"qa
83-215
zeiwlg zeil`ivpxt ze`eeyn
zepexzt -
3
milibxz
d`eeynd z` xezt
ux + uy = x − u
dlgzd i`pz mr
. u(x, 0) = e−x
2
The characteristics satisfy
dy
=1
dt
dx
=1,
dt
so
x = t + c1 ,
y = t + c2 .
We can choose c1 to be zero without loss of generality. Note that each characteristic
(the lines y = x + c2 ) intersects the line y = 0, on which the initial data is given,
exactly once, so the solution will be well-defined in the whole plane.
Solution on a characteristic: On a characteristic we have
du
=t−u
dt
with general solution
u = t − 1 + c3 e−t .
2
The initial condition is given on y = 0, i.e. at t = −c2 and then we have u = e−x =
2
e−c2 . Thus
2
e−c2 = −c2 − 1 + c3 ec2 ,
and
2
c3 = e−c2 1 + c2 + e−c2
The solution on the characteristic is
.
2
u = t − 1 + e−(t+c2 ) 1 + c2 + e−c2
.
It just remains to write the solution in terms of x and y: we have
(
so the solution is
x=t
y = t + c2
(
t=x
c2 = y − x
2
u = x − 1 + e−y 1 + (y − x) + e−(y−x)
.
It is a good idea to check!
.1
d`eeynd z` xezt
ux + xuy = y − 2xu
dlgzd i`pz mr
. u(0, y) = 1
The characteristics satisfy
dx
=1,
dt
dy
=x
dt
so
1
y = (t + c1 )2 + c2 .
2
We can choose c1 to be zero without loss of generality. Note that each characteristic
(the parabolas y = 12 x2 + c2 ) intersects the line x = 0, on which the initial data is
given, exactly once, so the solution will be well-defined in the whole plane.
x = t + c1 ,
Solution on a characteristic: On a characteristic we have
1
du
= t2 + c2 − 2tu .
dt
2
Writing this as
1
du
+ 2tu = t2 + c2
dt
2
2
we see there is an integrating factor et and
d t2 1 2
2
e u =
t + c2 et ,
dt
2
so
1 2
2
s + c2 es ds .
c3 +
u=e
2
The initial condition is given on x = 0, i.e. at t = 0 and then we have u = y = c3 .
Thus c3 should be taken to be 1 and the solution on the characteristic is
−t2
Z t
−t2
Z t
u=e
1+
1 2
2
s + c2 es ds
2
.
It just remains to write the solution in terms of x and y: we have
(
x=t
y = 21 t2 + c2
(
t=x
c2 = y − 21 x2
so the solution is
u = e−x
It is a good idea to check!
2
1+
Z
x
1 2
2
(s − x2 ) + y es ds
2
.
.2
d`eeynd z` xezt
xux + yuy = 1
dlgzd i`pz mr
. u(1, y) = y
?oexztd ly dxbdd
megz `ed dn
The characteristics satisfy
dx
=x,
dt
dy
=y
dt
so
x = c1 et ,
y = c2 et .
The characteristics are “half lines through (0, 0)” — et is always positive so the characteristics never reach (0, 0). (There is also a characteristic which is a single point —
the point (0, 0) corresponding to c1 = c2 = 0.) The initial data is specified on the line
x = 1. This intersects (once) with each of the characteristics in the half-plane x > 0,
so we expect a solution in this region. Given that x > 0 we can choose c1 = 1 without
loss of generality.
Solution on a characteristic: On a characteristic we have
du
= 1,
dt
with solution
u = t + c3 .
The initial data is specified at x = 1, i.e. t = 0. At t = 0 we have y = c2 and u = t3
so to impose the condition u(1, y) = y we must take c3 = c2 . Thus the solution on the
characteristic is
u = t + c2 .
It just remains to write the solution in terms of x and y: we have
(
x = et
y = c2 et
(
so the solution is
u = log x +
t = log x
c2 = xy
y
.
x
It is easy to check this satisfies both the equation and the initial condition. It is defined
everywhere except on the y axis (x = 0). But it is only the solution of our problem in
the region x > 0. No characteristics connect the initial data and the region x < 0.
.3
m`
.iynn
reaw
`ed
p
xy`k
xux + yuy = pu
u(x, y) = y
lbrnd lr
lk
oeayga
zgwl
yi
?
u(x, y)
d`eeynd
dtyd
ly
ezxbd
megz
z`
zniiwn
i`pz z` zniiwn
`ed
dn
u(x, y)
.
z`
u(x, y)
u(x, y)
divwpetd
mb
`vn
p
.
,
-y oezp
x2 + y 2 = 1
ly ixyt` jxr
The characteristics satisfy
dx
=x,
dt
dy
=y
dt
so
x = c1 et ,
y = c2 et .
The characteristics are “half lines through (0, 0)” — et is always positive so the characteristics never reach (0, 0). (There is also a characteristic which is a single point —
the point (0, 0) corresponding to c1 = c2 = 0.) Each half-line intersects (once) with
the circle x2 + y 2 = 1 so we expect a solution on the whole plane except at the point
(0, 0).
Solution on a characteristic: On a characteristic we have
du
= pu,
dt
with solution
u = c3 ept .
The initial data is specified on the circle x2 + y 2 = 1 which a characteristic intersects
when (c21 + c22 )e2t = 1, i.e. when et = (c2 +c12 )1/2 . On the circle we want to impose
1
2
u(x, y) = y i.e.
c3 ept = c2 et ⇔ c3 = c2 (c21 + c22 )(p−1)/2 .
Thus the solution on a characteristic is
u = c2 (c21 + c22 )(p−1)/2 ept .
It just remains to write the solution in terms of x and y: we have

√
(
 (c21 + c22 )1/2 et = x2 + y 2
x = c1 et
y
c2
 (c2 +c2 )1/2 = √ 2 2
y = c2 et
x +y
1
2
so the solution is
u = y(x2 + y 2)(p−1)/2 .
It is easy to check this satisfies both the equation and the initial condition.
What about the domain of definition of the solution? It certainly is defined on the
whole plane except possibly at (0, 0). At (0, 0) there are several possibilities:
( ) If p < 1 then the solution is not defined.
( ) If p = 1, 3, 5, . . . the solution is defined and the solution is infinitely differentiable
at (0, 0).
( ) If p takes other values bigger than 1 the solution is defined, but it is only differentiable a finite number of times (in fact not at all if 1 < p < 3).
`
a
b
.4
diral oexzt miiw m`d
?
yux − xuy = u
u(x, 0) = x ,
x>0
!wnp
The characteristics satisfy
dx
=y,
dt
dy
= −x
dt
so
x = c1 sin t ,
y = c1 cos t .
(There is another constant of integration which we can ignore.) The characteristics
are circles centered at (0, 0). We are given an initial condition on the positive xaxis (not including the origin). This intersects every characteristic once except for
the characteristic consisting of the single point (0, 0). But — the characteristics are
closed curves so this is not enough to guarantee a solution.
Solution on a characteristic: On a characteristic we have
du
= u,
dt
with solution
u = c2 et .
The initial data is specified on the half line y = 0, x > 0. The characteristic intersects
the initial line when t = 21 π, so we need to choose c2 such that c2 eπ/2 = c1 . In other
words, the solution on the characterstic is
u = c1 et−π/2 .
It just remains to write the solution in terms of x and y: we have
(
(
x = c1 sin t
y = c2 cos t
t = arctan xy
√
c1 = x2 + y 2
.
Thus the solution is
u=
x2 + y 2 earctan( y )− 2 π =
q
x
1
y
x2 + y 2 e−arctan( x ) .
q
It is easy to check this satisfies both the equation and the initial condition. But —
the definition of arctan means this formula is “multivalued”: as we go round the point
(0, 0) (in an anticlockwise direction) the arctan increases by 2π. This is a result of the
closed trajectories. In the normal sense of the word, there is no solution.
.5