010.141 PARTIAL DIFFERENTIAL EQUATIONS MODULE 4ocw.snu.ac.kr/sites/default/files/NOTE/6784.pdf ·...
Transcript of 010.141 PARTIAL DIFFERENTIAL EQUATIONS MODULE 4ocw.snu.ac.kr/sites/default/files/NOTE/6784.pdf ·...
ENGINEERING MATHEMATICS II
010.141
PARTIAL DIFFERENTIAL EQUATIONS
MODULE 4
2B.D. Youn2010 Engineering Mathematics II Module 4
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
22
2
2
ut
ux
= c
¶¶
¶¶
ut
ux
2 2
2 = c
2 2
2 2
u u + = 0x y¶ ¶¶ ¶
( )2 2
2 2
u u + f x, yx y¶ ¶¶ ¶
=
¶¶
¶¶
¶¶
2
22
2
2
2
2
ut
ux
uy
= c + æèç
öø÷
¶¶
¶¶
¶¶
2
2
2
2
2
2
ux
uy
uz
+ + = 0
A PDE is an equation involving one or more partial derivatives of a function.
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EXAMPLE OF WAVE EQUATION
Ø Problem, as a solution to the wave equation
( )( )
( )
( )
( )( )( )
14
14
2142
1 14 4
21 1
16 42
21 1 14 16 4
2
( , ) = sin 9t sin u = 9 cos 9t sin tu = 81 sin 9t sin
tu = sin 9t cos xu = sin 9t cos
x81 sin 9t sin = c sin 9t sin identically true,
if c = 16 81
u t x x
x
x
x
x
x x is
¶¶¶¶¶¶¶¶
×
- ×
×
× -
- × - ×
c = 36® ±
¶¶
¶¶
2
22
2
2
ut
ux
= c ( , )u t x
4B.D. Youn2010 Engineering Mathematics II Module 4
EXAMPLE OF HEAT EQUATION
22
2
4t
4t
4t
24t
2
4t 4t 2
2
u u = ct x
u = e cos 3xu = 4e cos 3xtu = 3e sin 3xxu = 9e cos 3x
x4e cos 3x = 9e cos 3x c
4if c = , an identity9
-
-
-
-
- -
¶ ¶¶ ¶
¶-
¶¶
-¶¶
-¶- - ×
5B.D. Youn2010 Engineering Mathematics II Module 4
EXAMPLE OF LAPLACE EQUATION
¶¶
¶¶
¶¶¶¶¶¶
¶¶
2
2
2
2
2
2
2
2
ux
uy
uxu
xuyu
y
+ = 0
= e sin y
= e sin y
= e sin y
= e cos y
= e sin y
e sin y e sin y = 0 0 = 0
u x
x
x
x
x
x x
-
-
6B.D. Youn2010 Engineering Mathematics II Module 4
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
7B.D. Youn2010 Engineering Mathematics II Module 4
DERIVATION OF WAVE EQUATION
Deflected string at fixed time t
T1, T2 = tension in string at point P and QT1 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
8B.D. Youn2010 Engineering Mathematics II Module 4
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 accelerationuIf u is the vertical position, = acceleration
T sin T
Thus
t
rr
¶¶
b
DD D
´
- ( )
( )
( )
2
2
22 1
22 1
+ x
usin = x
T sin T sin x u = tan tan = equation 2T cos T cos T
uAt P x is distance from origin , tan is slope =
uLikewise at Q, tan =
x
x
t
t
x
x
¶a r¶
b a r ¶b ab a ¶
¶a¶
¶b¶ D
D
D- -
9B.D. Youn2010 Engineering Mathematics II Module 4
DERIVATION OF WAVE EQUATION (Cont.)
Substituting and x: 1x
u u = T
u
x 0, L.H.S. becomes u
= T so that
u = c u
e 1- D Wave equationIf T or c
+ x
2
2
2
¸ -é
ëêê
ù
ûúú
®
¯
DD
D
D
¶¶
¶¶
r ¶
¶
¶
¶
r
¶
¶
¶
¶
r
x x t
as x
Let c
t xThis is th
x x
2
2
2
2
2
2
2
2
,
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Solution by Separating Variables
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Solution by Separating Variables
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Solution by Separating Variables
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Solution by Separating Variables
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Solution by Separating Variables
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Solution by Separating Variables
16B.D. Youn2010 Engineering Mathematics II Module 4
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
¶¶
¶¶
ut
ux
= c22
2
¶¶ srut
u = c c = k2 2Ñ2
17B.D. Youn2010 Engineering Mathematics II Module 4
CONVERT TOORDINARY 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 = 0Solution for F is:
F = A cos px +B sin px
Applying boundary conditions at x = 0 and x = L gives A = 0 andsin pL = 0, which results in pL = np, or:
p = (np)/L.Now
Fn = sin (npx/L)
let ln = (cnp/L), and proceed to time solution
18B.D. Youn2010 Engineering Mathematics II Module 4
TIME-DEPENDENT SOLUTION
Gn = Bnexp(-ln2t)
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 Bn must be the coefficients of the Fourier sine series:
( )untnx, t = B sin n x
L en
p l- 2
( ) ( )u xLn
, = B sin n x = f xn= 1
0¥
å p
( )BLn
L
= 2 f x sin dx0ò
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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, lnt, 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
20B.D. Youn2010 Engineering Mathematics II Module 4
VIBRATING MEMBRANE ANDTHE 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
21B.D. Youn2010 Engineering Mathematics II Module 4
GEOMETRY OF VIBRATING "DRUM"
Vibrating Membrane
22B.D. Youn2010 Engineering Mathematics II Module 4
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 ut
x x xD D D D D = T y u + x, y u , y + T x u x , y + y u , y2
x 1 x 2 y 1 y2 2- -
( ) ( ) ( ) ( )¶¶ r
2x 1 x 2 y 1 y = T u + x, y u , y
+ u x , y + y u , y
yu
tx x
xx
22D
D
D
D
- -é
ëêê
ù
ûúú
¶¶
¶¶
¶¶
22
2 2 = c + u
tu
xu
y2 2 2
æ
èç
ö
ø÷
T = c2
r
23B.D. Youn2010 Engineering Mathematics II Module 4
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
22u
tu = c2 Ñ
24B.D. Youn2010 Engineering Mathematics II Module 4
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.
( )&&GG
Fxxc = 1
F + F 2 yy = -n2
&&G G + = 02l
( )FG F G Gxx&& ; = c + F2
yy
F F xx + F + = 0yy n2
25B.D. Youn2010 Engineering Mathematics II Module 4
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 22
2 2Q = + d H d QH HQdx dy
næ ö
- ç ÷è ø
1 12
2
2
2Hd Hdx Q
d Qdy
Q = + = k2 2-æ
èç
ö
ø÷ -n
d Hdx
H d Qdy
Q2
2
2
20 0 + k + p2 2= =,