010.141 PARTIAL DIFFERENTIAL EQUATIONS MODULE 4

(1)

ENGINEERING MATHEMATICS II

010.141 PARTIAL DIFFERENTIAL EQUATIONS

MODULE 4

(2)

EXAMPLES OF PDE

One-D Wave equation

One-D Heat equation

Two-D Laplace equation

Two-D Wave Equation

Two-D Poisson equation

Three-D Laplace Equation

¶

2 2

2

u t

u = c x

¶

¶ u

t

u x

2 2

= c 2

2 2

u u

+ = 0

x y

¶ ¶

( )

2 2

u u

+ f x, y

x y

¶ ¶

¶ ¶ ⁼

¶

2 2

2

2 2

u t

u x

u = c æ + y

èç ö

ø÷

¶

2 2

u x

u y

u

+ + z = 0

A PDE is an equation involving one or more partial derivatives of a function.

(3)

EXAMPLE OF WAVE EQUATION

Ø Problem, as a solution to the wave equation

( ) ( )

( )

1 4

1 4 2

1 2 4

1 1

4 4

2

1 1

16 4

2

1 2 1 1

4 16 4

( , ) = sin 9t sin u = 9 cos 9t sin

t

u = 81 sin 9t sin t

u = sin 9t cos x

81 sin 9t sin = c sin 9t sin identically true,

u t x x

x

x x

x

x x is

¶

×

- ×

×

× -

- × - ×

¶

2 2

2

u t

u

= c x ^{u t x}^{( , )}

(4)

EXAMPLE OF HEAT EQUATION

2 2

2 4t

4t

2

4t 2

4t 4t 2

2

u u

t = c x u = e cos 3x

u = 4e cos 3x t

u = 3e sin 3x x

u = 9e cos 3x x

4e cos 3x = 9e cos 3x c if c = 4, an identity

9

-

- -

¶ ¶

¶ -

¶

¶ -

¶

¶ -

¶

- - ×

(5)

EXAMPLE OF LAPLACE EQUATION

¶

2 2

u x

u y u

x u x u y

u y

+ = 0

= e sin y = e sin y

= e sin y = e cos y

= e sin y

e sin y e sin y = 0

u ^x

x

x x

- -

(6)

VIBRATING STRING AND THE WAVE EQUATION

Ø General assumptions for vibrating string problem:

Ø mass per unit length is constant; string is perfectly elastic and no resistance to bending.

Ø tension in string is sufficiently large so that gravitational forces may be neglected.

Ø string oscillates in a plane; every particle of string moves vertically only;

and, deflection and slope at every point are small in terms of absolute value.

Deflected string at fixed time t

(7)

DERIVATION OF WAVE EQUATION

Deflected string at fixed time t

T₁, T₂ = tension in string at point P and Q

T₁ cos a = T₂ cos b = T, a constant (as string does not move in horizontal dir.)

Vertical components of tension:

– T₁ sin a and T₂ sin b

(8)

DERIVATION OF WAVE EQUATION (Cont.)

2 2

2 1

Let x = length PQ and = mass unit length x has mass x

Newtonns Law: F = mass acceleration

If u is the vertical position, u = acceleration

T sin T Thus

t r

r

¶

¶ b D

D D

´

-

( )

2 2

2

2 1

2

2 1

+ x

sin = x u

T sin T sin x u

= tan tan = equation 2

T cos T cos T

At P x is distance from origin , tan is slope = u Likewise at Q, tan = u

x

t

x

a r ¶

¶

b a r ¶

b a

b a ¶

a ¶

¶ b ¶

¶ _D

D

- - D

(9)

DERIVATION OF WAVE EQUATION (Cont.)

Substituting and x: 1 x

u u

= T

u

x 0, L.H.S. becomes u = T

so that

u

= c u e 1- D Wave equation

If T or c

+ x

2

¸ é -

ë êê

ù û úú

®

¯

D D

D

¶

r ¶

¶

¶ r

¶

¶ r

x x t

as

x Let c

t x

This is th

x x

2 2 2

2

2 2

,

(10)

Solution by Separating Variables

(11)

Solution by Separating Variables

(12)

Solution by Separating Variables

(13)

Solution by Separating Variables

(14)

Solution by Separating Variables

(15)

Solution by Separating Variables

(16)

From prior work the heat equation is:

In one dimension (laterally insulated):

Some boundary conditions at each end:

u(0, t) = u(L, t) = 0, for all t

Initial Condition:

u(x, 0) = f(x), specified 0 ≤ x ≤ L

HEAT EQUATION

¶

¶ u

t

u = c² x

2 2

¶

¶ sr

u

t = c₂ u c = ₂ k Ñ²

(17)

CONVERT TO

ORDINARY DIFFERENTIAL EQUATIONS

U(x,t) = F(x)G(t) FG' = c²F"G

¸ FGc²

G'/(c²G) = F"/F = -p² F" + p²F = 0, and

G' + (cp)²G = 0 Solution for F is:

F = A cos px +B sin px

Applying boundary conditions at x = 0 and x = L gives A = 0 and sin pL = 0, which results in pL = np, or:

p = (np)/L.

Now

F_n= sin (npx/L) let l_n= (cnp/L), and proceed to time solution

(18)

TIME-DEPENDENT SOLUTION

G_n= B_nexp(-l_n²t) Combined solution in terms of space and time:

This solution must satisfy the initial condition that u(x,0) equals f(x) Thus:

As before, the constants B_n must be the coefficients of the Fourier sine series:

( )

u_n x, t = B sin n x ⁿ^t L e

n

p _-_l²

( ) ( )

u x, = _n B sin n xL

= f x

n

= 1

0

¥

å

^p

( )

Bⁿ L

L

= 2

f x sin dx

0

ò

(19)

SOME OBSERVATIONS

Ø The exponential terms in u(x,t) approach zero for increasing time, and this should be expected, as the edge conditions (x = 0 and x = L) are at zero, and there is no internal heat generation.

Ø Since exp(-0.001785t) is 0.5, whenever the exponent for the nth term, l_nt, is -0.693, the temperature has decreased by 1/2.

Ø If the ends of the bar were perfectly insulated, then over time the bar temperature will approach some uniform value

(20)

VIBRATING MEMBRANE AND

THE TWO-DIMENSIONAL WAVE EQUATION

Three Assumptions:

Ø The mass of the membrane is constant, the membrane is perfectly flexible, and offers no resistance to bending;

Ø The membrane is stretched and then fixed along its entire boundary in the x-y plane. Tension per unit length (T) which is caused by stretching is the same at all points and in the plane and does not change during the motion;

Ø The deflection of the membrane u(x,y,t) during vibratory motion is small compared to the size of the membrane, and all angles of

inclination are small

(21)

GEOMETRY OF VIBRATING "DRUM"

Vibrating Membrane

(22)

NOW FOR NEWTON'S LAW

Divide by rDxDy:

Take limit as Dx ® 0, Dy ® 0

This is the two-dimensional wave equation

(^rD^{x y}^D ) ^¶_¶_t^u ^{= T y u}^D

[ (

^x+ x, y ^D

)

u

(

^x, y

) ]

+ T x u^D

[ ⁽

x , y + y ^D

⁾

u

⁽

^x , y

⁾ ]

2

x 1 x 2 y 1 y

2 - - 2

( ) ( ) ( ) ( )

¶

¶ r

2 x 1 x 2 y 1 y

= T u + x, y u , y

+ u x , y + y u , y y

u t

x x

x

2

D 2

D

D D

- -

é ë ê ê

ù û ú ú

¶

2

2 2

= c + u

t

u x

u

2 2 y2

æ è

ç ö

ø

÷ T = c²

r

(23)

WAVE EQUATION

Ø In Laplacian form:

where

c²= T/r

§ Some boundary conditions: u = 0 along all edges of the boundary.

§ Initial conditions could be the initial position and the initial velocity of the membrane

§ As before, the solution will be broken into separate functions of x,y, and t.

§ Subscripts will indicate variable for which derivatives are taken.

¶

2 2

u 2

t = c² Ñ u

(24)

SOLUTION OF 2-D WAVE EQUATION

Let

u(x, y, t) = F(x, y)G(t) substitute into wave equation:

divide by c²FG, to get:

This gives two equations, one in time and one in space. For time,

where l = cn, and, what is called the amplitude function:

also known as the Helmholtz equation.

( )

&&

G

G F_xx

c = 1

F + F

2 yy = -n²

&&

G + l²G = 0

( )

FG&& = c² F G _xx + F_yyG ;

F_xx + F + _yy n²F = 0

(25)

SEPARATION OF THE HELMHOLTZ EQUATIONS

Let F(x, y)=H(x)Q(y) and, substituting into the Helmholtz:

Here the variables may be separated by dividing by HQ:

Note: p² = n² – k²

As usual, set each side equal to a constant, -k². This leads to two ordinary linear differential equations:

2 2

2

2 Q = 2 +

d H d Q

H HQ

dx _{- ç}^æ dy n ^ö_÷

è ø

1 ² 1

2

H 2

d H

dx Q

d Q

dy Q

= - æ + ² = k² è

ç ö

ø

÷ -

n

d H H d Q

Q

2 2

0 0

+ k² = , + p² =