Partial Differential Equations Formation and Solutions

PARTIAL DIFFERENTIAL EQUATIONS Presented by Dr .M.Santhi Associated Professor Department of Mathematics Hajee Karutha Rowther Howdia college Uthamapalayam Class: III Mathematics Sub Code: 17UMAC54 Sub: Differential Equation

Formation of Partial Differential equations Partial Differential Equation can be formed either by elimination of arbitrary constants or by the elimination of arbitrary functions from a relation involving three or more variables . SOLVED PROBLEMS 1.Eliminate two arbitrary constants a and b from constant . ( ) ( ) 2 2 + + = 2 2 x a y b z R here R is known

(OR) Find the differential equation of all spheres of fixed radius having their centers in x y- plane. solution ( ) ( ) ( ) 1 2 2 + + = 2 2 ....... x a y b z R Differentiating both sides with respect to x and y z z = 2 2 ( ) x a x z = 2 2 ( ) z y b y z z = = , p q x y = = , x a pz y b qz

By substituting all these values in (1) + + R = 2 2 2 2 2 2 p z q z z R 2 = 2 z + + 2 2 1 p q or 2 R = 2 z 2 2 z z + + 1 x y

2. Find the partial Differential Equation by eliminating arbitrary functions from SOLUTION = 2 2 ( ) z f x y = 2 2 ( to )......... ) 1 .( z f x y . z . . . d w r xandy = ' 2 2 ( ) 2 ......( x ) 2 f x y x z = ' 2 2 ( ) 2 ......( y ) 3 f x y y

) 2 ( By ) 3 ( z x x = y z p y x = + = 0 py qx q y

3.Find Partial Differential Equation by eliminating two arbitrary functions from ( x yf z = + ) ( ) xg y SOLUTION = + ( ) ( )......( ) 1 z yf x xg y Differentiating both sides with respect to x and y z + = ( ) ( )........( ) 2 f y x g y x z = + ( ) ( )........( ) 3 f x x g y y

Again d . w .r. to x and yin equation (2)and(3) 2 z = + ( ) ( ) f x g y x y + ( 2 ) )...... 3 ( ... x y to get

z z + = x y yf x ( y + + + ) ( ) ( ( ) ( )) xg y x xy f x g y ( ) = + + z xy f g x 2 z z z + = + x y z xy x y y

Different Integrals of Partial Differential Equation 1. Complete Integral (solution) z z = = ( , , , , ) ( , , , , ) ......( 0 ) 1 F x y z F x y z p q Let be the Partial Differential Equation. The complete integral of equation (1) is given by x y = ( , , , , ) .......... 0 ) 2 ( x y z a b

2. Particular solution A solution obtained by giving particular values to the arbitrary constants in a complete integral is called particular solution . 3.Singular solution The eliminant of a , b between ) , , , , ( = b a = 0 x y z a b = , 0 0 when it exists , is called singular solution

4. General solution In equation (2) assume an arbitrary relation of the form . Then (2) becomes ) (a f b= = ( , , , , ( )) .........( 0 ) 3 x y z a f a Differentiating (2) with respect to a, + = ( ) .......... 0 ) 4 ( f a a b The eliminant of (3) and (4) if exists, is called general solution

Standard types of first order equations TYPE-I The Partial Differential equation of the form 0 ) , ( = q p f has solution with by ax z + + = c = ( , ) 0 f a b TYPE-II The Partial Differential Equation of the form = + + ( , ) z px qy f p q is called Clairaut sform of pde , it s solution is given by z = + + ( b a , ) ax by f

TYPE-III If the pde is given by = ( , , ) 0 f z p q then assume that z = + ( ) x ay = + u x ay = ( ) z u

z z u z dz = = = = 1 . p x u x u du z z u z dz = = = = a . q a y u y u du The given pde can be written as dz a dx this can dz = ( , , ) 0 f z .And also dy be integrated to get solution

TYPE-IV The pde of the form can be solved by assuming ( ) , ( a p x f = z dz + = ( , ) ( , ) f x p g y q = = , ) f x p g y q a = = ( , ) ( , ) p x a = z ( , ) ( , ) g y q a q y a = dx dy x ( y + = , ) ( , ) dz x a dx y a dy Integrate the above equation to get solution

SOLVED PROBLEMS q = 2 1 p 1.Solve the pde and find the complete and singular solutions Solution Complete solution is given by ax z + = + by c b = a 2 1 a b with = 2 1

= + + 2 ( ) 1 z ax a y c d.w.r.to. a and c then z = + 2 x ay a z = = 1 0 Which is not possible c Hence there is no singular solution + + = 0 pq p q 2.Solve the pde and find the complete, general and singular solutions

Solution The complete solution is given by ax z = + + by c + + b = 0 ab a b with b = a + 1 b = + + .......( c ) 1 z x by + 1 b

1 z = + = 0 x y ( ) 2 + b 1 b z = = 1 0 no singular solution To get general solution assume that ) (b g c = c From eq (1) b = + + ( ).......( 2 ) z x by g b + 1 b

1 z = + + ( ).......( 3 ) x y g b ( ) 2 + c 1 b Eliminate from (2) and (3) to get general solution 3.Solve the pde and find the complete and singular solutions = + + + + 2 2 1 z px qy p q Solution The pde is in Clairaut s form = + + + + 2 2 1 z px qy p q

complete solution of (1) is = + + + + 2 2 1 .......( 2 ) z ax by a b d.w.r.to a and b z a = + = 0 x a + + 2 2 1 a b ........( ) 3 z b = + = 0 y b + + 2 2 1 a b

From (3) 2 2 a b = = 2 2 , x y + + + + 2 2 2 2 1 1 a b a b + 2 2 a + b + + = 2 2 x y 2 2 1 a b 1 2 = + 2 2 1 ( ) x y + + 2 1 a b

2 a + = 0 ax + + 2 2 1 a b 2 b + = 0 by + + 2 2 1 a b 1 a + + + + = 2 2 1 0 ax by a b + + 2 2 1 b 1 a = = + 2 2 2 0 1 ( ) z z x y + + 2 2 1 b + + = 2 2 2 1 x y z is required singular solution

+ = 4.Solve the pde 1 ( ) 2 ( ) 3 x p y q z Solution pde + = 1 ( z ) ( 2 ) 3 x p + y q z = + 3 ( 2 ) px qy p q Complete solution of above pde is ax z + = p + q z ( 3 = 2 ) by 2 a b + 2 5.Solve the pde Solution Assume that = + ( ) z x ay

= + u x ay = (u ) z z z u z dz = = = = 1 . p x u x u du z z u z dz = = = = a . q a y u y u du From given pde 2 2 dz dz + = + = 2 2 2 2 p q z a z du du

2 dz z = + 2 1 du a 1 + dz z dz = = du + 2 1 du a z 2 1 a Integrating on both sides u + = + 2 z b 2 1 a + x ay = + 2 z b + 2 1 a

= + zpq p q 6. Solve the pde Solution Assume Substituting in given equation zpap + = 1 q = ap = + p ap + 1 a a = , p q az z z z = + dz dx dy x y + + 1 1 a a = + dz dx dy az z

= + + 1 ( )( ) zadz a dx ady Integrating on both sides a = 2 + + + 2 1 ( )( ) z a x ay b = pq ( xy 7.Solve pde z z = )( ) xy (or) x y Solution p= q x y

Assume that p y = = a x q y = = , p ax q a y = + = + dz pdx qdy axdx dy a Integrating on both sides 2 2 x 2 y 2 = + + z a b a

+ = + 2 2 p q x y 8. Solve the equation Solution p = = 2 2 x y q a = + = , p a x q y a = + = + + dz pdx qdy a x dx y a dy integrating 3 3 2 = + + + ( ) ( ) z a x y a b 2 2 3

Equations reducible to the standard forms xm yn ( ) ( ) p q (i)If and occur in the pde as in 0 ) , ( = q y p x F Or in = m n ( , , ) 0 F z x p y q m n n= 1 y Y m= 1 Case (a) Put and if ; m z x = x X 1 1 n z X z = = = m 1 ( ) p m x X x X z z Y z = = n 1 ( ) q n y y Y x Y

z = = m 1 ( ) 1 ( P ) x p m m X z = = n 1 ( ) 1 ( ) y q n Q n Y z z = = , P Q where X Y = = m n Then reduces to ( , ) 0 F P Q ( , ) 0 F x p y q = = m n ( , , ) 0 Similarly reduces to , , ( y p x z F ) 0 F z P Q q

case(b) If or put = = 1 Y = n 1 x m log = z , log 1 X y = = p px P X x 1 z = = q qy Q Y y zk ( q zk k k ( ) ) p ( , ) F z p z q (ii)If and occur in pde as in , ( 1 z x f Or in = k k ) ( , ) p f y z q 2

+ k= 1 z Z 1 k Case(a) Put if z z Z Z 1 ` 1 ` = = + = + k k 1 ( ) 1 ( ) z k z p k P x Z x x z z Z Z 1 ` 1 ` = = + = + k k 1 ( ) 1 ( ) z k z q k Q y Z y y Z Z = = , P Q where x y Given pde reduces to = ( , ) ( , ) f x P f y Q ( , ) F P Q and 1 2

= 1 k Case(b) if z = log Z z z Z Z = = = 1 z z p P x Z x x z z Z Z = = = 1 z z q Q y Z y y Solved Problems 1.Solve + = 2 4 2 4 2 p x q 2 y z 2 2 2 px qy Solution + = 1 .......( ) 1 z z

= = , 2 2 m n = 1 y k 1 z = = log Z Y 1 = x X z z Z X Z = = = = 2 2 p zx zx P x Z X x X z z Z Y Z = = = = 2 2 q zy zy Q y Z Y Z y Y Z = = , P Q where X Y

2 2 px qy = = , P Q z z (1)becomes ( P ) ( = ) 2 2 + = 1 P Q + 2 2 1 Q + = + Z aX bY c + = = 2 2 2 , 1 1 a b b a = + + 2 2 2 log 1 z ax a y c

+ = + 2 2 2 2 2 ( ) p q z x y 2. Solve the pde SOLUTION 2 2 p q + = + 2 2 z = Z ( ).....( ) 1 x y z z = z log z 1 Z k Z = = = 1 z z p P x Z x x z z Z Z = = = 1 z z q Q y Z y y

Eq(1) becomes + = + 2 2 2 2 ( ).....( = ) 2 P Q x y = 2 2 2 2 2 P x y Q a 2 a 2 x x 2 = + + + 1 2 2 log sinh ( ) z a x a 2 2 2 ( ) y y a a 2 y + 1 cosh b 2 a

Lagranges Linear Equation Def: The linear partial differenfial equation of first order is called as Lagrange s linear Equation. + = Pp Qq R This eq is of the form Where and are functions x,y and z Q P, R The general solution of the partial differential equation is Qq Pp = + = ( , ) 0 R F u v = F ( , , ) u x y z c Where is arbitrary function of and 2 ) , , ( c z y x v = 1

u = v = 1c 2c Here and are independent solutions dx dy dz = = of the auxilary equations R P Q Solved problems + = + 2 2 ( ) x p y q x y z 1.Find the general solution of Solution auxilary equations are dx dy dz + = = 2 2 ( ) x y x y z

dx dy = Integrating on both sides 2 2 x y ( ) = = 1 1 u x y c 1 dx dy dz + = 2 2 ( ) x y x y z ( ) d x y + dz + = ( )( ) ( ) x ( y x ) y x y z d x y ) dz = Integrating on both sides ( x y z

= + log( v = ) log = log x y z c 2 1 ( ) x y z c 2 = ( , ) 0 F = u v The general solution is given by y x F 1 1 1 ( ( , ) ) 0 x y z + = 2 2 2 ( ) ( ) ( ) x y z y z x q z x y 2.solve solution Auxiliary equations are given by dy z dx dz = = 2 2 2 ( ) ( ) ( ) x y y z x z x y

dy dx dz 2 y 2 2 x z = = ( ) dx ( ) ( ) y z z x x y dy dz + + 2 2 2 x y z + + ( dx ) ( ) ( ) y z z x x y dy dz + + = 0 2 2 2 x y z Integrating on both sides

1 1 1 = + + = u a x y y z 1 1 1 x dx dy z dz = = x dy ( ) dx ( ) ( ) x y z y z x z x y + + 1 1 1 y z dz + + ( ) ( ) ( ) x y z y z x z x y dx dy dz + + = 0 Integrating on both sides x y z = = v xyz b

The general solution is given by + + = 1 1 1 ( , ) 0 F x y z xyz HOMOGENEOUS LINEAR PDE WITH CONSTANT COEFFICIENTS Equations in which partial derivatives occurring are all of same order (with degree one ) and the coefficients are constants ,such equations are called homogeneous linear PDE with constant coefficient

n n n n z n z z z n + + + = ........ ( , ) a a a F x y 1 2 n 1 2 2 n n x x y x y y = = , . D D Assume that x y th n then order linear homogeneous equation is given by ( 2 1 D D a D D a D or ) , ( F z D D f = + + + + = 1 2 2 n n n n ......... ) ( , ) a D z F x y n ( , )......... ) 1 ( x y

The complete solution of equation (1) consists of two parts ,the complementary function and particular integral. The complementary function is complete solution of equation of z = ( , ) 0 f D D Rules to find complementary function Consider the equation + y x x or 2 2 2 z z z + = 0 k k 1 2 2 2 y + + = 2 2 ( ) .......... 0 ...( ) 2 D k D D k D z 1 2

The auxiliary equation for (A.E) is given by + + = 2 2 0 D k D D k D 1 2 = m = m , + 1 + D m D k And by giving = 2 ....( 0 ) 3 k The A.E becomes 1 2 Case 1 If the equation(3) has two distinct roots The complete solution of (2) is given by 1,m m 2 = + + + ( ) ( ) z f y m x f y m x 1 1 2 2