Weak Solutions to Partial Differential Equations Case study: - PowerPoint PPT Presentation

Weak Solutions to Partial Differential Equations Case study: Poisson’s Equation William Golding University of Maryland May 5, 2016 W. Golding (UMD) Weak Solutions May 2016 1 / 17

Outline Introduction 1 Weak formulation 2 Functional Analysis 3 Existence and Uniqueness 4 Regularity 5 W. Golding (UMD) Weak Solutions May 2016 2 / 17

Motivation Poisson’s Equation For u ∈ C 2 ( U ) we say that u satisfies Poisson’s equation if on U ⊂ R n ∆ u = f u = 0 on ∂ U Figure : A steady state of the heat equation W. Golding (UMD) Weak Solutions May 2016 3 / 17

Outline Introduction 1 Weak formulation 2 Functional Analysis 3 Existence and Uniqueness 4 Regularity 5 W. Golding (UMD) Weak Solutions May 2016 4 / 17

Weak Derivatives Definition For, α a multiindex and f ∈ L p ( U ) for some open set U , we say g = D α f is the weak derivative of f if � � g φ dx = ( − 1) | α | f D α φ dx U U for each φ ∈ C ∞ c ( U ). W. Golding (UMD) Weak Solutions May 2016 5 / 17

Example 1 Figure : Example of a weakly differentiable function f ( x ) = | x | � ∞ � 0 � ∞ � ∞ | x | φ ′ dx = − − x φ ′ dx − x φ ′ dx = − h ( x ) φ dx −∞ −∞ 0 −∞ for each φ ∈ C ∞ c ( R ) where h ( x ) is the weak derivative of f given by � 1 if x ≥ 0 h ( x ) = − 1 if x < 0 W. Golding (UMD) Weak Solutions May 2016 6 / 17

Example 2 � 1 if x ≥ 0 g ( 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) = − δ 0 φ dx −∞ 0 −∞ W. Golding (UMD) Weak Solutions May 2016 7 / 17

Sobolev Spaces Define the Sobolev spaces W k , p ( U ) = { f ∈ L p ( U ) | D α f ∈ L p ( U ) ∀| α | < k } . which we norm with � 1 / p � 1 / p � � � � D α f � p � f � p L p + �∇ f � p L p + · · · + �∇ k f � p � f � W k , p = = L p L p | α |≤ k In particular we denote the Hilbert space W k , 2 ( U ) = H k ( U ) W. Golding (UMD) Weak Solutions May 2016 8 / 17

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 ∈ H 1 ( U ) and integrate by parts to get the weak formulation, � � � v ∆ u dx = − ∇ v · ∇ u dx = fv dx U U U Definition We say that u is a weak solution to Poisson’s equation if ∀ v ∈ H 1 , it satisfies � � B [ u , v ] = − ∇ v · ∇ u dx = fv dx = f ( v ) U U W. Golding (UMD) Weak Solutions May 2016 9 / 17

Outline Introduction 1 Weak formulation 2 Functional Analysis 3 Existence and Uniqueness 4 Regularity 5 W. Golding (UMD) Weak Solutions May 2016 10 / 17

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 ] | ≤ α � u � H � v � H and β � u � 2 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. W. Golding (UMD) Weak Solutions May 2016 11 / 17

Energy Estimates for Poisson’s Equation First, by Holder’s inequality � | B ( u , v ) | ≤ |∇ u ||∇ v | dx ≤ �∇ u � L 2 �∇ v � L 2 ≤ � u � H 1 � v � H 1 U By a Poincare inequality, for some β > 0, � |∇ u | 2 dx = β | B ( u , u ) | � u � 2 L 2 ≤ β �∇ u � 2 L 2 = β U Thus, � u � 2 H 1 = � u � 2 L 2 + �∇ u � 2 L 2 ≤ ( β + 1) | B ( u , u ) | W. Golding (UMD) Weak Solutions May 2016 12 / 17

Outline Introduction 1 Weak formulation 2 Functional Analysis 3 Existence and Uniqueness 4 Regularity 5 W. Golding (UMD) Weak Solutions May 2016 13 / 17

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 ] = − ∇ v · ∇ u dx = fv dx = f ( v ) U U for each v ∈ H 1 . By Lax-Milgram, there exists a unique weak solution u to Poisson’s equation. W. Golding (UMD) Weak Solutions May 2016 14 / 17

Outline Introduction 1 Weak formulation 2 Functional Analysis 3 Existence and Uniqueness 4 Regularity 5 W. Golding (UMD) Weak Solutions May 2016 15 / 17

Recovering Classical Solutions Assume u ∈ C 2 ( U ) for u our weak solution Then, for each v ∈ H 1 , � � � − ∇ u · ∇ v dx = v ∆ u = fv dx U U U So, for each v ∈ H 1 , � (∆ u − f ) v dx = 0 U Therefore, ∆ u = f a.e. on U W. Golding (UMD) Weak Solutions May 2016 16 / 17

References [1] L. Evans Partial Differential Equations. American Mathematical Society (2010) W. Golding (UMD) Weak Solutions May 2016 17 / 17