WEBSITE : http://maths-minhthe.violet.vn BOUNDARY VALUE PROBLEMS OF DIFFERENTIAL EQUATIONS 5390 The Minh Tran --- 1... Draw the dividing waves shocks, rarefactions ,.... Draw the wa
Trang 1WEBSITE : http://maths-minhthe.violet.vn
BOUNDARY VALUE PROBLEMS OF DIFFERENTIAL EQUATIONS
(5390)
The Minh Tran
-
1 Solution
a) Because u (x)'' = 1 , so we have the genaral form of u(x) : u x ( ) x2 ax b
2
= + +
'
'
From the condition boundary : u(0) 2u (0) 1 ; u(1) 0
We have
a u(0) 2u (0) 1
2 1
a b u(1) 0
b 2 2
+ =
+ = −
=
b)
Solution :
We have the equation :
( ) ( ) ( )
1
N 0
u x = ∫ G x, f ξ ξ d ξ + u (x)
We see that
Trang 21
1
d
dx
We compute L to both sides of the equation so
−
−
1
0
2
write out as:
d
dx
∫
From the condition boundary, the Green’s function G(x, )ξ satisfies :
2
' 2
d
We have
G x, ( ) a bx , x
c dx , x
+ < ξ
ξ =
+ > ξ
The constant a, b,c,d are different
Where x< ξ :
'
G(0, ) 2G (0, ) 1ξ + ξ = ⇒ +a 2b= ⇒1 a= −1 2b
Where x> ξ :
G(1, )ξ =0⇒ + = ⇒ = − c d 0 c d
Hence,
( ) 1 2b bx , x
(1) G x,
d dx , x
− + < ξ
ξ =
− + > ξ
We continue to check at x= ξ
( )
From (1) and (2) we have :
From (2) and (3) we have :
− =
Trang 3Soving the equation system
⇒
Thus
1 2 x 2 , x
G x,
− − ξ − < ξ
ξ =
− ξ − > ξ
The representation :
( ) ( ) ( )
1
0
u x = ∫ G x, f ξ ξ d ξ − 2(1 x) −
- c) We have the equation :
( ) = ∫1 ( ξ ) ( ) ξ ξ
0
u x G x, f d with 0 < < x 1
We have the condition boundary :
,
( )
u (x) x u (0) u(0) 2u (0) 1
u(0) 2
1 3
2 2
= − = α
0
We have the function G x ( , ξ ) as the formula below :
1 2 x 2 , x
G x,
− − ξ − < ξ
ξ =
− ξ − > ξ
with u x f ξ
Trang 4( ) ( ) ( ) ( )
0
2
x
= ξ − ξ − − + ξ − − − −
∫
Thus, ( )
2
We see two solutions from the above are the same
-
2
( ) ( )
( ) ( )
2
2
2
2
We assum that u(x,t) x G t
where x is only a function of x and G t is a function of t
We will get first partial derivetive of u(x,t) with respect to t and the fourth
partial derivative with respect to x
dG t
u
x
u
x
= φ φ
∂
= φ
∂
∂
=
∂
( ) ( )
2
2
2 2
d
G(t)
dx
dG t d
φ
φ
Trang 5( )
2
We can separate var iable by dividing both sides of (1) by x G(t)
dG t
We can not solve by the method of separation of var iables
because the presence of a terms like sin 5x is forbidden
φ φ
"
2 n
n 1
'
n 1
2
n 1
2
∞
=
∞
=
∞
=
φ + λφ =
π
∑
∑
∑
n 1
5
n 1
n
2
2
We see that C (t)
∞
=
∞
=
π
=
∑
∑
The initial condition is u(x,0)=0=sin0= 2φ1(x)
π
The fourier expantion is
n 1
n
2
2
n 0
B (0)
∞
=
π
=
= π
∑
Trang 6t xx
n
From the equation u u sin5x and the formula from the above
The differential equation :
B (t) n B (t) C (t)
We can solve :
2
B (t) n B (t)
2
n 0 with B (0)
⇒
=
'
0
0 0
Case 1 : n 0
B (0) 0
2
B (0) 2
B (0)
=
π
=
π
'
5 5
Case 2 : n 5
2
25
B (0) 0
−
=
π
π
n
n
=
n 1
Finally ,the solution is :
25
∞
=
−
∑
-
3 Solve the following two types of problems :
a) − u = λ u subject to u 0 = 0 u 1 =
We have the equation form :
Trang 7( )
( )
( )
2
2s
0
We have the auxiliary equation : r 0 (*)
d
dx
A B 0 and so
dx
There are no
−
−
φ + λφ =
+ λ =
λ <
⇔ + λ = ⇔ = −λ > ⇔ = ± −λ = ±
φ
=
λ
( )
( )
( )
( )
2
d
d
dx
x A const is an eigenfunction
λ =
φ
=
φ
( )
( )
( )
Case 3 : 0
x A cos x B sin x
d
d
1 0
dx
when n 0 dim 2
λ >
⇔ + λ = ⇔ = −λ < ⇒ = ± λ
φ
=
φ
⇔ λ = π ∈ ⇒ λ = π
> ⇒ =
( ) ( )
b) − u = λ u subject to u 0 = 0 ; u 1 = 1
We have the equation form :
Trang 8( )
( )
( )
"
2
s 2s
0
We have the auxiliary equation : r 0 (*)
Case 1 : 0
d
e
s(e 1) d
1 1
dx
When 0 the solution is Ae Be , where A,B is the same as the a
−
−
−
φ + λφ =
+ λ =
λ <
⇔ + λ = ⇔ = −λ > ⇔ = ± −λ = ±
φ
=
− φ
( )
( )
( )
2
Case 2 : 0
x A Bx
d
0 0 A 0; B 0
dx
d
1 1 A 0; B 1
dx
There is no 0
λ =
φ
φ
λ =
( )
( )
( )
( )
x A cos x B sin x
d
dx
d
dx
1
A
sin
1
sin
1
sin
λ >
⇔ + λ = ⇔ = −λ < ⇒ = ± λ
φ
φ
−
=
−
−
Trang 9Condition :
The problem a) is an eigenvalue prolem
-
( ) ( )
( ) ( )
( ) ( ) ( ) ( )
tt
xx
1
2
1
2
4
We have u(x,t) e
and u cu
u cu
Re turn the problem which is given
c for x 0
We see that u cu where c(x)
c for x 0
k c for x 0
k c for x
−ω
−ω
−ω
=
=
<
>
<
ω =
>
ikx t
0
u (x,0) ( i)e g(x)
−ω
x
x
<
ω
ω
∫
∫
Trang 10( )
1
1
tki c it
1
1 x
x
c
−
−ω
x
x
2
2
For x 0
u(x,t) Q(x c t) P(x c t)
x c t : h(t) Q( c t) P(c t) where x 0
z Q(z) h( ) P( z) , z 0
c
x u(x,t) h(t
c
>
ω
ω
−
∫
∫
2
x
i t c
) e
=
( )
x
We continue to use the continuity of u at x 0
u (0 ,t) u (0 ,t)
Setting x 0 we have
f ( c t) h (t) f c t h (t)
2c f( c t)
h(t)
=
=
=
−
+
Trang 111 2 2
c
ik ( x c t ) c
2
Thus
x 0
x 0
u(x,t) f(x c t) f( c t x)
>
<
−
+
−
+
-
tt
i t
tt
i t
tt
2
2
5
u u , [0,L] [0,L] : the boundary
From the data is given
u(x, y,t) e u(x,y)
u (i ) e u(x,y) (1)
u e u(x, y) (2)
(1)& (2) u e u(x,y) e u(x,y) where i 1
u(x, y) u(x,y)
u(x,y) u(
∧ ω
∧ ω
∧ ω
=
2
mn
x,y)
u u , where
u 0 on
∧
mn
2
mn
Trang 12In physics ,the equation from above is called Maxwell ' s wave equation
E
t
∂
∂
-
6
2
s1
s2
s1
Answer :
From the quation , we have
Solving for the shock velocity
dx q(4) q(2) 16 4
6
dx q(2) q(1) 4 1
3
dx q(4) q(1) 16 1
5
<
ρ
ρ
−
−
−
s2
s3
The intersection of waves :
1 6t 3t c where t c 1 x 3t c 3t 1
3 1 6t 3t 1 t
3
Thus, the formulas for the shock waves
1
x 6t ; x 3t 1; x 5t
3
= + => =
Trang 13Draw the three shock waves
-
7
2
s1
Answer :
From the quation ,we have
Solving for the shock velocity
3
<
ρ
−
Trang 140
dx
dt
We have : (x,t)
<
>
2
s2
1
ln t 2 1
ln t
s2
2
x
q( ) q(2)
From derivativing :
−
−
< < ⇒ < ρ <
−
1
ln t 2
s2
s2
s2
s2
dx
dt
We have int er sec tion of the equations :
−
−
=
=
∫
Trang 15Draw the dividing waves ( shocks, rarefactions , ) on the x-t place
-
8
t
Answer :
=
Trang 162 0
Generally, we have two cases
d
dt
t (x,t) p(x ) p(x ) with (x,0) p(x) f(x)
2
ω
2
2 0
2
t
2
t Thus, (x,t) f(x )
2 d
t
2 t
x
2
d
(x,t) (x ,t)e where (x ,t) is the value of (x,t) at t 0
dt
t From the initial condition (x,0) f(x) (x ,0) f(x ) f(x )
2 Thus, (x,t) f(x
= + >
⇒ > −
ω
2 t
t )e 2 +
-
2 0
2
t
2 t
x
2
⇒ < −
t
2
2
0
2 2
0
t
d
(x,t) (x ,t)e where (x ,t) is the value of (x,t) at x 0;t t
dt
t
2
t
x
2
From the initial condition (0,t) t
(0,t) (x ,t)e t
ω
= − +
= −
( )
t 0
t t
(x ,t) t e (x,t) t e e t e , where t t 2x
−
−
−
=
Trang 17Draw the waves