Unformatted text preview: rtial integration
(since it contains derivatives with respect to each variable). So, we’ll use the
method of characteristics instead. To ﬁnd the “base characteristics”, we need
to solve the DE
dy
1
=.
dx
2
That’s easy: y = x/2+C . Expressing this in implicit form, we have x 2y = C2 ,
(these are the characteristics) which tells us that we should let one of our new
variables be x 2y . So, let’s let
⇠=x 2y, ⌘ = y. With this choice, we have
u x = u ⇠ ⇠ x + u ⌘ ⌘x = u ⇠
ˆ
ˆ
ˆ
u y = u ⇠ ⇠ y + u ⌘ ⌘y =
ˆ
ˆ 2 u⇠ + u⌘
ˆ
ˆ and this converts the original PDE into the form
2ˆ⇠ + ( 2u⇠ + u⌘ ) + u = 0,
u
ˆ
ˆ
ˆ
or simply
u⌘ =
ˆ u.
ˆ This is an equation we can solve! We just need to observe how similar it is
du
ˆ
to the ODE
= u. Since we know that the general solution to this is (by
ˆ
d⌘
inspection) u = Ce ⌘ , we know immediately that the solution to our PDE is
ˆ
u = f (⇠ ) e
ˆ ⌘ where f is an arbitrary function of one variable.
Returning to the original variables, we have
2y ) e y . u = f (x Applying the initial condition, we discover that
f (x) = cos x,
so the solution we seek is
u (x, y ) = e y cos (x 2y ) . AMATH 350 Page 9 Assignment #7 Solutions  Fall 2013 9. Solve the initial value problem
3 y 2 ux
on the domain x 2xuy = x, u(0, y ) = y
2 0. Solution:
We start by solving the equation
dy
=
dx
which is separable:
3 Z 2x
,
3y 2 2 y dy = 2 y3 = =) Z xdx x2 + C. Solving for C gives the characteristic curves: C = x2 + y 3 . We introduce a
change of coordinates which will map these curves to straight horizontal (or
vertical1 ) lines:
Let ⇠ = x2 + y 3 ,
Then @u
@ u @⇠
ˆ
@u
ˆ
=
= 2x
= 2x u⇠
ˆ
@x
@⇠ @ x
@⇠ ux =
and
uy = ⌘ = y. @u
@ u @⇠ @ u @⌘
ˆ
ˆ
@u @u
ˆ
ˆ
=
+
= 3y 2
+
= 3y 2 u⇠ + u⌘ .
ˆ
ˆ
@y
@ y @ y @⌘ @ y
@⇠
@⌘ These allow us to rewrite the given PDE:
3 y 2 ux
=) 3y 2 (2xu⇠ )
ˆ
=) 2xuy = x
2 x 3 y 2 u⇠ + u⌘ = x
ˆ
ˆ
u⌘ =
ˆ...
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 Fall '12
 DavidHamsworth
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