Mathematical Methods in the Engineering Sciences

Elena Beretta

Dipartimento di MatematicaPolitecnico di Milano (ITALY)

elena.beretta@polimi.it

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Main goals of the course

Many constitutive laws of physics are governed by partial differentialequations wThe qualitative study of solutions to these equations from amathematical viewpoint give an insight of the physical processes orsystems. In general additional information is necessary to select orpredict a unique solution. This information is often supplied in the formof initial and/or boundary data wWe will analyze the following fundamental equations: Laplaceequation,Poisson equation, Heat equation, Schroedinger equationanalyzing some boundary value problems and Cauchy problems. Inparticular we will investigate well-posedness (existence, uniqueness,stability) of such problems using several tools of functional analysis.Finally we will study two mathematical models, one arising in medicalimaging (Optical Tomography) and the second in quantum mechanics(Equilibrium of atoms)

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Part I

Prerequisits

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Banach Spaces

X vector space on R

A norm on X is a function ‖ · ‖ : X → [0,+∞) s.t.‖x‖ ≥ 0 and ‖x‖ = 0 if and only if x = 0 (positivity)‖λx‖ = |λ|‖x‖, ∀x ∈ X , ∀λ ∈ R (homogeneity)‖x + y‖ ≤ ‖x‖+ ‖y‖, ∀x , y ∈ X (triangular inequality)

(X , ‖ · ‖) is a normed space

(X , ‖ · ‖) is a metric space (X ,d) w.r.t. d induced byd(x , y) = ‖x − y‖, ∀x , y ∈ X

xn → x∗ in X if ‖xn − x∗‖ → 0 as n→ +∞(strong convergence)

A Banach space is a complete normed space(any Cauchy sequence is convergent in X )

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Examples

(Rn, ‖ · ‖n) and (Cn, ‖ · ‖n) are Banach spaces with respect to

‖x‖ =√∑n

i=1 x2i where x = (x1, . . . , xn).

C0(K ) = u : Ω ⊂ Rn → R continuous on K, K is compact‖u‖C0(Ω) = ‖u‖∞ = sup

x∈Ω

|u(x)|,

(C0(K ), ‖ · ‖∞) is a Banach space

Lp(Ω) = u :(∫

Ω |u(x)|pdx)1/p

< +∞, 1 ≤ p < +∞

‖u‖p =(∫

Ω |u(x)|pdx)1/p

(Lp(Ω), ‖ · ‖p) is a Banach space

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Examples

xk = (x1, x2, . . . ), xk ∈ R1 ≤ p < +∞,

`p =

x = xk :

∞∑k=1

|xk |p < +∞

(`p, ‖ · ‖) is a Banach space with ‖x‖p = (∑+∞

k=1 |xk |p)1/p

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Hilbert Spaces

X vector space on R

An inner product on X is a function (·, ·) : X × X → R s.t.(x , x) ≥ 0, ∀x ∈ X , (x , x) = 0 iff x = 0 (positivity)(y , x) = (x , y), ∀x , y ∈ X (symmetry)(λx + µy , z) = λ(x , z) + µ(y , z), ∀x , y , z ∈ X , ∀λ, µ ∈ R (bilinearity)

(X , (·, ·)) is an inner product space

If H is a vector space on C then symmetry is substituted by(x , y) = (y , x)

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Hilbert Spaces

(X , (·, ·)) is a normed space (X , ‖ · ‖) w.r.t. ‖ · ‖ induced by‖x‖ = (x , x)1/2, ∀x ∈ X

|(x , y)| ≤ ‖x‖‖y‖, ∀x , y ∈ X Cauchy-Schwarz inequality

‖x + y‖2 + ‖x − y‖2 = 2‖x‖2 + 2‖y‖2 Parallelogram law

A Hilbert space is a complete inner product space

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Examples of Hilbert Spaces

Rn (x , y) =∑n

i=1 xiyi ,Cn (x , y) =

∑ni=1 xiy i

L2(Ω) is a Hilbert space(u, v) =

∫Ω u(x)v(x)dx , (u, v) =

∫Ω u(x)v(x)dx

`2 is a Hilbert space

(x,y) =∞∑

k=1

xkyk

∞∑k=1

xkyk

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Orthogonal projections

V ⊂ H is closed in H if it contains all limit points of sequences inV .

V⊥ = u ∈ H : (u, v) = 0, ∀ v ∈ Vorthogonal complement of V ⊂ H, H Hilbert space

(Projection Theorem) (with proof) If V is a closed subspace of Hthen ∃! decompositionx = u + v , u ∈ V , v ∈ V⊥, ∀ x ∈ H

PV x = u orthogonal projection of x onto V

QV x = v orthogonal projection of x onto V⊥

‖x‖2 = ‖PV x‖2 + ‖x − PV x‖2, ‖PV x‖ ≤ ‖x‖

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Hilbert Spaces: separability

H Hilbert space, Y ⊂ H is dense if ∀ x ∈ H ∃ yn ⊂ Y :yn → x

A Hilbert space is separable if there exists a countable and denseY ⊂ X .

Examples of separable Hilbert spaces:

(Rn, ‖ · ‖n) and (Cn, ‖ · ‖n)

(L2(Ω), ‖ · ‖2)

(`2, ‖ · ‖2)

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Bases in Hilbert Spaces

ej : (ei ,ej) = δij orthonormal (countable) set in H

ej, orthonormal set, is a (countable) basis for H if

x =∞∑

j=1

(x ,ej)ej , ∀ x ∈ H

Generalized Fourier Series

cj = (x ,ej)

Fourier coefficients

‖x‖2 =∞∑

j=1

c2j

(follows from Pythagora’s Theorem)

H is separable iff H has a countable basis12 / 82

Examples of orthonormal bases

Rn, Cn

e1 = (1,0, . . . ,0), . . . ,e2 = (0,1,0 . . . ,0) . . .

L2(0,2π)e0 = 1√

2π,e1 = 1√

πcos x ,e2 = 1√

πsin x ,e3 = 1√

πcos 2x . . .

`2

e1 = (1,0, . . . ), e2 = (0,1, . . . ) . . .

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Linear operators on Hilbert spaces

L : (H1, ‖ · ‖H1)→ (H2, ‖ · ‖H2)

L is linear if L(λx + µz) = λLx + µLz, ∀ x , z ∈ H1

Kernel of L (N (L))

N (L) = x ∈ H1 : Lx = 0

Range of L (R(L))

R(L) = y ∈ H2 : ∃x ∈ H1 : Lx = y

L is bounded if ∃M > 0 : ‖Lx‖Y ≤ M‖x‖H1 , ∀ x ∈ H1

L is continuous if xn → x in H1 ⇒ Lxn → Lx in H2

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Linear operators on Hilbert spaces

L(H1,H2) set of all bounded linear operator from X to Y

L(H1,H2) vector space

‖L‖L(H1,H2) = supx 6=0

‖Lx‖H2

‖x‖H1

= sup‖x‖H1

=1‖Lx‖H2

‖Lx‖H2 ≤ ‖L‖L(H1,H2)‖x‖H1 , ∀ x ∈ H1

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Examples of linear bounded operators

L : Rn → Rm

Lx = Ax

where A is an m × n real valued matrix.

V ⊂ H, closed subset of a Hilbert space H.L1 : H → H with L1x = PV x and L2 : H → H with L1x = QV x

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Results on linear operators

TheoremL(H1,H2) is a Banach space

TheoremL : (H1, ‖ · ‖H1)→ (H2, ‖ · ‖H2), L linear.L is continuous ⇔ L is bounded.

(with proof)

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Linear functionals and dual spaces

L : (H, ‖ · ‖H)→ R bounded and linear: functional on H

L(H,R) is the dual space of H: H∗ or H ′

H∗ = L(H,R), ‖L‖H∗ = ‖f‖L(H,R) = supx 6=0

|Lx |‖x‖H

(H∗, ‖ · ‖H∗) is a Banach space (R is a Banach space)

H∗〈·, ·〉H : H∗ × H → R H∗〈L, x〉H = Lx is a bilinear form

H∗〈Lx〉H or 〈L, x〉 (duality)

|〈L, x〉| ≤ ‖L‖H∗‖x‖H

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Example of linear, bounded functionals on Hilbertspaces

H = Rn, L : Rn → R, let a ∈ Rn,

Lx = (a, x)

H = L2(Ω), Ω ⊂ Rn, L : L2(Ω)→ R, fix g ∈ L2(Ω)

Lf =

∫Ω

fg dx

Fix y ∈ H and let L : H → R

Lx = (x , y)

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Dual spaces of Hilbert spaces

TheoremRiesz representation TheoremLet H be a real Hilbert space. Then, ∀L ∈ H∗ ⇒ ∃! yL ∈ H :Lx = (x , yL),∀ x ∈ H and ‖L‖H∗ = ‖yL‖H

with proof

RemarkRiesz Theorem ⇒ we identify H and H∗ endowing H∗ with theinner product (L1,L2) = (yL1 , yL2).

Exercise: Prove that the kernel of a linear bounded operator is aclosed vector space.Exercise: Show that the Representation Theorem holds also incomplex Hilbert spaces.

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Bilinear Forms

DefinitionLet V be a real vector space. a : V × V → R is a bilinear form on V if∀y ∈ V the function x → a(x , y) is linear∀x ∈ V the function y → a(x , y) is linear

ExampleThe inner product on a real vector space is a bilinear form

In complex inner product spaces we define sesquilinear formssubstituting the linearity with respect to the second component withantilinearity

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Lax-Milgram Theorem

Problem (P):Let F ∈ H∗. Find u ∈ H such that

a(u, v) = 〈F , v〉, ∀v ∈ H

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TheoremLax-Milgram Theorem Let H be a real Hilbert space, a be a bilinearform such that

a is continuous i.e.∃M > 0:

|a(u, v)| ≤ M‖u‖‖v‖, ∀u, v ∈ H

a is coercive i.e. ∃α > 0:

a(v , v) ≥ α‖v‖2, ∀v ∈ H.

Then, there exists a unique solution u to P and

‖u‖ ≤ 1α‖F‖

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RemarkIf ∃M > 0:

|a(u, v)| ≤ M‖u‖‖v‖, ∀u, v ∈ H

then a(u, v) is continuous in H × H

RemarkLax-Milgram Theorem still holds in complex Hilbert spaces forsequilinear forms which are continuous and coercive.

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Minimization Problem

If the bilinear form a is symmetric then Problem (P) is equivalent to

minv∈H

E(v) =12

a(v , v)− 〈F , v〉

If the bilinear form a is symmetric ( antisymmetric in the complexcase)

(u, v)H = a(u, v)

is an inner product and Lax Milgram Theorem derives from RieszTheorem.Exercise: check that this is true.

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Galerkin Approximation Method

H Hilbert space, H = ∪Vk , Vk finite dimensional space for all k .Key ideas:

Project Problem (P)

a(u, v) = 〈F , v〉, v ∈ H

on Vk i.e. find solution uk of Problem (Pk )

a(uk , v) = 〈F , v〉, v ∈ Vk

Show‖uk − u‖H → 0

as k → +∞.

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Galerkin Approximation Method

PropositionAssume the assumptions of the Lax-Milgram Theorem are satisfiedand let u be the unique solution to Problem (P). If uk is the uniquesolution of Problem (Pk ) then

‖u − uk‖ ≤Mα

infv∈Vk‖u − v‖

Theorem

uk → u in V as k → +∞

with proof

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Weak and strong convergence

DefinitionH Hilbert space, a sequence xn of elements in H converges stronglyto an element x ∈ H

xn → x

if‖xn − x‖H → 0

as n→ +∞

DefinitionLet H be a Hilbert space, xn, x ∈ H.xn x (weak convergence) if (xn, y)→ (x , y) for any y ∈ H.

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Strong and weak convergence

TheoremH Hilbert space, xn, x ∈ H.xn → x (strong convergence) ⇒ xn x (weak convergence)

Proof.If xn → x strongly then

|(xn, y)−(x , y)| = |(xn−x , y)| ≤ ‖xn−x‖‖y‖ → 0, ∀y ∈ H ⇒ xn x

Exercise: Prove that if H finite dimensional Hilbert space, xn, x ∈ H.xn → x ⇔ xn x

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Counterexample

Example

H = `2 y ∈ `2

y ∈ `2 ⇒∞∑

k=1

y2k < +∞ ⇒ yk → 0

ek orthonormal basis

(ek ,y) = yk → 0, ∀y ∈ `2 ⇒ ek 0

ek 9 0 strongly in `2

(ek is not a Cauchy sequence: ‖ek − ej‖`2 =√

2, k 6= j )

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Uniqueness of weak limit

TheoremH Hilbert space, xn ∈ H.If xn x ∈ H then x is unique

Proof.Assume xn x . Then

(x − x , y) = (x − xn, y)− (x − xn, y)→ 0, ∀y ∈ H

Hence(x − x , y) = 0, ∀y ∈ H

which impliesx = x .

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Boundedness of weak convergent sequences

TheoremH Hilbert space, xn ∈ H.If xn x ∈ H then xn is bounded

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An estimate of the norm of the weak limit

TheoremH Hilbert space, xn ∈ H.If xn x ∈ H then ‖x‖ ≤ lim infn→∞‖xn‖

Proof.xn x ⇒

‖x‖2 = (x , x) = limn→∞(xn, x) = lim infn→∞(xn, x)

≤ lim infn→∞‖xn‖‖x‖ = ‖x‖lim infn→∞‖xn‖

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Compactness

E ⊂ X is compact if every open cover of E contains a finite subcover

E ⊂ X is compact if and only if every bounded sequence in E containsa convergent subsequence in E (sequentially compact ).

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Weak convergence and compactness Theorems

TheoremA sequentially compact subset E of a normed space is closed andbounded.

Exercise: Prove the theorem.

RemarkIf E is a subset of a finite dimensional normed space (Rn) then

E sequentially compact ⇐⇒ Eclosed and bounded

In particularB1 = x ∈ Rn : ‖x‖ ≤ 1

is compact.

TheoremA normed space is finite dimensional ⇐⇒ B1 is compact.

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Weak convergence and compactness Theorems

ExampleH infinite dimensional separable Hilbert space. and ej∞1 anothonormal basis. Then

‖ej‖H = 1 ∀j

so that ej ⊂ B1 but

‖en − em‖H =√

2 ∀n,m, n 6= m

so ej∞1 is not a Cauchy sequence.

TheoremAny bounded sequence in a Hilbert space H has a weak convergentsubsequence in H.

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Distributions

DefinitionGiven a continuous function ψ : Ω ⊂ Rn → R the support of ψ is the set

supp ψ = x ∈ Ω : ψ(x) 6= 0

DefinitionWe indicate by C∞0 (Ω) the set of functions in C∞(Ω) compactlysupported in Ω

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Distributions

We now endow C∞0 (Ω) with a suitable notion of convergence (notinducing any metric)

Definitionψk∞k=1 ⊂ C∞0 (Ω) , ψ ∈ C∞0 (Ω).

ψk → ψ in C∞0 (Ω) as k →∞

ifDαψk → Dαψ, ∀α uniformly in Ω

and there exists a compact set K ⊂ Ω which contains the support of allthe ψk ’s.

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Distributions

ExampleLet ak ⊂ R, ak → 0 as k →∞, let ψ ∈ C∞0 (Ω). Then

ψk = akψ → 0, k →∞

in the above sense.

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Distributions

C∞0 (Ω) endowed with this convergence will be denoted as D(Ω).

DefinitionA distribution f is a linear continuous functional on D(Ω) (f ∈ D′(Ω))i.e. such that

< f , αψ1 + βψ2 >= α < f , ψ1 > +β < f , ψ2 >, ∀ψ1, ψ2 ∈D(Ω), α, β ∈ R< f , ψk >→< f , ψ > as ψk → ψ in D(Ω) as k → +∞.

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Distributions

Examplesf ∈ C0(Ω) defines a distribution

< f , ψ >=

∫Ω

fψ

since|∫

Ωfψ| ≤ ‖f‖K sup

Ω|ψ|

f ∈ L2(Ω) defines a distribution

< f , ψ >=

∫Ω

fψ

since|∫

Ωfψ| ≤ ‖f‖L2(Ω)‖ψ‖L2(Ω)

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Distributions

Examplesf = δ0 is a distribution

< f , ψ >= ψ(0)

since|ψ(0)| ≤ sup

Ω|ψ|

f ∈ L1loc(Ω) is a distribution

< f , ψ >=

∫Ω

fψ

since|∫

Ωfψ| ≤ ‖f‖L1(K ) sup

Ω|ψ|

Exercise: Verify that for p < n the function f (x) = 1|x |p ∈ D

′(Rn)

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Distributions

TheoremThe map I : L2(Ω)→ D′(Ω) where

< I(f ), ψ >=

∫Ω

fψ dx

is injective

with proof

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Distributional derivatives

Let f ∈ C1(R) with and ψ ∈ D(R). Then∫R

f ′ψ dx = fψ|+∞−∞ −∫R

fψ′ dx = −∫R

fψ′ dx

Hence< f ′, ψ >= − < f , ψ′ >

DefinitionLet f ∈ D′(Ω) where Ω ⊂ Rn. The derivative ∂xi f is the distributiondefined by

< ∂xi f , ψ >= − < f , ∂xiψ >

Every distribution possesses (distributional) derivatives of any order

< Dαf , ψ >= (−1)|α| < f ,Dαψ > ∀α

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Examples of distributional derivatives

Examplesf = δ0

< ∂xi δ0, ψ >= −∂xiψ(0)

f = H (Heaviside function)

< H, ψ >= − < H, ψ >=

∫RHψ′ dx = −

∫ +∞

0ψ′ dx = ψ(0)

Hence< H′, ψ >= ψ(0) =< δ, ψ >

Exercise: Compute the derivative of f (x) = |x | ∈ D(R)

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Some important Sobolev spaces: H1(Ω)

DefinitionWe denote by H1(Ω)

H1(Ω) = v ∈ L2(Ω) : ∇v ∈ L2(Ω)

i.e.< ∂xi v , ψ >= −

∫Ω

v∂xiψ dx , ∀ψ ∈ D′(Ω)

TheoremH1(Ω) is a Hilbert space with inner product

(u, v)1 =

∫Ω

uv dx +

∫Ω∇u · ∇v dx

and

‖u‖1 =

√∫Ω|v |2 dx +

∫Ω|∇v |2 dx

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Some important Sobolev spaces: H10 (Ω)

DefinitionWe denote by H1

0 (Ω) the closure of D(Ω) in H1(Ω).

H10 (Ω) ⊂ H1(Ω) and elements of H1

0 (Ω) have zero trace on ∂Ω

TheoremPoincaré inequality. Let Ω ⊂ Rn. There exists a positive constant CPsuch that

‖u‖0 ≤ CP‖∇u‖0

We can choose(u, v)1 = (∇u,∇v)0

as inner product

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Some important Sobolev spaces: H−1(Ω)

DefinitionWe denote by H−1(Ω) the dual of H1

0 (Ω) with

‖F‖−1 = sup|Fv | : v ∈ H10 (Ω), ‖v‖1 ≤ 1

TheoremH−1(Ω) is the set of distributions of the form

F = f0 + div f

where f0 ∈ L2(Ω) and f ∈ L2(Ω)

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Part II

Elliptic and parabolic PDE’s

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Elliptic equations

Laplace equation

∆u = 0

Stationary diffusion equation, conductivity equation, equilibrium of anelastic membrane

Poisson equation

∆u = f

f external force or source e.g. density of electric charges, load

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Harmonic functions

DefinitionA solution of

∆u = 0

is called harmonic.

Examplesz = ρeiθ, <zm and =zm, m ∈ N are harmonic functionsx , y , xy , x2 − y2, x3 − 3xy2, . . . are harmonic functionsuζ(x) = eζ·x , ζ ∈ Cn : ζ · ζ = 0, x ∈ Rn

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Mean Value Theorem

TheoremIf u is harmonic in Ω ⊂ Rn, n ≥ 2. Then, for any ball BR(x) ⊂ Ω

u(x) =1

|BR(x)|

∫BR(x)

u(y)dy

u(x) =1

|∂BR(x)|

∫∂BR(x)

u(y)dσy

TheoremIf u ∈ C0(Ω) has the mean value property then u is C∞(Ω) and isharmonic in Ω.

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Maximum principle

TheoremIf u ∈ C0(Ω) has the mean value property then, if u attains a maximumor a minimum at x0 ∈ Ω then u is constant. In particular if Ω isbounded, u ∈ C0(Ω) and u is not constant then, ∀x ∈ Ω

min∂Ω

u < u(x) < max∂Ω

u

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Application of maximum principle

Uniqueness and continuous dependence of boundary value problems

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Boundary value problems

Dirichlet Problem∆u = f in Ω ⊂ Rn,n ≥ 2

u = g on ∂Ω.

Neumann Problem∆u = f in Ω ⊂ Rn,n ≥ 2∂u∂n = h on ∂Ω.

Robin Problem∆u = f in Ω ⊂ Rn,n ≥ 2

∂u∂ν + αu = h on ∂Ω.

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Well-posedness

DefinitionA problem is well posed if

there exists a solution (existence)the solution is unique (uniqueness)the solution depends continuously upon the data (stability)

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Uniqueness

TheoremΩ ⊂ Rn bounded and smooth. There exists a unique solutionu ∈ C2(Ω) ∩ C1(Ω) of the Dirichlet and of the Robin problem. In thecase of the Neumann problem the solution is unique up to an additiveconstant.

with proof

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Existence

Existence is much more complicated to prove. For particulargeometries of the domain Ω existence can be proved via method ofseparation of variables

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Existence via method of separation of variables: anexample

Let Ω = BR(P) ball centered at P of radius R. Then

TheoremPoisson formula The unique solution to the Dirichlet problem

∆u = 0 in BR(P) ⊂ R2

u = g on ∂BR(P).

is given by

u(x) =R2 − |x − P|2

2πR

∫∂BR(P)

g(y)

|x − y |2dσy

with proof

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Existence via singular solutions

u harmonic function. Then, for fixed y ∈ Rn

v(x) = u(x − y)

is harmonic (translation invariance of ∆)

v(x) = u(Mx)

for M orthogonal (MT = M−1) is harmonic (rotation invariance of ∆).

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The fundamental solution

Radially symmetric harmonic solutions: u = u(r), r = |x |.n = 2

∂2u∂r2 +

1r∂u∂u

= 0

Thenu(r) = C1 ln r + C2

n = 3∂2u∂r2 +

2r∂u∂u

= 0

Thenu(r) = C1

1r

+ C2

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The fundamental solution

The fundamental solution for the Laplacian

Φ(x) =

1

2π ln |x |, n = 2− 1

4π|x | , n = 3

is solution to∆Φ = δ0

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The representation formula

TheoremRepresentation FormulaΩ ⊂ Rn smooth and bounded, u ∈ C2(Ω). Then, ∀x ∈ Ω

u(x) =

∫Ω

Φ(x − y)∆u dy −∫∂Ω

Φ(x − y)∂u∂ν

dσy +

∫∂Ω

u∂Φ(x − y)

∂νdσy

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The Green’s function

DefinitionFix y ∈ Ω

∆G(·, y) = δy in ΩG(·, y) = 0 on ∂Ω.

ANSATZ:

G(x , y) = Φ(x − y)− ψ(x , y)

where ∆ψ(·, y) = 0 in Ωψ(·, y) = Φ(x − y) on ∂Ω.

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Properties of Green’s function

G < 0 (negativity)G(x , y) = G(y , x) (symmetry)|G(x , y)| → +∞ as x → y (singularity)

For particular geometries of Ω the Green’s function can be computedexplicitly.

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Representation formula

TheoremConsider

∆u = f in Ωu = g on ∂Ω.

Then,

u(x) =

∫Ω

G(x , y)f (y) dy +

∫∂Ω

u∂G(x , y)

∂νdσy

If u is harmonicu(x) =

∫∂Ω

u∂G(x , y)

∂νdσy

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Representation formula

TheoremConsider

∆u = f in Ω∂u∂ν = h on ∂Ω.

Then,

u(x) =

∫Ω

N(x , y)f (y) dy −∫∂Ω

hN(x , y) dσy

where N(x , y) is the Neumann function solution to∆N(·, y) = δy in Ω

∂N(·,y)∂ν = 1

|∂Ω| on ∂Ω.

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More general equations

Electrical Impedence Tomographydiv(σ∇u) = f in Ω ⊂ R3

u = g on ∂Ω,

σ conductivity, u electrostatic potential, g boundary potential, f externalsource.

Optical Tomography−div(k∇ψ) + µψ = f in Ω ⊂ R3

∂ψ∂n + αψ = h on ∂Ω,

k diffusion coefficient, µ absorption coefficient, ψ optical density, hillumination function.

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More general elliptic equations

DefinitionConsider Ω ⊂ Rn, A(x) = (aij(x))n

i,j=1, c(x) = (ci(x))ni=1, a0(x) and f (x)

scalar functions. An equation of the form

−div (A∇u) + c · ∇u + a0u = f in Ω

is elliptic in Ω if

A(x)ξ · ξ > 0, ∀x ∈ Ω,∀ξ ∈ Rn, ξ 6= 0

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Variational formulation: Dirichlet Problem

Consider the Dirichlet problem−div(a∇u) + a0u = f in Ω

u = 0 on ∂Ω.

Suppose a, a0 and f smooth and u ∈ C2(Ω) ∩ C0(Ω) is a solution. Letv ∈ C1

0(Ω), multiplying the equation by v and integrating by parts wehave ∫

Ωa∇u · ∇v + a0uv =

∫Ω

fv

Viceversa, integrating back last equation implies that u is solution ofthe Dirichlet problem.

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Variational formulation

Problem (DP)Find u ∈ H1

0 (Ω) such that∫Ω

a∇u · ∇v + a0uv =

∫Ω

fv , ∀v ∈ H10 (Ω)

LetB(u, v) :=

∫Ω

a∇u · ∇v + a0uv

and the linear functional

〈F , v〉 =

∫Ω

fv

Problem (DP) is equivalent to the abstract variational problem

B(u, v) = 〈F , v〉, ∀v ∈ H10 (Ω)

⇒ well-posedness follows from Lax-Milgram Theorem if

a ≥ λ, and a0 ≥ 0

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Well-posedness

TheoremLet f ∈ L2(Ω) and λ−1

0 ≤ a ≤ λ0, 0 ≤ a0(x) ≤ γ0 in Ω. Then, Problem(DP) has a unique solution u ∈ H1

0 (Ω). Moreover

‖∇u‖0 ≤ C‖f‖0

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Remarks

Problem (DP) is equivalent to the minimization in H10 (Ω) of the

functional (total energy)

E(u) =

∫Ω

a|∇u|2 dx +

∫Ω

a0u2 dx −∫

Ωfu dx

In the case of non homogeneous boundary conditions ifg ∈ H1(Ω) and u − g ∈ H1

0 (Ω) then we reduce to the previouscase considering w = u − g.

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Variational formulation: Neumann Problem

Consider the Neumann problem−div(a∇u) + a0u = f in Ω

a∂u∂ν = g on ∂Ω.

Suppose a0 and f smooth and u ∈ C2(Ω) ∩ C0(Ω) is a solution. Letv ∈ C1(Ω), multiplying the equation by v and integrating by parts wehave

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Variational formulation: Neumann Problem

∫Ω

a∇u · ∇v + a0uv =

∫Ω

fv +

∫∂Ω

gv , ∀v ∈ C1(Ω)

Assume last relation holds and integrate by parts we get∫Ω

(−div(a∇u) + a0u − f )v +

∫∂Ω

a∂u∂ν

v =

∫∂Ω

gv , ∀v ∈ C1(Ω)

Choose v ∈ C10(Ω) and we get

−div(a∇u) + a0u − f = 0

and we have ∫∂Ω

a∂u∂ν

v =

∫∂Ω

gv ∀v ∈ C1(Ω)

which gives

a∂u∂ν

= g

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Variational formulation: Neumann Problem

Problem (NP)Find u ∈ H1

0 (Ω) such that∫Ω

a∇u · ∇v + a0uv =

∫Ω

fv +

∫∂Ω

gv , ∀v ∈ H1(Ω)

LetB(u, v) :=

∫Ω

a∇u · ∇v + a0uv

and the linear functional

〈F , v〉 =

∫Ω

fv +

∫∂Ω

gv

Problem (DP) is equivalent to the abstract variational problem

B(u, v) = 〈F , v〉, ∀v ∈ H1(Ω)

⇒ well-posedness follows from Lax-Milgram Theorem if

a ≥ λ−10 > 0, a0 ≥ γ−1

0 > 0

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Well-posedness

TheoremLet f ∈ L2(Ω), g ∈ L2(∂Ω), 0 < λ−1

0 ≤ a(x) ≤ λ0 and0 < γ−1

0 ≤ a0(x) ≤ γ0 in Ω. Then, Problem (NP) has a unique solutionu ∈ H1

0 (Ω). Moreover

‖u‖1 ≤1

minλ−10 , γ−1

0 (‖f‖0 + C‖g‖L2(∂Ω))

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Well-posedness

RemarkThe condition a0 ≥ γ−1

0 > 0 is necessary for existence anduniqueness!

If a0 = 0 then uniqueness fails and a normalization condition is neededto restore uniqueness.Existence of solutions requires that the compatibility condition∫

Ωf +

∫∂Ω

g = 0

is satisfied.

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Eigenvalues

DefinitionLet Ω ⊂ Rn bounded. A nontrivial weak solution u ∈ H1

0 (Ω) to theproblem

−∆u = λu in Ωu = 0 on ∂Ω.

is called Dirichlet eigenfunction and λ the correspondingDirichlet eigenvalue

TheoremThere exists in L2(Ω) an orthonormal basis uk consisting of Dirichleteigenfunctions. The corresponding eigenvalues are an increasingsequence

0 < λ1 < λ2 ≤ · · · ≤ λk ≤ . . .

with λk → +∞. The sequence uk/λk is an orthonormal basis inH1

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Eigenvalues

RemarkIf u ∈ L2(Ω) then u =

∑ckuk and ‖u‖2 =

∑c2

k . So

‖∇uk‖2 = (∇uk ,∇uk )0 = λk (uk ,uk )0 = λk

Thus, u ∈ H10 (Ω) iff

‖∇u‖20 =∑

c2kλk <∞

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Eigenvalues

By‖∇u‖20 =

∑c2

kλk <∞

and monotonicity of eigenvalues we get

‖∇u‖20 ≥ λ1‖u‖20

and we derive the following variational problem

λ1 = min

∫Ω |∇u|2∫

Ω u2 : u ∈ H10 (Ω),u 6= 0

or equivalently

λ1 = min∫

Ω|∇u|2 : u ∈ H1

0 (Ω), ‖u‖0 = 1

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Eigenvalues

Let F (u) =∫

Ω |∇u|2 and G(u) =∫

Ω u2 − 1

TheoremLet u ∈ H1

0 (Ω) be a local extremum of the functional F subject to thecondition G(u) = 0, then u is an eigenfunction and λ = F (u) is thecorresponding eigenvalue.

Proof. Follows by using Lagrange multiplier Theorem

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Eigenvalues

TheoremThere exists a global minimum u ∈ H1

0 (Ω) of the functional F subject tothe condition G(u) = 0.

Main steps of the proof.

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