Mathquery

How To Multiply Fractions With Whole Numbers

2020-05-14T12:29:00.002+05:30

How To Multiply Fractions With Whole Numbers :

Multiplication of two fractional numbers or rational numbers is solved only in three steps i.e.

Before we discuss the method of how to multiply fractions we have to understand that what is a fractional number or a rational number?

So let us go through definition,

Definition Of Fraction or Rational Numbers :

A fraction or rational number is of the form p/q, where p and q are integers and q≠0.

Then here is the question arises that what is an integer?

To answer this question let us take examples,

The set of numbers like .............-3,-2,-1,0,1,2,3..........

are called as the set of integers.

Now come to our initial question of how to multiply fraction with mixed numbers?

Let us discuss the methods

(1) First of all multiply the numerators of two fractional numbers i.e.

for example 3 / 5 × 2 / 5

Here first we have to multiply 3 with 2.

Let us proceed another step

(2) Then multiply the same or different denominators of the two fractional numbers

for example 3 / 5 × 2 / 5

Here to multiply 5 of the fraction 3/5 with 5 of the fraction 2/5.

(3) Here is the next step to simplify if needed ,

for example 3 × 2 / 5 × 5 = 6 / 25

Here we know that there is nothing to simplify.

But there are some problems that needed simplification after multiplication for an accurate result.

For example 2 / 4 × 6 / 8 = 2 × 6 / 4 × 8

= 12 / 32

Here we need simplification, we have to simplify the fractional number till numerator or denominator is converted into prime numbers, i.e.

Let us take the above example that

12 / 32, where the numerator is 12 . Let us divide it with such number which results in a prime number like 2,3,5,7,11...etc.

Let us divide it by 4, then the result is 12/ 4 = 3 which is a prime number.

There is a rule of the fraction that if we divide or multiply some number with numerator we have to divide or multiply the same whole number or any integer with denominator.

So let us divide the denominator by 4 we get 32/4 = 8

Hence the simplification result is 3/8.

We reached a conclusion that

A rational number is of the form p/q, where p and q are integers and q ≠ 0 and p and q both are relatively prime.

One thing we have to consider that in case of complicated multiplication or multiply two large numbers is not so easy.

So, in that case, we have to simplify the fractions before multiply with each other,

For example :

374 / 462 × 564 / 126

Ans: In this case, we have to first simplify 374 / 462 to relatively prime numbers.

In case we can't find the exact number which if we divide by 374 it is converted into a prime number, then divide this number with a least positive divisor by which it is divisible.

So let us divide both 374 and 462 by 2 we get 187 / 231, so here we get the exact relatively prime numbers.

Let us do the same with a second fraction that 564 / 126.

Let us divide both 564 and 126 by 2 we get 282/63. Again let us divide the result by 3 we get 94/21. Here we get the relatively prime numbers.

So let us multiply

187/231 × 94/21

= 187 × 94 / 231 × 21

= 17578 / 4851

which is the required solution.
This is the solution to the problem How to multiply an improper fraction with a proper fraction.

Let us discuss the improper and proper fraction .

Improper Fraction:

The fraction of the form p/q is said to be an improper fraction if p> q .

for example :

3/2 , 4/3 , 5/4 , 7/6........etc.

Proper Fraction :

The fractional number of the form p/q is said to be a proper fraction if p<q .

for example : 3/4 , 11/12, 16/17, 49/50 , 60/61 ............etc .

Prime Number :

The number which is only divisible by 1 or itself is called a prime number.

Ex : 2,3,5,7,11,13,17,19,.......

Relatively Prime :

Two numbers are called relatively prime to each other if their greatest common divisor or g.c.d is 1.

This is all about How To Multiply Fraction.

References:-

(I). Multiplication And Fractions Math Games tough topics.

(II). Multiplication, division, the addition of fractions by OUP Oxford.

PLEASE LEAVE A COMMENT AND SHARE

Continuity And Differentiability In Mathematics | Mathquery

2020-05-01T19:36:00.004+05:30

Welcome To Mathquery

In mathematics continuity and derivability or differentiability plays an important role . So let us discuss it .

Definition Of Continuity Of A Function :-

Let f be a function defined on an interval [a,b] . We shall now consider the behaviour of f at points of [a,b] .

Continuity At A Point :

Definition(Continuity At An Internal Point) :

A function f is said to be continuous at a point c , a<c<b , if

lim f(x) = f(c)

x-->c

In other words , the function is continuous at c , if for each ε>0 , there exists δ>0 such that

|f(x) - f(c) | < ε , when |x-c|<δ

Here lim f(x) is called the limit of a function

x-->c

i.e. the function exists when x tends to c.

(i) A function f is said to be continuous from the left at c if

lim f(x) = f(c)

x-->c-0

(ii) Also f is continuity from the right at c if

lim f(x) = f(c)

x-->c+0

Clearly a function is continuous at c iff it is continuous from the left as well as from the right .

Definition(Continuity At An End Point):

A function f is defined on a closed interval [a,b] is said to be continuous at the end point a if it is continuous from the right at a , i.e.

lim f(x) = f(a)

x-->a+0

Also the function is continuous at the end point b of [a,b] if

lim f(x) = f(b)

x-->b-0

Thus a function f is continuous at a point c if

(i) lim f(x) exists , and

x-->c

(ii) limit equals the value of the function at x= c.

Let's discuss about discontinuity of a function .

Example :

Show that the function defined by

{ x sin 1/x , when x≠0

f(x) ={

{ 0 , when x= 0

is continuous at x = 0.

Solution :

Now

lim f(x) = lim (x sin 1/x) = 0

x-->0

so that

lim f(x) = f(0)

x-->0

Hence f is continuous at x = 0 . (Proved)

Discontinuous Function :

A function is said to be discontinuous at a point c of its domain if it is not continuous there at c. The point c is then called a point of discontinuity of the function .

Now let us discuss about Uniform continuous of a function in mathematics .

Definition Of Uniform Continuity:

A function f defined on an interval I is said to be uniformly continuous on I if to each ε>0 there exists a δ>0 such that

|f(x₂) - f(x₁)| < ε , for arbitrary points x₁,x₂ of I for which |x₁-x₂|<δ.

Example :

Prove that f(x) = sin x² is not uniformly continuous on [0,∞[ .

Solution :

Let ε = 1/2 and δ be any positive number such that for n>π/δ²

|√(nπ/2 ) - √(n+1)π/2| <δ

Therefore , taking x₁ = √(nπ/2 ) and x₂= √(n+1 )π/2 , as any two points of the interval [0,∞[.

|f(x₁) - f(x₂)| = | sin nπ/2 - sin (n+1)π/2 | = 1

>ε,

|x₁-x₂| <δ

Hence f(x) = sin x² is not uniformly continuous on [0,∞[ .

Differentiability At A Point :

Let f be a real valued function defined on an interval I = [a,b] ⊂ R . It is said to be derivable at an interior point c (where a<c<b) if

lim f(c+h) - f(c) / h

h-->0

or lim f(x) - f(c) / x-c exists .

x-->c

The limit in case it exists , is called the Derivative or the Differential Coefficient of the function f at x = c , and is denoted by f'(c) . The limit exists when the left - hand and the right - hand limits exists and are equal .

lim f(x) - f(c) / x-c

x-->c-0

is called the left - hand Derivative and is denoted by

f'(c-0) , f'(c⁻) or Lf'(c)

While lim f(x) - f(c) / x-c

x-->c+0

is called the right - hand Derivative and is denoted by

f'(c+0) , f'(c+) or R f'(c)

Thus , the derivative f'(c) exists when

Lf'(c) = Rf'(c)

Example :

Show that the function f(x) = x² is derivable on [0,1] .

Solution :

Let x₀ be any point of ]0,1[, then

f'(x₀) = lim x²-x₀² / x-x₀ = lim(x+x₀) = 2x₀

x-->x₀ x-->x₀

At the end points , we have

f'(1) = lim f(x)-f(0) / x-0 = lim x²/x = lim x =0

x-->0+0 x-->0+

f'(1) = lim f(x)-f(1) / x-1 = lim x²-1 / x-1

x-->1-0 x-->1⁻

= lim (x+1) =2

x-->1⁻

Thus , the function is derivable in the closed interval [0,1].

References :-

(I).

By Abhijit Debnath.

(II).

(Dover Books on Mathematics)

Leibnitz's Rule Statement And It's Proof

2020-04-29T14:42:00.002+05:30

WELCOME TO MATHEMATICS

In this mathematics session I shall prove that , under suitable conditions, ' the derivative of the integral and the integral of the derivative are equal' , and consequently , 'the two repeated integrals are equal for continuous functions'.

Leibnitz's Rule In Mathematics:

If f is defined and continuous on the rectangle R = [a,b;c,d] , and if

(i) fₓ(x,y) exists and is continuous on the rectangle R , and

(ii) g(x) = ∫ f(x,y) dy , for x∈ [a,b]

then g is differentiable on [a,b]and

g'(x) = ∫ fₓ(x,y) dy

d d

i.e., d/dx {∫ f(x,y) dy }=∫ ∂f(x,y)/∂x dy

c c

Proof Of Leibnitz's Rule In Mathematics :

Since fₓ (∂f/∂x) exists on R , therefore, for each y∈ [c,d] , and each h≠ 0 , it follows by the Lagrange's Mean Value Theorem , that

f(x+h , y) - f(x,y) = h fₓ(x+θh , y),

for some 0<θ<1

Now fₓ being continuous is integrable on [c,d] for each x∈ [a,b] , therefore g(x) is well defined function on [a,b].

g(x+h) - g(x) /h = 1/h ∫ {f(x+h ,y) - f(x,y)}dy

= ∫ fₓ(x+θh ,y) dy , 0<θ<1

c .......(1)

Let ε>0 be given . Then by continuity and uniform continuity of fₓ on R , ∃ δ>0 , such that if (x,y) ,(x',y') ∈ R with

|x - x'| <δ , |y - y'| <δ , then

|fₓ(x,y) - fₓ(x',y')| <ε/(d-c) .........(2)

From equations (1) and (2) , we obtain

| g(x+h) - g(x) / h - ∫ fₓ(x,y) dy |

≤ ∫ |fₓ(x+θh , y) - fₓ(x,y)| dy

< ε(d-c)/ (d-c) = ε , 0<|h|<δ
d
Here g'(x) = lim g(x+h) - g(x) /h ∫fₓ(x,y) dy
h-->0 c

(Proved)

To understand this rule we have to solve an example , so let us take an example .

Example Of Leibnitz's Rule In Mathematics :

Show that

π/2

∫ log(1-x²sin²θ) dθ

= π log(1+√(1-x²) - π log 2 , if |x|<1

Proof :

The function log (1-x² sin²θ) is well defined in the rectangle [-1,1;0,π/2] and satisfies the conditions of the Leibnitz's rule . π/2

Let g(x) = ∫ log (1-x² sin²θ) dθ , |x|<1

0 ..........(1)

By differentiating under the integral sign , w.r.t. x , we get

π/2

g'(x) = ∫ -2x sin² θ dθ/(1-x² sin²θ)

π/2

= 2/x ∫ (1-x²sin²θ -1) dθ /(1-x²sin²θ) dθ

0 x≠ 0

π/2

= π/x -2/x ∫ dθ/(1-x² sin²θ) , put cot θ = 1

∞

= π/x - 2/x ∫ dt /(1+t²-x²)

∞

= π/x - 2/x√(1-x²) tan⁻¹ t/√(1-x²) ]

= π/x - π/x√(1-x²)

Integrating w.r.t. x , we obtain

g(x) = π log x - π log {1-√(1-x² ) / x} + c

where c is an arbitrary constant

= π log {x²/(1-√(1-x² )} + c

= π log {x²(1+√(1-x² ) / 1-(1-x²)} + c

= π log (1+√(1-x²)) + c

But g(0) = 0 , by equation(1) , therefore

c = -π log 2

Hence , g(x) = π log (1+√(1-x² ) - π log 2,

for |x| < 1

(Proved)

About The Scientists:

Gottfried Wilhelm Leibniz was a prominent German polymath and one of the most important logicians, mathematicians and natural philosophers of the Enlightenment.Readmore.

Fubini's Theorem

2020-04-27T00:29:00.000+05:30

Fubini's Theorem :-

In the world of mathematics integration plays an important role . So we have discussed the advance form of integration which results the famous theorem Fubini's Theorem.

Statement:

If a double integral , I = ∫∫ f dx dy

exists over a rectangle R = [a,b;c,d] , and if

∫ f dy also exists , for each fixed x in [a,b] ,

c b d

then the iterated integral ∫ dx ∫ f dy exists

a c

and is equal to double integral I.

Proof :

Let ε be any positive number.

Since the upper integral , Iᵘ is the infimum of the upper sums , there exists a partition P of R such that

Σ Σ Mᵢⱼ ΔRᵢⱼ < Iᵘ + ε

i j

or Σ Σ Mᵢⱼ Δxᵢ Δyⱼ < Iᵘ + ε ...........(1)

i j

Again , since for any fixed value of x ∈ Δxᵢ
, Mᵢⱼ is upper bound (not necessarily the supremum) of f in Δyⱼ , therefore
_d
∫ f dy ≤ Σ Mᵢⱼ Δyⱼ , when x∈ Δxᵢ
c j ..........(2)

Again , from equation(2) , Σ Μᵢⱼ Δyⱼ is upper bound of the function
d
Φ(x) = ∫ f dy , x∈ Δxᵢ
c

therefore by the same reasoning as above , we get
b
∫̅ Φ(x) dx ≤ Σ(Σ Μᵢⱼ Δyⱼ)Δxᵢ
a i j

or
_b _d
∫ dx ∫ f dy ≤ Σ Σ Mᵢⱼ Δyⱼ Δxᵢ < Iᵘ +ε
a c i j [by(1)] ........(3)

Also , since by hypothesis ,
d _d d
∫ f dy = ∫ f dy = ∫ f dy
̅ c c c

and ε is an arbitrary positive number , we get from equation(3),
_b _d _b d
∫ dx ∫ f dy = ∫ dx ∫ f dy ≤ Iᵘ ........(4)
a c a c

By considering the lower integral Iₗ, we can similarly show that
b _d
∫ dx ∫ f dy ≥ Iₗ ..........(5)
̅ a a

Again , since Iₗ = Iᵘ = I as the double integral exists , from (4) and (5) , we get
b d _d
I ≤ ∫ dx ∫ f dy ≤ ∫ f dy ≤ I
̅a c a
b d _b d
⇒ ∫ dx ∫ f dy = ∫ dx ∫ f dy = ∫∫ f dx dy
̅ a c a c R
b d
Thus ∫ [∫ f dy ] dx exists and equals
a c

∫∫ f dx dy . (Proved)
R

Note :

Similarly , if ∫∫ f dx dy and ∫ f dx both exist

R a

then the double integral can be expressed as

d b

∫∫ f dx dy = ∫ dy ∫ f dx

R c a

2. Since ε is an arbitrary positive number , inequality (3) shows that

_b _d __

∫ dx ∫ f dy ≤ ∫∫ f dx dy

a c R

and similarly by considering the lower integrals ,

b d

∫ dx ∫ f dy ≥ ∫∫ f dx dy

̅ a ̅ c ̅ ̅ ̅ R

Remarks :

1. The theorem holds even if f has a finite number of discontinuities , or an infinite of discontinuities lying on a finite number of lines x = Cᵢ , i= 1,2,...m parallel to y-axis .

For , the function Φ(x) = ∫ f dy will be

discontinuous at a finite number of points C₁,C₂,.........,Cₘ only and will therefore be still integrable . A similar remark holds for discontinuities parallel to x - axis .

2. If a double integral exists , then the two repeated integrals cannot exist without being equal . However , if the double integral does not exist , nothing can be said about the repeated integrals ; they may or may not exist .

Also one of the repeated integrals may exist or even that both may exist and be equal and yet the double integral may not exist , i.e. the existence of one or of both of the repeated integrals is no guarantee for the existence of the double integrals .

However , if the two repeated integrals exist both are unequal , the double integral cannot exist .

Example :

A function is defined on a rectangle [0,1;0,1] as

{ 1/2 when y is rational

f(x,y) = {

{ x , when y is irrational

Prove that the iterated integral

1 1

∫ dy ∫ f dx = 1/2 , but the other iterated

0 0

integral does not exist .

Proof :

1 1

For y rational ∫ f dx = ∫ 1/2 dx = 1/2

0 0

For y irrational

1 1

∫ f dx = ∫ x dx = 1/2

0 0

1 1 1

∴ ∫ dy ∫ f dx = ∫ 1/2 dx = 1/2

0 0 0

Proceeding as for a function of a single

variable , we may show that the integral

1 1 1

∫ f dy does not exist and so ∫ dx ∫ f dy

0 0 0

also does not exist .

It was shown earlier that the double integral ∫∫ f dx dy does not exist .

(Proved)

About Scientist :-

Guido Fubini (19 January 1879 – 6 June 1943) was an Italian mathematician, known for Fubini's theorem and the Fubini–Study metric. Born in Venice, he was steered towards mathematics at an early age by his teachers and his father, who was himself a teacher of mathematics.For more

Recommended Books :

Measure And Integration by Springer for undergraduate studies.

Concise Introduction To Measure Theory for undergraduate students .

Cauchy's Integral Theorem | Mathquery

2020-04-25T07:55:00.000+05:30

Cauchy's Integral Theorem:

Statement :

The theorem is usually formulated for closed paths as follows :

Let U be an open subset of C which is simply connected . Let f: U-->C be a holomorphic function , and let γ be a rectifiable path in U whose start point is equal to its end point . Then

∮ f(z) dz = 0
γ

Proof :

Let us assume that the partial derivatives of a holomorphic function are continuous , the Cauchy Integral Theorem can be proved as direct sequence of Green's Theorem and the fact that the real and imaginary parts of f = u + i v must satisfy the Cauchy - Riemann equations in the region bounded by γ , and moreover in the open neighbourhood U of this region.

Cauchy provided this proof , but it was later proved by Goursat without requiring techniques from vector calculus , or the continuity of partial derivatives .

We can break the integrand f , as well as the differential dz into their real and imaginary components .

f = u + i v ..............(1)

dz = dx + i dy .............(2)

Then we have

∮ f(z) dz = ∮ (u+iv)(dx+idy)
γ γ

= ∮(u dx - v dy)+i∮(v dx + u dy)
γ γ ........(3)

By Green's Theorem , we then replace the integrals around the closed contour γ with an area integral throughout the domain D that is enclosed by γ as follows.

∮(u dx - v dy) = ∫∫(-∂v/∂x - ∂u/∂y) dx dy
γ D ............(4)

∮(v dx + u dy) = ∫∫(∂u/∂x - ∂v/∂y) dx dy
γ D ............(5)

But the real and imaginary parts of a function holomorphic in the domain D , u and v must satisfy the Cauchy-Riemann equations so ,

∂u/∂x = ∂v/∂y ..............(6)
∂u/∂y = -∂v/∂x .............(7)

We therefore find that both integrands are zero

∫∫(-∂v/∂x - ∂u/∂y) dx dy
D

= ∫∫(∂u/∂y - ∂u/∂y) dx dy =0
D ...........(8)

∫∫(∂u/∂x - ∂v/∂y) dx dy
D
= ∫∫(∂u/∂x - ∂u/∂x) dx dy
D .............(9)

This gives the desired result

∮ f(z) dz = 0 ...............(10)
γ (Proved)

Example :

Prove that ∫ eᶻ dz = 0

Solution :

Here f(z) = eᶻ = e^(x+iy) (∵ In complex

z = x+iy)

= e^x . e^ iy

= e^x (cos y + i sin y)

(∵ polar form of

complex function)

= e^x cos y + i e^x sin y

= u + iv

where u = e^x cosy v= e^x sin y

∂u/∂x = e^x cosy ∂v/∂x = e^x sin y

∂u/∂y = -e^x sin y ∂v/∂y = e^x cos y

∴ ∂u/∂x = ∂v/∂y and ∂u/∂y = - ∂v/∂x

⇒ f(z) = eᶻ satisfy C-R equation

⇒ eᶻ is analytic for all z .

So ∫ eᶻ dz = 0 by Cauchy's Integral Theorem

Which is the required solution .

About Scientist :-

Baron Augustin-Louis Cauchy (21 August 1789 – 23 May 1857) was a French mathematician, engineer, and physicist who made pioneering contributions to several branches of mathematics, including mathematical analysis and continuum mechanics .For More

Green's Theorem | Mathquery

2020-04-24T07:35:00.000+05:30

Green's Theorem:

Statement :-

If a domain E, regular with respect to both the axes , is bounded by a contour C , and f and g are two single - valued functions which along with their partial derivatives ∂f/∂y and ∂g/∂x are continuous on E , then

∫∫ (∂g/∂x - ∂f/∂y) dx dy = ∫ (f dx + g dy )

E C

where the line integral is taken in the positive direction .

Proof :-

Let us first consider a function f which , alongwith its partial derivative ∂f/∂y ,is continuous on a region E , regular with respect to y-axis . Let E be bounded by contour C , consisting of the curves y= φ(x) , y= ψ(x) , x = a , x = b , such that

φ(x) ≤ ψ(x) , ∀ x ∈ [a,b]

we have

∫∫ ∂f(x,y)/∂y dx dy

b ψ(x)

= ∫ dx ∫ ∂f(x,y)/∂y dy

a φ(x)

b b

= ∫ f(x,ψ(x)) dx - ∫ f(x,φ(x)) dx

a a

The two integrals on the right are line integrals along the contour from A to B and from C to D , respectively .(The portions BC and DA of the contour coincide with the lines x = a and x=b respectively .)

Also contour

C = C₁ + C₂ + C₃ + C₄

Now

∫ f(x,ψ(x)) dx = ∫ f(x,y) dx = -∫f(x,y)dx

a -c₁ c₁

∫ f(x,y) dx = 0= ∫ f(x,y) dx .

c₂ c₄

∫ f(x,φ(x)) dx = ∫ f(x,y) dx

a c₃

∴ ∫∫ ∂f/∂y dx dy =

-∫ f dx - ∫ f dx - ∫f dx -∫ f dx

c₁ c₂ c₃ c₄

= -∫ f dx

In order to extend the result to any region , let us first suppose the region E to be piecewise regular with regard to the y-axis , i.e., E can be split up into a finite number of sub-regions, E₁,E₂,....,Eₙ, the contour of each of which is cut in at most two points by a line parallel to y-axis .

Since result(1) holds for each sub-region , let us apply it to all the sub-regions and add. The double integrals add up to the double integral on the whole region E.

Since the line integrals along the partition lines cancel each other because along each line the integral is taken twice in opposite directions, therefore the line integrals (in positive sense ) along the contours of sub-regions also add up to the line integral along the contour C of the whole region E.

Hence for any region E, piecewise regular with respect to y-axis, we have

∫∫ ∂f/∂y dx dy = -∫ f dx .....(2)

E c

It can similarly be proved that for any region E , piecewise regular with respect to x-axis ,

∫∫ ∂g/∂x dx dy = ∫ g dy .......(3)

E c

The for any region E, which is regular with respect to both the axes , we have

∫∫ (∂g/∂x - ∂f/∂y) dx dy = ∫ f dx + g dy

E c

...............(4)

Hence the theorem .

Note :-

The theorem holds even when the domain is enclosed by only two curves , y= φ(x) and y = ψ(x), between x=a and x=b , the contours C₂ and C₄ reducing to zero.

Corollary:-

If f(x,y) = y , we see from (2) that the area of a domain which is piecewise regular with respect to y-axis

= -∫ y dx ..........(5)

Similarly putting g(x,y) = x, we see from(3) that the area of a domain , piecewise regular with respect to x-axis

= -∫ x dy .............(6) c

By adding the above results, or putting
g(x,y) = x and f(x,y) = -y , we see from(4) that

= 1/2 ∫(x dy - y dx) = ∫∫ dx dy = area of E
c E

Thus the area of a domain E (with contour C), regular with respect to both the axes

= 1/2 ∫(x dy - y dx) .................(7)
c

Example :-

With the help of Green's formula , compute the difference between the line integrals

I₁ = ∫ {(x+y)² dx - (x-y)² dy}

ACB

I₂ = ∫ {(x+y)² dx - (x-y)² dy}

ADB

where ACB and ADB are respectively the straight line y = x , and the parabolic are y = x² , joining the points A(0,0) and B(1,1) .

Solution :-

Now ,

I₁-I₂ = ∫ {(x+y)² dx - (x-y)² dy}

ADBCA

The line integral is along the closed contour ADBCA which encloses a domain E, bounded by y = x and y= x² from (0,0) to (1,1) and quadratic with respect to y-axis (in fact both axes) . Therefore by Green's Theorem ,

I₁-I₂ = ∫∫{-∂(x-y)²/∂x - ∂(x+y)²/∂y } dx dy

= -4 ∫∫ x dx dy

1 x

= -4 ∫ x dx ∫ dy = 1/3

0 x²

∴ |I₂-I₁| = 1/3

Hence the required solution .

About The Scientists :

George Green (mathematician) ... Green was the first person to create a mathematical theory of electricity and magnetism and his theory formed the foundation for the work of other scientists such as James Clerk Maxwell, William Thomson, and others. His work on potential theory ran parallel to that of Carl Friedrich Gauss. For more .

Cauchy Riemann Equations For Analytic Function

2020-04-19T15:22:00.000+05:30

Cauchy - Riemann Equations :-

Cauchy-Riemann Equation

Necessary Conditions For A Function To Be Analytic :

Cauchy-Riemann Equation

Statement:-

Cauchy-Riemann Equation

The necessary conditions for

w = f(z) = u(x,y) + i v(x,y) to be analytic (differentiable) at any point z = x + i y of its domain D is that the four patial derivatives ∂u/∂x , ∂u/∂y ,∂v/∂x ,∂v/∂y should exists and satisfy the C-R partial differential equations .

∂u/∂x = ∂v/∂y and ∂u/∂y = - ∂v/∂x

Proof :-

Cauchy-Riemann Equation

Let f(z) = u(x,y) + i v(x,y) be analytic at any point z of its domain , then

f'(z) = lim f(z+δz) - f(z) / δz

δx-->0

exists and is unique .

i.e it is independent of the path along which δz -->0

Let z = x+ iy ∴ δz = δx + i δy and as δz-->0

then δx , δy --> 0

∴ f'(z) =

lim [u(x+δx ,y+δy) + iv(x+δx,y+δy)]/δx+iδy

δx-->0

δy-->0

lim [u(x,y) + iv(x,y)] / δx+iδy

δx-->0

δy-->0

lim [u(x+δx , y+δy) - u(x,y)] / δx+iδy

δx-->0

δy-->0

lim i [v(x+δx , y+δy) - v(x,y)] / δx + i δy

δx-->0

δy-->0 .......(1)

Now let us consider two possible approaches in which δz -->0

Case-I:-

Cauchy-Riemann Equation

Let δz be purely real , then δz = δx , δy=0 and δx -->0

Hence from (1) , we get

f'(z) = lim [u(x+δx ,y) - u(x,y)]/δx

δx-->0

lim i[v(x+δx , y) - V(x,y)]/ δx

δx-->0

or f'(z) = ∂u/∂x + i ∂v/∂x ..........(2)

Since f'(z) exists , then above limit exists i.e. ∂u/∂x and ∂v/∂x exists .

Case-II:-

Cauchy-Riemann Equation

Let δz-->0 along the imaginary axis , then δz is purely imaginary and then δx = iδy , δx= 0 and δy-->0 . Hence from(1) we get

f'(z) = lim [u(x, y+iy) - u(x,y)]/iy

δy-->0

lim i [v(x,y+δy) - u(x,y)]/iδy
δy-->0

or, f'(z) = 1/i ∂u/∂y + ∂v/∂y

or, f'(z) = -i ∂u/∂y + ∂v/∂y .........(3)

Since f'(z) exists ,

then above limit exists i.e. ∂u/∂y abd ∂v/∂y exists .

Now , the two limits in (2) and (3) are identical and unique .

∂u/∂x + i ∂v/∂x = -i ∂u/∂y + ∂v/∂y

Equating real and imaginary parts we get

∂u/∂x = ∂v/∂y ........(4)

and ∂u/∂y = -∂v/∂x ........(5)

Equations (4) and (5) are called Cauchy - Riemann Partial Differential Equations .

Polar Form Of Cauchy - Riemann Equations :

Cauchy-Riemann Equation

If w = f(z) = u+iv is an analytic function, show that , in polar form , the C - R equations are :

∂u/∂r = 1/r ∂v/∂θ , ∂v/∂r = -1/r ∂u/∂θ

Proof :-

Cauchy-Riemann Equation

We know that C-R equations are

∂u/∂x = ∂v/∂y and ∂u/∂y = - ∂v/∂x ...(1)

Let x = r cos θ , y= r sin θ

So (x²+y²) = r² ⇒√(x²+y²) = r

Now y/x = tan θ ⇒θ = tan⁻¹(y/x)

Then ∂r/∂x = 1/2 .1/√(x²+y² ) .2x = x/√(x²+y²)

= x/r = r cos θ /r = cos θ

Similarly , ∂r/∂y = y/r = r sin θ /r = sin θ

Again θ = tan⁻¹(y/x)

∂θ/∂x = 1/(1+y²/x²) (-y/x²) = -y/x²+y²

= -r sin θ /r² = -sin θ /r

∂θ/∂y = 1/(1+y²/x²) (1/x) = x/x²+y²

= r cos θ / r² = cos θ /r

∴ ∂u/∂x = ∂u/∂r . ∂r/∂x + ∂u/∂θ . ∂θ/∂x

= cos θ . ∂u/∂r - sin θ / r .∂u/∂θ

∂v/∂y = ∂v/∂r . ∂r/∂y + ∂v/∂θ .∂θ/∂y

= sin θ . ∂v/∂r + cos θ /r . ∂v/∂θ

From (1) , we get

∂u/∂x = ∂v/∂y

⇒cos θ ∂u/∂r - sin θ /r ∂u/∂θ

= sin θ ∂v/∂r + cos θ /r ∂v/∂θ
..........(2)

∂v/∂x = ∂v/∂r . ∂r/∂x + ∂v/∂θ . ∂θ/∂x

= cos θ. ∂v/∂r - sin θ /r .∂v/∂θ

∂u/∂y = ∂u/∂r . ∂r/∂y + ∂u/∂θ . ∂θ/∂y

= sin θ. ∂u/∂r + cos θ / r .∂u/∂θ

Again from (1) , we get

-∂v/∂x = ∂u/∂y

⇒sin θ /r . ∂v/∂θ - cos θ .∂v/∂r

= sin θ .∂u/∂r + cos θ /r .∂u/∂θ
............(3)

Multiplying equation (2) by cos θ and equation (3) by sin θ , we get

cos² θ . ∂u/∂r - sin θ . cos θ /r . ∂u/∂θ

= sin θ .cos θ ∂v/∂r + cos² θ /r . ∂v/∂θ

..............(4)

sin² θ . ∂u/∂r + cos θ sin θ /r . ∂u/∂θ

= -cos θ .sin θ ∂v/∂r + sin² θ /r . ∂v/∂θ

...............(5)

Adding equation(4) with equation (5), we get

(cos² θ + sin² θ).∂u/∂r

= (cos² θ + sin² θ)/r . ∂v/∂θ

⇒∂u/∂r = 1/r . ∂v/∂θ .............(6)

Multiplying equation (2) by sin θ and equation (3) by cos θ , we get

cos θ . sin θ . ∂u/∂r - sin² θ /r . ∂u/∂θ

= sin² θ . ∂v/∂r + sin θ . cos θ /r . ∂v/∂θ

...........(7)

sin θ . cos θ . ∂u/∂r + cos² θ /r . ∂u/∂θ

= -cos² θ . ∂v/∂r + cos θ.sin θ /r . ∂v/∂θ

............(8)

Subtracting we get ,

-(sin² θ + cos² θ)/r . ∂u/∂θ

= (sin² θ + cos² θ).∂v/∂r

⇒-1/r . ∂u/∂θ = ∂v/∂r

⇒∂v/∂r = -1/r . ∂u/∂θ ................(9)

So , equation(6) and equation (8) are called C-R equations in polar form .

Cauchy-Riemann Equation

About Scientists :

Cauchy-Riemann Equation

Baron Augustin-Louis Cauchy (21 August 1789 – 23 May 1857) was a French mathematician, engineer, and physicist who made pioneering contributions to several branches of mathematics, including mathematical analysis and continuum mechanics.For more .

Cauchy-Riemann Equation

Georg Friedrich Bernhard Riemann was a German mathematician who made contributions to analysis, number theory, and differential geometry. In the field of real analysis, he is mostly known for the first rigorous formulation of the integral, the Riemann integral, and his work on Fourier series. For more.

Cauchy-Riemann Equation

Lioville's Theorem In Complex Analysis | Mathquery

2020-04-16T22:47:00.000+05:30

Lioville's Theorem:-

Statement :

If a function f(z) is analytic for all limit values of z and is bounded then f(z) is constant .

Proof :

Let z₁ , z₂ be any two point of the z-plane the contour C to be a large circle of radius R centred at origin and containing the point z₁ , z₂ . Therefore |r₁|<R and |z₂|<R . Also as f(z) is bounded therefore ∃ a positive M such that f|(z)|≤M for all z.

By Cauchy's Integral formula we have

f(z₁) = 1/2πi ∫ f(z) dz /z-z₁

f(z₂) = 1/2πi ∫ f(z) dz /z-z₂

∴ f(z₁) - f(z₂) = 1/2πi ∫ f(z) dz /z-z₁

- 1/2πi ∫ f(z) dz /z-z₂

= 1/2πi ∫ (z₁-z₂) f(z) dz /(z-z₁)(z-z₂)

|f(z₁)-f(z₂)|

= |1/2πi ∫(z₁-z₂)f(z) dz /(z-z₁)(z-z₂)|
c

≤ |1/2πi|∫|z₁-z₂||f(z)||dz|/|z-z₁||z-z₂|
c

≤ 1/2π |z₁-z₂|M ∫ |dz|/(|z|-|z₁|)(|z|-|z₂|)
c

as |f(z)|<M

= 1/2π |z₁-z₂|M ∫dz /(R-|z₁|)(R-|z₂|)
c
as |z|=R

Again |z|=R ⇒z= R exp(iθ)

∴ dz = R exp(iθ). i .dθ

∴ |dz| = R dθ
2π
= 1/2π |z₁-z₂|M .R ∫ dθ
0

/ R² (1-|z₁|/R)(1-|z₂|/R)

= 1/2π |z₁-z₂| M .2π
--->0 as R-->∞
/ R(1-|z₁|/R)(1-|z₂|/R)

∴ f(z₁) - f(z₂) = 0 or f(z₁) = f(z₂)

As z₁ and z₂ are arbitrary it follows that f(z) is constant .

About The Scientist :-

Liouville's Theorem

Joseph Liouville (24 March 1809 – 8 September 1882)was a French mathematician.For more.

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Morera's Theorem

2020-04-09T22:32:00.000+05:30

Morera's Theorem In Complex Analysis :-

Morera's Theorem

Statement :

If f(z) is continuous in a simple connected domain D and if

∫f(z) dz = 0

for every closed path in D, then f(z) is analytic in D .

Morera's Theorem

Proof :

Morera's Theorem

Let z₀ be a fixed point and z a variable point inside the domain D , then the value of the integral

∫ f(z) dz

z₀

is independent of the curve joining z₀ to z and is a function of the upper limit z . Then we have
z
f(z) = ∫ f(t) dt .............(1)
z₀
z+h
then F(z+h) = ∫ f(t) dt
z₀
z+h z
Now F(z+h) - F(z) = ∫ f(t) dt - ∫ f(t) dt
z₀ z₀
z₀ z+h
= ∫ f(t) dt + ∫ f(t) dt
z+h z₀
z+h
= ∫ f(t) dt ........(2)
z

Now the integral in (2) is independent of the path and may be taken along a straight line segment joining z to z+h .

Morera's Theorem

Hence [F(z+h) - F(z)]/ h - f(z)
z+h
= 1/h ∫ f(t) dt - f(z)/h . h
z
z+h z+h
= 1/h [ ∫ f(t) dt - f(z) ∫ dt ]
z z
z+h
= 1/h ∫ [f(t) - f(z)] dt ..........(3)
z

Since f(t) is continuous at z , so given ε>0 ∃ δ>0 such that |f(t) - f(z)|<ε for every t satisfying |t-z|<δ .

Let us choose h such that |h|<δ so that the relation |f(t) - f(z)| <ε is satisfied for every point t on the line segment joining z to z+h .

Morera's Theorem

From equation (3) we get

|F'(z+h) - F(h) / h - f(z) |
z+h
≤ 1/|h| ∫ |f(t) - f(z)||dt|
z
z+h
< 1/|h| . ε ∫ |dt|
z
= 1/|h| ε |h|

= ε

Since ε is arbitrary we get

lim F(z+h) - F(z) / h = f(z)
h--->0

⇒F'(z) exists for all values of z in D and F'(z) = f(z) .

Hence F(z) is analytic in D .

We know that derivative of an analytic function is analytic

⇒ f(z) is also analytic in D . (Proved)

About The Scientist :-

Giacinto Morera (18 July 1856 – 8 February 1909), was an Italian engineer and mathematician. He is known for Morera's theorem in the theory of functions of a complex variable and for his work in the theory of linear elasticity. For More

Fourier Transform Understanding- Mathquery

2020-01-26T08:07:00.000+05:30

Definition Of Fourier Transform:

The Fourier Transform of a function f denoted as f̂ is defined by

∞

f̂ (ξ) = 1/√(2π ) ∫ f(x) e ^ -iξx dx .....(1)

-∞

whenever the integral on the right exists . It is obvious that the integral on the right of (1) exists if
∞
∫ |f(x)| dx exists .
-∞

If the fourier transform f̂ of a function f is known the function f can be obtained by the following formula , known as the inversion formula :
∞
f(x) = 1/√(2π ) ∫ f̂(ξ) e^iξx dξ ..............(2)
-∞

Sometimes it is convenient to use the operator notation F and F⁻¹ for the Fourier transform and its inverse i.e.

F(f) = f̂ and F⁻¹(f̂) = f .

The following important properties of the Fourier transform can be easily verified :

(i) Linearity Of The Transform and its inverse

If f, g are any transformable functions and a and b are real numbers then

F(af + bg) = a F(f) + b F(g)

F⁻¹(af̂ + bĝ) = a F⁻¹(f̂) + b F⁻¹(ĝ) .

(ii) Transform Of nth Derivative

If fⁿ⁻¹(x) , fⁿ⁻²(x),.....f(x) all approach zero as x--> ±∞ , then

F(f)ⁿ(x) = (iξ)ⁿ f̂(ξ)

and F((-ix)ⁿf(x)) = f̂ ⁿ(ξ)

(iii) Shift Formula .

F(f(ax-b)) = 1/|a| e^ -iξb/a f̂(ξ/a)

and F(eⁱᵃˣ f(x)) = f̂(ξ-a)

(iv) Transform Of The Convolution .

Defining the Fourier convolution , f*g of f(x) and g(x) as

∞

f*g(-x₀) = ∫ f(x₀-x) g(x) dx

-∞

we have F(f*g) = f̂ * ĝ

Here * stands for function composition .

If f(x) is even or it is only defined over 0<x<∞ in which case we can extend it so as to be even , then we observe that

∞

∫ f(x) cos ξx dx = 0

-∞

Hence the Fourier transform reduces to

∞

f̂(ξ) = i/√(2π ) ∫ f(x) sin ξx dx

-∞

∞

= -i √(2/π ) ∫ f(x) sin ξx dx

Thus if f(x) is given for 0<x<∞ its Fourier sine transform is defined as

∞

Fₛ(f) = √(2/π ) ∫ f(x) sin ξx dx ,

and the inversion formula is

∞

f(x) = √(2/π ) ∫ sin (ξx) Fₛ(f) dξ

Similarly , the cosine transform is defined as

∞

Fₒ[f] = √(2/π ) ∫ f(x) cos ξx dx

and the inversion formula becomes

∞

f(x) = √(2/π) ∫ cos ξx Fₒ (f) dx

Sine and Cosine transforms are useful in treating problems with boundary condition at x= 0 and involving only derivatives of even order as is seen from the properties :

Fₛ[f''] = f(0) ξ - ξ² Fₒ[f]

Fₒ[f"] = -f'(0) - ξ²Fₒ[f] ,

Provided f(x) and f'(x) -->0 as x-->∞

Thus sine transform is useful when f(0) is given and cosine transform is usual when f'(0) is known .

Motivational Quote

" A Thought Is An Idea In Transit "

By Pythagoras

Fourier Series Of Even And Odd Functions

2020-01-25T00:23:00.000+05:30

Definition :

A function f is said to be an even function if f(x) = f(-x) for all x and it is odd if f(-x) = -f(x) .

Examples :

Sin kx , x,x³ and any power of x are all odd functions where as Cos kx , x ,1,x² and any even power of x are even functions .

The following properties of even and odd functions are easy to check

a a

(i) ∫ f(x) = 0 if f is odd = 2∫ f(x) dx

-a 0

if f is even .

(ii) The product of even and odd function are characterised by

even * odd = odd

even * even = even

odd * odd = even .

Now suppose f is an even periodic function with period 2π . Then since the sine function is odd , f(x) sin nx is odd and f(x) cos nx is even .

So by (i)

bₙ = 1/π ∫ f(x) sin nx = 0 ,

-π

a₀= 2/π ∫ f(x) dx

and

aₙ = 1/π ∫ f(x) cos nx dx , n= 1,2....

-π

= 2/π ∫ f(x) cos nx dx

Similar results hold for odd functions and we obtain the following results :

Theorem :

If f is even on internal -π<x<π then the Fourier series of f is given by

∞

f(x) = a₀/2 + Σ aₙ cos nx

n=1

where a₀ = 2/π ∫ f(x) dx ,

aₙ = 2/π ∫ f(x) cos nx dx , n=1,2,.... and if

f is odd on the interval -π<x<π then the Fourier Series of f is given by

∞

f(x) = Σ bₙ sin nx

n=1

where bₙ = 2/π ∫ f(x) sin nx dx

Fourier Series For A Function With Arbitrary Period :

Let f(x) be a periodic function with arbitrary period 2c . Then we can introduce a new variable z such that f(x) , as a function of z , has period 2π . If we set

x= c z/π so that z = πx/c .....(1)

then x = ±c corresponds to z = ±π . Thus the function f(x) of period 2c in (-c,c) is transformed to the function f(c z/π) of period 2π in (-π,π) . Hence f(c z/π) can be expressed as the Fourier Series

∞

f(x) = f(c z/π) = a₀/2 + Σ (aₙcos nz+bₙsin nz)

n=1 .........(2)

with coefficients
π
a₀ = 1/π ∫ f(cz/π) dz ,
-π

π
aₙ = 1/π ∫ f(cz/π) cos nz dz
-π ..........(3)
π
bₙ = 1/π ∫ f(cz/π) sin nz dz
-π

We could use these formulae directly , but change to the original variable x simplifies the calculation . Since

z = πx/c , we have dz = π/c dx ,

and the interval of integration corresponds to the interval -c≤x≤c and the Euler's formulae become
c
a₀ = 1/c ∫ f(x) dx
-c
c
aₙ = 1/c ∫ f(x) cos nπx/c dx
-c

The Fourier series (2) takes the form
∞
f(x) = a₀/2 + Σ (aₙ cos nπx/c + bₙ sin nπx/c)
n=1
.............(4)

Note :

1 . If f(x) is an even function , then

∞

f(x) = a₀/2 + Σ aₙ cos nπx/c .......(5)

n=1

c
with a₀ = 2/c ∫ f(x) dx ,
0
c
aₙ = 2/c ∫ f(x) cos nπx/c dx ........(6)
0

2. If f(x) is an odd function , then
∞
f(x) = Σ bₙ sin nπx/c ..........(7)
n=1

with
c
bₙ = 2/c ∫ f(x) sin nπx/c dx .........(8)
0

Motivational Quote

"A Mathematical Theory Is Not To Be Considered Complete Until You Have Made It So Clear That You Can Explain It To The First Man Whom You Meet On The Street".

By David Hilbert

Periodic Functions And Fourier Series

2020-01-11T00:38:00.001+05:30

Periodic Functions And Fourier Series :-

Definition :

A function f is said to be periodic with period T if

(i) f(x) is defined for all x and f(x+T) = f(x) for all x for some positive number T . For example , sin x is periodic with period 2π , since sin (x+2π) = sin x . A periodic function has many periods, for if f(x) = f(x+T) then

f(x) = f(x+T) = f(x+2T) = ....=f(x+nT),

where n is any integer. Hence when T is a period of f, nT is also a period of f , but while referring to period we mean the smallest one .

Since each of the functions sin x , cos x , sin 2x , cos 2x , .... are of period 2π , we may think of representing a given function f of period 2π by an infinite series of these function as

∞

f(x) = a₀/2 + Σ (aₙ cos nx + bₙ sin nx) ......(1)

n=1

Now two questions arise :

(i) Supposing the representation (1) is possible , how are the aₙ 's and bₙ 's determined ?

(ii) If the appropriate values are assigned to the coefficients , does the series converge to f(x) ?

To answer the first question we assume that the series (1) can be integrated term by term from x= -π to x= π . Then integrating both sides of (1) from -π to π , we have

π π ∞

∫ f(x) = ∫ [a₀/2 + Σ (aₙ cos nx + bₙ sin nx)]dx

-π -π n=1

= a₀/2 ∫ dx

-π

∞ π π

+ Σ (aₙ∫cos nx dx+bₙ∫sin nx dx)

n=1 -π -π

π π

= πa₀ [∵∫ cos nx dx = 0, ∫ sin nx dx =0]

-π -π

Hence a₀ = 1/π ∫ f(x) dx ...........(2)

-π

To find aₙ , we multiply each side of (1) by cos mx , where m is any fixed positive integer and integrate from -π to π . Then

π π

∫ f(x) cos mx dx = a₀/2 ∫ cos mx dx

-π -π

∞ π

+ Σ [aₙ ∫ cos nx cos mx dx

n=1 -π
π
+bₙ∫ sin nx cos mx dx]

-π

= πaₙ
π
[∵ ∫ cos nx cos mx dx = 0 when m≠n
-π

=π when m=n
π
∫sin nx cos mx dx = 0 for all m,n]
-π
π
Hence aₙ= 1/π ∫ f(x) cos nx dx ..........(3)
-π

Similarly to find bₙ , we multiply (1) by sin mx and integrating term wise from -π to π we get
π π
∫ f(x) sin mx dx = a₀/2 ∫ sin mx dx
-π -π
∞ π
+ Σ [aₙ ∫ cos nx sin mx dx
n=1 -π
π
+ bₙ ∫sin nx sin mx dx ]
-π

= πbₙ
π
[∵∫ cos nx sin mx dx = 0 for all m,n
-π
π
∫sin nx sin mx = 0 when m≠n
-π
π
∫ sin nx sin mx = π when m=n ]
-π
π
Hence bₙ = 1/π ∫ f(x) sin nx dx .......(4)
-π

The equations (2) , (3) and (4) are called Euler's formulae , a₀ ,aₙ and bₙ are called the Fourier Coefficients of f(x) and the series
∞
1/2 a₀ + Σ (aₙ cos nx + bₙ sin nx )
n=1

when aₙ's and bₙ's are determined by (2) , (3) and (4) is called the Fourier Series of f whether or not the series converges .

Motivational Quote

" The truth, he thought, has never been of any real value to any human being - it is a symbol for mathematicians and philosophers to pursue. In human relations kindness and lies are worth a thousand truths."

by Graham Greene

PFAFFIAN Differential Equations And It's Solutions

2020-01-10T00:50:00.002+05:30

PFAFFIAN Differential Equations And It's Solutions :-

The relation

Σ Fᵢ(x₁,x₂,......xₙ) dxᵢ = 0

i=1

where Fᵢ's (i=1,2,....n) are functions of all or some of the n independent variables x₁,x₂,.....xₙ is called a Pfaffian differential equation . Here we shall study Pfaffian differential equation in three independent variables which is of the form

P dx + Q dy + R dz = 0 ...........(1)

where P, Q , R are functions of x , y and z . Setting --> -->

X = (P,Q,R) and dr = (dx,dy,dz), equation (1) can be written in the vector notation as

--> -->

X . dr = 0 ..............(2)

If there exists a function μ(x,y,z) such that μ(P dx + Q dy + R dz) is an exact differential dΦ , the equation is said to be integrable and to possess an integrating factor μ(x,y,z) . In such case(1) becomes

dΦ = 0

which on integration yields

Φ(x,y,z) = c ..................(3)

The function Φ is called the primitive and the relation (3) is called the solution of the differential equation . In general all equations of the form (1) may not possess of equation (1) . This is contained in the following theorem .

Before going to state the theorem first we shall prove two Lemmas .

Lemma 1 :

If two functions u(x,y) and v(x,y) satisfy the relation ∂(u,v) / ∂(x,y) = 0 then there exists between them a relation F(u,v) =0 , not involving x or y explicitly .

Proof :

Eliminating y between the equations

u= u(x,y) , v = v(x,y) ............(4)

we obtain a relation

F(u,v,x) = 0 ....................(5)

Differentiating (6) with respect to x and y respectively we obtain ,

∂F/∂x + ∂F/∂u ∂u/∂x + ∂F/∂v ∂v/∂x = 0

and ∂F/∂u ∂u/∂y + ∂F/∂v ∂v/∂y = 0

Elimination of ∂F/∂v from these equations yields

∂F/∂x ∂v/∂y+

(∂u/∂x ∂v/∂y - ∂u/∂y ∂v/∂x) ∂F/∂u = 0

⇒ ∂F/∂x ∂v/∂y +∂(u,v)/∂(x,y) ∂F/∂u = 0

Using the given condition this equation reduces to

∂F/∂x ∂v/∂y = 0 ......................(6)

Since v is a function of both x and y , ∂v/∂y ≠ 0 . So equation(7) implies

∂F/∂x = 0

which shows that F does not contain x explicitly and hence (6) may be written as

F(u,v ) = 0 (Proved)

Lemma 2 :

-->

If X is a vector and μ is an arbitrary function of x,y,z then

--> -->

X . curl X = 0

--> -->

if and only if (μ X ) . curl (μ X ) = 0 .

Proof :

--> -->

Suppose first that X . curl X = 0 and

-->

left X = (P,Q,R) , then

--> -->

(μ X ) . curl (μ Χ ) = μP{∂(μR)/∂y) - ∂(μQ)/∂z}

+μQ{∂(μP)/∂z - ∂(μR)/∂x}

+ μR{∂(μQ)/∂x - ∂(μp)/∂y}

= μ²{P(∂R/∂y - ∂Q/∂z) + Q(∂P/∂z - ∂R/∂x)

+ R(∂Q/∂x - ∂P/∂y)}

+ {(PR ∂μ/∂y - PQ ∂μ/∂z)

+(QP ∂μ/∂z -QR ∂μ/∂x)+(RQ ∂μ/∂x - RP ∂μ/∂y)}

--> -->

= μ²( X . curl X ) [∵ the term in the second bracket vanishes identically ]

= 0
--> -->
Conversely , suppose (μ X . curl(μ X ) = 0

Then , by the above result ,
--> -->
(1/μ μ X ) . curl(1/μ μ X ) =0
--> -->
X . curl X = 0 (Proved)

The solution of the PFAFFIAN Differential Equations is the proof of the theorem given below .

Theorem :-

A necessary and sufficient condition that the PFAFFIAN Differential Equation
--> -->
X . dr = 0 is integrable is that
--> -->
X . curl X = 0 .................(7)

Proof :

Necessary Part :

Suppose that the differential equation

--> -->

X . dr = 0

i.e. P dx + Q dy + R dz = 0 ...........(1)

is integrable . Then there exists a function μ(x,y,z) such that μ(P dx + Q dy + R dz)

is an exact differential dΦ .

Hence we must have

μP = ∂Φ/∂x , μQ = ∂Φ/∂y , μR= ∂Φ/∂z

Now grad Φ = (∂Φ/∂x , ∂Φ/∂y , ∂Φ/∂z)

-->

= (μP , μQ , μR) = μX .

-->

So curl (μX ) = curl (grad Φ) = 0

--> -->

Hence (μΧ ) . curl (μΧ ) = 0

which implies by lemma 2 , that

--> -->

X . curl X = 0

Sufficient Part :

Let z be treated as a constant . Then the differential equation(1) reduces to

P(x,y,z)dx + Q(x,y,z)dy = 0 .......(8)

i.e. dy/dx = -P(x,y,z)/Q(x,y,z)

if Q(x,y,z) ≠ 0.

So , if P and Q are continuous functions of x and y , then equation (8) has a solution of the form

Φ(x,y,z) = c₁ ............(9)

where the constant c₁ may involve z . In differential form equation (9) is

∂Φ/∂x dx + ∂Φ/∂y dy = 0

which should be equivalent to (8) so that

∂Φ/∂x/P = ∂Φ/∂y/Q = μ(x,y,z) (say)

⇒ ∂Φ/∂x = μP , ∂Φ/∂y = μQ ................(10)

Multiplying equation(1) by μ and using equation (10) it can be written as

∂Φ/∂x dx + ∂Φ/∂y dy + ∂Φ/∂z dz +

(μR - ∂Φ/∂z) dz = 0

which is equivalent to

dΦ + Ψ dz = 0 ..........(11)

where Ψ = μR - ∂Φ/∂z ............(12)

--> -->

The given condition X . curl X = 0 implies , by lemma 2 ,

--> -->

(μΧ ) . curl (μΧ ) = 0

But as

-->

μΧ = (μP , μQ , μR)

= (∂Φ/∂x , ∂Φ/∂y , ∂Φ/∂z + Ψ) by (10) and (12)

= (∂Φ/∂x , ∂Φ/∂y ,∂Φ/∂z) + (0,0,Ψ)

= grad Φ + (0,0,Ψ) .
-->
curl (μΧ ) = curl (grad Φ) + curl(0,0,Ψ)

= 0 + (∂Ψ/∂y , -∂Ψ/∂x ,0)
--> -->
Hence 0= (μΧ ) . curl (μΧ )

= (∂Φ/∂x , ∂Φ/∂y , ∂Φ/∂z + Ψ) (∂Ψ/∂y , -∂Ψ/∂x , 0)

= ∂Φ/∂x ∂Ψ/∂y - ∂Φ/∂y ∂Ψ/∂x

= ∂(Φ,Ψ) / ∂(x,y)

It follows from lemma 1 that there exists a relation between Φ and Ψ not involving x and y explicitly but it may involve z . So Ψ may be written as a function Ψ(Φ,z) of Φ and z . The equation (11) assumes the form

dΦ/dz = - Ψ(Φ,z)

which possess a solution of the form

f(Φ,z) = c .............(13)

Substituting for Φ its expression in terms of x,y,z the solution (13) of (11) which is equivalent to (1) is of the form

F(x,y,z) = c

This shows that equation (1) is integrable .

(Proved)

Motivational Quote

"Mathematics Is The Art Of Giving The Same Name To Different Things "

by Henri Poincare

Green's Functions

2020-01-08T17:56:00.000+05:30

Green's Functions :-

The objective of this section to construct Green's (1793 - 1841) function and use it to solve the boundary value problem

L(y) + f(x) = 0, a≤x≤b .........(1)

a₁ y(a) + a₂ y'(a) = 0 , ................(2a)

b₁ y(b) + b₂ y'(b) =0, ..................(2b)

Where L is the differential operator defined by L(y) = (py')' + qy.p , p' and q are given functions continuous on [a,b] with p(x) ≠0 on [a,b] and at least one of a₁ and a₂ and one of b₁ and b₂ are non zero .

Definition :

The Green's function for L(y) =0 under given homogeneous boundary conditions(2) is the function G(x,ξ) satisfying the following conditions .

(i) G(x,ξ) is continuous for all values of ox but its first and second derivatives are continuous for all

x≠ ξ ; a≤x≤b , a<ξ<b ,

(ii) At x= ξ the first derivative of G(x,ξ) has a jump discontinuity given by

d G(x,ξ)/dx | x= ξ⁺ = G'(ξ⁺,ξ)-G'(ξ⁻,ξ) = -1/p(ξ)
x=ξ⁻ ......(3)

(iii) For fixed ξ , G(x,ξ) satisfies the given boundary conditions and is the solution of the associated homogeneous equation L(y) = 0 except at x= ξ .

i.e a₁ G(a,ξ) + a₂ G'(a,ξ) = 0 .............(4a)

b₁ G(b,ξ) + b₂ G'(b,ξ) = 0 ............(4b)

and L(G(x,ξ)) = 0 for all x except x= ξ ......(5)

Note :

In equation(3) G'(ξ⁺,ξ) =

lim G'(ξ+ε,ξ) and G'(ξ⁻,ξ)

ε-->0

= lim G'(ξ-ε, ξ)

ε-->0

Theorem _1 :-

If the equation L(y) = 0 satisfying the boundary condition (2a) and (2b) has only the trivial solution , then the Green's function exists and is unique .

Proof :

Let φ₁(x) be a non- trivial solution of L(y)=0 satisfying the condition (2a).

Also let φ₂(x) be a nontrivial solution of L(y)=0 satisfying the condition (2b). Then φ₁ and φ₂ are linearly independent , for if they are not linearly independent , then φ₁ = C φ₂ and φ₁ would satisfy both of the conditions (2a) and (2b) implying that φ₁ is non trivial solution of L(y) = 0 with (2a) and (2b) which contradicts the hypothesis .

Since G(x,ξ) satisfies (condition(iii) of the definition) the same conditions as φ₁(x) and φ₂(x) in the intervals [a,ξ) and (ξ,b] respectively , it follows that

{
C₁ φ₁(x) for x<ξ
G(x,ξ) = {
C₂ φ₂(x) for x>ξ ...........(5)
{

for some choice of C₁ and C₂ .

Since G(x,ξ) is continuous at x= ξ (condition (i) of the definition), we have from equation(5) ,

C₂ φ₂(ξ) - C₁ φ₁(ξ) = 0 ...............(6)

The discontinuity in the derivative of G at x= ξ , (condition(ii) of the definition) yields

C₂ φ₂'(ξ) - C₁ φ₁'(ξ) = -1/p(ξ) ..............(7)

Solving equations (6) and (7) for C₁ and C₂ we obtain ,

C₁ = -φ₂(ξ)/[p(ξ) W(φ₁,φ₂;ξ)]

C₂ = -φ₁(ξ)/[p(ξ) W(φ₁,φ₂;ξ)] ..........(8)

where W(φ₁,φ₂;ξ) is the Wronskian of φ₁ and φ₂ at ξ

i.e. W(φ₁,φ₂;ξ) = φ₁(ξ) φ₂'(ξ) - φ₂(ξ)φ₁'(ξ) .

Since φ₁ and φ₂ are linearly independent, the Wronskian W (φ₁,φ₂;ξ) ≠0 which implies in turn that p(ξ) W(φ₁,φ₂;ξ)≠0 so that C₁ and C₂ are well defined in equation(8) .

Since φ₁ and φ₂ are solutions of L(y)=0 we have

d/dx (p φ₁) + q φ₁ = 0 ,

and d/dx (p φ₂') + q φ₂= 0 ,

Multiplying the first equation by φ₂ and the second by φ₁ and subtracting , we obtain

φ₁ d/dx (p φ₂') - φ₂ d/dx (p φ₁') = 0,

which can be written in the form

d/dx [p(φ₁ φ₂' - φ₂ φ₁')] = 0

Integration of this equation yields

p(φ₁ φ₂' - φ₂ φ₁') = constant = A .......(9)

Hence from(5) , (8) and (9) the Green's function is uniquely determined as

{ - φ₁(x) φ₂(ξ)/A for x≤ ξ
G(x,ξ) = {
{-φ₂(x) φ₁(ξ)/A for x≥ξ ......(10)

Using the definition of the Green's function we can obtain a solution of the boundary value problem (1) and (2) .

Orthogonality Of Eigen Functions

2020-01-02T12:12:00.000+05:30

Orthogonality Of Eigen Functions:-

Definition :

Two distinct continuous functions f and φ on [a,b] are said to be orthogonal with respect to a continuous weight function γ if

∫ f(x) φ(x) γ(x) dx = 0 ...(14)

An infinite set of functions defined on [a,b] is to be an orthogonal system with respect to the weight function γ on [a,b] if every pair of distinct functions of the set are orthogonal with respect to γ .

Example :

The set of functions {φₙ} , where φₙ(x) = sin nx , n=1,2,.... on [0,π] is an orthogonal system with respect to the weight function having the constant value 1 on [0,π] , for

∫ (sin mx)(sin nx)(1) dx

=[sin(m-n)x / 2(m-n) - sin(m+n)x / 2(m+n)]

= 0 for m≠n

The infinite set of eigen functions of the Sturm - Liouville Problem (1) and (2) form an orthogonal system with respect to weight function γ , as is evident from the following theorem .

Theorem :-

If φₘ and φₙ are eigen functions of the Sturm - Liouville Problem(1) and (2) corresponding to the distinct eigen values λₘ and λₙ respectively , then they are orthogonal with respect to the weight function γ .

Proof :

Since φₘ is an eigen function corresponding to λₘ . It satisfies equation (1) with λ = λₘ . Hence we have,

d/dx [p(x)φ'ₘ(x)] + [q(x)+λₘ γ(x)] φₘ(x) = 0

Similarly for φₙ we get

d/dx [p(x)φₙ'(x)] + [q(x) + λₙ γ(x)]φₙ(x) =0

Multiply equation(15) by φₙ(x) and equation(16) by φₘ(x) and subtracting we obtain

(λₘ-λₙ) γ(x) φₘ(x)φₙ(x) =

d/dx [p(x)φₙ'(x)]φₘ(x)

- d/dx [p(x)φₘ'(x)]φₙ(x),

that is

(λₘ- λₙ) γ(x) φₘ(x) φₙ(x) =

d/dx [p(x){φ'ₙ(x) φₘ(x) - φₘ'(x) φₙ(x)}];

which on integration yields

( λₘ- λₙ) ∫ γ(x) φₘ(x) φₙ(x) dx

= p(b){φ'ₙ(b) φₘ(b) - φ'ₘ(b) φₙ(b)}

- p(a){φₙ'(a) φₘ(a) - φ'ₘ(a) φₙ(a)}

The orthogonality of φₙ and φₘ is established if we show that the right side of equation(17 ) vanishes .

From the first equation of (2) we get

a₁ φₘ(a) + a₂ φₘ'(a) =0 and a₁φₙ(a) + a₂φₙ'(a)

= 0

Elimination of a₂ from these two equations leads to

a₁[φₙ(a) φₘ'(a) - φₘ(a) φₙ'(a)] = 0

and if a₁ ≠ 0 then

φₙ(a) φ'ₘ(a) - φₘ(a) φ'ₙ(a)=0 ..........(18)

Similarly if b₁≠0 then from 2nd equation of (2) we get

φₙ(b) φ'ₘ(b) - φₘ(b) φₙ'(b) = 0 ...............(19)

By (18) and (19) the right side of (17) vanishes, establishing that φₘ and φₙ are orthogonal .

If either a₁= 0 or b₁= 0 or a₁= b₁= 0 it can be checked that the right side of(17) vanishes .

Corollary (Eigen function Expansion) :-

Let g(x) be a piece wise continuous function defined on [a,b] satisfying the boundary conditions (2) . Let φ₁,φ₂,......φₙ be the set of eigen functions of the Sturm - Liouville Problems (1) and (2) . Then

g(x) = c₁φ₁(x) + c₂φ₂(x) + ....+cₙφₙ(x)+... ...(20)

Where cₙ's are given by

b b

cₙ ∫ γ(x) φₙ²(x) dx = ∫γ(x) g(x) φₙ(x) dx

a a

n= 1,2,3.... ........(21)

Proof :

By the above theorem φ₁,φ₂,...φₙ form an orthogonal system with respect to γ on [a,b] . Hence

∫ γ(x) φₙ(x) φₘ(x) dx = 0 for n≠m .....(22)

From (20) we have

∫ g(x) γ(x) φₙ(x) dx =

b ∞

∫( Σ cₖ φₖ(x) ) γ(x) φₙ(x) dx

a k=1

∞ b

= Σ cₖ (∫ γ(x) φₖ(x) φₙ(x) dx )

k=1 a

= cₙ ∫ γ(x) φₙ²(x) dx , by (22)

Here we have assumed the uniform convergence of the series in the right side of (20) .

Sturm - Liouville Problem

2019-12-16T21:40:00.000+05:30

Sturm - Liouville Problem:

The Sturm - Liouville Problem is a special class of linear homogeneous boundary value problem given in the following definition.

Definition :

Consider a second order homogeneous linear differential equation of the form

d/dx [p(x) dy/dx ] + [q(x) + λ γ(x)]y=0, a≤x≤b

...............(1)

where p, q and γ are real - valued continuous functions on [a,b] and λ is real parameter . Further consider the following two sets of prescribed conditions

a1 y(a) + a2 y'(a) = 0}

}...................(2)

b1 y(b) + b2 y'(b) = 0}

y(a) = y(b) , y'(a) = y'(b) , p(a) = p(b) .........(3)

where the real constants a1 and a2 are not both zero and b1 and b2 are not both zero .

A boundary value problem consisting of equation(1) with either the boundary condition (2) or the boundary condition (3) is called Sturm - Liouville Problem (or Sturm - Liouville System ).

Example 1 :

The boundary - value problem

d²y/dx² + λy = 0, ..............(4)

y(0) = 0 , y'(π) = 0 ......................(5)

is a Sturm - Liouville Problem , since , equation (4) may be written as

d/dx [1. dy/dx ] + [0 + λ.1] y = 0

which is of the form (1) with p(x) =1 , q(x) =0 and γ(x) =1 and the boundary condition (5) is the form (2) with a1 = 1 , a2 = 0 , b1= 0 , b2= 1.

It is easy to see that the identically vanishing function φ i.e. φ(x) = 0 for all x in [a,b] is always a solution of a Sturm - Liouville problem . But this trivial solution is not very useful . So we seek for nontrivial solutions of the problems . That is , we seek for functions , identically zero , which satisfy both the differential equation(1) and the boundary conditions (2) . In the next example we shall see that the existence of nontrivial solutions depends upon the value of the parameter λ .

Example 2 :

Find nontrivial solutions of the Sturm - Liouville problem

d²y/dx² + λy = 0 , ..........(4)

y(0) =0, y'(π) =0 ...............(5)

Solution :

We shall discuss the problem under three separate cases : λ=0 , λ<0 and λ>0.

Case (i) :

λ=0 . In this case the differential equation(4) becomes

d²y/dx² = 0,

and hence the general solution is

y(x) = c1 x + c2 ...............(6)

Using y(0) = 0 in (6) we obtain c2= 0. The condition y'(π) = 0 yields (since v'(x) = c1)c1 =0 . Thus the solution of the given problem is y(x) =0 for all values of x . So when λ=0, the only solution of the given Sturm - Liouville problem is the trivial solution .

Case (ii) :

λ<0 . In this case the auxiliary equation of (4) is m² + λ =0 which has roots ±√(-λ). Since λ<0 , -λ is positive so that √(-λ) is real . Hence the general solution is

y(x) = c1 e^√(-λ)x + c2 e^-√(-λ )x ........(7)

Application of y(0) = 0 in (7) yields

c1 + c2 = 0 .................(8)

Since y'(x) = √(-λ)[c1 e^√(-λ)x - c2 e^-√(-λ )x]

the condition y'(π) = 0 gives

√(-λ)[c1 e^√(-λ)π - c2 e^-√(-λ)π]=0 .........(9)

Since √(-λ), e^√(-λ)π and e^-√(-λ) are all positive (8) and (9) yield c1=c2=0 . Thus when λ<0 , the only solution of the given problem is trivial solution y(x) =0 for all values of x .

Case (iii) :

λ>0 . Since λ>0 the roots ±√(-λ) of the auxiliary equation are conjugate complex numbers ±√(λ)i and the general solution is of the form

y(x) = c1 sin√(λ)x + c2 cos√(λ )x ...........(10)

Using the condition y(0) in (10) we obtain

0 = c1 sin 0 + c2 cos 0

and hence c2 = 0 . Since

y'(x) = √(λ) [c1 cos √(λ)x - c2 sin √(λ)x]

from the condition y'(π) =0 we obtain

0 = √(λ)[c1 cos √(λ)π - c2 sin √(λ)π]

Which reduces to c1 cos√(λ) π = 0, .........(11)

since c2 =0 and √(λ) ≠ 0 .

If c1 = 0 in (11) then we obtain from (10) the unwanted trivial solution y(x) . So for a nontrivial solution of the problem we must take

cos √(λ)π = 0

which gives √(λ)π = (2n - 1)π/2

i.e. λ = (2n - 1)²/4 , where n=1,2,3...

Thus for a nontrivial solution the parameter λ must have one of the values of the sequence

{(2n-1)²/4} , n=1,2,3.........

and the nontrivial solution , as obtained from (10) , corresponding to

λ = (2n-1)²/4 are

y= ci sin (2n-1)/2 x, n=1,2,3... .........(13)

where ci' s are nonzero constants.

The values of λ in (12) are called eigen values and the corresponding solution in (13) are called eigen functions . Thus the following definition is given

Definition :

The values of the parameter λ in (1) for which there exist nontrivial solutions of the Sturm - Liouville problem (1) and (2) are called the eigen values or characteristic values of the problem and the corresponding nontrivial solutions are called eigen functions or characteristic functions of the problem .

It is observed from the above example that the eigen values are uniquely determined by the problem , but a one - parameter family of eigen functions since ci is arbitrary for each n correspond to each eigen value and any two eigen functions corresponding to definite eigen value are non zero constant multiples of each other . Further it is seen that the eigen values form an increasing sequence of positive numbers that approach ∞ and the nth eigen function sin (2n-1)/2 x has exactly n-1 zeroes inside the interval [0,π] . [The zeros of sin (2n-1 / 2)x are given by (2n-1 / 2) x =K π i.e. x= K π/(n-1/2) , K is any integer . So the zeros which lie in 0<x<π are those for which K = 1,2,.....n-1]

Let us check the above observation is true or not for Sturm - Liouville problem using the following theorem .

Theorem :

If p(x)>0 and γ(x)>0 for all x on [a,b] the Sturm - Liouville problem consisting of the equation(1) and the boundary condition (2) has an infinite number of eigen values λn which can be arranged in a monotonic increasing sequence λ1<λ2<..... with no finite limit . Further corresponding to each eigen value there exists a one- parameter family of eigen functions such that any two eigen functions corresponding to the same eigen value λn has exactly (n-1) zeros in the open interval a<x<b.

Laplace Transform Of Periodic Functions

2019-11-19T08:08:00.002+05:30

Laplace Transform Of Periodic Functions :

Definition :

A function f(t) is said to be periodic , with a period 'l' , if it satisfies the functional equation

f(t±l) = f(t)

So, if t>0 , a periodic function f(t) can be written as

f(t) = f(t + nl) , n= 0,1,2.. ......(1)

For example sin t = sin(t+2πn) , n=0,1,2.... is a periodic function with period 2π .

In case a function f(t) is periodic , the Laplace transform can be expressed as an integral over one cycle of the function instead of an integral over an infinite range.

The transform of equation(1) is

∞

f̅(p) = ∫ e⁻ᵖᵗ f(t) dt

l 2l 3l

= [ ∫ + ∫ + ∫ .....]e⁻ᵖᵗ f(t) dt

0 l 2l

∞ (n+1)l

= Σ ∫ e⁻ᵖᵗ f(t) dt ............(2)

n=0 nl

Now let us substitute

t = τ + nl ,

Where τ is a new variable , so that

f(t) = f(τ+nl) = f(τ) .

We have dt = dτ. In the integration , when t= nl , we have τ = 0 and when t= (n+1)l , τ= l.

Hence from (2) we get

∞ l

f̅(p) = Σ ∫ f(τ+nl) e^ -p(τ+nl) dτ

n=0 0

∞ l

= Σ e⁻ᵖⁿᵗ ∫ f(τ) e^-pτ dτ .......(3)

n=0 0

Now

∞

Σ e⁻ᵖⁿᵗ = 1+e⁻ᵖᵗ + e⁻²ᵖᵗ+ e⁻³ᵖᵗ + ...

n=0

= 1/ 1-e⁻ᵖᵗ , since e⁻ᵖᵗ < 1 .......(4)

From the relations (3) and (4) , we have

f̅(p) = 1/ 1-e⁻ᵖᵗ ∫ e⁻ᵖᵗ f(t) dt .

We have thus established the theorem i.e.

L{f(t)} = 1/ 1-e⁻ᵖᵗ ∫ e⁻ᵖᵗ f(t) dt .

Which is known as Laplace Transform Of Periodic Function Theorem .

If you have any doubt in any mathematical problems then you can comment me and I will try my level best to solve your problem.

Unit Impulse Function

2019-10-21T08:07:00.000+05:30

Unit Impulse Function

The above figure shows a function which has a zero value when t is negative , rises instantaneously to a value 1/t₀ when 0<t<t₀ , and is zero thereafter.

If we let t₀--> 0 , this function tends towards infinity as t₀--> 0, but width decreases inversely with the magnitude ; hence area under the curve remains finite .

The unit impulse function is represented by δ(t) .

Among Physicists, the unit impulse function is referred to as the 'Dirac δ-function ' after the name of great Physicist Dirac , who is the first to use this function in systematic manner . The Dirac δ - function at the point t=a as in above figure is represented by δ(t-a) . Thus

{0 for t < a

δ(t-a) ={ 1/t₀ for a<t<a+t₀

{0 for t> a+t₀ .......(1)

Where t₀ --> 0 in the limit .

The Laplace transform of δ(t-a) is

∞

L{δ(t-a)} = ∫ e⁻ᵖᵗ δ(t-a) dt

a a+t₀

= ∫ e⁻ᵖᵗ δ(t-a) dt + lim ∫ e⁻ᵖᵗ δ(t-a) dt

0 t₀-->0 a

∞

+ ∫ e⁻ᵖᵗ δ(t-a) dt

a+t₀

= lim 1/t₀ ∫ e⁻ᵖᵗ dt

a+t₀

= lim 1/t₀ [-1/p e⁻ᵖᵗ ]

= e⁻ᵖᵃ/p lim 1-e⁻ᵖᵗ / t₀ = e⁻ᵖᵃ .....(2)

t₀-->0

If you have any doubt in any mathematical problems then you can comment me and I will try my level best to solve your problem.

Unit Step Function

2019-10-19T13:45:00.000+05:30

Unit Step Function :

A unit step function is a function which has a zero value for t≤t₀ and then rises instantaneously to a sustained unity value.

This is shown in the figure below. In electric circuit , and in mechanical system it could represent a force suddenly impressed on the system .

This unit step function starting at zero time will be designated by u(t) , and that starting at time t₀ is u(t-t₀) . Thus

{ 0 for 0<t≤t₀

u(t-t₀) ={

{ 1 for t>t₀ .....(1)

The Laplace transform of unit step function is

∞

L{u(t-t₀)} = ∫ e⁻ᵖᵗ u(t-t₀) dt

∞ ∞

= ∫ e⁻ᵖᵗ u(t-t₀) dt + ∫ e⁻ᵖᵗ u(t-t₀) dt

0 t₀

∞ ∞

= ∫ e⁻ᵖᵗ dt = [-1/p e⁻ᵖᵗ]

t₀ t₀

= 1/p e⁻ᵖᵗ₀ ......(2)

When t₀= 0 i.e., the function instantaneously takes the value unity at zero time , then

L{u(t)} = 1/p

Which is the function.

Example _1:

Express the square wave shown in the figure below in terms of unit step function , and obtain its Laplace transform .

Solution :

The wave can be written in terms of unit step functions as

f(t) = u(t)- 2u(t-a) + 2u(t-2a) - 2u(t-3a)+...

Assuming that term wise integration is permissible and using (2) , we obtain

f̅(p) = 1/p - 2e⁻ᵃᵖ/p + 2e⁻²ᵃᵖ/p - 2e⁻³ᵃᵖ/p +...

= 1/p [1-2e⁻ᵃᵖ(1-e⁻ᵃᵖ + e⁻²ᵃᵖ - e⁻³ᵃᵖ+...)]

= 1/p [1 - 2e⁻ᵃᵖ/1+e⁻ᵃᵖ]

the expression in parentheses () is a geometric series .

Hence f̅(p) = 1/p (1-e⁻ᵃᵖ / 1+e⁻ᵃᵖ)

= 1/p[ e^-ap/2 (e^ap/2 - e^-ap/2)/

e^-ap/2 (e^ap/2 + e^-ap/2)

= 1/p (e^ap/2 - e^-ap/2) /

(e^ap/2 + e^-ap/2)

= 1/p (2sinh ap/2) / (2cosh ap/2)

= 1/p tanh ap/2 . (a>0 , p>0),

Theorem _1:

If {f(t)} = f̅(p), then for a>0,

L{u(t-a)f(t-a)} = e⁻ᵃᵖ f̅(p) .

Proof :

By definition

∞

L{u(t-a) f(t-a)} = ∫ e⁻ᵖᵗ u(t-a) f(t-a) dt

a ∞

= ∫e⁻ᵖᵗ f(t-a).0dt + ∫ e⁻ᵖᵗ f(t-a).1dt

0 a

∞

= ∫ e⁻ᵖᵗ f(t-a) dt

In the last integral write x = t-a

Then

∞ ∞

∫ e⁻ᵖᵗ f(t-a) dt = ∫ e^-p(x+a) f(x) dx

a 0

∞

= e⁻ᵖᵃ ∫ e⁻ᵖˣ f(x) dx

= e⁻ᵖᵃ f̅(p)

Hence, L{u(t-a) f(t-a)} = e⁻ᵖᵃL{f(t)}=e⁻ᵃᵖ f̅(p)

Example_2 :

What function has Laplace transform

e⁻ᵖ/p² ?

Solution :

We know that L{t} = 1/p²

Hence by Theorem_1

e⁻ᵖ/p² = L{u(t-1)(t-1)} .

Which is the required solution ,

The Shifting Theorem

2019-10-13T22:01:00.000+05:30

The Shifting Theorem :

If L{f(t)}= f̅(p), then

L{eᵃᵗf(t)} = f̅(p-a) , p>a

Proof :

By definition

∞

f̅(p) = ∫ e⁻ᵖᵗ f(t) dt

∞

Therefore, f̅(p-a) = ∫ e^-(p-a)t f(t) dt

∞

= ∫ e⁻ᵖᵗ[eᵃᵗ f(t)] dt

= L{eᵃᵗ f(t)}.

Corollary :

L{e⁻ᵃᵗ f(t)} = f̅(p+a); (p>-a).

This follows immediately from the above theorem by writing -a for a.

Example _1:

Find the transform of eᵃᵗ tⁿ .

Solution :

Since L{tⁿ} = n!/pⁿ⁺¹

by using the above theorem , we find

L{eᵃᵗ tⁿ} = n!/(p-a)ⁿ⁺¹

which is the required solution .

Example_2 :

Find the Laplace Transformation of

f(t) = . e³ᵗ

Solution :

To find L{t^7/2 . e³ᵗ} first we have to find out the value of L{t^7/2}

Now , As we know that

L{tⁿ} = n!/pⁿ⁺¹

So L{t^7/2} = (7/2)!/p^[(7/2)+1]

⇒L{t^7/2} = (7/2)Γ(7/2) / p^(9/2)

= (7/2)Γ(5/2 + 1) / p^(9/2)

= (7/2)(5/2)Γ(5/2) / p^(9/2)

[∵Γ(n+1) = nΓ(n)]

= (7/2)(5/2)Γ(3/2 + 1) / p^(9/2)

= (7/2)(5/2)(3/2)Γ(3/2) / p^(9/2)

= (7/2)(5/2)(3/2)Γ(1/2 + 1)/ p^(9/2)

= (7/2)(5/2)(3/2)(1/2)Γ(1/2) / p^(9/2)

= (7/2)(5/2)(3/2)(1/2)√(π ) / p^(9/2)

[∵Γ(1/2) = √(π )]

⇒L{t^7/2} = 105√(π )/16 p^(9/2)= F(p)

∴ L{t^7/2 . e³ᵗ} = 105√(π )/16(p-a)^(9/2)

Which is the required solution .

Existence And Property Of Laplace Transform

2019-09-08T08:38:00.000+05:30

Existence And Property Of Laplace transforms :

Existence Of Laplace Transforms:

In order to discuss the theorem on existence of transform we need the following definition .

Definition :

A function f(t) is said to be piecewise continuous in a finite range a≤t≤b , if it is possible to divide the range into a finite number of sub-intervals such that f(t) is continuous inside each sub-interval and approaches finite values as t approaches either end of any interval from the interior.

Theorem :

Let f(t) be a function which is piecewise continuous on every finite interval in the range t≥0 and satisfies

|f(t)|≤Me^αt , for all t≥0 .........(1)
and for some constants α and M . Then the Laplace transform of f(t) exists for all p>α .

Proof :

Since f(t) is piecewise continuous , e⁻ᵖᵗf(t) is integrable over any finite interval on the t-axis ,

i.e ∫ e⁻ᵖᵗf(t) dt exists for all T.

Hence for all T ,

T T

∫ |e⁻ᵖᵗf(t)|dt ≤M∫ e⁻ᵖᵗ e^αt dt

0 0

= M[e^(α-p)Τ -1]/(α-p)

≤ M/(p-α) , for p>α

Thus lim ∫ |e⁻ᵖᵗ f(t)| dt ≤ M/(p-α)

T-->∞ 0
∞
for p>α which implies ∫ |e⁻ᵖᵗf(t) | dt
∞ 0
exists and hence ∫ e⁻ᵖᵗ f(t) dt exists for p>α,
0

that is L{f(t)} exists for all p>α .

From now on we shall assume that the functions under discussion satisfy conditions in above theorem so that the Laplace transform exists on a suitable interval .

Linearity Property Of Laplace Transform :

Theorem :

If f(t) = Σ Cᵢ fᵢ(t) ,

i=1

where Cᵢ are constants then

L{f(t)} = Σ Cᵢ L{fᵢ(t)} .

i=1

Proof :

By definition , we have

∞ n

L{f(t)} = ∫ e⁻ᵖᵗ Σ Cᵢ fᵢ(t) dt

0 i=1

∞

= ∫ e⁻ᵖᵗ[C₁f₁(t)+ C₂f₂(t)+....+Cₙfₙ(t)]dt

∞ ∞

= C₁∫ e⁻ᵖᵗ f₁(t) dt + C₂∫ e⁻ᵖᵗ f₂(t)dt

0 0

∞

+ ..........+Cₙ∫ e⁻ᵖᵗ fₙ(t) dt

= C₁L{f₁(t)} + C₂L{f₂(t)}+....+CₙL{fₙ(t)}

= Σ Cᵢ L{fᵢ(t)}

i=1

Example :

Find the transform of function

f(t) = 3 sin4t - 2 cos5t .

Solution :

Applying the above theorem , we have

L{f(t)| = 3L{sin 4t} - 2L {cos 5t}

= 3 . 4/p²+16 - 2.p/p²+25
by Laplace function (v) .

= 12/p²+16 - 2p/p²+25 .

Which is the required transformation .

The Laplace Transformation

2019-08-27T01:31:00.000+05:30

The Laplace Transformation:

In recent years , in the solution of differential equations much use has been made of what are known as "Operational Methods". Such methods have wide applications in science and engineering and represent a large field for advanced study .

This method consists of a procedure of solving differential equations where the boundary or initial conditions are automatically satisfied in the course of the solution . The Laplace transformation is but one of many possible operational methods of solving linear differential equations .

Definition:

Given a function f(t) of a real variable t>0 , if we multiply it by e⁻ᵖᵗ and with respect to t between the limits 0 and ∞, the result is a function of p, say f̅(p). This function f̅(p) is called the Laplace Transformation of f(t) , also written as L{f(t)}. Thus

∞

L{f(t)} ≡ f̅(p) = ∫ e⁻ᵖᵗf(t) dt ,

provided the integral exist. Here p, in general , is a complex variable. However, in special cases it may be either real or imaginary . We shall restrict p to assume only real values .

Laplace Transforms Of Some Functions:

(i) If f(t) = 1,

from definition we get

∞ ∞

f̅(p) = ∫ e⁻ᵖᵗ .1 dt = [-1/p e⁻ᵖᵗ] = 1/p

0 0

Hence L{1} = 1/p .

∞

(ii) L{t} = ∫ e⁻ᵖᵗ t dt

∞ ∞

= [-1/p e⁻ᵖᵗ t] + 1/p ∫ e⁻ᵖᵗ dt

0 0

= 1/p² .

∞

since , lim te⁻ᵖᵗ = 0 and ∫ e⁻ᵖᵗ dt = 1/p .

t-->∞ 0

∞

(iii) L{tⁿ} = ∫ e⁻ᵖᵗ tⁿ dt ,

where n is a positive integer

∞ ∞

= [-1/p e⁻ᵖᵗ tⁿ] + n/p ∫ e⁻ᵖᵗ tⁿ⁻¹ dt

0 0

= n/p L{tⁿ⁻¹}.

The successive application of the formula (iii) gives

L{tⁿ} = n!/pⁿ⁺¹ .

∞

(iv) L{eᵃᵗ}= ∫ e⁻ᵖᵗ eᵃᵗ dt

∞ ∞

= ∫ e⁻ᵖᵗ⁺ᵃᵗ dt = [-e⁻ᵖᵗ⁺ᵃᵗ/p-a]

0 0

= 1/p-a

∞

(v) L{cos at } = ∫ e⁻ᵖᵗ cos at dt

∞

= 1/2 ∫ e⁻ᵖᵗ[eⁱᵃᵗ+e⁻ⁱᵃᵗ]dt

∞ ∞

= 1/2 ∫ e⁻ᵖᵗ⁺ⁱᵃᵗ dt +1/2 ∫ e⁻ᵖᵗ⁻ⁱᵃᵗdt

0 0

= 1/2(p-ia) + 1/2(p+ia) = p/p²+a²

Similarly,

L{sin at } = p/p²+a² .

Integral Representation Of Confluent Hypergeometric Function

2019-08-25T08:21:00.000+05:30

Integral Representation Of Confluent Hypergeometric Function:

Theorem :

If γ>α>0 , then the function F(α;γ;x) can be expressed as

Γ(γ)/Γ(γ)Γ(γ-α) ∫ eˣᵗ t^(α-1)(1-t)^(γ-α-1) dt.

Proof :

We know that

B(α+n,γ-α)/B(α,γ-α) =

Γ(α+n)Γ(γ-α)/Γ(γ+n) / Γ(α)Γ(γ-α)/Γ(γ)

= Γ(α+n)/Γ(α) / Γ(γ+n)/Γ(γ)

But Γ(α+n)/Γ(α) = (α)ₙ and Γ(γ+n)/Γ(γ) = (γ)ₙ

Therefore,

(α)ₙ/(γ)ₙ = B(α+n,γ-α)/Β(α,γ-α)

= Γ(γ)/Γ(α)Γ(γ-α) ∫ t^(α+n-1) (1-t)^(γ-α-1) dt

∞

Now F(α;γ;x) = Σ (α)ₙxⁿ/n!(γ)ₙ

n=0

= Γ(γ)/Γ(α)Γ(γ-α)

∞ 1

Σ xⁿ/n! ∫ t^(α-1) tⁿ(1-t)^(γ-α-1) dt

n=0 0

Interchanging the order of summation and integration we have

F(α;γ;x) = Γ(γ)/Γ(α)Γ(γ-α)×

1 ∞

∫ t^(α-1) (1-t)^(γ-α-1) dt Σ (xt)ⁿ/n!

0 n=0

= Γ(γ)/Γ(α)Γ(γ-α) ∫ eˣᵗ t^(α-1) (1-t)^(γ-α-1) dt

0 ............(1)

We can use this integral representation to deduce an important relation satisfied by the function F(α;γ;x).

Putting t= 1-s in the above result we get

F(α;γ;x) = Γ(γ)/Γ(α)Γ(γ-α)×

∫ e⁻ˣˢ s^(γ-α-1) (1-s)^(α-1) (-ds)

= Γ(γ)eˣ/Γ(α)Γ(γ-α) ×

∫ s^(γ-α-1) (1-s)^(α-1)e⁻ˣˢ ds

Let γ-α = α₁; then γ-α₁ = α.

Hence

F(α;γ;x) = Γ(γ)eˣ/Γ(γ-α₁)Γ(α₁)×

∫ s^(α₁-1) (1-s)^(α-1) e⁻ˣˢ ds

= eˣ F(α₁;γ;-x)

= eˣ F(γ-α;γ;-x), .........(2)

Which is known as Kummer's result.

Whittaker Functions :

If we substitute y= x^-γ/2 e^x/2 W(x) in the equation

xy" + (γ-x)y' -αy = 0,

We find that the function W(x) satisfies the differential equation

d²W/dx² + [-1/4 + k/x + (1/4 - m²) /x²]W =0,

where k stands for γ/2 -α and m for

1/2 (1-γ). Solution of this equation are known as Whittaker's confluent hypergeometric functions .

Confluent Hypergeometric Function

2019-08-24T21:42:00.001+05:30

Confluent Hypergeometric Function:

Let us put x = z/β or z= βx in the hypergeometric equation . Then the equation assumes the form

z(1- z/β)d²y/dz² +

[γ-(1+α+β)z/β]dy/dz -αy = 0

There are three regular singularities , one at z= 0 , another given by z= β , and also at z=∞.

When β-->∞ , we get the equation

z = d²y/dz² + (γ-z)dy/dz -αy = 0 .......(1)

The equation is known as the Confluent Hypergeometric Equation following to confluence of two singularities β,∞ when β-->∞ .

Again putting x= z/β in the hypergeometric series we obtain

1+ α.βz/1.γβ + α(α+1).β(β+1)z²/1.2.γ(γ+1)β²+..

= 1+ αz/1.γ + α(α+1)(1+ 1/β)z²/1.2.γ(γ+1)+...

When β-->∞ this series reduces to

1+αz/1.γ + α(α+1)z²/2!γ(γ+1) + ....

which can be put as

∞

Σ (α)ₖ zᵏ/k! (γ)ₖ or simply F(α;γ;z) . ...(2)

k=0

This function is called Confluent Hypergeometric Function .

Note :

If we put γ= α and z= x in equation(2) , we have

∞ ∞

F(α;α;x) = Σ (α)ₖxᵏ/k!(α)ₖ = Σ xᵏ/k! = eˣ

k=0 k=0

From the definition of confluent hypergeometric function it can be shown that

d/dx [F(α;γ;x)] = α/γ F(α+1;γ+1;x) ......(3)
∞
Since F(α;γ;x) = 1+ Σ (α)ₖ xᵏ /k!(γ)ₖ .
k=1

differentiating with respect to x , we get
∞
d/dx F(α;γ;x) = Σ (α)ₖ xᵏ⁻¹/(k-1)!(γ)ₖ
k=1

Writing k-1 = n , we obtain
∞
d/dx F(α;γ;x) = Σ (α)ₙ₊₁ xⁿ / n!(γ)ₙ₊₁
k=1
∞
= Σ α(α+1)...(α+n)xⁿ/n!γ(γ+1)..(γ+n)
n=0
∞
= α/γ Σ (α+1)ₙxⁿ/n!(γ+1)ₙ
n=0

= α/γ F(α+1;γ+1;x).

Example:

Prove that

αF(α+1;γ+1;x) - γF(α;γ;x)

= (α-γ)F(α;γ+1;x).

Solution :

L.H.S =

∞

Σ [α(α+1)ₖxᵏ/k!(γ+1)ₖ- γ(α)ₖxᵏ/k!(γ)ₖ]

k=0

∞

= Σ (α)ₖxᵏ[(α+k)-γ(γ+1)ₖ/(γ)ₖ] / k!(γ+1)ₖ

k=0

Now γ(γ+1)ₖ/(γ)ₖ =

γ(γ+1)....(γ+k)/ γ(γ+1)..(γ+k-1) =γ+k.

∞

Hence the L.H.S = Σ (α)ₖxᵏ(α+k-γ-k)/k!(γ+1)ₖ

k=0

∞

= (α-γ)Σ (α)ₖxᵏ/k!(γ+1)ₖ

k=0

= (α-γ)F(α;γ+1;x) .(Proved)

Integral Formula For The Hypergeometric Series

2019-08-22T08:26:00.001+05:30

Integral Formula For The Hypergeometric Series :

Theorem :

The hypergeometric function F(α,β;γ;x) can be represented by

[Γ(γ)/Γ(β)Γ(γ-β)]

∫(1-t)^(γ-β-1) t^(β-1) (1-xt)^-α dt

Proof :

We know that

(β)ₙ = β(β+1)...(β+n-1) = Γ(β+n)/Γ(β)

Similarly (γ)ₙ = Γ(γ+n)/Γ(γ)

Again

B(m,n) = Γ(m)Γ(n)/Γ(m+n)

Consider,

B(β+n,γ-β) / B(β,γ-β)

= [Γ(β+n)Γ(γ-β)/Γ(γ+n)]/[Γ(β)Γ(γ-β)/Γ(γ)

= [Γ(β+n)/Γ(β)] / [Γ(γ+n)/Γ(γ)]

Thus (β)ₙ/(γ)ₙ = B(β+n, γ-β) / B(β, γ-β)

From the definition of Beta function , we have

B(p,q) = ∫ tᵖ⁻¹ (1-t)^(q-1) dt , t<1.

Therefore,

(β)ₙ/(γ)ₙ = 1/B(β, γ-β)

∫ t^(β+n-1) (1-t)^(γ-β-1) dt

= Γ(γ) / Γ(β)Γ(γ-β)

∫ t^(β+n-1) (1-t)^(γ-β-1) dt , n=0,1,2....

0 .........,(1)

∞

Now F(α,β;γ;x) = Σ (α)ₙ(β)ₙ xⁿ/n!(γ)ₙ

n=0

∞

= Σ (α)ₙxⁿ/n! . Γ(γ) / Γ(β)Γ(γ-β)

n=0

∫ t^(β+n-1) (1-t)^(γ-β-1) dt

Reversing the order of summation and integration which is justified for an absolute convergence argument , we see that

F(α,β;γ;x) = 1

Γ(γ)/[Γ(β) Γ(γ-β)] ∫ t^(γ-β-1) t^(β-1)

∞

{ Σ (α)ₙ (xt)ⁿ/n!} dt

n=0

...............(2)

Since t<1 , we have if |x|≤1

(1-xt)^-α = 1+α(xt)/1! + α(α+1)(xt)²/2!+.....

∞

= Σ (α)ₙ (xt)ⁿ/n!

n=0

Hence from equation(2)

F(α,β;γ;x) = Γ(γ)/Γ(β)Γ(γ-β)

∫ (1-t)^(γ-β-1) t^(β-1) (1-xt)^-α dt

γ>β>0 , |x|≤1 (Proved)

..........(3)

Corollary :

If we put x=1 in equation(3) , we get

F(α,β;γ;1) = Γ(γ)/Γ(β)Γ(γ-β)

∫ (1-t)^(γ-α-β-1) t^(β-1) dt

= Γ(γ)/ Γ(β)Γ(γ-β) Β(β,γ-α-β)

if β>0 , γ-α-β>0

= Γ(γ)/Γ(β)Γ(γ-β) × Γ(β)Γ(γ-α-β) / Γ(γ-α)

= Γ(γ)Γ(γ-α-β) / Γ(γ-α) Γ(γ-β)

Which is called Gauss's Theorem . ........(4)

Mathquery

How To Multiply Fractions With Whole Numbers

How To Multiply Fractions With Whole Numbers :

Definition Of Fraction or Rational Numbers :

Improper Fraction:

Proper Fraction :

Prime Number :

Relatively Prime :

References:-

Continuity And Differentiability In Mathematics | Mathquery

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Definition Of Continuity Of A Function :-

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Discontinuous Function :

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Example :

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Differentiability At A Point :

Example :

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References :-

Leibnitz's Rule Statement And It's Proof

Leibnitz's Rule In Mathematics:

Proof Of Leibnitz's Rule In Mathematics :

Example Of Leibnitz's Rule In Mathematics :

Proof :

About The Scientists:

Fubini's Theorem

Fubini's Theorem :-

Statement:

Proof :

Note :

Remarks :

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Cauchy's Integral Theorem | Mathquery

Cauchy's Integral Theorem:

Statement :

Proof :

Example :

Solution :

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Green's Theorem | Mathquery

Green's Theorem:

Statement :-

Proof :-

Note :-

Corollary:-

Example :-

Solution :-

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Cauchy Riemann Equations For Analytic Function

Cauchy - Riemann Equations :-

Necessary Conditions For A Function To Be Analytic :

Statement:-

Proof :-

Case-I:-

Case-II:-

Polar Form Of Cauchy - Riemann Equations :

Proof :-

About Scientists :

See Also

Example :-

Solution :-

Lioville's Theorem In Complex Analysis | Mathquery

Lioville's Theorem:-

Statement :

Proof :

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Morera's Theorem

Morera's Theorem In Complex Analysis :-

Statement :

Proof :

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