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

Concept: 

If $J(y)={\int }_0^1[F(x,y,y^{\prime })]dx,$ then its extremum is

${\displaystyle \frac{\partial F}{\partial y}}-{\displaystyle \frac{d}{dx}}({\displaystyle \frac{\partial F}{\partial y^{\prime }}})=0$

Explanation:

$J(y)={\int }_0^1[3{\left(y^{\prime }\right)}^2-xy]dx,$  y(0) = 0, y(1) = 1,  $y\in C^2[0,1]$

According to the question 

$F(x,y,y^{\prime })=3(y^{\prime }{)}^2-xy$

${\displaystyle \frac{\partial F}{\partial y}}-{\displaystyle \frac{d}{dx}}({\displaystyle \frac{\partial F}{\partial y^{\prime }}})$ = 0

⇒ $-x-{\displaystyle \frac{d}{dx}}(6y^{\prime })$ = 0

⇒ x + 6y'' = 0

⇒ $y^{″}=-{\displaystyle \frac{x}{6}}$

after integrating with respect to x, we get

$y^{\prime }=-{\displaystyle \frac{x^2}{12}}+c$

again integrating with respect to x, we get

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

now, $y(0)=d=0$

and $y(1)=-{\displaystyle \frac{1}{36}}+c=1\Rightarrow c={\displaystyle \frac{37}{36}}$

Hence our extremum would be

 $y(x)=-{\displaystyle \frac{x^3}{36}}+\frac{37}{36}x$

Hence option (2) is correct

Concept: 

If $J(y)=\in t_0^1[F(x,y,y^{\prime })]dx,$ then its extremum is

${\displaystyle \frac{\partial F}{\partial y}}-{\displaystyle \frac{d}{dx}}({\displaystyle \frac{\partial F}{\partial y^{\prime }}})=0$

Explanation:

Here F(x, y, y') = $y^{\prime 2}-k^2{y}^2$

Using 

${\displaystyle \frac{\partial F}{\partial y}}-{\displaystyle \frac{d}{dx}}({\displaystyle \frac{\partial F}{\partial y^{\prime }}})=0$

-2k²y - $\frac{d}{dx}(2y^{\prime })$ = 0

-2k2y - 2y'' = 0

y'' + k2y = 0 

General solution is

y = a cos kx + b sin kx

y(0) = 0 ⇒ a = 0

y(l) = 0 ⇒ b sin kl = 0 ⇒ sin kl = 0 (if b ≠ 0)

⇒  kl = nπ ⇒ $l=\frac{n\pi }{k}$ , n ∈ ℤ⁺

for k2 > 0 and l is a multiple of $\frac{\pi }{k}$, there are infinitely many extremals because n can take any positive integer value, leading to different sinusoidal solutions

The functional has infinitely many extremals for k2 > 0 if l is a multiple of $\frac{\pi }{k}$

(3) is true and (2) is false.

The functional is dependent on k, as k determines the frequency of the sinusoidal solutions.

(4) is false.

If k = 0 then F(x, y, y') = y'²

then

${\displaystyle \frac{\partial F}{\partial y}}-{\displaystyle \frac{d}{dx}}({\displaystyle \frac{\partial F}{\partial y^{\prime }}})=0$

0 - ${\displaystyle \frac{d}{dx}}(2y^{\prime })=0$

y'' = 0

y = ax + b

y(0) = y(l) = 0

⇒ b = 0 and al = 0 i.e., a = 0

so, y = 0

Hence, there are no non-trivial extremals..

(1) is true.

Concept:

For the functional I[y(x)] = ${\int }_{x_1}^{x_2}F(x,y,y^{\prime })dx$, y(x2) = y2 and y(x1) = y1

The Euler equation is

$\frac{\partial F}{\partial y}-\frac{d}{dx}(\frac{\partial F}{\partial y^{\prime }})$ = 0

Explanation:

Here F(x, y, y') = xyy'

Using Euler's equation,

xy' - $\frac{d}{dx}$(xy) = 0

⇒ xy' - y - xy' = 0

⇒ y = 0

It satisfies y(0) = 0 but does not satisfy y(1) = 1.

So, the functional has no extremal.

(4) is true.

Concept:

For the functional I[y(x)] = ${\int }_{x_1}^{x_2}F(x,y,y^{\prime })dx$, y(x2) = y2 and y(x1) = y1

The Euler equation is

$\frac{\partial F}{\partial y}-\frac{d}{dx}(\frac{\partial F}{\partial y^{\prime }})$ = 0

Explanation:

Here F(x, y, y') = $y(3x+y)$

Using Euler's equation,

3x + 2y - $\frac{d}{dx}$(0) = 0

⇒ y = -3x/2

Using

y(1) = 1, y(3) = 4

and y(1) = 1

⇒ 1 = -3/2 which is not true.

Hence the functional has no extremal.

Option (4) is true.

Concept:

For the functional I[y(x)] = ${\int }_{x_1}^{x_2}F(x,y,y^{\prime })dx$, y(x2) = y2 and y(x1) = y1

The Euler equation is

$\frac{\partial F}{\partial y}-\frac{d}{dx}(\frac{\partial F}{\partial y^{\prime }})$ = 0

Explanation:

Here F(x, y, y') = $\sqrt{1+[y^{\prime }(x){]}^2}$

Using Euler's equation,

0- $\frac{d}{dx}$($\frac{1}{2\sqrt{1+[y^{\prime }(x){]}^2}}.2y^{\prime }(x)$) = 0

⇒ $\frac{d}{dx}$($\frac{y^{\prime }(x)}{\sqrt{1+[y^{\prime }(x){]}^2}}$) = 0

⇒ $\frac{y^{″}(x)}{(1+[y^{\prime }(x){]}^2{)}^{3/2}}$ = 0

Option (2) is true.

Concept:

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

Explanation:

J(y(x)) = ${\displaystyle {\int }_0^1}$[(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  $\frac{\partial f}{\partial y}$ - $\frac{d}{dx}$($\frac{\partial f}{\partial y^{\prime }}$) = 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  $\frac{\partial f}{\partial y}$ - $\frac{d}{dx}$($\frac{\partial f}{\partial y^{\prime }}$) = 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 = $\frac{x^3}{12}+c_{1}x+c_2$

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 $J(y)={\int }_0^1[F(x,y,y^{\prime })]dx,$ then its extremum is

${\displaystyle \frac{\partial F}{\partial y}}-{\displaystyle \frac{d}{dx}}({\displaystyle \frac{\partial F}{\partial y^{\prime }}})=0$

Explanation:

$J(y)={\int }_0^1[2{\left(y^{\prime }\right)}^2+xy]dx,$  y(0)=0, y(1) = 1,  $y\in C^2[0,1]$

According to the question 

$F(x,y,y^{\prime })=2(y^{\prime }{)}^2+xy$

${\displaystyle \frac{\partial F}{\partial y}}-{\displaystyle \frac{d}{dx}}({\displaystyle \frac{\partial F}{\partial y^{\prime }}})$ = 0

${\displaystyle \frac{\partial (2(y^{\prime }{)}^2+xy)}{\partial y}}-{\displaystyle \frac{d}{dx}}({\displaystyle \frac{\partial (2(y^{\prime }{)}^2+xy)}{\partial y^{\prime }}})$ = 0

⇒ $x-{\displaystyle \frac{d}{dx}}(4y^{\prime })$ = 0

⇒ x - 4y'' = 0

$y^{″}={\displaystyle \frac{x}{4}}$

after integrating  with respect to x, we get

$y^{\prime }={\displaystyle \frac{x^2}{8}}+c$

again integrating with respect to x, we get

$y(x)={\displaystyle \frac{x^3}{24}}+cx+d$

where c and d are constants of integration

now, $y(0)=d=0$

and $y(1)={\displaystyle \frac{1}{24}}+c=1\Rightarrow c={\displaystyle \frac{23}{24}}$

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

Hence option (4) is correct

Concept:

Euler Equation:

Let us examine the functional 

$I\left(y\right)={\int }_{x1}^x\left(F(x,y,y^{\prime }\right))dx$

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

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

Calculation:

Given: F( x , y , y' ) = y'² -2xy

Now by Euler's equation 

$\frac{\partial F}{\partial y}$ - $\frac{d}{dx}$($\frac{\partial F}{\partial y^{\prime }}$) = 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) = ${\int }_a^{b}F(x,y,y^{\prime })dx$ subject to y(a) = y₁, y(b) = y₂, J(y) = ${\int }_a^{b}G(x,y,y^{\prime })dx=L$ then the extrema satisfy the Euler-Lagrange's equation

$\frac{\partial H}{\partial y}-\frac{d}{dx}(\frac{\partial H}{\partial y^{\prime }})$ = 0

where H = F + λH

Explanation:

${\displaystyle J[y]={\int }_0^1{\left(y^{\prime }\right)}^{2}dx}$, y(0) = 1, y(1) = 6,${\displaystyle {\int }_0^{1}ydx=3}$

Let H = $y^{\prime 2}+\lambda y$

Then using

$\frac{\partial H}{\partial y}-\frac{d}{dx}(\frac{\partial H}{\partial y^{\prime }})$ = 0

⇒ λ - $\frac{d}{dx}(2y^{\prime })$ = 0

⇒ 2y'' = λ

⇒ y'' = λ/2

Integrating both sides we get

y' = $\frac{\lambda }{2}$x + a

Again integrating

⇒ y = $\frac{\lambda x^2}{4}$ + 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 = $\frac{\lambda x^2}{4}$ + ax + 1

Also, ${\displaystyle {\int }_0^{1}ydx=3}$

⇒ ${\displaystyle {\int }_0^1(\frac{\lambda x^2}{4}+ax+1)dx}$ = 3

⇒ ${[\frac{\lambda x^3}{12}+\frac{a{x}^2}{2}+x]}_0^1$ = 3

⇒ $\frac{\lambda }{12}+\frac{a}{2}+1$ = 3

⇒ $\frac{\lambda }{12}+\frac{a}{2}$ = 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 $F-y^{\prime }(\frac{\partial F}{\partial y^{\prime }})$ = 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 

$\frac{y^2}{2}=\frac{1}{2}x$ ⇒ 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)] = ${\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\mathrm{f}(\mathrm{x},\mathrm{y},{\mathrm{y}}^{\prime })\mathrm{d}\mathrm{x}$, y(x₁) = y1, y(x2) = y1 attains 

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

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

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

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

Explanation:

I[y(x)] = ${\int }_{\mathrm{b}}^{\mathrm{a}}{\mathrm{y}}^{\prime 10}\mathrm{d}\mathrm{x}$, y(0) = 0, y(a) = b

So f = y'¹⁰

$f_{y^{\prime }}=10y^{\prime 9}$

So $f_{y^{\prime }{y}^{\prime }}=90y^{\prime 8}$ > 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) = ${\int }_a^{b}f(x,y,y^{\prime })dx$, y(a) = y₁, y(b) = arbitrary then satisfy Lagrange equation

$\frac{\partial f}{\partial y}-\frac{d}{dx}(\frac{\partial f}{\partial y^{\prime }})$ = 0 and $\frac{\partial f}{\partial y^{\prime }}{|}_{x=b}$ = 0

Explanation:

J(y) = ${\int }_2^3(y^{\prime 2}+2y^5{y}^{\prime }+y^{10})dx$,  y(0) = e2 . Then

$\frac{\partial f}{\partial y^{\prime }}{|}_{x=3}$ = 0

⇒ 2y' + 2y⁵ = 0

⇒ y' + y⁵ = 0

⇒ $\frac{dy}{y^5}=-dx$

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 $J(u)={\int }_{t_1}^{t_2}F(t,u,u^{\prime })dt$, u(t1) = u1, u(t2) = u2, will satisfy Euler equation

$\frac{\partial F}{\partial u}-\frac{d}{dt}(\frac{\partial F}{\partial u^{\prime }})$ = 0

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

Explanation:

J(u) = ${\int }_0^1{(u^{\prime }(t))}^{2}dt$, u(0) = 0 and $\{\underset{[0,1]}{max}|u|=1\}$ 

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

$\{\underset{[0,1]}{max}|u|=1\}$ i.e., at x = 1, u = 1

So, 1 = a × 1 ⇒ a = 1

Solution is u(t) = t ⇒ u'(t) = 1

Therefore infimum value

= ${\int }_0^1{(u^{\prime }(t))}^{2}dt$ = ${\int }_0^1{1}^{2}dt$ = [t]₀¹ = 1

So infimum value is 1

(3) is correct

Concept:

A typical isoperimetric problem is to find an extremum of 

I(y) = ${\int }_a^{b}F(x,y,y^{\prime })dx$ subject to y(a) = y₁, y(b) = y₂, J(y) = ${\int }_a^{b}G(x,y,y^{\prime })dx=L$ then the extrema satisfy the Euler-Lagrange's equation

$\frac{\partial H}{\partial y}-\frac{d}{dx}(\frac{\partial H}{\partial y^{\prime }})$ = 0

where H = F + λH

Explanation:

${\displaystyle J[y]={\int }_0^1{\left(y^{\prime }\right)}^{2}dx}$, y(0) = 2, y(1) = 5,${\displaystyle {\int }_0^{1}ydx=4}$

Let H = $y^{\prime 2}+\lambda y$

Then using

$\frac{\partial H}{\partial y}-\frac{d}{dx}(\frac{\partial H}{\partial y^{\prime }})$ = 0

⇒ λ - $\frac{d}{dx}(2y^{\prime })$ = 0

⇒ 2y'' = λ

⇒ y'' = λ/2

Integrating both sides we get

y' = $\frac{\lambda }{2}$x + a

Again integrating

⇒ y = $\frac{\lambda x^2}{4}$ + 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 = $\frac{\lambda x^2}{4}$ + ax + 2

Also, ${\displaystyle {\int }_0^{1}ydx=4}$

⇒ ${\displaystyle {\int }_0^1(\frac{\lambda x^2}{4}+ax+2)dx}$ = 4

⇒ ${[\frac{\lambda x^3}{12}+\frac{a{x}^2}{2}+2x]}_0^1$ = 4

⇒ $\frac{\lambda }{12}+\frac{a}{2}+2$ = 4

⇒ $\frac{\lambda }{12}+\frac{a}{2}$ = 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:

${\displaystyle f\star g(t)={\int }_0^{t}f(s)g(t-s)ds.}$

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

So, ${\displaystyle f\star g(t)={\int }_0^t{e}^{-s}\sin (t-s)ds}$

Let 

I = ${\displaystyle f\star g(t)={\int }_0^t{e}^{-s}\sin (t-s)ds}$

I = ${[-e^{-s}\sin (t-s)]}_0^t-{\displaystyle {\int }_0^t(-e^{-s})(-\cos (t-s))ds}$

I = 0 + sin t - ${\displaystyle {\int }_0^t{e}^{-s}\cos (t-s)ds}$

I = sin t - $\{{[-e^{-s}\cos (t-s)]}_0^t+{\displaystyle {\int }_0^t{e}^{-s}\sin (t-s)ds}\}$

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