G G G
EEEOOO---D D D
IIISSSAAASSSTTTEEERRRP P P
RRREEEVVVEEENNNTTTIIIOOONNNS S S
YYYSSSTTTEEEMMM 1/4G G G
EEEOOOS S S
CCCIIIEEENNNCCCEEE&&&G G G
EEEOOOE E E
NNNGGGIIINNNEEEEEERRRIIINNNGGG LLLAAABBB http://www.cm.nitech.ac.jp/maeda-lab/3. 微分方程式の導出 Introduction of PDE of 2
ndorder
3.1 一次元波動方程式の 入(Introduction to Wave Propagation in 1D)
・ 微分表記 Detonations for differential
x t
u
u ,
,x
u
xu
,t
u
tu
, 22
x
u
xxu
, 2 2t
u
ttu
・ 微 振幅 Infinitesimal Amplitude
0
⇒cos 1
,sin
,tan
:x
x
x
u
・ 釣合い式・運動方程式 Equilibrium / Motion Equation
cos 0
cos
T
xT dT
xdx x-direction sin 0
sin
dx u
ttT
xT dT
xdx y-direction↓
0
dT
x-direction 0
dx u
ttT u
xu
xdx y-direction↓
x t T x dx t
T , ,
x-direction0
dx x dx x xx
tt
dx
u
T u
T u
u
y-direction振動する弦の 片
x
,x x
と作用する力Infinitesimal element
x
,x x
of bowstring in 1D vibration and forcesT
(internal tension) applied to it.x x dx
dx
T
T d
T u x dx , t
x t
u ,
dx x
x x
u
x tan sin
xG G G
EEEOOO---D D D
IIISSSAAASSSTTTEEERRRP P P
RRREEEVVVEEENNNTTTIIIOOONNNS S S
YYYSSSTTTEEEMMM 2/4G G G
EEEOOOS S S
CCCIIIEEENNNCCCEEE&&&G G G
EEEOOOE E E
NNNGGGIIINNNEEEEEERRRIIINNNGGG LLLAAABBB http://www.cm.nitech.ac.jp/maeda-lab/4. 微分方程式の解析解 (Analytical solution of PDE)
4.1波動方程式のD’Alembert解 D’Alembert’s Solution for Differential Equations for Wave Propagation
Differential Equation
xx
tt
V u
u
2 , ( x
,0 t
) (1)Initial Conditions: We must two initial conditions to solve wave propagation with second differential equation
x
g
x
u
x
f
x
u
t
, 0
0
,
, ( x
) (2)D’Alembert’s Solution
1.- Replacement:
x , t ,
Vt
x
Vt
x
(3) 0
u
(4)
u
u
u
V
u
u
u
u
u
u
u
V
u
u
u
u
tt xx
t x
2
2
2
(5)
2.- Integration
,
u
; Arbitrary Function
as
(6) ,
u
(7)Arbitrary Function
as
d
G G G
EEEOOO---D D D
IIISSSAAASSSTTTEEERRRP P P
RRREEEVVVEEENNNTTTIIIOOONNNS S S
YYYSSSTTTEEEMMM 3/4G G G
EEEOOOS S S
CCCIIIEEENNNCCCEEE&&&G G G
EEEOOOE E E
NNNGGGIIINNNEEEEEERRRIIINNNGGG LLLAAABBB http://www.cm.nitech.ac.jp/maeda-lab/ 3.- Replacement: , x , t
x t x Vt x Vt
u ,
(8)4.- Initial Condition:
x t x Vt x Vt
u ,
x
g
x
u
x
f
x
u
t
, 0
0
,
x
g
x
V
x
V
x
f
x
x
(9)
xx
K
d
g
x
V
x
V
0
(10) x
x
K
d
V g
x
f
x
2
01
2
1
(11) x
x
K
d
V g
x
f
x
2
01
2
1
(12)
x VtVt x
d
V g
Vt
x
f
Vt
x
f
t
x
u
2
1
2
, 1
(13)u
x 0
V: propagation velocity
1 2
t=0 t=1/V t=2/V
φ ( )=C
u
x 0 -1
t=0 t=1/V
t=2/V
(ξ )=C
-2
G G G
EEEOOO---D D D
IIISSSAAASSSTTTEEERRRP P P
RRREEEVVVEEENNNTTTIIIOOONNNS S S
YYYSSSTTTEEEMMM 4/4G G G
EEEOOOS S S
CCCIIIEEENNNCCCEEE&&&G G G
EEEOOOE E E
NNNGGGIIINNNEEEEEERRRIIINNNGGG LLLAAABBB http://www.cm.nitech.ac.jp/maeda-lab/ - 特性曲線(Characteristic Curve)Example-1
Initial condition
0
0
,
0
,
x
g
x
u
x
f
x
u
t
(14)
x t f x Vt f x Vt
u
2
, 1
(15)characteristic curve
a a
a a
Vt
x
Vt
x
Vt
x
Vt
x
(16)Example-2
0
0
,
int
0
1
1
0 1
,
x
u
s
po
other
the
x x
f
x
u
t
(17)
Characteristic curve x = -1 or 1 , t=0
1
1
Vt
x
Vt
x
(18)t
x
u(x
a, t
a)
(x
a-Vt
a, 0) 0 (x
a+ Vt
a, 0)
x-Vt=x
a-Vt
ax + Vt=x
a+ Vt
ax-Vt=0
x + Vt=0
u
x
-1 0 1
1
u
x
-1 0 1
1
1/2 1/2
t
x -1
u=0 u=1/2
u=1/2 u=0
u=0 u=0
0 1
u=1
x-Vt=1 x-Vt=-1
x+Vt=1 x+Vt=-1