Separable Partial Differential Equation

7/29/2019 Separable Partial Differential Equation

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BITS PilaniPilani Campus

BITS Pilani

presentationDr. Padma Murali

Department of Mathematics


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BITS PilaniPilani Campus

Engineering Mathematics

IILecture 16


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BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Separable partial differential equation

Review: Partial differential equation(PDEs), like ordinary differential equations

(ODEs), are classified as linear or

nonlinear. Analogous to a linear ODE (see(6) of section 1.1) the dependent variable

and its partial derivatives appear only to

the first power in a linear PDE. In this

chapter we are concerned only with linear

partial differential equations.


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Linear Partial Differential Equation: If we

let u denote the dependent variable and x

and y the independent variables, then the

general form of a linear second-order

partial differential equation is given by

.2

22

2

2

GFuy

uE

x

uD

y

uC

yx

uB

x

uA


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Where the coefficients A,B,C,G areconstants or functions of x and y. When

G(x,y) = 0. equation (1) is said to be

homogeneous; otherwise it isnonhomogeneous. For example, the linear

equations

xyy

u

x

u

andy

u

x

u

2

2

2

2

2

2

0

are homogeneous and nonhomogeneous,

respectively.


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Separation of variables:

Although there are several methods that canbe tried to find particular solutions of a linear

PDE, in the method of separation of variables

we seek to find a particular solution of the formof a product of a function of x and a function of

y. yYxXyxu ,

With this assumption, it is sometimes possible

to reduce a linear PDE in two variables to two

ODEs. To this end we observe that


7/72BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

YXy

uYX

x

uYX

y

uYX

x

u

2

2

2

2

,,,

Where the primes denote ordinarydifferentiation.

Example: Using Separation of Variables

Find product solution of yu

xu

42

2

Solution: Substituting u(x,y) = X (x) Y(y) into

the partial differential equation yields



YXYX 4After dividing both sides by 4XY we have

separated the variables:

.4 Y

Y

X

X

Since the left hand side of the last equationis independent of y and is equal to the right

hand side, which is independent of x, we

conclude that



both sides of the equation are independentof x and y. In other words, each side of the

equation must be a constant. As a practical

matter it is convenient to write this realseparation constant as From the two equalities.

Y

Y

X

X

4

We obtain the two linear ordinary differential

equations.



004 YYandXX

For the three cases for

zero, negative, or positive; that is

0,00,0 22 whereand

the ODEs in (2) are, in turn

,00

YandX004 22 YYandXX

004 22 YYandXX



Case I: 0

The DEs in (3) can be solved by integration.

The solutions are X = c1 + c2x and Y=c3.

Thus a particular product solution of the

given PDE is

xBAcxccXYu

11321

Where we have replaced c1c3 and c2c3 by

A1 and B1, respectively



Case II 2 The general solutions of the DEs in (4) are

yecYandxcxcX2

654 2sinh2cosh

respectively. Thus another particular

product solution of the PDE is

yecxcxcXYu2

654 2sinh2cosh

OR

xeBxeAu yy 2sinh2cosh22

22

.652642 ccBandccA Where



Case III 2

Finally, the general solutions of the DEs in

(5) are:yecYandxcxcX

2

987 2sin2cos

respectively. These results give yet anotherparticular solution

xeBxeAu yy 2sin2cos22

33



Where 973 ccA And 983 ccB

It is left as an exercise to verify that (6), (7)

and (8) satisfy the given partial differential

equation

yxx uu 4

See problem 29 in exercises 13.1



Theorem 13.1 Superposition Principle

If kuuu ,........, 21

are solutions of a homogeneous linear

partial differential equation, then the linearcombination.

kkucucucu .......2211

Where the kici ,....,2,1,

are constants is also a solution.



Definition 13.1 Classification of Equations

The linear second order partial differentialequation

02

22

2

2

Fuy

u

Ex

u

Dy

u

Cyx

u

Bx

u

A

Where A,B,C,D,E and F are real

constants is said to be

Hyperbolic if 042 ACB

Parabolic if 042 ACB

Elliptic if042 ACB



Heat Equation

Introduction

Consider a rod of length L with an initial

temperature f(x) through-out and whoseends are held at temperature zero for all

time t>0. The temperature u(x,t) in the rod is

determined from the boundary value

problem

0,0,2

2

tLx

t

u

x

uk (1)



)3(...0,0,

)2(...0,0,,0,0

Lxxfxu

ttLutu

Solution of the BVP: Using the product

andtTxXtxu ,,as the separation constant, leads to

kT

T

X

X(4)

OR



0 XX (5)0 TkT (6)

Now the boundary conditions in (2) becomes

0,00,0 tTLXtLuandtTXtuSince the last equalities must hold for all

time t, we must have X(0) = 0 and X(L) = 0.

These homogeneous boundary conditionstogether with the homogeneous ODE(5)

constitute a regular Sturm-Liouville

problem:



0,00,0 LXXXX (7)

The general solution of the DEs are 0,21 ifxccxX

(8)

0,sinhcosh2

21

ifxcxcxX(9)

0,sincos 221 ifxcxcxX (10)



Recall, when the boundary condition X(0) =0 and X (L) = 0 are applied to (8) and (9)

these solutions yield only X(x) = 0 and so

we are left with the unusable result u = 0.Applying the first boundary condition X(0) =

0 to the solution in (10) gives c1 = 0.

Therefore xcxX

sin2

The second boundary condition X(L) = 0

now implies

0sin

2 LcLX



If c2 = 0, then X = 0 so that u=0. But (11)can be satisfied for 02 c when

.0sin L

This last equation implies that ,/LnL where n = 1,2,3,.. Hence (7) possesses

nontirivial solution when

,....3,2,1,/

2222

nLnnn The values n

and the corresponding solutions

,.....3,2,1,sin2 nxL

ncxX


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are the eigenvalues and eigenfunctions,respectively, of the problem in (7). The

general solution of (6) is

,

222 /

3

tLnk

ecT

and so xL

neAtTxXu tLnknn

sin222 / (13)

where we have replaced the constant c2c3 by

An. The products un(x,t) given in (13) satisfythe partial differential equation (1) as well as

the boundary condition (2) for each value of

the positive integer n.


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However, in order of the functions in (13)

to satisfy the initial condition (3), we

would have to choose the coefficient An

in such a manner that

xL

nAxfxu nn

sin0, (14)


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is not a solution of the problem given in (1)

(3). Now by the superposition principlethe function

In general, we would not expect

condition (14) to be satisfied for an

arbitrary, but reasonable, choice of f.

Therefore we are forced to admit that txun ,


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1 1

/ sin,222

n n

tLnk

nn xL

neAutxu

(15)

Must also, although formally, satisfy

equation (1) and the conditions in (2). If we

substitute t = 0 into (15) then

1

.sin0,n

n xL

nAxfxu


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The last expression is recognized as thehalf-range expansion of f in a sine series.

If we make the identification

........3,2,1, nbA nnit follows from (5) to section 12.3 that

L

n dxx

L

nxf

L

A0

.sin2

(16)

We conclude that a solution of the

boundary-value problem described in (1),

(2) and (3) is given by the infinite series


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1

/

0

sinsin2

,223

n

tLnkL

xL

nedxx

L

nxf

Ltxu

(17)

In the special case when the initial

temperature is 1,,1000, kandLxu you should verify that the coefficient (16)

are given by

n

A

n

n

11200

And that the series (17) is

nxen

txu tn

n

n

sin11200

,2

1


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Wave Equation

Introduction

The vertical displacement u(x,t) of a string

of length L that is freely vibrating in thevertical plane is determined from

0,0,2

2

2

2

2

tLxt

u

x

ua (1)


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0,0,0,0 tLutu (2)

.0,,0,0

Lxxgt

uxfxu

t

(3)

Solution of the BVP with the usual

assumption that u(x,t) = X (x) T(t), separating

variables in (1) gives

Ta

T

X

X2


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So that0 XX

02 TaT

(4)

(5)

As in Section 13.3 the boundary condition(2) translate into X(0) = 0 and X(L) = 0.

The ODE in (4) along with these

boundary-conditions is the regular Sturm-Liouville problem

0,00,0 LXXXX (6)


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Of the usual three possibilities for the parameter

.0,0,0: 22 and

only the last choice leads to nontirivalsolutions. Corresponding to 0,2

the general solution of (4) is

xcxcX sincos 21 X(0) = 0 and X (L) = 0 indicate that c1 = 0

and 0sin2

Lc


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The last equation again implies that./LnornL

The eigenvalues and corresponding

eigenfunctions of (6) are ,.....3,2,1,sin/ 2

222 nxL

ncxXandLnn

the general solution of the second order

equation (5) is then

.sincos 43 tL

anct

L

anctT


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By rewriting c2c3 and An and c2c4 as Bn

solutions that satisfy both the wave equation

(1) and boundary conditions (2) are

xL

n

L

anBt

L

anAu nnn

sinsincos

(7)

and

1

.sinsincos,n

nn xL

nt

L

anBt

L

anAtxu


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Setting t = 0 in (8) and using the initial

condition u(x,0) = f(x) gives

1 .sin0, nn xL

n

Axfxu

Since the last series is a half-range

expansion for f in a sine series, we can write

An = bn.

L

n dxxL

nxf

LA

0

.sin2

(9)


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To determine Bn we differentiate (8) with

respect to t and then set t = 0.

1sincossin

n

nn xL

ntL

an

L

anBtL

an

L

anAt

u

10

.sin

n

n

t

xL

n

L

anBxg

t

u

In order for this last series to be half

range sine expansion of the initial velocity g

on the interval, the total coefficient

LanBn /


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must be given by the form bn in (5) of section

12.3 that is,


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L

n xdxLnxg

LLanB

0

sin2

From which we obtain

L

n dxxL

nxgan

B0

sin2

(10)

The solution of the boundaryvalue problem

(1) (3) consists of the series (8) with

coefficients An and Bn defined by (9) and

(10), respectively.


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We note that when the string is released

from rest, then g(x) = 0 for every x in the

interval

Lx 0 and consequently

0

nB


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Separable Partial Differential Equation