1a. We need to solve the characteristic equations
x′ (s) = x,
y ′ (s) = 1,
u′ (s) = u
subject to the initial conditions
x(0) = r,
y(0) = 0,
u(0) = f (r).
Solving this system of equations, one easily finds that
x = x(0)es = res ,
u = u(0)es = f (r)es .
y = s,
Once we now eliminate the auxiliary variables r and s, we may conclude that
u = f (r)es = f (xe−s )es = f (xe−y )ey .
1b. 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 y2
along the initial curve. Let us then impose the condition u(0, r) = f (r), where the
function f is arbitrary. The associated characteristic equations are
x′ (s) = 1,
y ′ (s) = y 2 ,
u′ (s) = u
and we need to solve these equations subject to the initial conditions
x(0) = 0,
y(0) = r,
u(0) = f (r).
Solving this system of equations, one easily finds that
x = s,
y=
r
,
1 − rs
u = u(0)es = f (r)es .
In particular, r = y(1 − rs) = y − rxy and this implies that
y
s
r(1 + xy) = y =⇒ u = f (r)e = f
ex .
1 + xy
Since u(x, 0) = f (0)ex and the function f is arbitrary, there exists an infinite number
of solutions such that u(x, 0) = ex .
1c. This is a fully nonlinear PDE with F = p2 + q − y and characteristic equations
x′ = 2p,
u′ = 2p2 + q,
y ′ = 1,
p′ = 0,
q ′ = 1.
Three of these equations can be integrated directly to give
y = y0 + s,
p = p0 ,
q = q0 + s
and this allows us to determine the remaining two variables
s2
+ u0 .
2
u = (2p20 + q0 )s +
x = 2p0 s + x0 ,
We note that y0 = 0 and u0 = x0 by assumption. When it comes to p0 , we have
u(x, 0) = x
=⇒
ux (x, 0) = 1
=⇒
p0 = 1,
so we can use the given PDE to also find
q0 = y0 − p20 = −1.
In particular, the solution that we obtained above can be expressed in the form
x = 2s + x0 ,
y = s,
u=s+
s2
+ x0 ,
2
p = 1,
q = s − 1.
Once we now eliminate x0 and s, we may finally conclude that
s + x0 = x − s = x − y
=⇒
u=
y2
s2
+x−y =
+ x − y.
2
2
2a. This problem is a special case of the first part of the extended maximum principle.
Were the set B = {(x, y) ∈ A : u(x, y) > 0} nonempty, we would have
uxx + uyy ≥ −cu ≥ 0 in B
=⇒
max u = max u ≤ 0
B∪∂B
∂B
by the standard maximum principle. This implies that u ≤ 0 within B, contrary to
the definition of B. In particular, the set B is empty and so u ≤ 0 within A.
2b. Consider the difference w = u − v. This is easily seen to satisfy the equation
wxx + wyy = uxx + uyy − vxx − vyy = u3 − v 3
= (u − v)(u2 + uv + v 2 ) = −cw,
where c = −(u2 + uv + v 2 ). To see that c ≤ 0 at all points, we note that
v 2 3v 2
2
2
−c = u + uv + v = u +
≥ 0.
+
2
4
We may thus apply the result of part (a). Since w = u − v ≤ 0 on the boundary of A,
we must have w ≤ 0 at all points and this means that u ≤ v at all points.
2c. Consider the function u(x, y) = (sin x)(sin y). It is easily seen to satisfy
uxx + uyy = −(sin x)(sin y) − (sin x)(sin y) = −2u,
while u = 0 on the boundary of the set A = (0, 2π) × (0, 2π). Since u attains positive
values within A, the result of part (a) does not hold in the case that c = 2.
3a. Since y(x) is an extremal of J(y), the directional derivative
Z b
′
Ly ϕ + Ly′ ϕ′ dx
J (y)ϕ =
a
must be zero for all functions ϕ. Integrating by parts, we conclude that
h
ib Z b d
′
J (y)ϕ = Ly′ ϕ +
Ly − Ly′ ϕ dx = 0
dx
a
a
for all functions ϕ. When it comes to test functions ϕ, this gives
Z b
d
d
Ly ′
Ly − Ly′ ϕ dx = 0 =⇒ Ly =
dx
dx
a
by the fundamental lemma of variational calculus. In particular, we must have
h
ib
′
J (y)ϕ = Ly′ ϕ = 0
a
for any function ϕ, so it easily follows that Ly′ = 0 when x = a, b.
3b. First of all, we need to solve the Euler-Lagrange equation
d
Ly ′ = Ly
dx
=⇒
2y ′′ (x) + 4y ′ (x) = 4y ′ (x) − 2y(x) + 6
=⇒
y ′′ (x) + y(x) = 3.
This is a second-order linear equation with constant coefficients whose solution is
y(x) = c1 sin x + c2 cos x + 3.
To ensure that the natural boundary conditions hold, we need to ensure that
Ly′ = 2y ′ (x) + 4y(x) = (4c1 − 2c2 ) sin x + (2c1 + 4c2 ) cos x + 12
vanishes when x = 0 and also when x = π/3. In other words, we need to have
√
√
2c1 + 4c2 = −12 = (1 + 2 3)c1 + (2 − 3)c2 .
Solving this system for c1 and c2 , we conclude that the only critical point is
√
√
4 3+6
2 3 − 12
y(x) = −
sin x +
cos x + 3.
5
5
3c. We check the sufficient condition for a local minimum. First, we note that
Z
π/3 Lyy ϕ2 + 2Lyy′ ϕϕ′ + Ly′ y′ (ϕ′ )2 dx
0
Z
Z π/3
Z π/3
′
2
ϕ(x)ϕ (x) dx + 2
ϕ(x) dx + 8
= −2
′′
J (y)ϕ =
0
0
π/3
ϕ′ (x)2 dx.
0
When it comes to the first integral, we can use the Poincaré inequality to get
Z
π/3
(π/3)2
ϕ(x) dx ≤
2
2
0
Z
π/3
π2
ϕ (x) dx =
18
2
′
0
Z
π/3
ϕ′ (x)2 dx.
0
In particular, the second variation of J(y) satisfies the inequality
′′
J (y)ϕ ≥
π2
2−
9
2
Z
π/3
0
π/3
′
2
ϕ (x) dx + 8
Z
π/3
ϕ(x)ϕ′ (x) dx
0
iπ/3
18 − π
′
2
2
=
ϕ (x) dx + 4ϕ(x)
9
0
0
2 Z π/3
18 − π
ϕ′ (x)2 dx
=
9
0
Z
h
for any test function ϕ. This proves the sufficient condition for a local minimum.