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)] = \(\rm \int_{x_1}^{x_2} f(x, y, y')d x\), 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)] = \(\rm \int_0^1\left(5x^2 + 3y^4 - 2y^{\prime 2}\right) d x\), y(1) = 1

So f = \(5x^2 + 3y^4 - 2y^{\prime 2}\)

\(f_{y'}=-4y'\)

So \(f_{y'y'}=-4\) < 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] = \(\int_a^bf(x, y, y')dx\) be a functional such that y(a) = c, y(b) = d then y(x) satisfy Euler-Lagrange's equation

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

Explanation:

J[y] = \(\rm \int_0^1[(y^{\prime})^2+2 y] d x\) subject to y(0) = 0, y(1) = 1

here f = y'² + 2y

Using Euler-Lagrange's equation we get

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

⇒ 2 - \(d\over dx\)(2y') = 0

⇒ y'' - 1 = 0

⇒ y'' = 1

Integrating both sides

y' = x + c₁

Integrating again

⇒ y = \({x^2\over 2}+c_1x+c_2\)

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) = \({x^2\over 2}+\frac x2\) so, y' = x + 1/2

Therefore

Inf[J] = \(\rm \int_0^1[(y^{\prime})^2+2 y] d x\)

         = \(\rm \int_0^1[(x+\frac 12)^2+2({x^2\over 2}+\frac x2)] d x\)

        = \(\rm \int_0^1(x^2+x+\frac 14+{x^2}+x) d x\)

       = \(\rm \int_0^1(2x^2+2x+\frac 14) d x\)

      = \(\left[\frac23x^3+x^2+\frac x4\right]_0^1\) = \(\frac{23}{12}\)

Hence option (1) is correct

Concept:

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

Explanation:

\(J(x)=\int_0^1\left[x^2(t)+\dot{x}^2(t)\right] d t \)

F(t, x, x') = [x²(t) + x'²(t)]

\(\frac{\partial F}{\partial y}- \frac{d}{dx}(\frac{\partial F}{\partial y'})=0\) ⇒ 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) = \(\int_a^bF(x, y, y')dx\) subject to y(a) = y1, y(b) = y2 then the extrema satisfy the Euler-Lagrange's equation

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

where H = F + λH

Explanation:

\(\displaystyle J[y]=\int_{-1}^0\left(12 x y-\left(y^{\prime}\right)^2\right) d x\), y(0) = 0, y(-1) = 1

Here F = 12xy - \((y')^2\)

Then using

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

⇒ 12x - \({d\over dx}(-2y')\) = 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 \(I(y)=\int_a^b F\left(x, y, y^{\prime}\right) d x\), y(a) = y1, y(b) = y2, will satisfy Euler equation

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

Explanation:

\(I(y)=\int_1^2 \frac{\sqrt{1+\left(y^{\prime}(x)\right)^2}}{x} d x\), y(1) = 0, y(2) = 1

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

Then from Euler equation we get

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

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

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

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

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

⇒ \(y'^2=c^2x^2(1+y'^2)\)

⇒ \(y'^2(1-c^2x^2)=c^2x^2\)

⇒ \(y'^2=\frac{c^2x^2}{1-c^2x^2}\)

⇒ y' = \(\frac{cx}{\sqrt{1-c^2x^2}}\)

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

Integrating both sides

⇒ y = \(-\frac1c\sqrt{1-c^2x^2}\) + b....(i)

Given y(1) = 0, y(2) = 1

y(1) = 0 ⇒ 0 = \(-\frac1c\) + b ⇒ b = 1/c

y(2) = 1 ⇒ 1 = \(-\frac1c\sqrt{1-4c^2}\) + b

⇒ 1 = \(-\frac1c\sqrt{1-4c^2}+\frac1c\) (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) = \(-\frac1c\sqrt{1-c^2x^2}\)

⇒ (y - b)² = \(\frac1{c^2}({1-c^2x^2})\)

⇒ (y - b)2 + x² =  \(\frac1{c^2}\)

Putting the values

Extremize function is

\((y-\frac52)^2+x^2=(\frac52)^2\) which is an equation of a circle

(3) is correct

Concept:

The extremal of the functional \(J(u)=\int_a^b F\left(x, u, u^{\prime}\right) d x\), u(a) = u1, u(b) = u2, will satisfy Euler equation

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

Explanation:

\(\rm J(u)=\int_0^1\left[u_x^2+4 \frac{u^2}{x^2}\right] x d x\), u(0) = 0 and u(1) = 1

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

Then from Euler equation we get

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

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

⇒ \(\frac{4u}{x}-u_x-xu_{xx}\) = 0

⇒ \(x^2u_{xx}+xu_x-4u\) = 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 = \(c_1x^2+c_2x^{-2}\)

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

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

Option 4: \(\rm u(x)=\frac{1}{4} x^2\) 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] = \(\rm \int_0^1[720 x^2 y-\left(y^{\prime \prime}\right)^2] d x\)

Using

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

720x2 - 0 + \(\frac{d^2}{dx^2}\)(-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:

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

Explanation:

f(x, y, y') = y²(1-y'²)

so \(\frac{\partial f}{\partial y}\)= 2y(1-y'2), \(\frac{\partial f}{\partial y'}\) = -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 ⇒ \(y =\sqrt{-x^2+π x}\) ,which is a circle of centre (π/2, 0) and radius π / 2

To find the local maxima or minima

 y' = \(\frac{-2x+π}{2\sqrt{-x^2+π x}}\)

Stationary point is given by

y' = 0 ⇒  x = π / 2

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

Now y(π / 2) = \(\sqrt{-π^2/4+π^2/2}\) = 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|| = \(\sqrt{y^Ty}=\sqrt{-x^2+\pi x}\) 

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

So option (1) is correct.

Concept:

The functional \(\int_{x_0}^{x_1}f(x, y, y')dx\), \(y(x_0)=y_0, y(x_1)=y_1\) satisfy 

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

Explanation:

here f = xy + y²y'

Using

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

⇒ x + 2yy' - \({d\over dx}(y^2)\) = 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 \(F(x, y, y') = \frac{1}{2} \left( [y'(x)]^2 + [y(x)]^2 \right) \),

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

Explanation:

\(\min \frac{1}{2} \int_{-1}^{1} \left( [y'(x)]^2 + [y(x)]^2 \right) dx\)

subject to the conditions:

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

2. \(\int_{-1}^{1} x y(x) dx \)= 0

3. \(y(-1) = y(1) = 1\)

Euler-Lagrange Equation

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

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

\(\frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) = y''(x)\)
 

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 :

\(A e^1 + B e^{-1} = 1\)

⇒\(A e + \frac{B}{e} = 1 \tag{1}\)

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

⇒\(\frac{A}{e} + B e = 1 \tag{2}\)

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

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

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

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

Thus, \(A + B = \frac{2}{e + \frac{1}{e}} = \frac{2}{\frac{e^2 + 1}{e}} = \frac{2e}{e^2 + 1} \tag{3}\)

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

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

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

Thus, \((A - B) \left( e - \frac{1}{e} \right) = 0\)

Since, \(e - \frac{1}{e} \neq 0\) , it must be that\(A = B \tag{4}\)

Substitute A = B into equation (3):

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

Solve for A :

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

Thus, \( A = B = \frac{e}{e^2 + 1} .\)

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

This matches option 3).