MA342H: Homework #1 solutions
1. Find all separable solutions u(x, y) = F (x)G(y) of the equation xux + uy = 3y 2 u.
To say that u = F (x)G(y) is a solution is to say that
xF ′ (x)G(y) + F (x)G′ (y) = 3y 2 F (x)G(y).
Let us divide this equation by F (x)G(y) and then rearrange terms to get
xF ′ (x)
G′ (y)
= 3y 2 −
.
F (x)
G(y)
Here, the left hand side is a function of x and the right hand side is a function of y. Thus,
the two sides are equal if and only if they are both constant, say
xF ′ (x)
G′ (y)
= λ = 3y 2 −
.
F (x)
G(y)
When it comes to the unknown function F (x), separation of variables gives
λ
F ′ (x)
=
F (x)
x
=⇒
log F (x) = λ log x + C1
=⇒
F (x) = C2 xλ .
When it comes to the unknown function G(y), one similarly finds that
G′ (y)
= 3y 2 − λ
G(y)
=⇒
log G(y) = y 3 − λy + C3
=⇒
G(y) = C4 ey
3 −λy
.
In particular, every separable solution of the given equation must have the form
u(x, y) = F (x)G(y) = C5 xλ ey
3 −λy
.
2. (a) Show that ux + yuy = 1 has no solutions such that u(x, 0) = ex .
(b) Show that ux + yuy = 1 has infinitely many solutions such that u(x, 0) = x.
An initial condition of the form u(r, 0) = f (r) is not suitable because it gives
1 1
1 1
=0
= det
det
0 0
0 y
along the initial curve. Let us impose the condition u(0, r) = f (r), where the function f is
arbitrary. We must then solve the characteristic equations
x′ = 1,
y ′ = y,
u′ = 1
subject to the initial conditions
x(0) = 0,
y(0) = r,
u(0) = f (r).
These equations can be easily integrated to give
x(s) = s,
y(s) = y(0)es = res ,
u = s + f (r)
and one may eliminate the auxiliary variables r, s to conclude that
u(x, y) = s + f (r) = x + f (ye−s ) = x + f (ye−x ).
(∗)
In view of this computation, every solution of the PDE must have the form (∗) for some
function f , so every solution of the PDE satisfies
u(x, 0) = x + f (0).
In particular, no solution satisfies u(x, 0) = ex and part (a) follows. On the other hand,
the initial condition of part (b) is satisfied by any function f such that f (0) = 0 and this
obviously includes an infinite number of choices.
3. Solve xux + (x + y)uy = x subject to u(x, 0) = f (x) for each x > 0.
In this case, the characteristic equations are given by
x′ (s) = x,
y ′ (s) = x + y,
u′ (s) = x
and we need to solve these equations subject to the initial conditions
x(0) = r,
y(0) = 0,
u(0) = f (r).
Noting that x(s) = res , one easily finds that
u′ (s) = x = res
=⇒
u(s) = r(es − 1) + u(0) = res − r + f (r).
The remaining equation y ′ = x + y is first-order linear, so we have
(ye−s )′ = y ′ e−s − ye−s = xe−s = r
=⇒
ye−s = rs
=⇒
This implies y/x = s, hence r = xe−s = xe−y/x and we finally get
u(x, y) = res − r + f (r) = x − xe−y/x + f (xe−y/x ).
y = rses .
4. Solve u2x − u2y = 8u subject to u(x, x) = x2 .
This is a fully nonlinear PDE with F = p2 − q 2 − 8u and characteristic equations
x′ = 2p,
y ′ = −2q,
u′ = 2p2 − 2q 2 = 16u,
p′ = 8p,
q ′ = 8q.
Note that the last three equations can be easily integrated to give
u = u0 e16s ,
p = p0 e8s ,
q = q0 e8s .
Employing this fact in the first two equations, one now finds that
x′ = 2p0 e8s
=⇒
y ′ = −2q0 e8s
=⇒
p0 8s
(e − 1) + x0 ,
4
q0
y = − (e8s − 1) + y0 .
4
x=
Next, we use the initial condition to determine p0 and q0 . On one hand, we have
u(r, r) = r2
=⇒
d
u(r, r) = 2r
dr
=⇒
ux + uy = 2r
=⇒
p0 + q0 = 2r.
(E1)
On the other hand, we must also have p2 − q 2 = 8u at all times, hence
(p0 + q0 )(p0 − q0 ) = 8r2
=⇒
2r(p0 − q0 ) = 8r2
=⇒
p0 − q0 = 4r.
Solving the system (E1)-(E2) gives p0 = 3r and q0 = −r, so we end up with
x=
r
3r 8s
(e − 1) + r = (3e8s + 1),
4
4
y=
r 8s
r
(e − 1) + r = (e8s + 3).
4
4
To finally express the solution u = r2 e16s in terms of x and y, we note that
3x − y =
r
· 8e8s = 2re8s
4
=⇒
u(x, y) = (re8s )2 =
(3x − y)2
.
4
(E2)