Uniqueness in the magnetic shaping problem without high frequency assumption

Applied Mathematics Letters 102 (2020) 106106

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Applied Mathematics Letters www.elsevier.com/locate/aml

Uniqueness in the magnetic shaping problem without high frequency assumption Jaemin Shin Department of Mathematical Sciences & Institute for Applied Mathematics and Optics, Hanbat National University, Daejeon 34158, Republic of Korea

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Article history: Received 23 August 2019 Received in revised form 18 October 2019 Accepted 18 October 2019 Available online 6 November 2019

abstract In this work, we study an inverse source problem arising from the magnetic casting process. Specifically, we discuss the mathematical model without the high frequency assumption on the current in inductors. We derive an associated transmission problem and provide a proof of uniqueness by using the Bedrosian identity. © 2019 Elsevier Ltd. All rights reserved.

Keywords: Inverse shaping problem Electromagnetic casting Inverse source problem Bedrosian identity

1. Introduction and mathematical model There are several engineering processes in liquid metallurgical industry by which a liquid material such as a molten metal or a ferrofluid is shaped by magnetic fields. In this work, we focus on the case of a vertical column of liquid material falling into a magnetic field, B e = ⟨B1e , B2e , 0⟩ created by vertical inductors. The magnetic field produces a surface pressure that forces the liquid material to change its shape until it reaches an equilibrium state at the boundary between the magnetic pressure and the surface tension. We are mainly interested in a distribution of inductors or current density, j = ⟨0, 0, j⟩, such that the horizontal cross-section of the liquid material, denoted by Ω , has a prescribed shape. It is known (see, e.g., [1–3]) that the magnetic induction B e satisfies the following system of equations: ∇ × B e = µ0 j

in R2 \ Ω ,

(1.1a)

∇ · Be = 0

in R2 \ Ω ,

(1.1b)

|B e | = O(|x|

−1

(1.1c)

E-mail address: [email protected]. https://doi.org/10.1016/j.aml.2019.106106 0893-9659/© 2019 Elsevier Ltd. All rights reserved.

)

as |x| → ∞,

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J. Shin / Applied Mathematics Letters 102 (2020) 106106

where µ0 is the vacuum permeability, and j is compactly supported in R2 \ Ω such that ∫ jdx = 0.

(1.1d)

R2 \Ω

If A1. the frequency of the applied magnetic field is assumed to be sufficiently high that the field does not penetrate into the material, and A2. the perturbation of the cylinder that occurs under gravity is ignored, then B e satisfies the boundary conditions on Γ := ∂Ω ;

σκ +

Be · ν = 0

on Γ ,

(1.2a)

1 2 |B e | = P0 2µ0

on Γ ,

(1.2b)

where ν is the unit normal vector, σ is the surface tension of the liquid material, κ is the curvature of Γ , and P0 is a constant that represents the difference between the interior and exterior pressures. It is also assumed that the liquid material is incompressible so the area of the cross-section is fixed; i.e., for a given S0 , ∫ dx = S0 . (1.3) Ω

The direct problem is a free boundary problem to determine the shape of the liquid material for the given magnetic field; i.e., to solve (1.1)–(1.3) for B e and Ω for a given j. On the other hand, the inverse problem is formulated to seek j for a given Ω . These problems have been studied intensively, analytically and numerically. For example, the existence and uniqueness of j and stability were studied in [3–7]. Several numerical methods for direct and inverse problems using the variational approach, topological derivatives, the simple pole detection algorithm, and the level set method were proposed as well. We refer the reader to [2,3,8–13] and references therein. From the practical point of view, however, assumption A1 may not be valid for certain materials that have relatively low conductivity. To achieve a small skin depth such that assumption A1 is effective requires too high an electric current, which is expensive or even impossible to generate in a laboratory. For this reason, we must reformulate the mathematical model, especially, for a low conductive material. Without the high frequency assumption, we need to consider the magnetic field H i = µB i inside the material. Here, we assume that the material is a linear medium, and µ denotes the permeability of the material. Then, Maxwell’s equations give ∇ × Hi = 0

in Ω ,

(1.4a)

∇ · Hi = 0

in Ω .

(1.4b)

Here, we neglect the free current density as we assume that the conductivity of the material is low. The boundary conditions at the interface between the interior and exterior regions of the material are given by the following (see, e.g., [14,15]): (B i − B e ) · ν = 0, i

e

(H − H ) × ν = 0,

(1.5a) (1.5b)

J. Shin / Applied Mathematics Letters 102 (2020) 106106

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where the indexes i, e denote the interior and exterior region of the material, respectively. The equilibrium condition (1.2b) is replaced by the following Young–Laplace equation together with the no-gravity assumption A2: 1 1 2 2 σκ − (µ − µ0 )|H i | − (µ − µ0 )2 |H i · ν| = P1 on Γ , (1.5c) 2 2µ0 where P1 is an unknown constant. For the derivation of (1.5c), we refer the reader to [15–17]. We are interested in the shaping problem without the high frequency assumption A1, which is to seek j for a given Ω satisfying (1.1), (1.4), and (1.5). Here, we assume that the inductor consists of single solid-core wires with a negligible cross-sectional area; i.e., j is the finite sum of Dirac delta functions. The main result in this work is the following uniqueness theorem: Theorem 1.

Suppose that Ω is a simply connected bounded domain with a smooth boundary Γ , and µ > µ0 .

If j=

K ∑

βk δ(x − xk ),

βk ∈ R,

xk ∈ R2 \ Ω ,

β ′ ∈ R,

x′k ∈ R2 \ Ω

k=1

and

′

′

j =

K ∑

βk′ δ(x − x′k ),

k=1

are solutions to the shaping problem, (1.1), (1.4), and (1.5), then, either j = j ′ or j = −j ′ . In Section 2, we derive a transmission problem for the scalar potentials of the magnetic fields and compute the unknown constant P1 of (1.5c). Proof of Theorem 1 is provided in Section 3. 2. Transmission problem We define scalar potential functions ue and ui such that Hi = ⟨

∂ui ∂ui ,− , 0⟩, ∂y ∂x

He = ⟨

∂ue ∂ue ,− , 0⟩. ∂y ∂x

(2.1)

Then, (1.1), (1.3)–(1.5) can be rewritten as follows: −∆ui = 0 in Ω , ⏐ ∂ui ⏐2 ⏐ ∂ui ⏐2 2 ⏐ ⏐ ⏐ ⏐ (σκ − P1 ) ⏐ ⏐ + µr ⏐ ⏐ = ∂ν ∂τ µ − µ0 ∫ ui dS(x) = 0,

(2.2a) on Γ ,

(2.2b) (2.2c)

Γ

−∆ue = j

in R2 \ Ω ,

(2.2d)

as |x| → ∞,

(2.2e)

e

(2.2f)

ue = O(1) i

µu = µ0 u

on Γ ,

∂ue ∂ui = on Γ , ∂ν ∂ν ∫ jdx = 0, R2 \Ω

(2.2g) (2.2h)

J. Shin / Applied Mathematics Letters 102 (2020) 106106

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where ∂/∂τ denotes the tangential derivative in the counterclockwise direction, and µr = µ/µ0 is the relative permeability. The condition (2.2c) is invoked for the uniqueness of the potentials defined in (2.1). It is known [4] that the unknown constant P0 in (1.2b) can be computed by P0 = σ max κ, which is a necessary condition for a feasible domain. The constant P1 in (1.5c) or (2.2b) may be computed in a similar manner. Let u ˜i be a harmonic conjugate function of ui . Then, for any z0 ∈ C, f (z) = ui + i˜ ui + z0 is a holomorphic function in Ω . Then, the maximum modulus theorem gives 2

∂|f | ∂ui ∂˜ ui = 2(ui + Re z0 ) + 2(˜ ui + Im z0 ) =0 ∂τ ∂τ ∂τ at some point on the boundary Γ . As z0 is an arbitrary constant, we have Since the Cauchy-Riemann equation gives ∂ui ∂˜ ui = , ∂ν ∂τ we conclude that

∂ui ∂τ

˜ ui = ∂∂τ = 0 at a point on Γ .

σκ − P1 ≥ 0, and the equality must hold. It follows that P1 = σ min κ

(2.3)

as long as µ > µ0 . 3. Proof of uniqueness The uniqueness of j (up to the sign) in Theorem 1 follows from the uniqueness of ui . Indeed, if ui is given, then the transmitting boundary conditions, (2.2f) and (2.2g) give the Cauchy data of ue at Γ , which uniquely determines j assuming that it is a superposition of Dirac delta functions; i.e., Theorem 2. Let Ω be a simply connected bounded domain with a smooth boundary Γ . Suppose that ue and ue ′ solve (2.2d), (2.2e) with the source terms j=

K ∑

βk δ(x − xk ),

k=1

and j =

K ∑ k=1

e′

respectively. If ue and u

βk = 0,

xk ∈ R2 \ Ω ,

βk′ = 0,

x′k ∈ R2 \ Ω ,

k=1

′

′

K ∑

′

βk′ δ(x

−

x′k ),

K ∑ k=1

have the same Cauchy data on Γ , then j = j′.

One may find a proof of Theorem 2 in [7]. Thus, it suffices to show the uniqueness of ui satisfying (2.2a)–(2.2c). Without loss of generality, we may rewrite (2.2a)–(2.2c) for some g ∈ L2 (∂B) by means of a conformal transformation (see, e.g., [18]) together

J. Shin / Applied Mathematics Letters 102 (2020) 106106

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with (2.3) as follows: −∆U = 0 in B, ⏐ ∂U ⏐2 ⏐ ∂U ⏐2 ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ + µr ⏐ ⏐ =g ∂ν ∂τ ∫ U dS(x) = 0.

(3.1a) on ∂B,

(3.1b) (3.1c)

∂B

We then have the following uniqueness result for ui or U . Theorem 3. Suppose that g ∈ L2 (∂B) and 0 < µr < 1 or µr > 1. Let U1 and U2 be real solutions to (3.1). Then, either U1 = U2 , or U1 = −U2 . Let U1 , U2 be solutions to (3.1). Define w1 = U1 − U2 , w2 = U1 + U2 . Then, wi (i = 1, 2) are harmonic functions in B, and we deduced from (3.1b) the following: ∂w1 ∂w2 ∂w1 ∂w2 + µr =0 ∂ν ∂ν ∂τ ∂τ

on ∂B,

or

∂w ˜1 ∂ w ˜2 ∂w1 ∂w2 + µr = 0 on ∂B. ∂τ ∂τ ∂τ ∂τ Here, we denote the harmonic conjugate of wi with the mean zero by w ˜i . ∫ 2 2 Let H be the Hilbert transform defined in L0 := {ϕ ∈ L (∂B) : ∂B ϕdS(x) = 0} such that ∞ ∑

H(

an einθ ) =

n=−∞

∞ ∑

(3.2)

−i sgn(n)an einθ .

n=−∞

As the Hilbert transform converts the boundary value of the harmonic function to the boundary value of its harmonic conjugate, and commutes with the derivative (see, e.g., [19,20]), we deduced from (3.2) the following: ∂w2 ⏐⏐ ∂w1 ⏐⏐ ∂w2 ⏐⏐ ∂w1 ⏐⏐ H( ⏐ )H( ⏐ ) + µr ⏐ ⏐ = 0. ∂τ ∂B ∂τ ∂B ∂τ ∂B ∂τ ∂B i If ∂w ∂τ = 0 on ∂B, (3.1c) implies wi = 0 on ∂B. Then, the maximum principle implies wi ≡ 0 in B. Thus, Theorem 3 follows from the Lemma below.

Lemma 4. Let H be the Hilbert transform on the circle and 0 < µr < 1 or µr > 1. Suppose that ϕ, ψ are real-valued functions in L20 satisfying H(ϕ)H(ψ) + µr ϕψ = 0. (3.3) Then, either ϕ = 0 or ψ = 0 almost everywhere. Proof . For ϕ ∈ L20 , we define the following: ϕ = ϕ+ + ϕ− :=

∞ ∑ n=1

ϕn einθ +

∞ ∑ n=1

The Hilbert transform H then gives H(ϕ) = −iϕ+ + iϕ− .

ϕ−n e−inθ .

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J. Shin / Applied Mathematics Letters 102 (2020) 106106

Substitution of these representations into (3.3) yields (1 − µr )(ϕ+ ψ + + ϕ− ψ − ) − (1 + µr )(ϕ+ ψ − + ϕ− ψ + ) = 0. Since we assume that ϕ and ψ are real, ϕ± = ϕ ∓ ,

ψ± = ψ∓ ,

(3.4)

where ϕ is the complex conjugate of ϕ. It follows that (1 − µr )(ϕ+ ψ + + ϕ+ ψ + ) − (1 + µr )(ϕ+ ψ + + ϕ+ ψ + ) = 0.

(3.5)

+ + + + We split ϕ+ and ψ + into the real and imaginary parts as ϕ+ = ϕ+ R + iϕI , ψ = ψR + iψI . Then, (3.5) gives + + + ϕ+ I ψI + µr ϕR ψR = 0.

(3.6)

Since ϕ+ and ψ + are in the Hardy space H 2 and have homomorphic extensions in B, we have the following: + ϕ+ I = H(ϕR ),

+ ψI+ = H(ψR ).

Rewrite (3.6) as + + + H(ϕ+ R )H(ψR ) + µr ϕR ψR = 0. + As the Fourier coefficients of ϕ+ R and ψR are zeros for n ≤ 0, we can apply the Bedrosian identity [21] to obtain + + + + + H(ϕ+ R ψR ) = ϕR H(ψR ) = H(ϕR )ψR , + + + + 2 2 + H(H(ϕ+ R )H(ψR )) = H(ϕR )H (ψR ) = H (ϕR )H(ψR ).

Together with H2 = −I, we have + (1 − µr )ϕ+ R H(ψR ) = 0,

or by the Bedrosian identity again, + (1 − µr )ϕ+ R ψR = 0.

Since we assume that µr ̸= 1, it follows that + ϕ+ R ψR = 0.

(3.7)

On the other hand, the Bedrosian identity also gives + + + + + + + (ϕ+ R + iH(ϕR ))(ψR + iH(ψR )) = 2(ϕR ψR + iH(ϕR ψR )), + + + which vanishes thanks to (3.7). As ϕ+ R + iH(ϕR ) and ψR + iH(ψR ) can be extended to holomorphic functions in B, the product has a holomorphic extension in B. Using the maximum modulus theorem, we conclude that + + + ϕ+ = ϕ+ or ψ + = ψR + iH(ψR ) = 0. R + iH(ϕR ) = 0,

Hence, ϕ = 0 or ψ = 0 due to (3.4). □ Remark. We notice that µr ̸= 1 (i.e. µ ̸= µ0 ) is essential in Theorem 3. Indeed, if µr = 1 is assumed, then ∫ ˜ of U satisfying ˜ dS(x) = 0 solves (3.1) as well, since the harmonic conjugate U U ∂B ⏐ ∂U ⏐2 ⏐ ∂U ⏐2 ⏐ ∂ U ⏐ ⏐ ˜ ⏐2 ⏐ ⏐ ⏐ ⏐ ˜ ⏐2 ⏐ ∂ U ⏐ ⏐ ⏐ +⏐ ⏐ =⏐ ⏐ +⏐ ⏐ ⏐ ∂ν ∂τ ∂τ ∂ν

on ∂B.

As stated in Theorem 3, 0 < µr < 1 or µr > 1 is a sufficient condition for the uniqueness (up to the sign) for (3.1). However, P1 in (2.2b) should be evaluated to define g in (3.1b), in which the condition, µr > 1, is required in our approach. For this reason, µr = µ/µ0 > 1 should be assumed in Theorem 1.

J. Shin / Applied Mathematics Letters 102 (2020) 106106

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Acknowledgement This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (NRF-2015R1D1A3A01018164). References [1] J.A. Shercliff, Magnetic shaping of molten metal columns, Proc. R. Soc. Lond. Ser. A 375 (1981) 455–473. [2] M. Pierre, J.-R. Roche, Computation of free surfaces in the electromagnetic shaping of liquid metals by optimization algorithms, Eur. J. Mech. B Fluids 10 (5) (1991) 489–500. [3] A. Canelas, J.R. Roche, J. Herskovits, The inverse electromagnetic shaping problem, Struct. Multidiscip. Optim. 38 (4) (2009) 389–403. [4] A. Henrot, M. Pierre, Un probl` eme inverse en formage des m´ etaux liquides, RAIRO Mod´ el. Math. Anal. Num´ er. 23 (1) (1989) 155–177. [5] M. Dambrine, M. Pierre, About stability of equilibrium shapes, M2AN Math. Model. Numer. Anal. 34 (4) (2000) 811–834. [6] A. Novruzi, C 2,α Existence result for a class of shape optimization problems, SIAM J. Control Optim. 43 (1) (2004) 174–193 (electronic). [7] J. Shin, J.P. Spicer, J.A. Abell, Inverse and direct magnetic shaping problems, Struct. Multidiscip. Optim. 46 (2) (2012) 285–301. [8] M. Crouzeix, Variational approach of a magnetic shaping problem, Eur. J. Mech. B Fluids 10 (5) (1991) 527–536. [9] A. Novruzi, J.R. Roche, Second Order Derivatives, Newton Method, Application to Shape Optimization, Research Report RR-2555, INRIA, 1995. [10] A. Canelas, J.R. Roche, J. Herskovits, Inductor shape optimization for electromagnetic casting, Struct. Multidiscip. Optim. 39 (6) (2009) 589–606. [11] A. Canelas, A.A. Novotny, J.R. Roche, Topology design of inductors in electromagnetic casting using level-sets and second order topological derivatives, Struct. Multidiscip. Optim. 50 (6) (2014) 1151–1163. [12] T.J. Machado, A. Canelas, A.A. Novotny, J.R. Roche, Fast solution of the inverse electromagnetic casting problem, Struct. Multidiscip. Optim. 57 (6) (2018) 2447–2455. [13] A. Canelas, J.R. Roche, Solution to a three-dimensional axisymmetric inverse electromagnetic casting problem, Inverse Probl. Sci. Eng. 27 (10) (2019) 1451–1467. [14] J.D. Jackson, Classical Electrodynamics, second ed., John Wiley & Sons Inc., New York, London, Sydney, 1975. [15] R. Rosensweig, Ferrohydrodynamics, Dover Books on Physics, Dover Publications, 2013. [16] B. Berkovski˘ı, V. Medvedev, M. Krakov, Magnetic Fluids: Engineering Applications, Oxford Science Publications, Oxford University Press, 1993. [17] O. Lavrova, Numerical Methods for Axisymmetric Equilibrium Magnetic-Fluid Shapes (Ph.D. thesis), Universit¨ atsbibliothek, 2006. [18] L.V. Ahlfors, Complex Analysis, third ed., in: International Series in Pure and Applied Mathematics, McGraw-Hill, New York, NY, 1979. [19] S.R. Bell, The Cauchy Transform, Potential Theory and Conformal Mapping, second ed., Chapman & Hall/CRC, Boca Raton, FL, 2016. [20] F.W. King, Hilbert Transforms, Vol. 1, in: Encyclopedia of Mathematics and its Applications, vol. 124, Cambridge University Press, Cambridge, 2009. [21] E. Bedrosian, A product theorem for Hilbert transforms, Proc. IEEE 51 (5) (1963) 868–869.