Calculus of Variations MCQ Quiz in मराठी - Objective Question with Answer for Calculus of Variations - मोफत PDF डाउनलोड करा

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)] = ${\int }_0^1(5{\mathrm{x}}^2+3{\mathrm{y}}^4-2{\mathrm{y}}^{\prime 2})\mathrm{d}\mathrm{x}$, y(1) = 1

So f = $5x^2+3y^4-2y^{\prime 2}$

$f_{y^{\prime }}=-4y^{\prime }$

So $f_{y^{\prime }{y}^{\prime }}=-4$ < 0 for all y'

Therefore functional achieves a strong maxima on all its extremals.

Option (2) is correct

Concept:

Let If I(Y) is an extremum of the functional J[y] = ${\int }_a^{b}f(x,y,y^{\prime })dx$ be a functional such that y(a) = c, y(b) = d then y(x) satisfy Euler-Lagrange's equation

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

Explanation:

J[y] = ${\int }_0^1[({\mathrm{y}}^{\prime }{)}^2+2\mathrm{y}]\mathrm{d}\mathrm{x}$ subject to y(0) = 0, y(1) = 1

here f = y'² + 2y

Using Euler-Lagrange's equation we get

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

⇒ 2 - $\frac{d}{dx}$(2y') = 0

⇒ y'' - 1 = 0

⇒ y'' = 1

Integrating both sides

y' = x + c₁

Integrating again

⇒ y = $\frac{x^2}{2}+c_{1}x+c_2$

Given y(0) = 0, y(1) = 1

y(0) = 0 ⇒ c₂ = 0

y(1) = 1 ⇒ 1 = 1/2 + c₁ ⇒ c₁ = 1/2

So, y(x) = $\frac{x^2}{2}+\frac{x}{2}$ so, y' = x + 1/2

Therefore

Inf[J] = ${\int }_0^1[({\mathrm{y}}^{\prime }{)}^2+2\mathrm{y}]\mathrm{d}\mathrm{x}$

         = ${\int }_0^1[(\mathrm{x}+\frac{1}{2}{)}^2+2(\frac{{\mathrm{x}}^2}{2}+\frac{\mathrm{x}}{2})]\mathrm{d}\mathrm{x}$

        = ${\int }_0^1({\mathrm{x}}^2+\mathrm{x}+\frac{1}{4}+{\mathrm{x}}^2+\mathrm{x})\mathrm{d}\mathrm{x}$

       = ${\int }_0^1(2{\mathrm{x}}^2+2\mathrm{x}+\frac{1}{4})\mathrm{d}\mathrm{x}$

      = ${[\frac{2}{3}{x}^3+x^2+\frac{x}{4}]}_0^1$ = $\frac{23}{12}$

Hence option (1) is correct

Concept:

For ${\int }_{x_1}^{x_2}F(x,y,y^{\prime })dx$ extremal are given by $\frac{\partial F}{\partial y}-\frac{d}{dx}(\frac{\partial F}{\partial y^{\prime }})=0$

Explanation:

$J(x)={\int }_0^1[x^2(t)+{\dot{x}}^2(t)]dt$

F(t, x, x') = [x²(t) + x'²(t)]

$\frac{\partial F}{\partial y}-\frac{d}{dx}(\frac{\partial F}{\partial y^{\prime }})=0$ ⇒ 2x - $\frac{d}{dt}$(2x') = 0

⇒ x'' - x = 0 ⇒ x(t) = A$e^t$ + B$e^{-t}$

Now, x(0) = 0 ⇒ A+B = 0 ........... (1)

x(1) = 1 ⇒ Ae + Be^-1  = 1 ........... (2)

Solving (1) & (2) we get

B = $\frac{e}{1-e^2}$ & A = - $\frac{e}{1-e^2}$

x(t) = - $\frac{e}{1-e^2}$ $e^t$ + $\frac{e}{1-e^2}$ $e^{-t}$

x($\frac{1}{2}$) = $\frac{e}{1-e^2}$[$e^{-\frac{1}{2}}$ - $e^{\frac{1}{2}}$] = $\frac{\sqrt{e}}{1+e}$

Hence, Option (1) is true

Concept:

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

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

where H = F + λH

Explanation:

${\displaystyle J[y]={\int }_{-1}^0(12xy-{\left(y^{\prime }\right)}^2)dx}$, y(0) = 0, y(-1) = 1

Here F = 12xy - $(y^{\prime }{)}^2$

Then using

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

⇒ 12x - $\frac{d}{dx}(-2y^{\prime })$ = 0

⇒ 12x + 2y'' = 0

⇒ y'' = -6x

Integrating both sides we get

y' = - 3x² + a

Again integrating

 y = - x3 + ax + b

Given y(0) = 0, y(-1) = 1

⇒ b = 0

and

1 = 1 - a + 0

⇒ a = 0

Hence extremal is

y = - x3 

Hence, Option (3) is correct

Concept:

The extremal of the functional $I(y)={\int }_a^{b}F(x,y,y^{\prime })dx$, y(a) = y1, y(b) = y2, will satisfy Euler equation

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

Explanation:

$I(y)={\int }_1^2\frac{\sqrt{1+{(y^{\prime }(x))}^2}}{x}dx$, y(1) = 0, y(2) = 1

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

Then from Euler equation we get

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

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

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

⇒ $\frac{y^{\prime }}{x\sqrt{1+y^{\prime 2}}}$ = c (Integrating)

⇒ $\frac{y^{\prime 2}}{x^2(1+y^{\prime 2})}$ = c² (squaring both sides)

⇒ $y^{\prime 2}=c^2{x}^2(1+y^{\prime 2})$

⇒ $y^{\prime 2}(1-c^2{x}^2)=c^2{x}^2$

⇒ $y^{\prime 2}=\frac{c^2{x}^2}{1-c^2{x}^2}$

⇒ y' = $\frac{cx}{\sqrt{1-c^2{x}^2}}$

⇒ dy = $\frac{cx}{\sqrt{1-c^2{x}^2}}$dx

Integrating both sides

⇒ y = $-\frac{1}{c}\sqrt{1-c^2{x}^2}$ + b....(i)

Given y(1) = 0, y(2) = 1

y(1) = 0 ⇒ 0 = $-\frac{1}{c}$ + b ⇒ b = 1/c

y(2) = 1 ⇒ 1 = $-\frac{1}{c}\sqrt{1-4c^2}$ + b

⇒ 1 = $-\frac{1}{c}\sqrt{1-4c^2}+\frac{1}{c}$ (as b = 1/c)

⇒ c = 1 -  $\sqrt{1-4c^2}$

⇒ (1 - c)² = 1 - 4c²

⇒ 1 - 2c + c² = 1 - 4c²

⇒ 5c² = 2c

⇒ c = 2/5 as c can't be zero

So b = 1/c = 5/2 

Now, from (i)

(y - b) = $-\frac{1}{c}\sqrt{1-c^2{x}^2}$

⇒ (y - b)² = $\frac{1}{c^2}(1-c^2{x}^2)$

⇒ (y - b)2 + x² =  $\frac{1}{c^2}$

Putting the values

Extremize function is

$(y-\frac{5}{2}{)}^2+x^2=(\frac{5}{2}{)}^2$ which is an equation of a circle

(3) is correct

Concept:

The extremal of the functional $J(u)={\int }_a^{b}F(x,u,u^{\prime })dx$, u(a) = u1, u(b) = u2, will satisfy Euler equation

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

Explanation:

$\mathrm{J}(\mathrm{u})={\int }_0^1[{\mathrm{u}}_{\mathrm{x}}^2+4\frac{{\mathrm{u}}^2}{{\mathrm{x}}^2}]\mathrm{x}\mathrm{d}\mathrm{x}$, u(0) = 0 and u(1) = 1

F(x, u, u_x) = $(u_x^2+4\frac{u^2}{x^2})x$ = $x{u}_x^2+4\frac{u^2}{x}$

Then from Euler equation we get

$\frac{\partial F}{\partial u}-\frac{d}{dx}(\frac{\partial F}{\partial u_x})$ = 0

⇒ $\frac{8u}{x}-\frac{d}{dx}(2x{u}_x)$ = 0

⇒ $\frac{4u}{x}-u_x-x{u}_{xx}$ = 0

⇒ $x^2{u}_{xx}+x{u}_x-4u$ = 0

⇒ (x²D² +xD - 4)u = 0 where D = $\frac{d}{dx}$

Then auxillay equation will be

m² + (1 - 1)m - 4 = 0

⇒ m² - 4 = 0 ⇒ m = ± 2

General solution is

u = $c_1{x}^2+c_2{x}^{-2}$

Given u(0) = 0 and u(1) = 1

u(0) = 0 will satisfy only when c₂ = 0

and u(1) = 1 implies c₁ = 1

hence extremal is u(x) = x2

(1) is correct

Alternate Method Given u(0) = 0 and u(1) = 1 

Option 2: $\mathrm{u}(\mathrm{x})=\frac{1}{\sqrt{2}}{\mathrm{x}}^2$ not satisfying u(1) = 1, so (2) can't be a solution

Option 3: $\mathrm{u}(\mathrm{x})=\frac{1}{2}{\mathrm{x}}^2$ not satisfying u(1) = 1, so (3) can't be a solution

Option 4: $\mathrm{u}(\mathrm{x})=\frac{1}{4}{\mathrm{x}}^2$ not satisfying u(1) = 1, so (4) can't be a solution

Option 1: u(x) = x2, satisfying u(0) = 0 and u(1) = 1 so this can be extremal solution

(1) is correct

Explanation:

J[y] = ${\int }_0^1[720{\mathrm{x}}^2\mathrm{y}-{\left({\mathrm{y}}^{\prime \prime }\right)}^2]\mathrm{d}\mathrm{x}$

Using

$\frac{\partial f}{\partial y}$ - $\frac{d}{dx}(\frac{\partial f}{\partial y^{\prime }})$ + $\frac{d^2}{d{x}^2}(\frac{\partial f}{\partial y^{″}})$ = 0

720x2 - 0 + $\frac{d^2}{d{x}^2}$(-2y'') = 0

yiv = 360x2

y''' = 120x3 + a

y'' = 30x4 + ax + b

y' = 6x5 + $\frac{a}{2}$x2 + bx + c 

Using 

y'(0) = 0  we get c = 0

So y' = 6x5 + $\frac{a}{2}$x2 + bx 

y = x6 + $\frac{a}{6}$x3 + $\frac{b}{2}$x2 + d

using y(0) = 0 we get d = 0

y = x6 + $\frac{a}{6}$x3 + $\frac{b}{2}$x2 

Again using y(1) = 0 and y'(1) = 6 we get

$\frac{a}{6}$ + $\frac{b}{2}$ = -1 ⇒ a + 3b = -6

and $\frac{a}{2}$ + b = 0

Solving them we get a = 12, b = -6

Hence y = x6 + 2x3 - 3x2

Hence, (1) correct

Concept:

Euler Lagrange equation:

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

Explanation:

f(x, y, y') = y²(1-y'²)

so $\frac{\partial f}{\partial y}$= 2y(1-y'2), $\frac{\partial f}{\partial y^{\prime }}$ = -2y²y'

Using Euler-Lagrange equation

2y(1 - y'2) - $\frac{d}{dx}$( - 2y2y') = 0

y(1 - y'2) + $\frac{d}{dx}$(y2y') = 0

y - yy'² + y2y'' + 2yy'² = 0

y + yy'2 + y2y'' = 0

1 + y'2 + yy'' = 0

 $\frac{d}{dx}$(yy') = -1

Integrating both sides

yy' = - x + c

y$\frac{dy}{dx}$ = - x + c

ydy = (-x + c)dx

Again integrating both sides again

y²/2 = -x²/2 + cx + d

Putting y(0) = 0 we get

0 = d

And putting y(π) = 0 we get

0 = - π² / 2 +cπ ⇒ c = π / 2

Hence we get

 y2 / 2 = - x2 / 2 + πx / 2

y2 = - x2 + πx ⇒ $y=\sqrt{-x^2+\pi x}$ ,which is a circle of centre (π/2, 0) and radius π / 2

To find the local maxima or minima

 y' = $\frac{-2x+\pi }{2\sqrt{-x^2+\pi x}}$

Stationary point is given by

y' = 0 ⇒  x = π / 2

Also we have the endpoints x = 0, x = π 

Now y(π / 2) = $\sqrt{-{\pi }^2/4+{\pi }^2/2}$ = π  / 2, y(π) =0, y(0) = 0.

So y has local maxima at x = π / 2 and local minima at x = 0, x = π 

Now ||y|| = $\sqrt{y^{T}y}=\sqrt{-x^2+\pi x}$ 

so same as y,  ||y(π)|| =0, ||y(0)|| = 0 and sup ||y|| = π  / 2.

So option (1) is correct.

Concept:

The functional ${\int }_{x_0}^{x_1}f(x,y,y^{\prime })dx$, $y(x_0)=y_0,y(x_1)=y_1$ satisfy 

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

Explanation:

here f = xy + y²y'

Using

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

⇒ x + 2yy' - $\frac{d}{dx}(y^2)$ = 0

⇒ x + 2yy' - 2yy' = 0

⇒ x = 0

Putting  y(1) = 2 we get

1 = 0 which is not true.

Hence this functional has no extremal.

Option (4) is true.

Concept:

To minimize the functional, we use the Euler-Lagrange equation for the functional $F(x,y,y^{\prime })=\frac{1}{2}([y^{\prime }(x){]}^2+[y(x){]}^2)$,

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

Explanation:

$min\frac{1}{2}{\int }_{-1}^1([y^{\prime }(x){]}^2+[y(x){]}^2)dx$

subject to the conditions:

1. $y\in C^1[-1,1]$ (i.e., y(x) is continuously differentiable).

2. ${\int }_{-1}^{1}xy(x)dx$= 0

3. $y(-1)=y(1)=1$

Euler-Lagrange Equation

$\frac{\partial F}{\partial y}=y(x)$

$\frac{\partial F}{\partial y^{\prime }}=y^{\prime }(x)$

$\frac{d}{dx}\left(\frac{\partial F}{\partial y^{\prime }}\right)=y^{″}(x)$
 

Substitute into the Euler-Lagrange equation

$y(x)-y^{″}(x)=0$

This is a second-order linear differential equation,

$y^{″}(x)=y(x)$

The general solution to this differential equation is:

$y(x)=A{e}^x+B{e}^{-x}$

where A and B are constants to be determined using the boundary conditions.

 Apply y(1) = 1 :

$A{e}^1+B{e}^{-1}=1$

⇒$Ae+\frac{B}{e}=1$

2. Apply y(-1) = 1 :   $A{e}^{-1}+B{e}^1=1$

⇒$\frac{A}{e}+Be=1$

From equations (1) and (2), we can solve for A and B . Add both equations

$Ae+\frac{B}{e}+\frac{A}{e}+Be=2$

⇒ $A(e+\frac{1}{e})+B(e+\frac{1}{e})=2$

⇒ $(A+B)(e+\frac{1}{e})=2$

Thus, $A+B=\frac{2}{e+\frac{1}{e}}=\frac{2}{\frac{e^2+1}{e}}=\frac{2e}{e^2+1}$

Now subtract equation (2) from equation (1):

$Ae-\frac{A}{e}+Be-\frac{B}{e}=0$

⇒ $A(e-\frac{1}{e})+B(e-\frac{1}{e})=0$

Thus, $(A-B)(e-\frac{1}{e})=0$

Since, $e-\frac{1}{e}\ne 0$ , it must be that$A=B$

Substitute A = B into equation (3):

$2A(e+\frac{1}{e})=2$

Solve for A :

$A=\frac{1}{e+\frac{1}{e}}=\frac{e}{e^2+1}$

Thus, $A=B=\frac{e}{e^2+1}.$

$y(x)=\frac{e}{e^2+1}(e^x+e^{-x})$

This matches option 3).