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Quiz 7 (Notes, books, and calculators are not authorized)
Show all your work in the blank space you are given on the exam sheet. Always justify your answer. Answers with
no justification will not be graded.
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2
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Question 1: Consider the equation −∂x (x 2 ∂x u(x)) − x− 2 u(x) = 1 + x for all x ∈ ( π4 , π 2 ) with u( π4 ) = 0 and
u(π 2 ) = 0. Use the Fredholm alternative to determine whether the above problem has a solution. (Hint: The
√
√
1
1
general solution to −∂x (x 2 ∂x φ(x)) − x− 2 φ(x) = 0 is φ(x) = c1 cos(2 x) + c2 sin(2 x).)
2
2
1
Let us compute the null space of the operator L : {v ∈ C 2 ( π4 , π 2 ) : v( π4 ) = 0, v(π 2 ) = 0} 3 u 7−→ −∂x (x 2 ∂x u(x)) −
2
1
x− 2 u(x) ∈ C 0 ( π4 , π 2 ). Let N (L) be the null space and consider v ∈ N (L). By using the hint, we infer that
√
√
v(x) = c1 cos(2 x) + c2 sin(2 x)
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The boundary condition v( π4 ) = √
0 implies c1 = 0. The second boundary condition v(π 2 ) = 0 is automatically satisfied.
In conclusion N (L) = span{sin(2 x}. We are in the second case of Fredholm’s alternative. The problem has
√ solutions
∗
∗
∗
x)}. Let
(infinitely many)
if
the
right-hand
side
is
orthogonal
to
N
(L
).
But
L
=
L
,
i.e.,
N
(L
)
=
span{sin(2
√
v(x) = λ sin(2 x), λ ∈ R, then
Z
π2
π2
4
π2
Z
(1 + x)v(x)dx = λ
The above integral is nonzero unless λ = 0 since
any solution.
R π2
π2
4
π2
4
√
(1 + x) sin(2 x)dx.
√
(1 + x) sin(2 x)dx > 0. As a result, the problem does not have
Question 2: Use the Fredholm alternative to determine whether the following problem has a solution: Let f be
a smooth function in [0, 1], find u such that
u − ∂xx u = f (x),
x ∈ (0, 1),
∂x u(1) + u(1) = 0, −∂x u(0) + u(0) = 0
Let us compute the null space of the operator L : {v ∈ C 2 (0, 1) : ∂x v(1) + v(1) = 0, −∂x v(0) + v(0) = 0} 3 u 7−→
u(x) − ∂xx u(x) ∈ C 0 (0, 1). Let N (L) be the null space. Let v ∈ N (L). The energy argument implies that
Z
1
2
2
(v + (∂x v) )dx −
0=
0
[v∂x v]10
Z
=
1
(v 2 + (∂x v)2 )dx + v(1)2 + v(0)2 .
0
This proves that v = 0, i.e., N (L) = {0}. Fredholm’s alternative implies that the above problem has a unique solution.
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Quiz 7, April 1, 2014
Question 3: Let Ω = {(t, x) ∈ R2 : t > 0, x ≥ −t}. Let Γ be defined by the following parameterization
Γ = {x = xΓ (s), t = tΓ (s), s ∈ R}, with xΓ (s) = s and tΓ (s) = −s if s ≤ 0, xΓ (s) = s and tΓ (s) = 0 if s ≥ 0. Solve
the following PDE (give the implicit and the explicit representations):
(
1 if x > 0
for all (x, t) in Γ.
ut + 2ux + u = 0, in Ω,
u(x, t) = uΓ (x, t) :=
2 if x < 0
We define the characteristics by
dx(t, s)
= 2, x(tΓ (s), s) = xΓ (s).
dt
This gives x(t, s) = xΓ (s) + 2(t − tΓ (s)). Upon setting φ(t, s) = u(x(t, s), t), we observe that ∂t φ(t, s) + φ(t, s) = 0,
which means
φ(t, s) = ce−t .
The initial condition implies φ(tΓ (s), s) = uΓ (xΓ (s), tΓ (s)); as a result c = uΓ (xΓ (s), tΓ (s))etΓ (s) .
φ(t, s) = uΓ (xΓ (s), tΓ (s))etΓ (s)−t .
The implicit representation of the solution is
u(x(t, s), t) = uΓ (xΓ (s), tΓ (s))etΓ (s)−t ,
x(t, s) = xΓ (s) + 2(t − tΓ (s)).
Now we give the explicit representation.
Case 1: If s ≤ 0, xΓ (s) = s, tΓ (s) = −s, and uΓ (xΓ (s), tΓ (s)) = 2. This means x(t, s) = s + 2(t + s) and we obtain
s = 31 (x − 2t), which means
1
u(x, t) = 2e− 3 (x−2t)−t ,
if x − 2t < 0.
Case 2: If s ≥ 0, xΓ (s) = s, tΓ (s) = 0, and uΓ (xΓ (s), tΓ (s)) = 1. This means x(t, s) = s + 2t and we obtain s = x − 2t,
which means
u(x, t) = e−t ,
if x − 2t > 0.