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Weak Solutions to Partial Differential Equations

Case study: Poisson’s Equation William Golding

University of Maryland

May 5, 2016

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Outline

1

Introduction

2

Weak formulation

3

Functional Analysis

4

Existence and Uniqueness

5

Regularity

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Motivation

Poisson’s Equation

For u ∈ C 2(U) we say that u satisfies Poisson’s equation if ∆u = f

• n U ⊂ Rn

u = 0

• n ∂U

Figure : A steady state of the heat equation

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Outline

1

Introduction

2

Weak formulation

3

Functional Analysis

4

Existence and Uniqueness

5

Regularity

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Weak Derivatives

Definition

For, α a multiindex and f ∈ Lp(U) for some open set U, we say g = Dαf is the weak derivative of f if

• U

gφ dx = (−1)|α|

• U

f Dαφ dx for each φ ∈ C ∞

c (U).

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Example 1

Figure : Example of a weakly differentiable function f (x) = |x|

− ∞

−∞

|x|φ′ dx = −

−∞

−xφ′ dx − ∞ xφ′ dx = ∞

−∞

h(x)φ dx for each φ ∈ C ∞

c (R) where h(x) is the weak derivative of f given by

h(x) =

• 1

if x ≥ 0 −1 if x < 0

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Example 2

g(x) =

• 1

if x ≥ 0 if x < 0

Figure : Example of a non-weakly differentiable function g(x) the Heavyside function

− ∞

−∞

g(x)φ′ dx = − ∞ φ′ dx = φ(0) = ∞

−∞

δ0φ dx

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Sobolev Spaces

Define the Sobolev spaces W k,p(U) = {f ∈ Lp(U) | Dαf ∈ Lp(U) ∀|α| < k}. which we norm with f W k,p =

|α|≤k

Dαf p

Lp

1/p =

• f p

Lp+∇f p Lp+· · ·+∇kf p Lp

1/p In particular we denote the Hilbert space W k,2(U) = Hk(U)

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Weak Formulation for Poisson’s Equation

Now, for the classical formulation of Poisson’s equation we have ∆u = f We multiply by a test function v ∈ H1(U) and integrate by parts to get the weak formulation,

• U

v∆u dx = −

• U

∇v · ∇u dx =

• U

fv dx

Definition

We say that u is a weak solution to Poisson’s equation if ∀v ∈ H1, it satisfies B[u, v] = −

• U

∇v · ∇u dx =

• U

fv dx = f (v)

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Outline

1

Introduction

2

Weak formulation

3

Functional Analysis

4

Existence and Uniqueness

5

Regularity

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Energy Estimates/Lax Milgram

Theorem

Let H be a Hilbert space, and B : H × H → R be a bilinear form satisfying the energy estimates for some constants α, β > 0, |B[u, v]| ≤ αuHvH and βu2

H ≤ |B[u, u]|.

Then, if f : H → R be a bounded linear functional, there exists a unique u ∈ H such that B[u, v] = f (v) for each v ∈ H.

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Energy Estimates for Poisson’s Equation

First, by Holder’s inequality |B(u, v)| ≤

• U

|∇u||∇v| dx ≤ ∇uL2∇vL2 ≤ uH1vH1 By a Poincare inequality, for some β > 0, u2

L2 ≤ β∇u2 L2 = β

• U

|∇u|2 dx = β|B(u, u)| Thus, u2

H1 = u2 L2 + ∇u2 L2 ≤ (β + 1)|B(u, u)|

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Outline

1

Introduction

2

Weak formulation

3

Functional Analysis

4

Existence and Uniqueness

5

Regularity

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Existence and Uniqueness for Poisson’s Equation

Definition

We say u is a weak solution of Poisson’s equation if u satisfies B[u, v] = −

• U

∇v · ∇u dx =

• U

fv dx = f (v) for each v ∈ H1. By Lax-Milgram, there exists a unique weak solution u to Poisson’s equation.

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Outline

1

Introduction

2

Weak formulation

3

Functional Analysis

4

Existence and Uniqueness

5

Regularity

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Recovering Classical Solutions

Assume u ∈ C 2(U) for u our weak solution Then, for each v ∈ H1, −

• U

∇u · ∇v dx =

• U

v∆u =

• U

fv dx So, for each v ∈ H1,

• U

(∆u − f )v dx = 0 Therefore, ∆u = f a.e. on U

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References

[1] L. Evans Partial Differential Equations. American Mathematical Society (2010)

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