# Last name: name: 1 Quiz 1 (Notes, books, and calculators are not authorized)

```Last name:
name:
1
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
Question 1: Let φ(x, y) = cos(x + sin(x − y)). Compute ∂x φ(x, y) and ∂y φ(x, y). (Do not try
to simplify the results).
We apply the chain rule repeatedly
∂x φ(x, y) = − sin(x + sin(x − y))(1 + cos(x − y)).
∂y φ(x, y) = − sin(x + sin(x − y))(− cos(x − y)).
Question 2: Consider the heat equation ∂t T − k∂xx T = f (x), x ∈ [a, b], t &gt; 0, with f (x) = kx,
where k &gt; 0. Compute the steady state solution (i.e., ∂t T = 0) assuming the boundary
conditions: −k∂n T (a) = 0, T (b) = 0 (∂n is the normal derivative)
At steady state, T does not depend on t and we have ∂xx T (x) = −x, which implies ∂x T (x) =
α − 21 x2 , and T (x) = β + αx − 16 x3 , where α, β ∈ R. The two constants α and β are determined
by the boundary conditions. 0 = −∂n T (a) = ∂x T (a) = α − 21 a2 and 0 = T (b) = β + αb − 16 b3 .
We conclude that α = 12 a2 and β = −αb + 16 b3 = − 12 a2 b + 61 b3 . In conclusion
1
1
1
1
1
1
T (x) = − a2 b + b3 + a2 x − x3 = − a2 (b − x) + (b3 − x3 ).
2
6
2
6
2
6
2
Quiz 1, September 3, 2013
Question 3: Consider the equation ∂t c(x, t) − ∂xx c(x, t) = x, where x ∈ [0, L], t &gt; 0, with
c(x, 0) = f (x), −∂n c(0, t) = 6, −∂n c(L, t) = 5, (∂n is the normal derivative). Compute E(t) :=
RL
c(ξ, t)dξ.
0
We integrate the equation with respect to x over [0, L]
Z
L
Z
∂t c(ξ, t)dξ −
RL
0
Z
∂t c(ξ, t)dξ = dt
RT
0
L
ξdξ.
∂ξξ c(ξ, t)dξ =
0
0
0
Using that
infer that
L
c(ξ, t)dξ together with the fundamental theorem of calculus, we
1 2
L .
2
The boundary conditions ∂x c(0, t) = −∂n c(0, t) = 6, −∂x c(L, t) = −∂n c(L, t) = 5 give
dt E(t) − ∂x c(L, t) + ∂x c(0, t) =
dt E(t) + 5 + 6 =
1 2
L .
2
We now apply the fundamental theorem of calculus with respect to t
Z t
1
E(t) − E(0) =
∂τ E(τ )dτ = ( L2 − 11)t.
2
0
In conclusion
Z
E(t) =
0
L
1
f (ξ)dξ + ( L2 − 11)t.
2
Question 4: Let φ = x4 − y 4 (a) ComputeR∆φ(x, y). (b) Consider the square Ω = [0, 1]&times;[0, 1]
and let Γ be the boundary of Ω. Compute Γ ∂n φdΓ.
(a) The definition ∆φ = ∂xx φ + ∂yy φ implies that
∆φ = ∂xx φ + ∂yy φ = 12x2 − 12y 2 = 12(x2 − y 2 ).
(b)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&middot;∇φdΓ =
div(∇φ)dΩ =
∆φdΩ =
12(x2 − y 2 )dxdy = 0.
Γ
Γ
Ω
Ω
Ω
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