numerical method PDE

MATH572 Spring 2020 Assignment 3

Due: Wednesday March 11 Solve any set of problems for 100 points.

Problem 1: (30 points) Let Ω ⊂ R2 and u ∈ H1(Ω). Prove the following inequality∫ ∂Ω

u2dx ≤ C‖u‖21 (1)∫ Ω

u2dx ≤ C [∫

Ω

|∇u|2dx+ ∫ ∂Ω

u2dx

] (2)

Problem 2: (30 points) Consider the following B.V.P for elliptic equation in R2.

−∆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 bilin- ear 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

‖u‖2L2(Ω) ≤ C

( ‖∇u‖2L2(Ω) +

(∫ Ω

u dx

)2) .

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

‖v‖2L2(∂τ) ≤ C ( h−1τ ‖v‖2L2(τ) + hτ‖∇v‖2L2(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 + a20x2 + a21x2y + a12xy2 + a02y2};

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 x2y2 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.

2

 

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numerical method PDE was first posted on March 8, 2020 at 8:13 pm.
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