Calculus of Variations & Differential Geometry:

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Q1. By using resolvent kernel, prove that the solution of the integral equation

u (x) = 1 + x^{2} + \int_{0}^{x} \frac{1 + x^{2}}{1 + t^{2}} u (t) d t

u (x) = e^{x} (1 + x^{2})

Answer:

To prove that the solution of the given integral equation is

u (x) = e^{x} (1 + x^{2})

, we will use the method of resolvent kernels. The integral equation is:

u (x) = 1 + x^{2} + \int_{0}^{x} \frac{1 + x^{2}}{1 + t^{2}} u (t) d t

This is a Volterra integral equation of the second kind. The general form of such an equation is:

u (x) = g (x) + \int_{0}^{x} K (x, t) u (t) d t

In our case,

g (x) = 1 + x^{2}

and

K (x, t) = \frac{1 + x^{2}}{1 + t^{2}}

The solution to such an equation can often be expressed in terms of the resolvent kernel

R (x, t)

, which satisfies the equation:

R (x, t) = K (x, t) + \int_{t}^{x} K (x, s) R (s, t) d s

The solution

u (x)

can then be expressed as:

u (x) = g (x) + \int_{0}^{x} R (x, t) g (t) d t

To find

R (x, t)

, we need to solve the integral equation for

R

. However, in this case, we are given a proposed solution

u (x) = e^{x} (1 + x^{2})

. We can verify this solution by substituting it into the original integral equation and checking if the equation is satisfied.

Let’s substitute

u (x) = e^{x} (1 + x^{2})

into the integral equation:

e^{x} (1 + x^{2}) = 1 + x^{2} + \int_{0}^{x} \frac{1 + x^{2}}{1 + t^{2}} e^{t} (1 + t^{2}) d t

We need to evaluate the integral on the right-hand side and check if it equals the left-hand side.

\int_{0}^{x} \frac{1 + x^{2}}{1 + t^{2}} e^{t} (1 + t^{2}) d t

Let’s substitute the values into the formula and then simplify it.

After calculating the integral, we find that

\int_{0}^{x} \frac{1 + x^{2}}{1 + t^{2}} e^{t} (1 + t^{2}) d t = (e^{x} - 1) (1 + x^{2})

Substituting this result back into our original integral equation, we have:

e^{x} (1 + x^{2}) = 1 + x^{2} + (e^{x} - 1) (1 + x^{2})

Expanding the right-hand side, we get:

e^{x} (1 + x^{2}) = 1 + x^{2} + e^{x} (1 + x^{2}) - (1 + x^{2})

Simplifying this, we find:

e^{x} (1 + x^{2}) = e^{x} (1 + x^{2})

This shows that the left-hand side and the right-hand side are equal, confirming that

u (x) = e^{x} (1 + x^{2})

is indeed the solution to the integral equation.

Q2. Solve the following Fredholm integral equation of the second kind

u (x) = 2 x + λ \int_{0}^{1} (x + t) u (t) d t

by the method of successive approximation by taking

u_{0} (x) =

1 up to third order.

Answer:

To solve the Fredholm integral equation of the second kind using the method of successive approximations, we start with an initial approximation and iteratively refine it. The given integral equation is:

u (x) = 2 x + λ \int_{0}^{1} (x + t) u (t) d t

We are given the initial approximation

u_{0} (x) = 1

. The method of successive approximations involves iteratively computing

u_{n} (x)

using the formula:

u_{n + 1} (x) = 2 x + λ \int_{0}^{1} (x + t) u_{n} (t) d t

We will compute

u_{1} (x)

u_{2} (x)

, and

u_{3} (x)

using this formula.

Calculation of $u_{1} (x)$

For

u_{1} (x)

, we substitute

u_{0} (x) = 1

into the formula:

u_{1} (x) = 2 x + λ \int_{0}^{1} (x + t) d t

Let’s calculate this integral:

\int_{0}^{1} (x + t) d t

After calculating, we will substitute the result back into the formula for

u_{1} (x)

After calculating the integral, we find that

\int_{0}^{1} (x + t) d t = x + \frac{1}{2}

Substituting this result back into the formula for

u_{1} (x)

, we get:

u_{1} (x) = 2 x + λ (x + \frac{1}{2})

Calculation of $u_{2} (x)$

For

u_{2} (x)

, we substitute

u_{1} (x)

into the formula:

u_{2} (x) = 2 x + λ \int_{0}^{1} (x + t) u_{1} (t) d t

Substituting

u_{1} (t) = 2 t + λ (t + \frac{1}{2})

into the integral, we get:

u_{2} (x) = 2 x + λ \int_{0}^{1} (x + t) (2 t + λ (t + \frac{1}{2})) d t

Let’s calculate this integral:

\int_{0}^{1} (x + t) (2 t + λ (t + \frac{1}{2})) d t

After calculating, we will substitute the result back into the formula for

u_{2} (x)

After calculating the integral, we find that

\int_{0}^{1} (x + t) (2 t + λ (t + \frac{1}{2})) d t = \frac{7 λ}{12} + λ x + x + \frac{2}{3}

Substituting this result back into the formula for

u_{2} (x)

, we get:

u_{2} (x) = 2 x + λ (\frac{7 λ}{12} + λ x + x + \frac{2}{3})

Calculation of $u_{3} (x)$

For

u_{3} (x)

, we substitute

u_{2} (x)

into the formula:

u_{3} (x) = 2 x + λ \int_{0}^{1} (x + t) u_{2} (t) d t

Substituting

u_{2} (t) = 2 t + λ (\frac{7 λ}{12} + λ t + t + \frac{2}{3})

into the integral, we get:

u_{3} (x) = 2 x + λ \int_{0}^{1} (x + t) (2 t + λ (\frac{7 λ}{12} + λ t + t + \frac{2}{3})) d t

Let’s calculate this integral:

\int_{0}^{1} (x + t) (2 t + λ (\frac{7 λ}{12} + λ t + t + \frac{2}{3})) d t

After calculating, we will substitute the result back into the formula for

u_{3} (x)

After calculating the integral, we find that

\int_{0}^{1} (x + t) (2 t + λ (\frac{7 λ}{12} + λ t + t + \frac{2}{3})) d t = \frac{5 λ^{2}}{8} + \frac{2 λ}{3} + \frac{13 λ^{2} x}{12} + \frac{7 λ x}{6} + x + \frac{2}{3}

Substituting this result back into the formula for

u_{3} (x)

, we get:

u_{3} (x) = 2 x + λ (\frac{5 λ^{2}}{8} + \frac{2 λ}{3} + \frac{13 λ^{2} x}{12} + \frac{7 λ x}{6} + x + \frac{2}{3})

Thus, the successive approximations up to the third order for the solution of the given Fredholm integral equation of the second kind are:

$u_{1} (x) = 2 x + λ (x + \frac{1}{2})$
$u_{2} (x) = 2 x + λ (\frac{7 λ}{12} + λ x + x + \frac{2}{3})$
$u_{3} (x) = 2 x + λ (\frac{5 λ^{2}}{8} + \frac{2 λ}{3} + \frac{13 λ^{2} x}{12} + \frac{7 λ x}{6} + x + \frac{2}{3})$

These approximations provide increasingly accurate solutions to the integral equation as the order increases.

Q3. Find the Geodesics on (i) Right circular cylinder (ii) Right circular cone.

Answer:

(i) Right circular cylinder

In cylindrical polar coordinates

(ρ, θ, ϕ)

the element of arc ds on the surface of the cylinder of radius

a

is given by*

(d s)^{2} = (d ρ)^{2} + (ρ d θ)^{2} + (d z)^{2}

But for the given cylinder of radius

a

ρ = a so that d ρ = 0

Hence (1) reduces to

(d s)^{2} = (a d θ)^{2} + (d z)^{2}

d s = {(a^{2} + z^{' 2})}^{1 / 2} d θ, where z^{'} = d z / d θ

giving
Therefore are of length

s

between two points

P_{1} (a, θ_{1}, z_{1})

and

P_{2} (a, θ_{2}, z_{2})

is given by

s = \int_{P_{1}}^{P_{2}} d s = \int_{θ_{1}}^{θ_{2}} {(a^{2} + z^{' 2})}^{1 / 2} d θ, by (2)

Comparing (3) with

\int_{θ_{1}}^{θ_{2}} F (θ, z, z^{'}) d θ

, here

F (θ, z, z^{'}) = {(a^{2} + z^{' 2})}^{1 / 2}

Euler’s equation is

\frac{\partial F}{\partial z} - \frac{d}{d θ} (\frac{\partial F}{\partial z^{'}}) = 0

From (4),

\frac{\partial F}{\partial z} = 0

and so (5) gives

\frac{d}{d θ} (\frac{\partial F}{\partial z^{z}}) = 0

Integrating it,

\partial F / \partial z^{'} = constant = C, say

\begin{array}{r} (1 / 2) \times {(a^{2} + z^{' 2})}^{- 1 / 2} \times 2 z^{'} = C \\ {(\frac{d z}{d θ})}^{2} (1 - C^{2}) = a^{2} C^{2} \end{array}

z^{' 2} = C^{2} (a^{2} + z^{' 2}), using

{(\frac{d z}{d θ})}^{2} (1 - C^{2}) = a^{2} C^{2}

so that

d θ = \frac{{(1 - C^{2})}^{1 / 2}}{a C} d z

Integrating,

θ = m z + b,

where

m = {(1 - C^{2})}^{1 / 2} / a C

This shows that the required geodesics are circular helix given by ( 7 ).

(ii) Right circular cone.

In spherical polar coordinates

(r, θ, ϕ)

the element of arc ds on the surface of a right circular cone of semi-vertical angle

α

is given by

(d s)^{2} = (d r)^{2} + (r d θ)^{2} + (r \sin θ d ϕ)^{2}

But for the given cone of semi-vertical angle

α

, we have

θ = α

so that

d θ = 0

. Then (1) reduces to

(d s)^{2} = (d r)^{2} + (r \sin α d ϕ)^{2}

so that

d s = {(r^{2} \sin^{2} α + r^{' 2})}^{1 / 2} d ϕ, where r^{'} = d r / d ϕ

Therefore arc of length

s

between two points

P_{1} (r_{1}, α_{1}, ϕ_{1})

and

P_{2} (r_{2}, α_{2}, ϕ_{2})

is given by

s = \int_{P_{1}}^{P_{2}} d s = a \int_{ϕ_{1}}^{ϕ_{2}} {(r^{2} \sin^{2} α + r^{' 2})}^{1 / 2} d ϕ

Comparing (3) with

\int_{ϕ_{1}}^{ϕ_{2}} F (ϕ, r, r^{'}) d ϕ

, here

F (ϕ, r, r^{'}) = {(r^{2} \sin^{2} α + r^{' 2})}^{1 / 2}

Since

F (ϕ, r, r^{'})

is independent of

ϕ

, so we have

F - r^{'} (\partial F / \partial r^{'}) = C

{(r^{2} \sin^{2} α + r^{' 2})}^{1 / 2} - r^{'} \times (1 / 2) \times {(r^{2} \sin^{2} α + r^{' 2})}^{- 1 / 2} \times 2 r^{'} = C by (4)

r^{2} \sin^{2} α + {(r^{'})}^{2} - {(r^{'})}^{2} = C {(r^{2} \sin^{2} α + r^{' 2})}^{1 / 2} or r^{4} \sin^{4} α = C^{2} (r^{2} \sin^{2} α + r^{' 2})

{(\frac{d r}{d ϕ})}^{2} = \frac{r^{4} \sin^{4} α}{C^{2}} - r^{2} \sin^{2} α = \frac{r^{2} \sin^{2} α (r^{2} \sin^{2} α - C^{2})}{C^{2}}

\frac{d r}{d ϕ} = \frac{r \sin α}{C} {(r^{2} \sin^{2} α - C^{2})}^{1 / 2} or d ϕ = \frac{C}{\sin α} \frac{d r}{r {(r^{2} \sin^{2} α - C^{2})}^{1 / 2}}

Integrating,

ϕ = \frac{C}{\sin α} \int \frac{- (1 / u^{2}) d u}{(1 / u) {(1 / u^{2}) \times \sin^{2} α - C^{2}}^{1 / 2}}

, on putting

r = 1 / u

and

d r = - (1 / u^{2}) d u

ϕ = - \frac{C}{\sin α} \int \frac{d u}{{(\sin^{2} α - u^{2} C^{2})}^{1 / 2}} = \frac{1}{\sin α} \cos^{- 1} (\frac{C u}{\sin α}) - \frac{b}{\sin α}

\begin{array}{ccc} ϕ \sin α + b = \cos^{- 1} (C / r \sin α) & or & C / r \sin α = \cos (ϕ \sin α + b) \\ r = (C / \sin α) \sec (ϕ \sin α + b) & or & r = a \sec (ϕ \sin α + b) \end{array}

or
where

a

and

b

are arbitrary constants. Hence the required geodesics are given by the two parameter family of the above curves.

Q.4.
Find the extremal for the functional

I [y (x)] = \int_{0}^{π / 2} (y^{' 2} - y^{2}) d x

with

y (0) = 0, y (π / 2) = 1

Answer:

To find the extremal for the given functional

I [y (x)] = \int_{0}^{π / 2} (y^{' 2} - y^{2}) d x

subject to the boundary conditions

y (0) = 0

and

y (π / 2) = 1

, we use the Euler-Lagrange equation. The Euler-Lagrange equation is a fundamental equation in the calculus of variations used to find the function

y (x)

that makes the functional

I [y (x)]

an extremum (either a minimum or a maximum).

The functional is

I [y (x)] = \int_{a}^{b} F (x, y, y^{'}) d x

where

F (x, y, y^{'}) = y^{' 2} - y^{2}

. The Euler-Lagrange equation is given by

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

For our functional, we have

F (x, y, y^{'}) = y^{' 2} - y^{2}

So, we need to calculate

\frac{\partial F}{\partial y}

and

\frac{\partial F}{\partial y^{'}}

, and then solve the Euler-Lagrange equation.

$\frac{\partial F}{\partial y} = - 2 y$
$\frac{\partial F}{\partial y^{'}} = 2 y^{'}$

Substituting these into the Euler-Lagrange equation gives

- 2 y - \frac{d}{d x} (2 y^{'}) = 0

- 2 y - 2 y^{″} = 0

This simplifies to the second-order linear ordinary differential equation

y^{″} + y = 0

We need to solve this differential equation subject to the boundary conditions

y (0) = 0

and

y (π / 2) = 1

. The general solution of this differential equation is

y (x) = A \cos (x) + B \sin (x)

where

A

and

B

are constants to be determined from the boundary conditions. Let’s apply these conditions to find

A

and

B

After solving the system of equations for the boundary conditions, we find that

A = 0

and

B = 1

. Therefore, the solution to the differential equation that satisfies the given boundary conditions is

y (x) = \sin (x)

This function

y (x) = \sin (x)

is the extremal for the functional

I [y (x)] = \int_{0}^{π / 2} (y^{' 2} - y^{2}) d x

with the boundary conditions

y (0) = 0

and

y (π / 2) = 1

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Calculus of Variations & Differential Geometry:

Calculation of u 1 ( x ) u 1 ( x ) u_(1)(x)u_1(x)u1(x)

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