PFAFFIAN Differential Equations And It's Solutions

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

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