010.141 partial differential equations module 4ocw.snu.ac.kr/sites/default/files/note/6784.pdf ·...
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![Page 1: 010.141 PARTIAL DIFFERENTIAL EQUATIONS MODULE 4ocw.snu.ac.kr/sites/default/files/NOTE/6784.pdf · 2018. 1. 30. · PARTIAL DIFFERENTIAL EQUATIONS MODULE 4. 2 B.D. Youn 2010 Engineering](https://reader036.vdocuments.net/reader036/viewer/2022071409/610241b76dc5ce6a4214ffbb/html5/thumbnails/1.jpg)
ENGINEERING MATHEMATICS II
010.141
PARTIAL DIFFERENTIAL EQUATIONS
MODULE 4
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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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3B.D. Youn2010 Engineering Mathematics II Module 4
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
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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
-
-
-
-
- -
¶ ¶¶ ¶
¶-
¶¶
-¶¶
-¶- - ×
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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
-
-
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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
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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
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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- -
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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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10B.D. Youn2010 Engineering Mathematics II Module 4
Solution by Separating Variables
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11B.D. Youn2010 Engineering Mathematics II Module 4
Solution by Separating Variables
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12B.D. Youn2010 Engineering Mathematics II Module 4
Solution by Separating Variables
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13B.D. Youn2010 Engineering Mathematics II Module 4
Solution by Separating Variables
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14B.D. Youn2010 Engineering Mathematics II Module 4
Solution by Separating Variables
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15B.D. Youn2010 Engineering Mathematics II Module 4
Solution by Separating Variables
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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
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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
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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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19B.D. Youn2010 Engineering Mathematics II Module 4
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
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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
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21B.D. Youn2010 Engineering Mathematics II Module 4
GEOMETRY OF VIBRATING "DRUM"
Vibrating Membrane
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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
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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
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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
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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= =,