2.1 Given that
(1) =⇒ x = f1 (s)et ;
ux + uy = 1
(2) =⇒
u(x, 0) = f (x)
∂y
∂y
− y = x =⇒
− y = et
∂t
∂t
The above is a linear ODE =⇒ use an integrating factor
then the characteristic equations are
∂x
=1
∂t
∂y
=1
∂t
∂u
=1
∂t
∂ −t e y =1
∂t
=⇒
with
x(0, s) = s
with
y(0, s) = 0
with
u(0, x) = f (s)
f1 (s) = 1 =⇒ x = et
Thus, upon integrating, we get that
e−t y = t + f2 (s)
y = t et + f2 (s)et ;
f2 (s) = s
= (t + s)et
Thus
x =
y
=

t + f1 (s) 
t + f (s) 
initial conditions =⇒
2
x
= t+s
y
= t
(3) =⇒ u = t + f3 (s);
Thus we have that
a) Projection of characteristics curves onto the (x, y) plane.
y = t,
f3 (s) = s =⇒ u = t + s
x
t = x − s =⇒ y = x − s
y
= et
t


= (t + s)e 
=⇒
y
y
= t + s =⇒ u =
x
x
Note: The solution is not defined at x = 0
These are straight lines with slope, one for each s value.
2.3 Given that
b) Solutions:
x ux + y uy = P u
u = t + f3 (s)
thus with the given initial conditions we get that
a) The characteristic curves are given by

∂x

= x 

x = f1 (s) et
∂t
=⇒

∂y
y = f2 (s) et

= y 
∂t
u = t + f (s) = y + f (x − y)
2.2 Solve
x
= h(s)
y
b)
x ux + (x + y) uy = 1
∂u
= 4u =⇒ u = f3 (s) e4t
∂t
= 1, find a solution. Parameterize the unit circle
P = 4 =⇒
u(1, y) = y
Given that ux2 +y2 =1
The characteristic equations are
∂x
=x
∂t
∂y
=x+y
∂t
∂u
=1
∂t
=⇒
with x = cos s, y = sin s. We then have the initial conditions:
with
x(0, s)= 1
(1)
with
y(0, s)= s
(2)
with
u(0, s)= s
(3)
x(0, s) = cos s
y(0, s) = sin s
u(0, s) = 1
since x0 (s)2 + y0 (s)2 = 1. Upon applying the initial conditions, we
Therefore
determine that
x(t, s)
=
y(t, s)
=
u(t, s)
=

cos s e 
sin s et 
x = f1 (s) et
with
x(0, s)
= 1
y = f2 (s) et
with
y(0, s)
= s
t
=⇒ x2 + y 2 = e2t
u = f3 (s) e2t
e4t
with
u(0, s) = f (s); f (0) = 1
Thus
Therefore
u(x, y) = x2 + y 2
2
x = et
y = s et
c)
P = 2 =⇒
∂u
= 2u =⇒ u = f3 (s) e2t
∂t
Find a solution such that u(x, 0) = x2 .
u = f (s)e2t
from which we see that
Initial Conditions:
y
=s
x
x(0, s) = s =⇒ x = s et
x2 = e2t
y(0, s) = 0 =⇒ y = 0
Therefore
u(0, s) = s2 =⇒ u = s2 e2t
u = x2 f
Therefore
u = x2
a
J = ∂x0
∂s
b
∂y0
∂s
x y s et
=
=
1 0 1
0 =0
0 x
and the Jacobian of the transformation is
x y et s et
=
J =
1 0 0
1
Check:
ux = 2xf
d) Therefore if we have one solution, existence and uniqueness =⇒ infinitely many solutions.
uy = xf
c) (continued)
y
′
y
x
y
− yf ′
= et 6= 0
y
x
x
Thus,
Construction of another solution: Choose P0 (x0 , y0 , u0 ) = (1, 0, 1) and
x ux + y uy = 2x2 f
solve
y
and
x ux + y uy = 2u
u(1, y) = f (y);
u(x, 0) = x2
f (0) = 1
provided f (0) = 1.
x
= 2u
2.10 Given that
uτ − y 2 − 1 ux = 0
u(x, y, 0) = ey ex
2
Note: We may treat the variable y as if it were a parameter since there
are no derivatives with respect to it.
a) The characteristic equations are:
∂τ
=1
∂t
∂x
= − y2 − 1
∂t
∂u
=0
∂t
with
τ (0, s) = 0
with
x(0, s) = s
with
u(0, x) = ey e−s
2
Thus
τ
x
u
=
t + f1 (s)
=
− y 2 − 1 t + f2 (s)
=
f3 (s)





initial conditions =⇒




τ
=
t
x
=
− y2 − 1 t + s
u
=
ey e−s
Therefore t = τ and s = x + y 2 − 1 τ and thus the solution is
u(x, y, τ ) = ey e−(x+(y
2
−1)τ )
2
2