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Solutions to problems from the applied analysis and numerical analysis qualifying exam. The problems cover topics such as convergence of sequences of functions and distributions, properties of linear operators, and finite element methods. Students preparing for qualifying exams in applied analysis or numerical analysis may find this document useful for understanding the concepts and techniques required to solve similar problems.
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January 8, 2012
Cover Sheet – Applied Analysis Part
Policy on misprints: The qualifying exam committee tries to proofread exams as carefully as possible. Nevertheless, the exam may contain a few misprints. If you are convinced a problem has been stated incorrectly, indicate your interpretation in writing your answer. In such cases, do not interpret the problem so that it becomes trivial.
Name
Combined Applied Analysis/Numerical Analysis Qualifier Applied Analysis Part January 8, 2012
Instructions: Do any 3 of the 4 problems in this part of the exam. Show all of your work clearly. Please indicate which of the 4 problems you are skipping.
Problem 1. Let D be the set of compactly supported functions defined on R and let D′^ be the corresponding set of distributions.
(a) Define convergence in D and D′. (b) Consider a function f ∈ C(1)(R) such that both f and f ′^ are in L^1 (R), and
R f^ (x)dx^ = 1. Define the sequence of functions {Tn(x) := n^2 f ′(nx) : n = 1, 2 ,.. .}. Show that, in the sense of distributions — i.e., in D′^ —, Tn converges to δ′.
Problem 2. Let M : C[0, 1] → C[0, 1] be defined by M (u) :=
0 (2 +^ st^ +^ u(s)
(^2) )− (^1) ds. Let
‖ · ‖ := ‖ · ‖C[0,1]. Let Br := {u ∈ C[0, 1] | ‖u‖ ≤ r}.
(a) Show that M : B 1 → B 1 / 2 ⊂ B 1. (b) Show that M is Lipschitz continuous on B 1 , with Lipschitz constant 0 < α < 1 – i.e., ‖M [u] − M [v]‖ ≤ α‖u − v‖. (c) Show that M has a fixed point in B 1. State the theorem you are using to show that the fixed point exists.
Problem 3. Let Lu = − d
(^2) u dx^2 ,^ −π^ ≤^ x^ ≤^ π, with the domain of^ L^ given by DL := {u ∈ L^2 [−π, π] : u′′^ ∈ L^2 [π, π], u(−π) = −u(π), u′(−π) = −u′(π)}. (a) Show that L is self adjoint on D(L). (b) Find the Green’s function G(x, y) for the problem Lu = f , u ∈ DL. (c) Show that Ku :=
∫ (^) π −π G(·, y)u(y)dy^ is a compact self-adjoint operator. (d) Without actually finding them, show that the eigenfunctions of L form a complete, or- thogonal set for L^2 [−π, π]. (Hint: Relate the eigenfunctions of L to those of K. Use compactness.)
Problem 4. Let T be a (possibly unbounded) linear operator on a Hilbert space H, defined on the domain DT.
(a) Define these: the resolvent set of T , ρ(T ); the discrete spectrum, σd(T ); the continuous spectrum, σc(T ); and the residual spectrum, σr(T ). (b) Assume T is bounded. Show that the set {λ ∈ C : |λ| > ‖T ‖} ⊆ ρ(T ). (Hint: Use a Neumann series expansion.) (c) Let H = ℓ^2 , with the usual inner product. Define T to be the shift operator T (x 1 , x 2 ,.. .) = (0, x 1 , x 2 ,.. .). Show that every |λ| > 1 is in ρ(T ), that λ = 1 is in σc(T ), and that λ = 0 is in σr(T ).
Combined Applied Analysis/Numerical Analysis Qualifier Numerical analysis part January 8, 2012
Problem 1:
Let Ω = (0, 1) × (0, 1), f ∈ C^0 (Ω) and q ∈ R with q ≥ 0. Consider the boundary value problem
−∆u + qu = f in Ω; u = 0 on ∂Ω.
We are interested in approximating the quantity α :=
∂Ω n·∇u^ where^ n^ is the outward unit normal of Ω.
v ∈ C^0 (Ω) ∩ V | ∀T ∈ Th, v|T is linear
Write the weak formulation satisfied by the finite element approximation uh ∈ Vh of u. Prove that the function uh exists and is unique.
‖v − vh‖V ≤ Ch‖v‖H (^2) (Ω).
Problem 2:
Let K be a polyhedron in Rd, d ≥ 1. Let h = diam(K) and define
Kˆ = {ˆx = x/diam(K), x ∈ K}.
Show that there exists a constant c solely depending on Kˆ such that for any v ∈ H^1 (K),
‖v‖L (^2) (∂K) ≤ c
h−^1 /^2 ‖v‖L (^2) (K) + h^1 /^2 ‖∇v‖L (^2) (K)
Problem 3:
Let u 0 : (0, 1) → R be a given smooth initial condition and T > 0 be a given final time. Let u : [0, T ] × Ω → R be a smooth function satisfying u(t, 0) = u(t, 1) = 0 for any t ∈ [0, T ] and
∀v ∈ C c∞ ([0, T ) × (0, 1)) :
−
0
0
u(t, x) vt(t, x) dx dt −
0
u 0 (x)v(0, x) dx
0
0
ux(t, x) vx(t, x) dx dt +
0
0
u(t, x) v(t, x) dx dt = 0.
Here C∞ c ([0, T ) × (0, 1)) is the space of functions belonging to C∞([0, T ] × [0, 1]) and compactly supported in [0, T ) × (0, 1).
