Mathematics homework help. MATH572 Spring 2020

Assignment 3

Due: Wednesday March 11

Solve any set of problems for 100 points.

Problem 1: (30 points) Let Ω ⊂ R

2

and u ∈ H1

(Ω). Prove the following inequality

Z

∂Ω

u

2

dx ≤ Ckuk

2

1

(1)

Z

Ω

u

2

dx ≤ C

Z

Ω

|∇u|

2

dx +

Z

∂Ω

u

2

dx

(2)

Problem 2: (30 points) Consider the following B.V.P for elliptic equation in R

2

.

−∆u + q(x)u = f(x, y), (x, y) ∈ Ω = (0, 2) × (0, 1)

and

∂u

∂ν = g, (x, y) ∈ Γ,

where Γ is its boundary, and ν is the outward unit normal vector to Γ. Here q = 1 in (0, 1) × (0, 1)

and q = 0 in the remaining part of the domain.

Derive the weak formulation for this problem and show the coercivety of the corresponding bilinear form in H(Ω)− norm.

Problem 3: (50 points) Consider the τ to be a tetrahedron in the (x, y, x)− space determined

by its vertexes P1, P2, P3, P4. In order these four points to form a tetrahedron we assume that they

are not in a plane . Let Σ = {v(P1), v(P2), v(P3), v(P4)} be the set of values of a function v at

the vertices. Find a nodal basis for the space of linear functions over τ by using homogeneous

(baracentric) coordinates (λ1, λ2, λ3, λ4). Compute the element mass matrix.

Problem 4: (20 points) Let Ω be the square (0, 1) × (0, 1). Prove the Poincare inequality

kuk

2

L2(Ω) ≤ C

k∇uk

2

L2(Ω) +

Z

Ω

u dx2

!

.

Problem 5: (20 points) Let τ be a shape regular square in 2 − D with a side hτ . If ∂τ is the

1

boundary of τ show that there is a constant C independent of hτ such that

kvk

2

L2(∂τ) ≤ C

h

−1

τ kvk

2

L2(τ) + hτk∇vk

2

L2(τ)

.

Problem 6: (20 points) Consider (τ,P, Σ), where

τ = {rectangle (xi−1, xi) × (yj−1, yj ) with vertices P1, P2, P3, P4};

P = {v : v(x, y) = a00 + a10x + a01y + a11xy + a20x

2 + a21x

2

y + a12xy2 + a02y

2

};

Σ = {v(P1), v(P2), v(P3), v(P4), v(P12), v(P23), v(P34), v(P41)}

where Pij is the mid point of the edge joining Pi and Pj

. Show that the set Σ is P− unisolvent.

Note that the term x

2

y

2

is missing in the polynomial set and the center of the rectangle is allso

missing from the set of points values so that dimP = 8 and the number of degrees of freedom is 8