

Hence in both case we get

y'' = x/2

Integrating 

y' = \(\frac{x^2}{4}+c_1\)

Integrating again

y = \(\frac{x^3}{12}+c_1x+c_2\)

Using  y(0) = 0, y(1) = 0 we get

c₂ = 0 and 0 = \(\frac{1}{12}\) + c₁ ⇒ c1 = - \(\frac{1}{12}\)

Hence solution is

y = \(\frac{x^3}{12}-\frac{x}{12}\)

Option (3) is correct

Concept: If \(J(y)=\int_0^1[F(x,y,y')] d x,\) then its extremum is

\(\dfrac{\partial F}{\partial y}-\dfrac{d}{dx}(\dfrac{\partial F}{\partial y'})=0\)

Explanation:

\(J(y)=\int_0^1\left[2\left(y^{\prime}\right)^2+x y\right] d x,\)  y(0)=0, y(1) = 1,  \(y\in C^2[0,1]\)

According to the question 

\(F(x,y,y')=2(y')^2+xy\)

\(\dfrac{\partial F}{\partial y}-\dfrac{d}{dx}(\dfrac{\partial F}{\partial y'})\) = 0

\(\dfrac{\partial (2(y')^2+xy)}{\partial y}-\dfrac{d}{dx}(\dfrac{\partial (2(y')^2+xy)}{\partial y'})\) = 0

⇒ \( x-\dfrac{d}{dx}(4y')\) = 0

⇒ x - 4y'' = 0

\(y''=\dfrac{x}{4}\)

after integrating  with respect to x, we get

\(y'=\dfrac{x^2}{8}+c\)

again integrating with respect to x, we get

\(y(x)=\dfrac{x^3}{24}+cx+d\)

where c and d are constants of integration

now, \(y(0)=d=0\)

and \(y(1)=\dfrac{1}{24}+c=1⇒c=\dfrac{23}{24}\)

Hence our extremum would be \(y(x)=\dfrac{x^3}{24}+\dfrac{23x}{24}\)

Hence option (4) is correct

Concept:

Euler Equation:

Let us examine the functional 

\(I\left( y \right) = \int_{ x1}^x {\left( {{F(x , y,y'}} \right))} \,dx\)

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

then \(\frac{\partial F}{\partial y}\) - \(\frac{d}{dx}\)( \(\frac{\partial F}{\partial y'}\)) = 0 is called the Euler equation

Calculation:

Given: F( x , y , y' ) = y'² -2xy

Now by Euler's equation 

\(\frac{\partial F}{\partial y}\) - \(\frac{d}{dx}\)(\(\frac{\partial F}{\partial y'}\)) = 0

-2x - 2y'' = 0 

y'' = - x 

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

y' = \(\frac{-x^2}{2}\) + A 

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

y( x) = \(\frac{-x^3}{6}\)+ 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) = \(\frac{-1}{6}\)+ A + B 

1 = \(\frac{-1}{6}\)+ A +B 

A + B = \(\frac{7}{6}\) . . . . . . (2)

y( -1) =  \(\frac{1}{6}\) - A + B

-1 = \(\frac{1}{6}\) - A + B

- A + B = \(\frac{-7}{6}\) . . . . . . (3)

Adding equations 2 and 3, we get

B = 0 

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

A = \(\frac{7}{6}\)

Now from equation (1), 

y(x)  =  \(\frac{-x^3}{6}\)+\(\frac{7}{6}\)x

Option (3) is correct

Concept:

A typical isoperimetric problem is to find an extremum of 

I(y) = \(\int_a^bF(x, y, y')dx\) subject to y(a) = y₁, y(b) = y₂, J(y) = \(\int_a^bG(x, y, y')dx=L\) then the extrema satisfy the Euler-Lagrange's equation

\({\partial H\over \partial y}-{d\over dx}({\partial H\over \partial y'})\) = 0

where H = F + λH

Explanation:

\(\displaystyle J[y]=\int_0^1\left(y^{\prime}\right)^2 d x\), y(0) = 1, y(1) = 6,\(\displaystyle \int_0^1 y d x=3\)

Let H = \(y'^2+λ y\)

Then using

\({\partial H\over \partial y}-{d\over dx}({\partial H\over \partial y'})\) = 0

⇒ λ - \({d\over dx}(2y')\) = 0

⇒ 2y'' = λ

⇒ y'' = λ/2

Integrating both sides we get

y' = \({\lambda \over2}\)x + a

Again integrating

⇒ y = \({\lambda x^2\over 4}\) + 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 = \({\lambda x^2\over 4}\) + ax + 1

Also, \(\displaystyle \int_0^1 y d x=3\)

⇒ \(\displaystyle \int_0^1 \left({\lambda x^2\over 4}+ax+1\right)d x\) = 3

⇒ \(\left[{\lambda x^3\over 12}+{ax^2\over 2}+x\right]_0^1 \) = 3

⇒ \({\lambda\over 12}+{a\over 2}+1\) = 3

⇒ \({\lambda\over 12}+{a\over 2}\) = 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) = \(\int_0^a\) F dx where F is function of y and y' only then F satisfy \(F-y'(\frac{\partial F}{\partial y'})\) = c ...(1)

Explanation:

Here F = \(y^2\left(\frac{dy}{dx}\right)^2\) = 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 

\(\frac{y^2}{2}=\frac12x\) ⇒ 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)] = \(\rm \int_{x_1}^{x_2} f(x, y, y')d x\), y(x₁) = y1, y(x2) = y1 attains 

(i) strong maxima if \(f_{y'y'}\) < 0 for all y'

(ii) weak maxima if \(f_{y'y'}\) < 0 for some y'

(iii) strong minima if \(f_{y'y'}\) > 0 for all y'

(iv) weak minima if \(f_{y'y'}\) > 0 for some y'

Explanation:

I[y(x)] = \(\rm \int_b^a y^{\prime 10} d x\), y(0) = 0, y(a) = b

So f = y'¹⁰

\(f_{y'}=10y'^9\)

So \(f_{y'y'}=90y'^8\) > 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) = \(\int_a^bf(x, y, y')dx\), y(a) = y₁, y(b) = arbitrary then satisfy Lagrange equation

\(\frac{\partial f}{\partial y}-\frac{d}{dx}(\frac{\partial f}{\partial y'})\) = 0 and \({\partial f\over \partial y'}\Big|_{x=b}\) = 0

Explanation:

J(y) = \(\int_2^3\left(y^{\prime 2}+2 y^5 y^{\prime}+y^{10}\right) dx\),  y(0) = e2 . Then

\({\partial f\over \partial y'}\Big|_{x=3}\) = 0

⇒ 2y' + 2y⁵ = 0

⇒ y' + y⁵ = 0

⇒ \({dy\over y^5}=-dx\)

Integrating we get -

\(\frac{1}{y^4}= 4x - 4c\)

Now, y(0) = e² ⇒ -1/4e⁸ = c

Hence \(\frac{1}{y^4}= 4x + \frac{1}{e^8}\) ⇒ \(1= y^4( 4x + \frac{1}{e^8})\)

(3) is correct

Concept:

The extremal of the functional \(J(u)=∫_{t_1}^{t_2} F\left(t, u, u^{\prime}\right) d t\), u(t1) = u1, u(t2) = u2, will satisfy Euler equation

\(\frac{\partial F}{\partial u}-\frac d{dt}(\frac{\partial F}{\partial u'})\) = 0

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

Explanation:

J(u) = \(\int_0^1\left(u^{\prime}(t)\right)^2 d t\), u(0) = 0 and \(\left\{ \max _{[0,1]}|u|=1\right\}\) 

F(t, u, u') = \(u^{\prime 2}\) independent of t and u

So u = at + b is the general solution

u(0) = 0 implies b = 0

Hence u(t) = at

\(\left\{ \max _{[0,1]}|u|=1\right\}\) i.e., at x = 1, u = 1

So, 1 = a × 1 ⇒ a = 1

Solution is u(t) = t ⇒ u'(t) = 1

Therefore infimum value

= \(\int_0^1\left(u^{\prime}(t)\right)^2 d t\) = \(\int_0^11^2 d t\) = [t]₀¹ = 1

So infimum value is 1

(3) is correct

Concept:

A typical isoperimetric problem is to find an extremum of 

I(y) = \(\int_a^bF(x, y, y')dx\) subject to y(a) = y₁, y(b) = y₂, J(y) = \(\int_a^bG(x, y, y')dx=L\) then the extrema satisfy the Euler-Lagrange's equation

\({\partial H\over \partial y}-{d\over dx}({\partial H\over \partial y'})\) = 0

where H = F + λH

Explanation:

\(\displaystyle J[y]=\int_0^1\left(y^{\prime}\right)^2 d x\), y(0) = 2, y(1) = 5,\(\displaystyle \int_0^1 y d x=4\)

Let H = \(y'^2+λ y\)

Then using

\({\partial H\over \partial y}-{d\over dx}({\partial H\over \partial y'})\) = 0

⇒ λ - \({d\over dx}(2y')\) = 0

⇒ 2y'' = λ

⇒ y'' = λ/2

Integrating both sides we get

y' = \({\lambda \over2}\)x + a

Again integrating

⇒ y = \({\lambda x^2\over 4}\) + 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 = \({\lambda x^2\over 4}\) + ax + 2

Also, \(\displaystyle \int_0^1 y d x=4\)

⇒ \(\displaystyle \int_0^1 \left({\lambda x^2\over 4}+ax+2\right)d x\) = 4

⇒ \(\left[{\lambda x^3\over 12}+{ax^2\over 2}+2x\right]_0^1 \) = 4

⇒ \({\lambda\over 12}+{a\over 2}+2\) = 4

⇒ \({\lambda\over 12}+{a\over 2}\) = 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:

\(\displaystyle f ⋆ g(t)=\int_0^t f(s) g(t-s) d s .\)

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

So, \(\displaystyle f ⋆ g(t)=\int_0^t e^{-s} \sin(t-s) d s\)

Let 

I = \(\displaystyle f ⋆ g(t)=\int_0^t e^{-s} \sin(t-s) d s\)

I = \(\left[-e^{-s}\sin(t-s)\right]_0^t-\displaystyle\int_0^t(-e^{-s})(-\cos(t-s))ds\)

I = 0 + sin t - \(\displaystyle\int_0^te^{-s}\cos(t-s)ds\)

I = sin t - \(\left\{\left[-e^{-s}\cos(t-s)\right]_0^t+\displaystyle\int_0^te^{-s}\sin(t-s)ds\right\}\)

I = sin t - e^-t - cos t - I 

2I = sin t - e-t - cos t

I = \(\frac{1}{2}[\exp (-t)+\sin (t)-\cos (t)]\)

Hence f⋆g(t) = \(\frac{1}{2}[\exp (-t)+\sin (t)-\cos (t)]\)

Hence, option (1) is correct