Introduction to PML in time domain Alexander Thomann Introduction to PML in time domain - Alexander Thomann – p.1
Overview 1 Introduction PML in one dimension • Classical absorbing layers 2 • One-dimensional PML ’s • Approach with complex change of variables PML in two dimensions • PML for a general linear system 3 • Accoustic waves • Discretization and numerical problems Introduction to PML in time domain - Alexander Thomann – p.2
Introduction Task • Solution of wave scattering problem. • Interesting region is bounded. • The problem has to be solved numerically. Introduction to PML in time domain - Alexander Thomann – p.3
Introduction Task Problem • Solution of wave • Need to discretize scattering problem. space. • Interesting region is = ⇒ • Finite elements/finite bounded. differences. • The problem has to be ⇒ Need to bound the area of computation. solved numerically. Introduction to PML in time domain - Alexander Thomann – p.3
Introduction Task Problem • Solution of wave • Need to discretize scattering problem. space. • Interesting region is = ⇒ • Finite elements/finite bounded. differences. • The problem has to be ⇒ Need to bound the area of computation. solved numerically. ⇓ Idea • Construct artificial boundary. • Transparent for the solution. • Totally absorbs incom- ing waves, no reflec- tions. Introduction to PML in time domain - Alexander Thomann – p.3
Introduction Task Problem • Solution of wave • Need to discretize scattering problem. space. • Interesting region is = ⇒ • Finite elements/finite bounded. differences. • The problem has to be ⇒ Need to bound the area of computation. solved numerically. ⇓ Solution Idea • Absorbing Boundary • Construct artificial Conditions: Differential boundary. equations at the • Transparent for the boundary. ⇐ = solution. • "Classical" absorbing • Totally absorbs incom- layers. ing waves, no reflec- • Perfectly Matched tions. Layers . Introduction to PML in time domain - Alexander Thomann – p.3
Absorbing Layers in 1D Consider the 1D wave equation with velocity 1: ∂ 2 u ∂t 2 − ∂ 2 u ∂x 2 = 0 , x ∈ R , t > 0 . • As a first illustrative example we restrict the computational domain to x < 0 . • We therefore have to impose an Absorbing Boundary Condition at x = 0 . • In fact we dispose of a very simple and even local condition: ∂u ∂t + ∂u ∂x = 0 , x = 0 , t > 0 . Introduction to PML in time domain - Alexander Thomann – p.4
Absorbing Layers in 1D Consider the 1D wave equation with velocity 1: ∂ 2 u ∂t 2 − ∂ 2 u ∂x 2 = 0 , x ∈ R , t > 0 . • As a first illustrative example we restrict the computational domain to x < 0 . • We therefore have to impose an Absorbing Boundary Condition at x = 0 . • In fact we dispose of a very simple and even local condition: ∂u ∂t + ∂u ∂x = 0 , x = 0 , t > 0 . = ⇒ No exact local analogue in higher dimensions! Let us therefore find a transparent condition through an absorbing layer, infinite first and then in the interval [0 , L ] . Introduction to PML in time domain - Alexander Thomann – p.4
Classical Absorbing Layers In order to damp waves through a physical mechanism, we can add two terms to the wave equation, • fluid friction: ν ∂u ∂t , ν ≥ 0 , ∂x∂t ) , ν ∗ ≥ 0 . ∂x ( ν ∗ ∂ 2 u • viscous friction: − ∂ We then obtain the equation „ ∂u « ∂ 2 u ∂x + ν ∗ ∂ 2 u ∂t 2 + ν ∂u ∂t − ∂ = 0 . ∂x ∂x∂t The solution is k ( ω ) 2 = ω 2 − iων u ( x, t ) = Ae i ( ωt − k ( ω ) x ) + Be i ( ωt + k ( ω ) x ) , 1 + iων ∗ , ℑ k ( ω ) ≤ 0 . A natural choice thus would be ν ( x ) = 0 , ν ∗ ( x ) = 0 , x < 0 , ν ( x ) > 0 , ν ∗ ( x ) > 0 , x > 0 . Introduction to PML in time domain - Alexander Thomann – p.5
„ ∂u « „ ∂u « ∂ 2 u ∂ 2 u ∂x + ν ∗ ∂ 2 u ∂t 2 − ∂ ∂t 2 + ν ∂u ∂t − ∂ = 0 , = 0 . ∂x ∂x ∂x ∂x∂t x=0 The larger ν and ν ∗ , the smaller can we later on choose the length L of the absorbing layer. Introduction to PML in time domain - Alexander Thomann – p.6
„ ∂u « „ ∂u « ∂ 2 u ∂ 2 u ∂x + ν ∗ ∂ 2 u ∂t 2 − ∂ ∂t 2 + ν ∂u ∂t − ∂ = 0 , = 0 . ∂x ∂x ∂x ∂x∂t x=0 The larger ν and ν ∗ , the smaller can we later on choose the length L of the absorbing layer. But consider 8 e iω ( t − x ) + R ( ω ) e iω ( t + x ) , > x < 0 , < T( ω ) 1 u ( x, t ) = > : R( ω ) T ( ω ) e i ( ωt − k ( ω ) x ) , x > 0 . x=0 u (0 − ) u (0 + ) , We impose the right bound- = ary conditions, ∂x + ν ∗ ∂ 2 u ∂u ( ∂u ∂x (0 − ) ∂x∂t )(0 + ) . = Introduction to PML in time domain - Alexander Thomann – p.6
ω − k ( ω )(1 + iων ∗ ) This leads to R ( ω ) = ω + k ( ω )(1 + iων ∗ ) , ν →∞ | R ( ω ) | = lim ν ∗ →∞ | R ( ω ) | = 1 lim T ( ω ) = 1 + R ( ω ) , Introduction to PML in time domain - Alexander Thomann – p.7
ω − k ( ω )(1 + iων ∗ ) This leads to R ( ω ) = ω + k ( ω )(1 + iων ∗ ) , ν →∞ | R ( ω ) | = lim ν ∗ →∞ | R ( ω ) | = 1 lim T ( ω ) = 1 + R ( ω ) , The more a layer is absorbing, the more it is also reflecting! Reflection at a visco-elastic layer. On the right side the absorption and therefore the reflection is stronger (Joly). Introduction to PML in time domain - Alexander Thomann – p.7
Perfectly Matched Layers in 1D This was not satisfactory. In order to suppress reflections we want perfect adaption. For that reason, we return to the wave-equation with variable coefficients. With ρ, µ > 0 we have „ « ρ ( x ) ∂ 2 u ∂t 2 − ∂ µ ( x ) ∂u = 0 , ∂x ∂x and define p • the velocity of propagation c ( x ) = µ ( x ) /ρ ( x ) , p • the impedance z ( x ) = µ ( x ) ρ ( x ) . We impose u ( x ) = e i ( ωt − kx ) + R ( ω ) e i ( ωt + kx ) , ω k = c ( x ) , c ( x ) = c, z ( x ) = z, x < 0 , u ( x ) = T ( ω ) e i ( ωt − k ( ω ) x ) , ω k = c ( x ) , c ( x ) = c ∗ , z ( x ) = z ∗ , x > 0 . Introduction to PML in time domain - Alexander Thomann – p.8
With the right boundary conditions, u (0 − ) u (0 + ) , = µ (0 − ) ∂u µ (0 + ) ∂u ∂x (0 − ) ∂x (0 + ) , = we find R = z − z ∗ 2 z z + z ∗ , T = z + z ∗ . • It is obvious that R = 0 if z = z ∗ . • We thus need impedance-matching . • But how can we make the layer absorbing at the same time? • For that reason we change to frequency-space. Then we arrive at the Helmholtz-equation „ « ρ ( x, ω ) ω 2 u − ∂ µ ( x, ω ) ∂u − b b = 0 , ρ, b b µ > 0 . ∂x ∂x Introduction to PML in time domain - Alexander Thomann – p.9
The Idea The idea is simple but effective: We choose d ( ω ) ∈ C and ρ ( x, ω ) ≡ ρ, b µ ( x, ω ) ≡ µ, b x < 0 , ρ ρ ( x, ω ) = b d ( ω ) , µ ( x, ω ) = µ · d ( ω ) , b x > 0 . This then actually leads to z ( x > 0) = √ ρµ z ( x < 0) = b b = ⇒ we have impedance-matching, p b c ( x < 0) = µ/ρ = c, b c ( x > 0) = c · d ( ω ) ∈ C = ⇒ we can make the layer absorbing. • It must be possible to return to time domain. • Then the equation needs to be constructed out of differential operators. ⇒ A crucial condition is thus that d ( ω ) is a rational funtion in the variable iω with real coefficients. Introduction to PML in time domain - Alexander Thomann – p.10
Analysis of the Solution Writing d ( ω ) − 1 = a + ib , we have the solutions u ( x ) = e iω ( t ± ax c ) ∓ ω bx c , ωb < 0 , with • phase velocity c/a , c • that decay with penetration depth l ( ω ) = | ωb | in the direction of propagation. Possible choice: a = 1 , b = − σ ω , where σ is called the coefficient of absorption. Then we have the simple case where • l = c σ : absorption does not depend on the frequency, • the phase velocity remains c , iω • d ( ω ) = iω + σ . Introduction to PML in time domain - Alexander Thomann – p.11
In frequency domain the wave-equation becomes „ « ρ ( σ + iω ) u − ∂ µ ( σ + iω ) − 1 ∂u = 0 , ∂x ∂x | {z } v which corresponds in time domain to the differential equation ∂ 2 u ∂t + σ 2 u − c 2 ∂ 2 u ∂t 2 + 2 σ ∂u ∂x 2 = 0 , or as a first order system, describing a PML, „ ∂u « − ∂v ρ ∂t + σu ∂x = 0 , „ ∂v « − ∂u µ − 1 ∂t + σv ∂x = 0 . Introduction to PML in time domain - Alexander Thomann – p.12
In 1D we have the energy identity „ 1 Z « Z d ( ρ | u | 2 + µ − 1 | v | 2 ) dx σ ( ρ | u | 2 + µ − 1 | v | 2 ) dx = 0 . + dt 2 As one can see, • we do not only have dissipation in space but • we additionally have proof for temporal dissipation! All solutions to the 1D-equation are decaying! There will be NO such proof in higher dimensions! Introduction to PML in time domain - Alexander Thomann – p.13
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