SAMANTA
MAINAR
mmat 2328
1)
(x 2)ux + 2yny
+
&x
y Ly
=
2n
=
In(a + 2)
+
Thy
u+2
=
=
Hence we have
=
y
=
=
+ + c
=
c =
=
d'ER
2t + d ,
def
,
du
=
x+ 2
y
det
=
cet
=
,
d
k(x+ 2)"
=
c =
,
=
e
ed
ke
,
A
-
+
v
& (x k(x 2)2)
+
,
Choose the char
:
xEI] gives the family of characteristic
curve
.
& u(x , a(x+ 2)2)
S-(x
(x +
=
da
,
a (x + 2)2)
2a(x + 2)y
Thus a(x =2)
,
as
the
-
=
y(t)
du
=
=
=
a
fixed
a
.
of the PDE
Then
=y =
2
is
ODE
On along withe
parameter
M(s) (( (s)
x(t)
+
+
for
curves
cet
-
delt
2u
,
:
s>
03
2,MEC
u(d =
n
=
Set = (x+ 2)
=
2
Ty(x+ 2)
,
y
x-
I
2)
Kun
+
yuy
=
2n
x) 0
,
,
y0
ancee
at
=
u
=
&
N
:
Ex Rx xERYkor gives the family of characteristic curves of the PDE
:
,
a M(s)
[(Si 5)
=
:
seR150b 3
= - transversality holds
n
&
=
=
2n
,
=
&
=
=
2n
y ,
1
-
u
=
[Cross sins)
,
1lossing
An
=
e2t
y(0-5-y-get
y ,
b) T(s)
u(0)
,
u(0
y(0)
=
=
-u
sins
-
y
Thus
.
x
x(0) Su
=
,
v =
=
Se
xy , x)0 y>0
,
se[0 2n)}
,
Franversality hda
10
=
:
=
Sh a
=
=
et
sinset
.
=
Thus
x
,
n =
x0
=
c
x+ y2 ;
x ,
=
cost
y>0
6) T(s)
&(s es)/SER3
=
,
1ses
3a)
So
PDE
xux
+
=
x
Thus
y
yuy
=
kn
,
x + +
cet ,
=
,
k
So
.
a
if sol transversality condition fail
,
solved in
cannot be
kx
=
lates
=
general
.
y 2x)0 ; u(x a)
=
,
,
0
=
yy
=
F(x)
det ,
*) 0
=
#
characteristic
↑ (s)
=
[ (s x)
,
= kn
=y
,
,
ud
y(0)
-
Se}
:
curves
=
F(s)u F(s)ekt
,
=
n
det
ulay)
=
-
y
zwinital curve
>
=
.
Thus
=
=
x
,
x( =
F(q)( *"
x
=
e
x = R ,
y5x0
3b) you-sny
=y
xy"
=>
u(x 0)
>
x2
=
,
dyy
,
=
0
=
C
Ce[0, 0)
,
C
as
=
(0 , 0) has
solutions .
no
↑
>
[
characteristic
↑ (s)
-
[ (s 0) SER3
=
:
,
10)
=
5
,
810)
0
=
sinn-yuy
&
=>
=
x
a
+
+
0
=
=
=
d
+
=
.
x+y'=s"
Thus
36)
curves
=
=
U(l y)
:
++
-
n
=
,
it
stinia
I
0
*
U(x,y)
=
n
=
x y2
:
Fly)
i -y t
=
=
t+ d
where
e
&
&
characteristic
curves
ad
unique 10th on a to
un
T(s)
[(l s)/se]
=
1
,
I 1 2)
10
=
=
Transversality
hd o
= x( === -y y(
du
3d)
1-t
=
=
yuu
+
0
,
ulol
any
=
= =+
y
,
=
F(s) =>
;
u(0 , y)
0
u(xy)
=
=
F(s)
=
1
=
S
s
=
,
t
initial curve
>
=
y
=
E
=y
Fay/aryy
ey
=y ,= c ,
A
..
characteristic
curves
un
↑ (s)
=
[ (0 5)
,
1y Of
Au
x10)
=
=
=
0
0
,
,
y
-
u(0)
:
sett
unique
=
3
sol
on y
en =
y(0) = S
:
xky"--52
.
Thus xy)
=
Chey-
initial curve
4a)
UUx +
My
d
1
=
0
=
u(x, 8)
,
=
y
,
x(0 5)
:
0 , u(0)
=
&x
u(t)
=
,
5
4b)
y(0)
Una +
My
x(0)
s
=
0
=
u
x()
=
1
=
=
u(0)
Transversality condition
= 1 y(y
=
&
x = n
=>
x
=
,
s =
x(
l
As
:
y (0)
a =
=
=
<2
u(y 2)
-
-
,
=
=
+ 0
,
y > -1
=t
we
&
=
t
-
,
y +1
=
y) 10y)
sette
,
0
Se (0 1
as
,
=
+
=
t +
2n
-
S
,
y
10
2(y u)
-
=
=
u(y
-
2)
-
2 zy
+
E =E - = Oy
must have
22 2y
+
y
(0 1]
=
-
=
=
x
++ 1
&
He
,
(yu)(2n y) +
()
S
S/2
===
+
=
satisfied.
is
( % / E-1
:
<ly-u)
(2xy)
!)
a
4/2
=
,
We have
40, 5)
,
sdt , =
=
Thus
,
,
0
-1 FO
=
(f
U(x x)
,
y(0)
dx
.
=
,
S
=
:
yy-1
,
y(0 s)
,
/S')
=
=
S
:
Transversality condition
Since
=
,
Transversality condition
du
x
t + k (s)
=
Snition condition
at
=
y <2
= u(x , y)
=
(k +
-
2y)/(y 2) my(2
-
,
xe
(0 , 1)