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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 answer. Answers with no justification will not be graded. 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 > 0, with f (x) = kx, where k > 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 > 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]×[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·∇φdΓ = div(∇φ)dΩ = ∆φdΩ = 12(x2 − y 2 )dxdy = 0. Γ Γ Ω Ω Ω