PARTIAL DIFFERENTIAL EQUATIONS

Partial differential equation are those equations which involves Partial derivatives.

  • Notation: Let p=\frac{\partial z}{\partial x},q=\frac{\partial z}{\partial y},r=\frac{\partial^2 z}{\partial x^2},s=\frac{\partial^2z}{\partial x \partial y},t=\frac{\partial^2z}{\partial y^2}

RULE FOR FORMATION OF PARTIAL DIFFERENTIAL EQUATIONS BY ELIMINATION OF ARBITRARY CONSTANTS

  • If the number of arbitrary constants to be eliminated is equal to the number of independent variables we get a first order partial differential equation. If the number of arbitrary constants is more than the number of independent variables, we get a partial differential equations of higher order.

RULE FOR FORMATION OF PARTIAL DIFFERENTIAL EQUATIONS BY ELIMINATION OF ARBITRARY FUNCTIONS

  • If the number of arbitrary functions to be eliminated is equal to one then the required partial differential equations will be first order otherwise it will be of higher order.
  • When an equation involves one arbitrary function the required partial differential equations always reduce to a Lagranges the partial differential equations of the form Pp+Q_{{q}}=R.
  • Any solution which contains the same number of arbitrary constants as the number of independent variables is called a complete integral or solution.
  • Any solution which is got by giving particular values to the arbitrary constants in a complete integral is called a particular integral.
  • Let  Φ (x, y, z, a, b ) = 0   →( 1 ) be the complete integral of ƒ (x, y, z, p, q ) = 0 →( 2 ) where a and b are arbitrary constants.
  • To eliminate a and b let us partial differentiate ( 1 ) . write equation a and b and equating it to zero, we get
  • \frac{\partial \Phi }{\partial a} = 0 \rightarrow (3)  and \frac{\partial \Phi }{\partial b} = 0 \rightarrow (4)
  •  from equations (1),  (3),  (4) eliminate  ‘ a ‘ and ‘ b ‘ we  obtain the singular integral of ( 2 )
  • In the complete integral ( 1 )  put b = ƒ(a) then equations (1) becomes 

          Φ ( x, y, z, , ƒ(a) ) = 0  → ( 5 )

  •   Now partially differentiating ( 1 )  write the equations we have 

          \frac{\partial \Phi }{\partial a}+\frac{\partial \Phi }{\partial b} f^{'} (a) = 0  \rightarrow(6)

  •  Eliminating between equations ( 5 ) and (6), we get general integral of ƒ ( x, y, z, p, q ) = 0  

FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

  • An equation which involves partial derivatives \frac{\partial z}{\partial x}  and \frac{\partial z}{\partial y}  only is known as First order partial differential equations. The general first order partial differential equation is  ƒ ( x, y, z, p, q ) = 0 where p = \frac{\partial z}{\partial x} and q = \frac{\partial z}{\partial y} .
 TYPE  (1) :   F(p ,q) = 0
  • A partial differential equations which involves p and q only and the variables x, y, z do not  occur explicitly. Let  z = ax + by + c be the solution of the equations.
 Type (2) CLAIRAUT’S FORM
  • It is the form  z = px + qy + ƒ (p , q ). To find complete solution for clairaut’s type, replace p by a and q by b.
  • Complete solution in the form  is z = ax + by + ƒ (a ,b ) where a, b are arbitrary constants.
 Type (3) EQUATIONS OF TYPE F (z,p,q) = 0
  • In these types of equations x and y do not appear explicitly. Let z =( fx + ay) be a trail solution. Let p = \frac{dz}{du} , q = a \frac{dz}{du} substituting p and q in F (z,p,q) = 0 which is a first order ordinary differential equations on the integrating we get the complete solution.
  Type (4) F1 (x,p) = F2(y,q)
  • A first order partial differential equation is separable if it can be written as F1 (x , p) = F2 (y,q) (x , p). Let F_{{1}} (x,p) = F_{2} (y,q) = K







(say)    Express x in  terms of p and K and y in terms of q and K . Substituting p and q in dz= pdx + qdy and integrating we get the complete integral . The singular and general integrals can be found out as usual.
  Type (5) EQUATIONS REDUCIBLE TO THE STANDARD FORM

Case 1 

  • An equation of the form F (x^m p , y^n q ) = 0  where m and n are constants can be reduced to by the type (1) case by using the substituting x^1-m= X and  y^1-n = Y, m\neq 1 and n\neq 1.

Case 2

  • An equation of the formF (z,  x^{m} p,  y^{n} q) = 0  can be transformed to the type (3) by using the substitution x^1-m= X and y^1-n = Y if m\neq 1 ; and n\neq 1.

Case 3

  • Put X = log x and Y = log y if m =1  and n = 1 in the above two cases. 

Case 4 

  • An equation of the form F (z^{k} p,  z^{k} q) = 0 where K is any constant can be transformed into the type (1) by proper substitution.                                If  K \neq  -1, put Z =  z^{k+1} and if  K = -1 put Z  = log z

Case 5 :

  •  An equation of the form F(x^{m}   z^{k} p, )=0 can be reduced into type ( 1 ) by putting  ,x^1-m= X,y^1-n = Y, z=  z^{k+1} if  K \neq  -1, m\neq 1n\neq 1 or by putting X = log x, Y = logy, Z = log z  if m =1, n =1 and K = -1.
LAGRANGE’ S LINEAR EQUATIONS
  • A linear partial differential equation of first order is known as Lagrange’ s equations. Lagrange’ s linear equation is in the form  Pp+Q_{{q}}=R where P, Q and R are functions x, y, z.
  • Elimination of the arbitrary function from the relation \Phi ( u , v )= 0 give us the lagrange’ s linear equation Pp+Q_{{q}}=R. We can find two functions  u , v such as that the eliminating of \Phi from  \Phi ( u , v )=0  Gives us the Langrange’ s linear partial differential equations , then \Phi (u,v) =0 is the general solution where u and v are functions of x,y,z. 

WORKING RULE FOR SOLVING LAGRANGE’ S PARTIAL DIFFERENTIAL EQUATIONS

  • Put the partial differential equations of the first order in the formPp+Q_{{q}}=R,
  • Write the Lagrange’ s auxiliary equation \frac{dx}{p} = \frac{dy}{q} = \frac{dz}{r}
  •   Solve (2) and let u (x,y,z)= c_{1} and v(x,y,z)=c_{2} be two independent solutions.
  • The solution are written in the form \Phi ( u , v )=0 or u= \Phi (v) or v=\Phi (u).
METHOD OF GROUPING
  • Given\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}   we can solve it by variable separable method which gives us two independent solution u (x,y,z) =c_{1} and v(x,y,z)=c_{2} substituting u and v in \Phi ( u , v )=0 we get the equation for the Lagrange’s equation.
METHOD OF MULTIPLIERS
  • Given the subsidiary equations \frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}  we can solve it as follows . We known that if \frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R} =\frac{l dx + mdy + ndz}{lP +mQ +nR}=\frac{l^' dx + m^'dy + n^'dz}{l^'P +m^'Q +n^'R}
  • Where the two sets of multipliers  (l,m,n) , (l’, m’, n’) may be constants or functions of x,y,z. Choosing (l,m,n) such that lP +mQ+nR=0. We have l dx + m dy + n dz=0. Hence integrating we get u= u (x,y,z)=c_{1} as one of the solutions similarly we have  l’dx + m’dy + n’dz=0. Which on integrating we give as v=v(x,y,z)=c_{2} as the other solution the general  solution is \Phi (u , v)=0.
PARTIAL HOMOGENEOUS EQUATIONS OF HIGHER ORDER
  • Linear partial differential equations of higher order with constants coefficients are 
  1.  Homogeneous linear partial differential equations in the form \frac{\partial^3 z}{\partial x^3}+2\frac{\partial^3 z}{\partial x^2 
 \partial y}+3\frac{\partial^3 z}{\partial x \partial y^2}-4\frac{\partial^3 z}{\partial y^3} = f(x,y)
  2.  Non homogeneous linear differential equation is in the form \frac{\partial^3 z}{\partial x^3}+4\frac{\partial^2 z}{\partial y^2}-2\frac{\partial z}{\partial x }+z= x^2
  • If the roots of an arbitrary equation is real or complex and the different solution is z=f_{{1}}(y+m_{1}x)+f_{{2}}(y+m_{2}x)
  • If the roots are real and equal to the solution is  z=f_{{1}}(y+m_{1}x)+xf_{{2}}(y+m_{2}x)
NON HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS
  • If the polynomial expression f(D , D’) is not homogeneous in f(D,D’) z = f (x,y) then the equation is known as the non homogeneous linear equation.
  • Assume that z=c e^{hx+hy} where c,h,k are arbitrary constants then the solution is z=c. e^{hx+hy} in f(D D’)z=0

         c.f(h,k)  e^{hx+hy}=0\Rightarrowf(h,k)=0

  • If f(D D’) is of degree r in D’ then f(h,k)=0 will be the rth degree in k . solving for k from c f (h,k) e^{hx+ky} in terms of h we get k_{{s}} = f_{{s}} (h) where s =1,2,3,…….r
  • Hence z=c_{{s}} e^{hx+f_{s}(h)y} where s=1,2,…..r are separates solutions of f(D D’) z=0
  • The arbitrary values of c_{{s}} and h we get z=\Sigma c_{{1}} e^{hx+f_{1}(h)y} , \Sigma c_{{2}} e^{hx+f_{2}(h)y}

          to be the solution of f(D D’)z=0 . Hence the most general solution of

          z=\Sigma c_{{1}} e^{hx+f_{1}(h)y} +\Sigma c_{{2}} e^{hx+f_{2}(h)y}+...............+\Sigma c_{{r}} e^{hx+f_{r}(h)y}

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