Partial Differential Equation - Notes

Partial Differential Equations

Introduction

Partial Differential Equations(PDE) arise when the functions involved ordepend on two or more independent variables. Physical and Engineeringproblems like solid and fluid mechanics, heat transfer, vibrations, electro-magnetic theory and other areas lead to PDE.

Partial Differential Equations are those which involve partialderivatives with respect to two or more independent variables.

E.g.:

∂2u

∂x2+∂2u

∂y2= 0 Two-dimensional Laplace equation ..(1)

∂2u

∂t2= c2

∂2u

∂x2One-dimensional Wave equation ..(2)

∂2u

∂x2+∂2u

∂y2+∂2u

∂z2= 0 Three-dimensional Laplace equation ..(3)

∂u

∂t= c2

∂2u

∂x2One-dimensional heat equation ..(4)

∂z

∂x+∂z

∂y= 1 ..(5)

The ORDER of a PDE is the order of the highest derivatives in theequation.

• NOTE: We restrict our study to PDE’s involving one dependent vari-able z and only two independent variables x and y.

1


• NOTATIONS:∂z

∂x= p,

∂z

∂y= q,

∂2z

∂x2= r,

∂2z

∂x∂y= s and

∂2z

∂y2= t

Ex. 1 Show that u = sin9t sin(x4) is a solution of a one dimensional wave

equation.

Sol. Here u = sin9t sin(x4).

∂2u

∂t2= c2

∂2u

∂x2.. (1) is one dimensional wave equation.

∂u

∂t=

∂2u

∂t2=

∂u

∂x=

∂2u

∂x2=

Substitute above values in eq. (1), we get

2 Dept. of Mathematics, AITS - Rajkot


c =

∴ For c = , u = sin9t sin(x4) is a solution of a wave equation.

Ex. 2 Verify that the function u = excosy is the solution of the Laplace

equation∂2u

∂x2+∂2u

∂y2= 0.

Sol.

Dept. of Mathematics, AITS - Rajkot 3


Ex. 3 Verify that the function u = x3 + 3xt2 is the solution of the wave

equation∂2u

∂t2= c2

∂2u

∂x2for a suitable value of c.

Sol.



Exercise 1.1

Q.1 Verify the following functions are the solutions of the Laplace’s equation∂2u

∂x2+∂2u

∂y2= 0.

(a) u = log(x2 + y2)

(b) u = sinxsinhy

(c) u = tan−1(yx

)Q.2 Verify the following functions are the solutions of the wave equation∂2u

∂t2= c2

∂2u


(a) u = x3 + 3xt2

(b) u = cosctsinx

Q.3 Verify the following functions are the solutions of the heat equation∂u

∂t= c2

∂2u


(a) u = e−2tcosx



Formation of PDE

Partial Differential equations can be formed either by the elimination of(1) arbitrary constants present in the functional relation present in the rela-tion between the variable(2) arbitrary functions of these variables.

By elimination of arbitrary constants

Consider the function f(x, y, z, a, b) = 0 ..(i)

where a and b are two independent arbitrary constants. To eliminate twoconstants, we require two more equations.

Differentiating eq. (i) partially with respect to x and y in turn, we obtain

∂f

∂x+∂f

∂z

∂z

∂x= 0 ⇒ ∂f

∂x+ p

∂f

∂z= 0 .. (ii)

∂f

∂y+∂f

∂z

∂z

∂y= 0 ⇒ ∂f

∂y+ q

∂f

∂z= 0 .. (iii)

Eliminating a and b from the set of equations (i), (ii) and (iii) , we get aPDE of the first order of the form F (x, y, z, p, q) = 0. .. (iv)

Ex. 1 Eliminate the constants a and b form z = (x+ a)(y + b).

Sol. Given that z = (x+ a)(y + b) ... (1)

Differentiating partially eq. (1) with respect to x and y respectively,we get

∂z

∂x= p = ..(2)

∂z

∂y= q = ..(3)

Eliminating a and b from equation (1), (2) and (3), we get



is the required partial differential equation.

Ex. 2 Find the differential equation of the set of all spheres whose centreslie on the z-axis.Sol. The equation of the spheres with centres on the z-axis is

..(1)

Differentiating partially eq. (1) with respect to x and y respectively,we get

Eliminating arbitrary constant from (2) and (3), we get

is the required partial differential equation.



By elimination of arbitrary functions

Consider a relation between x, y and z of the type φ(u, v) = 0 ..(v)

where u and v are known functions of x, y and z and φ is an arbitraryfunction of u and v.

Differentiating equ. (v) partially with respect to x and y, respectively,we get

∂φ

∂u

(∂u

∂x+∂u

∂zp

)+∂φ

∂v

(∂v

∂x+∂v

∂zp

)= 0 .. (vi)

∂φ

∂u

(∂u

∂y+∂u

∂zq

)+∂φ

∂v

(∂v

∂y+∂v

∂zq

)= 0 .. (vii)

Eliminating∂φ

∂uand

∂φ

∂vfrom the eq. (vi) and (vii), we get the

equations

∂(u, v)

∂(y, z)p +

∂(u, v)

∂(z, x)q =

∂(u, v)

∂(x, y).. (viii)

which is a PDE of the type (iv).

since the power of p and q are both unity it is also linear equation,whereas eq. (iv) need not be linear.

Ex. 1 Eliminate the arbitrary function from the equation

z = xy + f(x2 + y2).

Sol. Let x2 + y2 = u.

∴ z = xy + f(u) .. (1)

differentiating eq. (1) with respect to x and y respectively, we get



p = ..(2)

q = ..(3)

eliminating from (2) and (3), we get

Ex. 2 Eliminate arbitrary function from the equationf(x+ y + z, x2 + y2 + z2) = 0.Sol. Let x+ y + z = u, x2 + y2 + z2 = v then

f(u, v) = 0 .. (1)

Differentiating (1) with respect to x and y, we get

∂f

∂u

(∂u

∂x+∂u

∂zp

)+∂f

∂v

(∂v

∂x+∂v

∂zp

)= 0 .. (2)

∂f

∂u

(∂u

∂y+∂u

∂zq

)+∂f

∂v

(∂v

∂y+∂v

∂zq

)= 0 .. (3)

Now

∂u

∂x=

∂v

∂x=

∂u

∂y=

∂v

∂y=

∂u

∂z=

∂v

∂z=



substituting above values in eq. (2) and (3), we get

eliminating and from above eq. (4) and (5), we get

is the required partial differential equation of the first order.



Exercise 1.2

Q.1 Form the partial differential equation by eliminating the arbitraryconstants from the following:

1. z = (x2 + a)(y2 + b)

2. (x− a)2 + (y − b)2 + z2 = 1

Q.2 Form the partial differential equations by eliminating the arbitraryfunctions from the following:

1. z = f(x2 − y2)

2. z = x+ y + f(xy)

3. f(xy + z2, x+ y + z) = 0

4. f(x2 + y2 + z2, xyz) = 0



Integrals of Partial Differential Equation

• A solution or integral of a partial differential equation is a relationbetween the dependent and the independent variables that satisfiesthe differential equation.

• A solution which contains a number of arbitrary constants equal tothe independent variables, is called a complete integral.

• A solution obtained by giving particular values to the arbitraryconstants 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)

eliminating a, b between eq. (2) and∂f

∂a= 0,

∂f

∂b= 0

if it exists, is called a singular integral.

• General Integral

In eq. (2), if we assume b = φ(a), then (2) becomes

f [x, y, z, a, φ(a)] = 0 .. (3)

differentiating (2) partially with respect to a,

∂f

∂a+∂f

∂bφ′(a) = 0 .. (4)

eliminating a between theses two equations (3) and (4), if it exists, iscalled the general integral of eq. (1).



Solutions of PDE by the method of Direct

Integration

This method is applicable to those problems, where direct integration ispossible.

Ex. 1 Solve∂2z

∂x∂y= x3 + y3

Sol. Given that∂

∂x

(∂z

∂y

)= x3 + y3

Integrating w.r.t to keeping constant, we get

∴∂z

∂y=

Now integrating w.r.t. to keeping as constant

∴ z =



Ex. 2 Solve∂2u

∂x∂t= e−tcosx

Sol. Given that∂

∂x

(∂u

∂t

)= e−tcosx


Now integrating w.r.t. to keeping as constant



Ex. 3 Solve∂3z

∂x2∂y= cos(2x+ 3y)

Sol. Given that∂2

∂x2

(∂z

∂y

)= cos(2x+ 3y)


∴∂

∂x

(∂z

∂y

)=

Integrating w.r.t. to keeping as constant

Finally integrating w.r.t. to keeping as constant



Lagrange’s Equation

We have∂(u, v)

∂(y, z)p+

∂(u, v)

∂(z, x)q =

∂(u, v)

∂(x, y)This can be expressed in the form Pp+Qq = R, where P, Q and R arefunctions of x, y and z. This partial differential equation is known asLagrange’s equation.

Method to solve Pp+Qq = R

In order to solve the equation Pp+Qq = R

1 Form the subsidiary (auxiliary ) equation

dx

P=dy

Q=dz

R

2 Solve these subsidiary equations by the method of grouping or by themethod of multiples or both to get two independent solutions u = c1and v = c2.

3 Then φ(u, v) = 0 or u = f(v) or v = f(u) is the general solution ofthe equation Pp+Qq = R.

Ex. 1 Solvey2z

x

∂z

∂x+ xz

∂z

∂y= y2.

Sol. The given equation can be written as ( in for of p and q)

y2zp+ x2zq = xy2

comparing with Pp+Qq = R, we get

P = , Q = and R =

The subsidiary equations are

dx

P=dy

Q=dz

R

Taking first two, we have



Now taking first and third, we have

Ex. 2 Find the general solution of the differential equationx2p+ y2q = (x+ y)zSol. Comparing with Pp+Qq = R, we get

P = , Q = and R =

The subsidiary equations are

dx

P=dy

Q=dz

R



Taking first two, we have

Ex. 3 Solve (y + z)p+ (z + x)q = x+ y

Ex. 4 Solve pz − qz = z2 + (x+ y)2

Ex. 5 Solve x2(y − z)p+ y2(z − x) = z2(x− y)

Ex. 6 Solve (z2 − 2yz − y2)p+ (xy + zx)q = xy − zx

Exercise

1. xp+ yq = z

2. xp− yq = xy

3. (1− x)p+ (2− y)q = 3− z

4. zxp− zyq = y2 − x2

5. (y − z)p+ (z − x)q = x− y

6. p− 2q = 2x− ey + 1

7. x(y2 − z2)p+ y(z2 − x2)q = z(x2 − y2)



Special type of Nonlinear PDE of the first

order

A PDE which involves first order derivatives p and q with degree more thanone and the products of p and q is called a non-linear PDE of the firstorder.There are four standard forms of these equations.

1. Equations involving only p and q

2. Equations not involving the independent variables

3. Separable equations

4. Clairaut’s form

Standard Form 1. Equations involving only p and q

Such equations are of the form f(p,q)=0

z = ax+ by + c ... (1)

is a solution of the equation f(p, q) = 0

provided f(a, b) = 0 (means put p = a and q = b) ... (2)

solving (2) for b, b = F (a)

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

where a and c are arbitrary constants.

Ex. 1 Solve pq = p+ qSol. Here the give DE involves on p and q (so standard form 1)

The solution is

put p = and q =

therefore,

b =



Hence the complete solution is

Exercise

Solve the following equations

1. pq = 1

2. p = q2

3. p2 + q2 = 4

4. pq + p+ q = 0



Standard Form 2. Equations not involving theindependent variables

Such equations are of the form f(z,p,q)=0

Consider f(z, p, q) = 0 .. (1)

since z is a function of x and y

dz =∂z

∂xdx+

∂z

∂ydy = pdx+ qdy

let q = ap

then eq (1) becomes f(z, p, ap) = 0 .. (2)

solving (2) for p i.e. p = φ(z, a)

∴ dz = φ(z, a) dx + a φ(z, a) dy

∴dz

φ(z, a)= dx+ ady

integrating, we get∫ dz

φ(z, a)= x+ ay + b which is a complete integral.

Ex. 1 Solve: p2z2 + q2 = 1Sol. Given equation involves only p, q and z (standard form 2)

Therefore putting q = ap in given equation, we get

Now dz = pdx+ qdy, substituting above values in this equation andintegrate



Ex. 2 Solve: p(1 + q) = qzEx. 3 Solve: z = p2 + q2

Exercise

1 q(p2z + q2) = 4

2 pq = z2

3 p+ q = z

4 zpq = p+ q

5 z2 = 1 + p2 + q2



Standard Form 3. Separable equations

Such equations are of the form f1(x,p) = f2(y,q)

A first order PDE is seperable if it can be written in the form

f1(x, p) = f2(y, q)

We assume each side equal to an arbitrary constant a.

solving f1(x, p) = a ⇒ p = φ1(x, a)

solving f2(y, q) = a ⇒ q = φ2(y, a)

∴ z =∫φ1(x, a)dx+

∫φ2(y, a)dy which is complete integral

Ex. 1 Solve: p− x2 = q + y2

Sol. Writing equation in the form

p− x2 = q + y2 = a

∴ p = a+ x2 and q = a− y2

Now dz = pdx+ qdy

dz = dx+ dy

Integrating both sides, we get

z =

Ex. 2 Solve: p2 + q2 = x2 + y2

Sol. Writing equation in the form

p2 − x2 = y2 − q2

Let p2 − x2 = a and y2 − q2 = a

Hence, p2 = x2 + a, q2 = y2 − a



∴ p = and q =

Now dz = pdx+ qdy

dz =

Exercise

1. q − p+ x− y = 0

2. p+ q = x+ y

3. P 2 + q2 = x+ y

4. pq = xy

5. py + qx = pq

6. p+ q = sinx+ siny



Standard Form 4. Clairaut’s form

A first-order PDE is said to be of Clairaut type if it can be written in theform,

z = px+ qy + f(p, q)

substitute p = a and q = b in f(p, q)

The solution of the the equation is z = ax+ by + f(a, b)

Ex. 1 Solve: z = px+ qy +√

1 + p2 + q2

Sol. The complete integral is obtained

z = px+ qy +√

1 + a2 + b2

Ex. 2 Solve: (p+ q)(z − xp− yq) = 1.Sol.

Exercise

Solve the following equations.

1. z = px+ qy + p2q2

2. z = px+ qy + 2√pq

3. z = px+ qy +q

p− p

4. (1− x)p+ (2− y)q = 3− z

5.z

pq=x

q+y

p+√pq



Classification of the PDE of second order

Consider the equation of the second order as

A∂2u

∂x2+B

∂2u

∂x∂y+ C

∂2u

∂2y+ f

{x, y, u,

∂u

∂x,∂u

∂y

}= 0 (1)

where A is positive

Now the equation (1) is

(i) elliptic if B2 − 4AC < 0

(ii) hyperbolic if B2 − 4AC > 0

(iii) parabolic if B2 − 4AC = 0.

Exercise

Classify the following differential equations.

1 2∂2u

∂t2+ 4

∂2u

∂x∂t+ 3

∂2u

∂x2= 0

2 4∂2u

∂t2− 9

∂2u

∂x∂t+ 5

∂2u

∂x2= 0

3∂2u

∂t2+ 4

∂2u

∂x∂t+ 4

∂2u

∂x2= 0


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Partial Differential Equation - Notes