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Quiz 1 (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.
Question 1: Let
R x v : [0, ∞) −→ R be a continuous function with bounded derivative, and let w : [0, ∞) −→ R be such
that w(x) = x1 0 (v(t) − v(x))dt. (a) Show that |w(x)| ≤ M2x where M = supx∈[0,∞) |∂x v(x)|.
Rt
Any time we see a quantity like v(t)−v(x) we must think of the fundamental theorem of calculus, i.e., v(t)−v(x) = x ∂x v(z)dz.
Hence, we have
Z
Z Z
Z Z
Z Z
1 x t
1 x t
1 x x
1 x
∂z v(z)dzdt ≤
∂z v(z)dz dt ≤
(v(t) − v(x))dt = |∂z v(z)|dzdt
|w(x)| = x 0
x 0 x
x 0 x
x 0 t
Z xZ x
Z x
M
M
M 2 1 2
M
≤
dzdt =
(x − t)dt =
(x − x ) =
x.
x 0 t
x 0
x
2
2
Hence |w(x)| ≤
Mx
2
for all x ∈ [0, ∞).
(b) Estimate w(0).
The estimate |w(x)| ≤
Mx
2
shows that |w(0)| ≤ 0, meaning that w(0) = 0.
(c) Show that ∂t (tw(t)) = −t∂t v(t) .
Rt
Upon observing that tw(t) = 0 (v(z) − v(t))dz and recalling that the fundamental theorem of calculus implies that
t
Z
∂t
f (z)dz
= f (t),
0
we have
Z
t
Z
(v(z) − v(t))dz = ∂t
∂(tw(t)) = ∂t
t
v(z)dz − ∂t (v(t)t) = v(t) − v(t) − t∂t v(t) = −t∂t v(t).
0
0
Hence ∂t (tw(t)) = −t∂t v(t).
(d) Prove that v(x) − v(0) = −w(x) −
by parts.)
Rx
0
w(t)
t dt.
(Hint: observe that v(x) − v(0) =
Rx
1
0 t
We follow the hint
Z x
1
1
(t∂t v(t)) dt = −
∂t (tw(t))dt
0 t
0 t
x
Z x
1
1
=
∂t
tw(t)dt − tw(t)
t
t
0
0
Z x
1
=−
tw(t)dt − w(x) + w(0),
2
0 t
Z
x
v(x) − v(0) =
thereby proving that v(x) − v(0) = −
Rx
1
w(t)dt
0 t
− w(x).
(t∂t v(t)) dt, use (c), and integrate
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Question 2: Consider the equation ∂t c(x, t)+∂x (x2 −xL)c(x, t) −∂x (1+x2 )∂x c(x, t) = 6x/L2 , where x ∈ [0, L], t > 0,
RL
1
with c(x, 0) = f (x), −∂n c(0, t) = 2, −∂n c(L, t) = 1+L
2 , (∂n is the normal derivative). Compute E(t) := 0 c(ξ, t)dξ.
We integrate the equation with respect to x over [0, L]
Z
L
Z
∂t c(ξ, t)dξ +
0
Using that
RL
0
L
∂x (x2 − xL)c(x, t) dξ −
Z
0
∂t c(ξ, t)dξ = dt
RT
0
0
L
6
∂ξ (1 + ξ 2 )∂ξ c(ξ, t) dξ = 2
L
Z
L
ξdξ.
0
c(ξ, t)dξ together with the fundamental theorem of calculus, we infer that
dt E(t) − (1 + L2 )∂x c(L, t) + ∂x c(0, t) = 3.
The boundary conditions ∂x c(0, t) = −∂n c(0, t) = 2, −∂x c(L, t) = −∂n c(L, t) =
1
1+L2
give
dt E(t) + 1 + 2 = 3.
We now apply the fundamental theorem of calculus with respect to t
Z t
∂τ E(τ )dτ = 0.
E(t) − E(0) =
0
In conclusion
Z
E(t) =
L
f (ξ)dξ,
∀t ≥ 0.
0
Question 3: Let φ = sin(x) cosh(y) + 2x2 + 3y 2 (a) Compute ∆φ(x, y).
The definition ∆φ = ∂xx φ + ∂yy φ implies that
∆φ = ∂xx φ + ∂yy φ = − sin(x) cosh(y) + sin(x) cosh(y)4 + 6 = 10.
(b)Let Ω be the disk of radius 1 centered at (0, 0) and let Γ be the boundary of Ω. Compute
R
Γ
∂n φdΓ.
The definition ∆φ = div(∇φ) and the fundamental theorem of calculus (also known as the divergence theorem) imply that
Z
Z
Z
Z
Z
∂n φdΓ =
n·∇φdΓ =
div(∇φ)dΩ =
∆φdΩ = 10
dΩ = 10π,
Γ
because the surface of Ω,
R
Ω
Γ
dΩ, is equal to π.
Ω
Ω
Ω