# Partial Differential Equations

1 Introduction 2 Formation Of Partial Differntial Equations 3 Solutions Of Partial Differential Equations 4 Lagrange’s Linear Equations 5 Partial Differential Equations Of Higher Order With Constant Co-Effecients 6 Non-Homogenous Linear Equations

PARTIAL DIFFERENTIAL EQUATIONS

This unit covers topics that explain the formation of partial differential equations and the solutions of special types of partial differential equations.

1 INTRODUCTION

2 FORMATION OF PARTIAL DIFFERNTIAL EQUATIONS

3 SOLUTIONS OF PARTIAL DIFFERENTIAL EQUATIONS

4 LAGRANGE’S LINEAR EQUATIONS

5 PARTIAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER WITH CONSTANT CO-EFFECIENTS

6 NON-HOMOGENOUS LINEAR EQUATIONS

1 INTRODUCTION

A partial differential equation is one which involves one or more partial derivatives. The order of the highest derivative is called the order of the equation. A partial differential equation contains more than one independent variable. But, here we shall consider partial differential only equation two independent variables x and y so that z = f(x,y). We shall denote A partial differential equation is linear if it is of the first degree in the dependent variable and its partial derivatives. If each term of such an equation contains either the dependent variable or one of its derivatives, the equation is said to be homogeneous, otherwise it is non homogeneous.

2 Formation of Partial Differential Equations

Partial differential equations can be obtained by the elimination of arbitrary constants or by the elimination of arbitrary functions.

By the elimination of arbitrary constants

Let us consider the function

f( x, y, z, a, b  ) = 0  -------------  (1)

where a & b are arbitrary constants

Differentiating equation (1) partially      w.r.t x & y, we get Eliminating a and b from equations (1), (2) and (3), we get a partial differential equation of the first order of the form f (x,y,z, p, q) = 0

Example 1

Eliminate      the arbitrary constants a & b        from  z = ax + by + ab

Consider      z  = ax + by + ab     ____________ (1)

Differentiating        (1) partially w.r.t     x & y, we get Using (2)      & (3)  in (1), we get

z        = px +qy+ pq

which is the required partial differential equation.

Example 2

Form the partial differential equation by eliminating the arbitrary constants a and b from

z = ( x2 +a2 ) ( y2 + b 2)

Given z = ( x2 +a2 ) ( y2 + b2)         ……..(1)

Differentiating        (1) partially w.r.t     x & y , we get

p        = 2x   (y2 + b2 )

q        = 2y   (x  + a  )

Substituting the values of p and q in (1), we get

4xyz = pq

which is the required partial differential equation.

Example 3

Find the partial differential equation of the family of spheres of radius one whose centre lie in the xy - plane.

The equation of the sphere is given by

( x –a )2 + ( y- b) 2 +  z2  = 1    _____________ (1)

Differentiating (1) partially w.r.t x & y , we get

2        (x-a ) + 2 zp   =       0

2 ( y-b ) + 2 zq  =    0

From these equations we obtain

x-a = -zp _________ (2)

y -b = -zq _________ (3)

Using (2) and (3) in (1),  we get

z2p2 + z2q2 + z 2    = 1

or  z2 ( p2  + q2  + 1) = 1

Example 4

Eliminate the arbitrary constants a, b & c  from and form the partial differential equation.

The given equation is or  -zp + xzr + p2x = 0

By the elimination of arbitrary functions

Let   u and v   be   any two   functions arbitrary function. This relation can be expressed as

u = f(v)  ______________ (1)

Differentiating  (1)  partially w.r.t    x &  y and eliminating    the arbitrary functions from these  relations, we get  a partial differential equation  of the first  order of the form

f(x, y, z, p, q )  = 0.

Example 5

Obtain the partial differential equation  by eliminating „f„from  z = ( x+y ) f ( x2 -  y2 )

Let us now consider the equation

z = (x+y ) f(x2- y2) _____________ (1)

Differentiating (1) partially w.r.t x & y , we get

p  = ( x + y ) f ' ( x2 -  y2 ) . 2x  +  f ( x2 -  y2 )

q  =  ( x + y ) f ' ( x2 -  y2 ) . (-2y)     + f ( x2 -  y2 ) i.e, py - yf( x2 - y2 ) = -qx +xf ( x2 - y2 )

i.e, py +qx   = ( x+y ) f ( x2 -  y2 )

Therefore, we have by(1),  py +qx  = z

Example 6

Form the partial differential equation by eliminating the arbitrary function f

from

z = ey f (x + y)

Consider  z  = ey f ( x +y )  ___________  ( 1)

Differentiating  (1) partially  w .r. t  x & y, we get

p     = ey f ' (x  + y)

q     = ey f '(x  + y) + f(x + y). ey

Hence, we have

q = p + z

Example 7 Exercises:

1. 1. Form the partial differential equation by eliminating the arbitrary constants „a‟ & „b‟ from the following equations. 2.                 Find the PDE of the family of spheres of radius 1 having their centres lie on the xy plane{Hint: (x –a)2 + (y –b)2 + z2 = 1}

3.                 Find the PDE of all spheres whose centre lie on the (i) z axis (ii) x-axis

4.                 Form the partial differential equations by eliminating the arbitrary functions in the following cases.

(i)                z = f (x + y)

(ii)             z = f (x2 –y2

(iii)           z = f (x2 + y2 + z2)

(iv)           f(xyz, x + y + z) = 0

(v)             F (xy + z2, x + y + z) = 0

(vi)           z = f (x + iy) +f (x –iy)

(vii)        z = f(x3 + 2y) +g(x3 –2y)

SOLUTIONS OF A PARTIAL DIFFERENTIAL EQUATION

A solution or integral of a partial differential equation is a relation connecting the dependent and the independent variables which satisfies the given differential equation. A partial differential equation can result both from elimination of arbitrary constants and from elimination of arbitrary functions as explained in section 1.2. But, there is a basic difference in the two forms of solutions. A solution containing as many arbitrary constants as there are independent variables is called a complete integral. Here, the partial differential equations contain only two independent variables so that the complete integral will include two constants.A solution obtained by giving particular values to the arbitrary constants in a complete integral is called a particular integral.

Singular Integral

Let  f (x,y,z,p,q) = 0 ----------   (1)

be the partial differential equation whose complete integral is

f (x,y,z,a,b) = 0-----------          (2)

where          „a‟   and   „b‟   are   arbitrary   constants.

Differentiating (2) partially w.r.t. a  and b, we obtain The eliminant of „a‟ and „b‟ from the equations (2), (3) and (4), when it exists, is called the singular integral of (1).

General Integral

In the complete integral (2), put b = F(a), we get

f (x,y,z,a, F(a) ) = 0       ----------       (5)

Differentiating (2), partially w.r.t.a,  we get The eliminant of „a‟ between (5) and (6), if it exists, is called the general integral of (1).

SOLUTION OF STANDARD TYPES OF FIRST ORDER PARTIAL

DIFFERENTIAL EQUATIONS.

The first order partial differential equation can be written as

f(x,y,z, p,q) = 0,

where p = z/x and q = z / y. In this section, we shall solve some standard forms of equations by special methods.

Standard I : f (p,q) = 0. i.e, equations containing p and q only.

Suppose that z = ax + by +c is a solution of the equation f(p,q) = 0, where f (a,b)

= 0.

Solving this for b, we get b = F (a).

 Hence the complete integral is z = ax + F(a) y +c------------ (1)

Now, the singular integral is obtained by eliminating a & c between

z = ax + y F(a) + c 0 = x + y F'(a)

0 = 1.

The last equation being absurd, the singular integral does not exist in this case.

To obtain the general integral, let us take c = F (a).

Then, z = ax + F(a) y + F  (a)  -------------- (2)

Differentiating (2) partially w.r.t. a, we get

0 = x + F'(a). y + F '(a)  --------------- (3)

Eliminating‟between„a  (2)   and   (3),   we   get   the   gene

Example 8

Solve pq = 2

The given equation is of the form  f (p,q) = 0

The solution is  z = ax + by +c, where ab = 2.

Solving,      b =  2/b

The complete integral is

Z =  ax  + 2/a y + c        ---------- (1)

Differentiating      (1)   partially   w.r.t   „c‟,   we

0 = 1,

which is absurd. Hence, there is no singular integral.

To find the general integral, put c =  (a) in (1), we get

Z = ax + 2/a y + F (a)

Differentiating      partially   w.r.t   „a‟,   we   get

0 = x – 2/ a2 y + F ‘(a)

Eliminating „a‟ between these equations gives the general integral.

Example 9

Solve pq + p +q = 0

The given equation is of the form f (p,q) = 0.

The solution is  z = ax + by +c, where ab + a + b = 0.

Solving, we get Differentiating (1) partially w.r.t. „c‟, we get

0 = 1.

The above equation being absurd, there is no singular integral for the given partial differential equation.

To find the general integral, put c = F (a) in (1), we have Differentiating (2) partially w.r.t a, we get Example 10

Solve p2 + q2 = npq

The solution of this equation is  z = ax + by + c, where a2 + b2 = nab.

Solving, we get Differentiating (1) partially w.r.t c, we get  0 = 1, which is absurd. Therefore, there is no singular integral for the given equation.

To find the general Integral,    put C = F (a), we get The eliminant of „a‟ between these equations gives the general integral

Standard II : Equations of the form f (x,p,q) = 0, f (y,p,q) = 0 and f (z,p,q) = 0. i.e, one of the variables x,y,z occurs explicitly.

(i)                Let us consider the equation f (x,p,q) = 0.

Since z is a function of x and y, we have or        dz = pdx + qdy

Assume that q = a.

Then the given equation takes the form f (x, p,a ) = 0

Solving, we get   p = F(x,a).

Therefore,        dz = F(x,a) dx + a dy.

(ii) Let us consider the equation f(y,p,q) = 0. Assume that p = a.

Then the equation becomes f (y,a, q) = 0 Solving, we get q = F (y,a).

Therefore, dz = adx + F(y,a) dy.

Integrating, z = ax + òF(y,a) dy + b, which is a complete  Integral.

(iii) Let us consider the equation f(z, p, q) = 0.

Assume that q = ap.

Then the equation becomes f (z, p, ap) = 0

Solving, we get p = F(z,a). Hence dz = F(z,a) dx + a F(z, a) dy. Example 11

Solve q = xp + p2

Given q = xp + p2   -------------(1)

This is of the form f (x,p,q) = 0.

Put q = a in (1), we get

a = xp + p2

i.e, p2 + xp –a = 0.

Therefore, Example 12

Solve q = yp2

This is of the form f (y,p,q) = 0

Then, put p = a.

Therfore, the given equation becomes q = a2y.

Since dz = pdx + qdy, we have

dz = adx + a2y dy

Integrating, we get z = ax + (a2y2/2) + b

Example 13

Solve 9 (p2z + q2) = 4

This is of the form f (z,p,q) = 0

Then, putting       q = ap, the given  equation becomes

9 (p2z + a2p2) = 4 or      (z + a2)3/2 = x + ay + b.

Standard III : f1(x,p) = f2 (y,q). ie, equations in which ‘z’ is absent and the variables are separable.

Let us assume as a trivial solution that

f(x,p) = g(y,q) = a (say).

Solving for p and q, we get p = F(x,a) and q = G(y,a). Hence dz = pdx + qdy = F(x,a) dx + G(y,a) dy

Therefore, z = òF(x,a) dx + òG(y,a) dy + b , which is the complete integral of the given equation containing two constants a and b. The singular and general integrals are found in the usual way.

Example 14

Solve pq = xy

The given equation can be written as Example 15

Solve p2 + q2 = x2 + y2

The given equation can be written as p2 –x2 = y2 –q2 = a2 (say)

p2 –x2 = aImplies p = Ö(a2 + x2)

andy2 –q2 = aImplies q = Ö(y2 –a2)

But   dz = pdx + qdy Standard IV (Clairaut’s) form

Equation of the type z = px + qy + f (p,q) ------(1) is   known   as   Clairaut‟s

Differentiating (1) partially w.r.t x and y, we get p = a and q = b.

Therefore, the complete integral is given by

z = ax + by + f (a,b).

Example 16

Solve z = px + qy +pq

The given equation is   in   Clairaut‟s.   form

Putting p = a and q = b, we have

z = ax + by + ab                                         -------- (1)

which is the complete integral.

To find the singular integral, differentiating (1) partially w.r.t a and b, we get

0 = x + b

0 = y + a

Therefore we have, a = -y and b= -x.

Substituting the values of a & b in (1), we get

z = -xy –xy + xy

or      z + xy = 0,  which is the singular integral.

To get the general integral, put b = F(a) in (1).

Then z = ax + F(a)y + a F(a)                              ---------- (2)

Differentiating (2) partially w.r.t a, we have

0 = x + F'(a) y + aF'(a) + F(a)                  ---------- (3)

Eliminating „a‟ between (2) and (3), we get the general integral.

Example 17

Find the complete and singular solutions of  Exercises

Solve the following Equations

1.     pq = k

2.     p + q = pq

3.     Öp +Öq = x

4.     p = y2q2

5.     z = p2 + q2

6.     p + q = x + y

7.     p2z2  + q2 = 1

8.     z = px + qy - 2Öpq

9.     {z –(px + qy)}2 = c2 + p2 + q2

10.                        z = px + qy + p2q2

EQUATIONS REDUCIBLE TO THE STANDARD FORMS

Sometimes, it is possible to have non –linear partial differential equations of the first order which do not belong to any of the four standard forms discussed earlier. By changing the variables suitably, we will reduce them into any one of the four standard forms.

Type (i) : Equations of the form F(xm p, ynq) = 0 (or) F (z, xmp, ynq) = 0.

Case(i) : If m ¹1 and n ¹1, then put x1-m = X and y1-n = Y. Hence, the given equation takes the form F(P,Q) = 0 (or) F(z,P,Q) = 0.

Case(ii) : If m = 1 and n = 1, then put log x = X and log y = Y. Example 18

Solve x4p2 + y2zq = 2z2

The given equation can be expressed as

(x2p)2 + (y2q)z = 2z2

Here m = 2, n = 2

Put X = x1-m = x -1  and  Y = y 1-n = y -1.

We have xmp = (1-m) P  and ynq = (1-n)Q

i.e, x2p = -P and  y2q = -Q.

Hence the given equation becomes

P2 –Qz = 2z----------(1)

This equation is of the form f (z,P,Q) = 0.

Let us take Q = aP.

Then equation (1) reduces to

P2 –aPz =2z2 Example 19

Solve x2p2 + y2q2  = z2

The given equation can be written as

(xp)2 + (yq)2  = z2

Here m = 1, n = 1.

Put X = log x  and Y = log y.

Then xp = P  and yq = Q.

Hence the given equation becomes

P2 + Q2  = z2   ------------- (1)

This equation is of the form F(z,P,Q) = 0.

Therefore, let us assume that Q = aP.

Now, equation (1) becomes,

P2 + a2 P2 = z2 Integrating, we get

Ö(1+a2) log z = X + aY + b.

Therefore, Ö(1+a2) log z = logx + alogy + b, which is the complete solution.

Type (ii) : Equations of the form F(zkp, zkq) = 0 (or) F(x, zkp) = G(y,zkq).

Case (i) : If k ¹-1, put Z = zk+1, Example 20

Solve z4q2 –z2p = 1

The given equation can also be written as

(z2q)2 –(z2p) =1

Here k = 2. Putting Z = z k+1 = z3, we get i.e,   Q2 –3P –9 = 0,

which is of the form F(P,Q) = 0.

Hence its solution is Z = ax + by + c, where b2 –3a –9 = 0.

Solving for b,  b = ± Ö(3a +9)

Hence the complete solution is

Z = ax  + Ö(3a +9) . y + c

or      z3 = ax  + Ö(3a +9) y + c 4 Lagrange’s   Linear   Equation

Equations of the form Pp + Qq  = R ________ (1), where P, Q and R are functions   of x,   y,   z,   are   known   as   Lagrang solve this equation, let us consider the equations u = a  and v = b, where a, b are arbitrary constants and u, v are functions of x, y, z. Equations (5) represent a pair of simultaneous equations which are of the first order and of first degree.Therefore, the two solutions of (5) are u = a and v = b. Thus, f( u, v ) = 0 is the required solution of (1).

Note :

To solve the Lagrange‟s equation,we have to form the subsidiary or auxiliary  equations which can be solved either by the method of grouping or by the method of multipliers.

Example 21

Find the  general solution of px + qy = z.

Here, the subsidiary equations are Integrating, log x = log y + log c1

or x = c1 y i.e, c1 = x / y

From the last two ratios, Integrating, log y = log z  + log c2

or  y = c2 z

i.e,  c2 = y / z

Hence the required  general solution is

Φ(   x/y,=  0,y/z)where Φ is   arbitrary

Example 22

Solve  p tan x + q tan y  = tan z

The subsidiary equations are Hence the required  general solution is where Φ is   arbitrary

Example 23

Solve (y-z) p + (z-x) q  = x-y

Here the subsidiary equations are Example 24

Find the general solution of  (mz  - ny) p  + (nx- lz)q = ly - mx.   Exercises

Solve the following equations

1.     px2 + qy2 = z2

2.     pyz + qzx = xy

3.     xp –yq = y2 –x2

4.     y2zp + x2zq = y2

5.     z (x –y) = px2 –qy2

6.     (a –x) p + (b –y) q = c –z

7.  (y2z p) /x  + xzq = y2

8.     (y2 + z2) p –xyq + xz = 0

9.     x2p + y2q = (x + y) z

10.                      p –q = log (x+y)

11.                      (xz + yz)p + (xz –yz)q = x2 + y2

12.                      (y –z)p –(2x + y)q = 2x + z

5 PARTIAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER WITH CONSTANT COEFFICIENTS.

Homogeneous Linear Equations with constant Coefficients.

A homogeneous linear partial differential equation of the nth order is of the form homogeneous because all its terms contain derivatives of the same order.

Equation (1) can be expressed as As in the case of ordinary linear equations with constant coefficients the complete solution of (1) consists of two parts, namely, the complementary function and the particular integral.

The complementary function is the complete solution of f (D,D') z = 0-------(3), which must contain n arbitrary functions as the degree of the polynomial f(D,D'). The particular integral is the particular solution of equation (2).

Finding the complementary function

Let us now consider the equation f(D,D') z = F (x,y)

The auxiliary equation of (3) is obtained by replacing D by m and D' by 1. Solving equation (4) for „m‟, we get „n‟ roots. Depending upon the nature of the roots, the Complementary function is written as given below: Finding the particular Integral Expand [f (D,D')]-1 in ascending powers of D or D' and operate on xm yn term by term.

Case (iv) : When F(x,y) is any function of x and y. into partial fractions considering f (D,D') as a function of D alone.

Then operate each partial fraction on F(x,y)  in such a way that where c is replaced by y+mx after integration

Example 26

Solve(D3 –3D2D' + 4D'3) z = ex+2y

The auxillary equation is m=m3 –3m2 + 4 = 0

The roots are m = -1,2,2

Therefore the C.F is f1(y-x) + f2 (y+ 2x) + xf3 (y+2x). Hence, the solution is z  = C.F. + P.I Example 27

Solve (D2 –4DD' +4 D' 2) z  = cos (x –2y)

The auxiliary equation is m2 –4m + 4 = 0

Solving, we get m = 2,2.

Therefore the        C.F is f1(y+2x) + xf2(y+2x). Example 28

Solve (D2 –2DD') z = x3y + e5x

The auxiliary equation is m2 –2m = 0.

Solving, we get     m = 0,2.

Hence the    C.F    is       f1 (y) + f2 (y+2x).  Example 29 The auxiliary equation is m2 + m –6 = 0.

Therefore,   m = –3, 2.

Hence the C.F is   f1(y-3x) + f2(y + 2x). =   ò(c + 3x) d(–cosx) –2òcosx  dx

= (c + 3x) (–cosx) –(3) ( - sinx) –2 sinx

=   –y cosx + sinx

Hence the complete solution is

z = f1(y –3x) + f2(y + 2x) –y cosx + sinx

Example 30

Solve r –4s + 4t = e 2x +y i.e, (D2 –4DD' + 4D' 2  ) z = e2x + y

The auxiliary equation is m2 –4m + 4 = 0.

Therefore, m = 2,2

Hence the C.F is f1(y + 2x) + x f2(y + 2x). Since D2 –4DD'+4D'2 = 0 for D = 2  and D' = 1, we have to apply the general rule. 6 Non –Homogeneous Linear Equations

Let us consider the partial differential equation

f (D,D') z = F (x,y)------- (1)

If f (D,D') is not homogeneous, then (1) is a non–homogeneous linear partial differential equation. Here also, the complete solution = C.F + P.I.

The methods for finding the Particular Integrals are the same as those for homogeneous linear equations.

But for finding the C.F, we have to factorize f (D,D') into factors of the form D –mD' –c.

Consider now the equation

(D –mD' –c) z = 0 -----------  (2).

This equation can be expressed as

p –mq = cz ---------(3),

which is in Lagrangian form.

The subsidiary equations are The solutions of (4) are y + mx = a and z = becx.

Taking b =  f (a), we get z = ecx f (y+mx) as the solution of (2).

Note:

1.  If   (D-m1D' –C1) (D –m2D'-C2)   …… –m(Dn'-Cn) z = 0 is the partial

differential equation, then its complete solution is

z = ec1x f1(y +m1x) + ec2x f2(y+m2x) + . . . . . + ecnx fn(y+mnx)

2.  In the case of repeated factors, the equation (D-mD' –C)nz = 0 has a complete

solution z = ecx f1(y +mx) + x ecx f2(y+mx) + . . . . . +x n-1 ecx fn(y+mx).

Example 31

Solve (D-D'-1) (D-D' –2)z = e 2x –y

Here  m1 = 1, m2 = 1, c1 = 1, c2 = 2.

Therefore, the C.F is ex f1 (y+x) + e2x f2 (y+x). Example 32

Solve (D2 –DD' + D' –1) z = cos (x + 2y)

The given equation can be rewritten as

(D-D'+1) (D-1) z = cos (x + 2y)

Here m1 = 1, m2 = 0, c1 = -1, c2 = 1.

Therefore, the C.F = e–x f1(y+x) + ex f2 (y) Example 33

Solve [(D + D'–1) (D + 2D' –3)] z = ex+2y + 4 + 3x +6y

Here m1 = –1, m2 = –2 , c1 = 1, c2 = 3.

Hence the C.F is z = ex f1(y –x) + e3x f2(y –2x).  It is the complete solution.

Exercises

(a) Solve the following homogeneous Equations. 6. (D2 + 4DD' –5D'2) z = 3e2x-y + sin (x –2y)

7. (D2 –DD' –30D'2) z  = xy + e6x+y

8. (D2 –4D' 2) z = cos2x. cos3y

9. (D2 –DD'  –2D'2) z = (y –1)ex

10.4r + 12s + 9t =  e3x –2y

(b)Solve the following non –homogeneous equations.

1.     (2DD' + D' 2 –3D') z = 3 cos(3x –2y)

2.     (D2 + DD' + D' –1) z = e-x

3.     r –s + p = x2 + y2

4.     (D2 –2DD' + D'2 –3D + 3D' + 2)z = (e3x + 2e-2y)2

(D2 –D'2 –3D + 3D') z = xy + 7.

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Mathematics (maths) - Partial Differential Equations : Partial Differential Equations |

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