Setting of the problem
DtN on elementary solutions
Boundary control
The Sommerfeld radiation condition
Transparent boundary conditions for the
Helmholtz equation with an infinite interface
October 15, 2015
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transparent boundary conditions, Study Guides, Projects, Research of Mathematical Physics
we investigate the non trivial case of transparent boundary conditions in the case of infinite heterogeneities and the analytical methods that can solve this problem
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DtN on elementary solutions Boundary control The Sommerfeld radiation condition
Transparent boundary conditions for the
Helmholtz equation with an infinite interface
October 15, 2015
DtN on elementary solutions Boundary control The Sommerfeld radiation condition
Outline
(^1) Setting of the problem
(^2) DtN on elementary solutions
(^3) Boundary control
(^4) The Sommerfeld radiation condition
DtN on elementary solutions Boundary control The Sommerfeld radiation condition
The Helmholtz equation.
The domain, the interface and the compact perturbation
DtN on elementary solutions Boundary control The Sommerfeld radiation condition
The Helmholtz equation.
The scattering problem reads
∆(fscatt + fin) + k^2 (fscatt + fin) = 0
where ∆fin + k mod^2 fin = 0
then interior problem : ∆fscatt + k^2 fscatt = known 2 nd^ member exterior problem : ∆fscatt + k^2 fscatt = 0
DtN on elementary solutions Boundary control The Sommerfeld radiation condition
What we know.
up to a change of function and some manipulations, we can show that our problem is equivalent to
H 2 (r , θ)u+^ + g(r )(θ)u−^ = 0
H 2 (r , θ)u−^ + g(r )(θ)u+^ = 0
it seems simpler but it is not formulation with mutually opposed functional spaces no simple separating variables solutions dirac mesure when eliminating the system
DtN on elementary solutions Boundary control The Sommerfeld radiation condition
Computation of elementary solutions
We want to find elementary solutions, not to far from the circular elementary solutions of the homogeneous case. Knowing that
(aJn(k 1 r ) + bYn(k 1 r )).(c.cos(nθ) + d.sin(θ))
is a local C^1 -solution in the upper half plane, find its C^1 continuation in the whole plane. We will need : analytic continuation type approach establishing a new multiplication theorem manipulating integral representations pray for good propeties of the solutions at the end
DtN on elementary solutions Boundary control The Sommerfeld radiation condition
The Bessel function of the first kind
What is a multiplication theorem?
Jn(k 1 r ) =
am(r )Jm(k 2 r )
already exist for polynomial am (bad !) no parity relation between n and m (bad !) we cannot make the θ-reconstruction
DtN on elementary solutions Boundary control The Sommerfeld radiation condition
The Bessel function of the first kind
2 π(− 1 )nJ 2 n(rk 1 )cos( 2 nθ) if θ ∈ [ 0 ; π] 4 J 0 (rk 2 )
0 √^ T^2 n(x) 1 −x^2
dx+
m≥ 1 (−^1 )
mJ 2 m(rk 2 )[∫^1 0
T 2 m ( k k^12 x)T 2 n(x) √ 1 −x^2
dx]cos( 2 mθ) if θ ∈ [π; 2 π]
DtN on elementary solutions Boundary control The Sommerfeld radiation condition
The Bessel function of the second kind
Yn(kr ) = 1 π
∫ (^) π
0
sin(kr .sinθ − nθ)dθ −
π
0
(ent^ + (− 1 )ne−nt^ )e−kr^ .sinh(t)dt
Yn(kr ) = 1 π
∫ (^) π
0
sin(kr .sinθ − nθ)dθ −
π
0
(ent^ − (− 1 )ne−nt^ )e−kr^ .sinh(t)dt
π
(− 1 )n
0
e−nt^ e−kr^ .sinh(t)dt,
DtN on elementary solutions Boundary control The Sommerfeld radiation condition
The Bessel function of the second kind
when k 1 > k 2 , the C^1 -continuation of
Y 0 (k 1 r ), θ ∈ [ 0 ; π]
is Y 0 (k 2 r ) + (^) π^2 J 0 (k 2 r ) ∫^0 argcosh (k^1 /k^2 ) 1 du π^4 ∑ m≥ 1 J^2 m^ (k^2 r^ )cos(^2 mθ)^ ∫^0 argcosh (k^1 /k^2 )(−^1 )m^ cosh(^2 mu)du
− (^) π^1 ∫^ argcosh∞ ( k 1 k 2 )
1 − k k^222 √√^1 √√ k^22 k^21 cosh^2 (u)−^1 (^
k k 2 1 sinh(u)+
√√√ √ k k (^222) 1
cosh^2 (u)− 1 )
exp(ik 2 r .cosh(u − iθ))du
− (^) π^1 ∫^ argcosh∞ ( k 1 k 2 )
1 − k k^222 √√^1 √√ k^22 k^21 cosh^2 (u)−^1 (^
k k 2 1 sinh(u)+
√√√ √ k k (^222) 1
cosh^2 (u)− 1 )
exp(−ik 2 r .cosh(u + iθ))du, θ ∈ [−π; 0 ]
DtN on elementary solutions Boundary control The Sommerfeld radiation condition
The Bessel function of the second kind
for n > 0, we show that
(ent^ − (− 1 )ne−nt^ )e−kr^ .sinh(t)dt = 2 cosh(u)e−r^ .sinhu^ ∗ [ (^) ∑n− 2 1 ]
m= 0
C n^2 m +^1 ( 1 +
k^2
sinh^2 (u))
kn−^2 m−^1
sinhn−^2 m−^1 (u)du
it will lead to a form
Y 2 n(k 1 r )cos( 2 nθ) = ∑ αmJ 2 m(k 2 r )cos( 2 mθ) +
finite
βmY 2 m(k 2 r )cos( 2 mθ) + R
DtN on elementary solutions Boundary control The Sommerfeld radiation condition
Summary
each elementary function has a known DtN simple expression at least on the half plane, parity preserving we have redundant solutions but it is not a problem, it could help the cases k 1 < k 2 and k 2 < k 1 , simple expression in the upper or lower half planes do not give the same results
DtN on elementary solutions Boundary control The Sommerfeld radiation condition
Imposing the boundary value
Suppose that f is the sum of upper elementary functions
f =
anw n+ (r , θ)
we can show that in order to impose f = f 0 on the boundary there exists an linear in < f 0 , w± > complicated and infinite system to solve ...
DtN on elementary solutions Boundary control The Sommerfeld radiation condition
Imposing the boundary value
Suppose that f is the sum of a finite number of upper class elementary functions and a finite number of lower class elementary functions
f =
n= 0 ..N+
anw n+ (r , θ) +
n= 0 ..N−
bnw n− (r , θ) = f 1 + f 2
this representation is more symmetric because there are as many cos and sin used in the upper and lower half planes to represent the solutions then we know how to express the DtN of f 1 and f 2 resp. as a function of f 1 |∂Ω and f 2 |∂Ω but not necessarily the DtN of (f 1 + f 2 ) as a function of f 1 + f 2 |∂Ω!