Laplace and Poisson equations: physical modellingljk.imag.fr/membres/Eric.Blayo/PDE-course/slides-chapter3.pdf · Laplace and Poisson equations: physical modelling ... Laplace and

Laplace and Poisson equations: physical modelling

Many physical phenomena are governed by diffusion (heat distribution,chemical or biological concentration, electric potential...)

Steady state solution: for any region ω ⊂ Ω, balance betweenI fluxes through the boundary ∂ωI sinks/sources within ω∫∂ω

Φ.n ds =

∫ω

f (x) dx

=⇒ div Φ(x) = f (x) ∀x ∈ Ω

A common law: (Fourier’s law, Fick’s law) Φ = −k∇u where u is thedensity (temperature, concentration...) and k is the diffusivity coefficient

−div (k(x)∇u(x)) = f (x)




Φ.n ds =

∫ω

f (x) dx =⇒ div Φ(x) = f (x) ∀x ∈ Ω


−div (k(x)∇u(x)) = f (x)




Φ.n ds =

∫ω

f (x) dx =⇒ div Φ(x) = f (x) ∀x ∈ Ω


−div (k(x)∇u(x)) = f (x)

Laplace and Poisson equations

−div (k(x)∇u(x)) = f (x)

I If k depends on u: nonlinear equation

I If k does not depend on u: linear equation

I If k is a constant: −∆u =1k f Poisson equation

I If k is a constant and f = 0: ∆u = 0 Laplace equation,harmonic functions

Some harmonic functions in R2

I Polynomials:

u(x , y) = a(x2 − y2) + b xy + cx + dy + e

I Separation of variables in cartesian coordinates u(x , y) = v(x)w(y)

elementary solutions:

u1λ(x , y) = (a cosλx + b sinλx)(ceλy + deλy )

u2λ(x , y) = (aeλx + beλx )(c cosλy + d sinλy)

∀λ 6= 0

hence any convergent sum of these functions


I Polynomials:

u(x , y) = a(x2 − y2) + b xy + cx + dy + e





∀λ 6= 0



I Polynomials:

u(x , y) = a(x2 − y2) + b xy + cx + dy + e





∀λ 6= 0



I Separation of variables in polar coordinates u(r , θ) = v(r)w(θ)

∆u(r , θ) =∂2u∂r2 +

1r∂u∂r +

1r2∂2u∂θ2

elementary solutionson R2 r (0, 0)

u0(r , θ) = c0 ln r + d0

un(r , θ) = (an cos nθ + bn sin nθ)

(cnrn +

dn

rn

)∀n ∈ N∗


u0 u2 u4 u5


I Separation of variables in polar coordinates u(r , θ) = v(r)w(θ)

∆u(r , θ) =∂2u∂r2 +

1r∂u∂r +

1r2∂2u∂θ2

elementary solutionson R2 r (0, 0)

u0(r , θ) = c0 ln r + d0


(cnrn +

dn

rn

)∀n ∈ N∗


u0 u2 u4 u5


I Polynomials

u(x , y , z) = ax2+by2−(a+b)z2+terms in 1, x , y , z , xy , xz , yz , xyz

I Extension of 2D harmonic functions

u(x , y , z) = uλ1 (x , y)(a1z+b1)+uλ2 (x , z)(a2y+b2)+uλ3 (y , z)(a3x+b3)

I Radial harmonic functions

∆u =∂2u∂r2 +

2r∂u∂r

+1

r2 sin θ∂

∂θ

(sin θ

∂u∂θ

)+

1r2 sin2 θ

∂2u∂ϕ2

u(r , θ, ϕ) = u(r) =ar

+ b on R3 r (0, 0, 0)


I Polynomials





∆u =∂2u∂r2 +

2r∂u∂r

+1

r2 sin θ∂

∂θ

(sin θ

∂u∂θ

)+

1r2 sin2 θ

∂2u∂ϕ2

u(r , θ, ϕ) = u(r) =ar

+ b on R3 r (0, 0, 0)


I Polynomials





∆u =∂2u∂r2 +

2r∂u∂r

+1

r2 sin θ∂

∂θ

(sin θ

∂u∂θ

)+

1r2 sin2 θ

∂2u∂ϕ2

u(r , θ, ϕ) = u(r) =ar

+ b on R3 r (0, 0, 0)


I Polynomials





∆u =∂2u∂r2 +

2r∂u∂r

+1

r2 sin θ∂

∂θ

(sin θ

∂u∂θ

)+

1r2 sin2 θ

∂2u∂ϕ2

u(r , θ, ϕ) = u(r) =

ar

+ b on R3 r (0, 0, 0)


I Polynomials





∆u =∂2u∂r2 +

2r∂u∂r

+1

r2 sin θ∂

∂θ

(sin θ

∂u∂θ

)+

1r2 sin2 θ

∂2u∂ϕ2

u(r , θ, ϕ) = u(r) =ar

+ b on R3 r (0, 0, 0)

Mean value property and maximum principle

I Mean value property:

u(x) =1

|B(x, r)|

∫B(x,r)

u(y) dy =1

|∂B(x, r)|

∫∂B(x,r)

u(σ) dσ ∀B(x, r) ⊂ Ω

An harmonic function is the average of its values over everysurrounding ball and sphere.

I Maximum principle: Let u an harmonic function. If u ∈ C2(Ω) andu ∈ C0(Ω), then u has no extreme values in Ω.

Harmonic functions in bounded domainsReminder: uniqueness of the solution of Laplace or Poisson problems inΩ ⊂ Rn: cf exercise session

div (k(x)∇u(x)) = f (x) in Ω

+ one of the following boundary conditions on ∂Ω

I Dirichlet: u = g on ∂Ω

I Neumann:∂u∂n

= h on ∂Ω

I Mixed: u = g on Γ0

∂u∂n

= h on Γ1where Γ0 ∪ Γ1 = ∂Ω and Γ0 ∩ Γ1 = ∅

−→ Uniqueness for Dirichlet and mixed BCs−→ No solution for Neumann BCs if

∫∂Ω

h(s)ds 6=∫

Ω

f (x)dx.No unique solution otherwise.

Harmonic functions in Ω = (0, Lx ) × (0, Ly ) ⊂ R2

∆u(x , y) = 0 in (0, Lx ) × (0, Ly )

u(0, y) = h(y), u(Lx , y) = u(x , 0) = u(x , Ly ) = 0

Elementary solutions using separation of variables u(x , y) = v(x)w(y)u1λ(x , y) = (a cosλx + b sinλx)(ceλy + deλy )


∀λ 6= 0

Using BCs: (cf exercise session)

u(x , y) =∑k≥1

αk(

eλk x − e2λk Lx e−λk x)

sin(λky)

with λk =kπLy

, αk =2

1− e2λk Lx

∫ Ly

0h(y) sin(λky) dy

Harmonic functions in Ω = Disk(O,R) ⊂ R2

∆u(r , θ) = 0 in Ω

u(R, θ) = g(θ) on ∂Ω (r = R)

Elementary solutions on R2 r (0, 0) using separation of variables:u0(r , θ) = c0 ln r + d0


(cnrn +

dn

rn

)∀n ∈ N∗

Given Ω: c0 = 0 and dn = 0 ∀n.

Finally: (cf exercise session)

u(r , θ) = a0 +∑n≥1

(an cos nθ + bn sin nθ)( r

R

)n

with a0 =1

2π

∫ 2π

0g(θ) dθ, an =

1π

∫ 2π

0g(θ) cos nθ dθ , bn =

1π

∫ 2π

0g(θ) sin nθ dθ


∆u(r , θ) = 0 in Ω

u(R, θ) = g(θ) on ∂Ω (r = R)

Can be transformed into u(r , θ) = K(r , θ) ∗ g(θ) =1

2π

∫ 2π

0K(r , θ − α) g(α) dα

where K(r , θ) =R2 − r2

R2 + r2 − 2rR cos θ

K u(r, θ)

−→ Illustration of the mean value property and of the maximum principle


∆u(r , θ) = 0 in Ω

u(R, θ) = g(θ) on ∂Ω (r = R)

Can be transformed into u(r , θ) = K(r , θ) ∗ g(θ) =1

2π

∫ 2π

0K(r , θ − α) g(α) dα

where K(r , θ) =R2 − r2


K u(r, θ)

−→ Illustration of the mean value property and of the maximum principle

Some properties of harmonic functions

I already seen: mean value property and maximum principle

I regularity In the 2-D case Ω = Disk(O,R):

u(r , θ) = K(r , θ) ∗ g(θ) with K(r , θ) =R2 − r2


g ∈ C1(∂Ω) =⇒ u ∈ C∞(Ω)

(because (a ∗ b)′ = a ∗ b′ = a′ ∗ b for a, b ∈ C1 and K ∈ C∞)

I global influence of boundary values In the 2-D case Ω = Disk(O,R):

u(r , θ) = K(r , θ) ∗ g(θ) =1

2π

∫ 2π

0K(r , θ − α) g(α) dα

Any change, even very local, in the boundary data gwill modify the solution everywhere in Ω.

Poisson kernels in R2 and R3

Reminder: K(r) =

1

2πln r in R2 r (0, 0)

−14πr

in R3 r (0, 0, 0)

are harmonic functions.

K (r) is called Poisson kernel in R2 or R3.

It will be of great importance for solving Poisson problems.

Distributions for dummies

Distributions

Laurent Schwartz (1915 - 2002)

Test functions D(Ω) = ϕ ∈ C∞(Ω), with compact support

Theorem: D(Ω) is dense in H1(Ω)

Distributions Dual space D′(Ω), i.e. linear forms on D(Ω).T : D(Ω) −→ R

ϕ −→ T (ϕ), also denoted < T , ϕ >

Distributions

Laurent Schwartz (1915 - 2002)

Test functions D(Ω) = ϕ ∈ C∞(Ω), with compact support

Theorem: D(Ω) is dense in H1(Ω)

Distributions Dual space D′(Ω), i.e. linear forms on D(Ω).T : D(Ω) −→ R

ϕ −→ T (ϕ), also denoted < T , ϕ >

Examples of distributionsDirac distribution (“delta function”)

δ : D(Ω) −→ Rϕ −→ ϕ(0)

Heaviside distribution

H : D(R) −→ R

ϕ −→∫R

H(x)ϕ(x) dx =

∫ +∞

0ϕ(x) dx

Usual functions can be identified to distributions (f ⇐⇒ Tf )

∀f ∈ L1loc , < Tf , ϕ >=

∫Ω

f (x)ϕ(x) dx

−→ generalization of the notion of function.


δ : D(Ω) −→ Rϕ −→ ϕ(0)


H : D(R) −→ R

ϕ −→∫R

H(x)ϕ(x) dx =

∫ +∞

0ϕ(x) dx


∀f ∈ L1loc , < Tf , ϕ >=

∫Ω

f (x)ϕ(x) dx



δ : D(Ω) −→ Rϕ −→ ϕ(0)


H : D(R) −→ R

ϕ −→∫R

H(x)ϕ(x) dx =

∫ +∞

0ϕ(x) dx


∀f ∈ L1loc , < Tf , ϕ >=

∫Ω

f (x)ϕ(x) dx


Operations on distributions

Derivative of a distribution by analogy with usual functions:

< Tf ′ , ϕ >=

∫R

f ′(x)ϕ(x) dx = −∫R

f (x)ϕ′(x) dx = − < Tf , ϕ′ >

Hence: T ′ is defined such that < T ′, ϕ >= − < T , ϕ′ > ∀ϕ ∈ D(Ω)

Example: H ′ ?

< H ′, ϕ >= − < H, ϕ′ >= −∫ +∞

0ϕ′(x) dx = ϕ(0) =⇒ H ′ = δ

Convolution of two distributions by analogy with usual functions:

< T1 ∗ T2, ϕ > = < T1(x), < T2(y), ϕ(x + y) >>

Theorem: T ∗ δ = T



< Tf ′ , ϕ >=

∫R

f ′(x)ϕ(x) dx = −∫R

f (x)ϕ′(x) dx = − < Tf , ϕ′ >


Example: H ′ ?

< H ′, ϕ >= − < H, ϕ′ >= −∫ +∞

0ϕ′(x) dx = ϕ(0) =⇒ H ′ = δ


< T1 ∗ T2, ϕ > = < T1(x), < T2(y), ϕ(x + y) >>




< Tf ′ , ϕ >=

∫R

f ′(x)ϕ(x) dx = −∫R

f (x)ϕ′(x) dx = − < Tf , ϕ′ >


Example: H ′ ?

< H ′, ϕ >= − < H, ϕ′ >= −∫ +∞

0ϕ′(x) dx = ϕ(0) =⇒ H ′ = δ


< T1 ∗ T2, ϕ > = < T1(x), < T2(y), ϕ(x + y) >>

Theorem: T ∗ δ =

T



< Tf ′ , ϕ >=

∫R

f ′(x)ϕ(x) dx = −∫R

f (x)ϕ′(x) dx = − < Tf , ϕ′ >


Example: H ′ ?

< H ′, ϕ >= − < H, ϕ′ >= −∫ +∞

0ϕ′(x) dx = ϕ(0) =⇒ H ′ = δ


< T1 ∗ T2, ϕ > = < T1(x), < T2(y), ϕ(x + y) >>


Poisson problem in R2 and R3

I Poisson kernel K (r) =

1

2π ln r in R2 r (0, 0)

−14πr in R3 r (0, 0, 0)

I Theorem 1 ∆K = δ in D′(R2) or D′(R3)

I Theorem 2 Let the Poisson equation ∆u = f in R2 or R3.

u = K ∗ f is a solution to this equation.

I Generalization to a bounded domain Ω: cf lecture notes

Poisson problem in R2 and R3

I Poisson kernel K (r) =

1

2π ln r in R2 r (0, 0)

−14πr in R3 r (0, 0, 0)

I Theorem 1 ∆K = δ in D′(R2) or D′(R3)

I Theorem 2 Let the Poisson equation ∆u = f in R2 or R3.

u = K ∗ f is a solution to this equation.

I Generalization to a bounded domain Ω: cf lecture notes

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Laplace and Poisson equations: physical modellingljk.imag.fr/membres/Eric.Blayo/PDE-course/slides-chapter3.pdf · Laplace and Poisson equations: physical modelling ... Laplace and