On a free boundary problem of magnetohydrodynamics in multi-connected domains

Σχετικά έγγραφα
Partial Differential Equations in Biology The boundary element method. March 26, 2013

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

2 Composition. Invertible Mappings

EE512: Error Control Coding

Example Sheet 3 Solutions

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3

Uniform Convergence of Fourier Series Michael Taylor

Other Test Constructions: Likelihood Ratio & Bayes Tests

4.6 Autoregressive Moving Average Model ARMA(1,1)

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

Reminders: linear functions

Math 446 Homework 3 Solutions. (1). (i): Reverse triangle inequality for metrics: Let (X, d) be a metric space and let x, y, z X.

D Alembert s Solution to the Wave Equation

Matrices and Determinants

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

Homework 3 Solutions

Areas and Lengths in Polar Coordinates

Every set of first-order formulas is equivalent to an independent set

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

5. Choice under Uncertainty

Homework 8 Model Solution Section

Statistical Inference I Locally most powerful tests

Areas and Lengths in Polar Coordinates

Concrete Mathematics Exercises from 30 September 2016

6.3 Forecasting ARMA processes

Math221: HW# 1 solutions

6.1. Dirac Equation. Hamiltonian. Dirac Eq.

The Pohozaev identity for the fractional Laplacian

L p approach to free boundary problems of the Navier-Stokes equation

ES440/ES911: CFD. Chapter 5. Solution of Linear Equation Systems

Generating Set of the Complete Semigroups of Binary Relations

Inverse trigonometric functions & General Solution of Trigonometric Equations

Section 8.3 Trigonometric Equations

ECE Spring Prof. David R. Jackson ECE Dept. Notes 2

12. Radon-Nikodym Theorem

Numerical Analysis FMN011

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

( ) 2 and compare to M.

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

Tridiagonal matrices. Gérard MEURANT. October, 2008

Congruence Classes of Invertible Matrices of Order 3 over F 2

C.S. 430 Assignment 6, Sample Solutions

Lecture 26: Circular domains

ORDINAL ARITHMETIC JULIAN J. SCHLÖDER

ω ω ω ω ω ω+2 ω ω+2 + ω ω ω ω+2 + ω ω+1 ω ω+2 2 ω ω ω ω ω ω ω ω+1 ω ω2 ω ω2 + ω ω ω2 + ω ω ω ω2 + ω ω+1 ω ω2 + ω ω+1 + ω ω ω ω2 + ω

SCITECH Volume 13, Issue 2 RESEARCH ORGANISATION Published online: March 29, 2018

Solvability of Brinkman-Forchheimer equations of flow in double-diffusive convection

Nonlinear Fourier transform for the conductivity equation. Visibility and Invisibility in Impedance Tomography

Local Approximation with Kernels

Solution Series 9. i=1 x i and i=1 x i.

Second Order Partial Differential Equations

Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in

Finite Field Problems: Solutions

b. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!

derivation of the Laplacian from rectangular to spherical coordinates

Lecture 34 Bootstrap confidence intervals

The Simply Typed Lambda Calculus

2. Let H 1 and H 2 be Hilbert spaces and let T : H 1 H 2 be a bounded linear operator. Prove that [T (H 1 )] = N (T ). (6p)

Space-Time Symmetries

ST5224: Advanced Statistical Theory II

Main source: "Discrete-time systems and computer control" by Α. ΣΚΟΔΡΑΣ ΨΗΦΙΑΚΟΣ ΕΛΕΓΧΟΣ ΔΙΑΛΕΞΗ 4 ΔΙΑΦΑΝΕΙΑ 1

Optimal Parameter in Hermitian and Skew-Hermitian Splitting Method for Certain Two-by-Two Block Matrices

A Lambda Model Characterizing Computational Behaviours of Terms

Problem Set 3: Solutions

Global nonlinear stability of steady solutions of the 3-D incompressible Euler equations with helical symmetry and with no swirl

k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +

Testing for Indeterminacy: An Application to U.S. Monetary Policy. Technical Appendix

1 String with massive end-points

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

On the Galois Group of Linear Difference-Differential Equations

Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

Srednicki Chapter 55

A General Note on δ-quasi Monotone and Increasing Sequence

Eulerian Simulation of Large Deformations

The semiclassical Garding inequality

Spherical Coordinates

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

( y) Partial Differential Equations

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007

Second Order RLC Filters

Quadratic Expressions

Higher Derivative Gravity Theories

Strain gauge and rosettes

Lecture 21: Properties and robustness of LSE

Iterated trilinear fourier integrals with arbitrary symbols

Fourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics

Fractional Colorings and Zykov Products of graphs

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 6/5/2006

MA 342N Assignment 1 Due 24 February 2016

DERIVATION OF MILES EQUATION FOR AN APPLIED FORCE Revision C

Forced Pendulum Numerical approach

SOME PROPERTIES OF FUZZY REAL NUMBERS

Memoirs on Differential Equations and Mathematical Physics

Abstract Storage Devices

THE SECOND ISOMORPHISM THEOREM ON ORDERED SET UNDER ANTIORDERS. Daniel A. Romano

Pg The perimeter is P = 3x The area of a triangle is. where b is the base, h is the height. In our case b = x, then the area is

Transcript:

On a free boundary problem of magnetohydrodynamics in multi-connected domains V.A.Solonnikov June 212, Frauenhimsee 1 Formulation of the problem Domains 1t : variable domain filled with a liquid, 2t : vacuum region surrounding 1t, 3 : fixed domain where the electric current j(x, t) is given; = 1t 3 2t, Γ t = 1t : free surface, S 3 : the boundary of 3, S: the boundary of a perfect conductor. Equations of magnetohydrodynamics. Navier-Stokes equations: { vt + (v )v (T (v, p) + T M (H)) = f(x, t), v(x, t) =, x 1t, t >, Maxwell equations: µh t = rote, H =, x 1t 2t 3, roth = α(e + µ(v H)), x 1t, roth =, H =, E =, x 2t, roth = αe + j(x, t), x 3. (1.1) (1.2) v, p: velocity and pressure, f: mass forces, H(x, t), E(x, t): magnetic and electric fields, µ: piecewise constant positive function equal to µ i in i α: piecewise constant function positive in 1t, 3 and α = in 2t, T (v, p) = pi + νs(v): the viscous stress tensor, S(v) = v + ( v) T : the doubled rate-of-strain tensor, T M (H) = µ(h H 1 2 H 2 I): the magnetic stress tensor. 1

Boundary and jump conditions: H n =, E τ =, x S, [µh n] =, [H τ ] =, [E τ ] =, x S 3, ( T (v, p) + [TM (H)] ) n =, V n = v n, [µh n] =, [H τ ] =, n t [µ]h τ + [n x E] =, x Γ t, (1.3) H τ = H n(n H), E τ = E n(n E): tangential components of H and E, [F ]: jump of the vector field F (x) on Γ t and S 3, V n : the velocity of evolution of Γ t in the direction of the exterior normal n, n = (n x, n t ): normal to the surface x Γ t, t (, T ) in R 4. Initial and orthogonality conditions v(x, ) = v (x), x 1, H(x, ) = H (x), x 1 2 3, (1.4) S k E nds =. (1.5) S k : connected components of the boundary of 2t, i = i, i = 1, 2, 3 Compatibility conditions v (x) =, x 1, H (x) =, x 1 2 3, roth (x) =, x 2, [(S(v )n) τ ] =, [H τ ] =, [µh N] =, x Γ, [H τ ] =, [µh n] =, x S 3, H (x) n =, x S. (1.6) 2 Transformation of the problem and formulation of main result. Lagrangian coordinates ξ 1 : x = ξ + t u(ξ, τ)dτ X(ξ, t), X : 1 1t, Γ Γ t. u(ξ, t) = v(x(ξ, t), t), u (x, t) : the extension of u from 1 in such that u (ξ, t) = in 3 and near S 3 S. x = ξ + t u (ξ, τ)dτ X(ξ, t), X : i it, i = 1, 2, 3 3, ( x ) L(u), L(u) = detl, L = LL 1 ; ξ 2

L = 1 for ξ 1, L T A(u): co-factors matrix, T means transposition. Let P = L T L/L, q(ξ, t) = p(x, t), h(ξ, t) = LH(X, t), e(ξ, t) = LE(X, t), i.e., H(X, t) = L L h, E(X, t) = L L e. The mapping X converts (1.1)-(1.5) in { ut ν 2 uu + u q u T M (Lh) = f(x, t), u u =, ξ 1, t >, µ(h t L T L t L h L T (u u ) L h) = rotp(ξ, t)e, L ξ i, i = 1, 2, 3, ProtPh = α(pe + µ(l 1 u h)), h =, ξ 1, rotph =, h =, e =, ξ 2, roth = αe + j(ξ, t), h =, ξ 3, (2.1) (2.2) { h n =, eτ =, ξ S, [µh n] =, [h τ ] =, [e τ ] = ξ S 3, [µh n ] =, [h τ ] = ( AT An An 2 n ))[h n ], [n Pe] = (u An )[µ]h τ, (T u (u, q) + [T M (Lh)])An =, ξ Γ, u(ξ, ) = v (ξ), ξ 1, h(ξ, ) = H (ξ), ξ i, i = 1, 2, 3, e(ξ, t) n(ξ)ds =, S k (2.3) (2.4) (2.5) where u = L T is the transformed gradient w.r.to x and is the gradient w.r.to ξ, n(x) = An (ξ)/ An (ξ), n : the exterior normal to Γ. T u = q + νs u (u): the transformed stress tensor; S u (w) = u w + ( u w) T : the transformed rate-of-strain tensor. We find the solution of (2.1)-(2.5) in anisotropic Sobolev-Slobodetskii spaces W r,r/2 2, r >. Theorem 1. Let Γ W l+3/2 2, u W2 l+1 ( 1 ), H W2 l( 1) W2 l( 2), W2 l( 3) with 3/2 < l < 2, and let the compatibility conditions (1.6) be satisfied. Assume also that f, f W l,l/2 2 (R 3 (, T )), j W (l 1)/2 2 (, T ; W 1 2 ( 3 )) L 2 (, T ; W l 2( 3 ), j(x, t) n(x) x S3 =. (2.6) 3

Then the problem (2.1)-(2.5) has a unique solution defined in a certain (small) time interval (, T ), T T with the following regularity properties: u W 2+l,1+l/2 2 (Q 1 T ), q W l,l/2 2 (Q 1 T ), q W l+1/2,l/2+1/4 2 (G T ), h W l+1,(l+1)/2 2 (Q i T ), e L 2 (, T ; W l 2( i )) W (l 1)/2 2 (, T ; W 1 2 ( i )), where i = 1, 2, 3,, Q i T = (, T ), G T = Γ (, T ). The solution satisfies the inequality u l+2,l/2+1 W 2 (Q 1 T ) + q W l,l/2 2 (Q 1 T ) + q l+1/2,l/2+1/4 W 2 (G T ) ( + h W l+1,l/2+1/2 2 (Q i T ) + e L 2 (,T ;W2 l( i)) ) + e (l 1)/2 W ( c + 2 (,T ;W 1 2 (Γ )) f W l,l/2 2 (Q 1 T ) + u W l+1 2 ( 1 ) H W l 2 ( i ) + j L 2 (,T ;W2 l( 3)) ) + j (l 1)/2 W 2 (,T ;W2 1(. 3)) (2.7) Exclusion of e. Let the test function ψ(ξ) satisfy the conditions ψ(ξ) =, ξ i, i = 1, 2, 3, rotψ(ξ) =, ξ 2, [µψ n ] =, [ψ τ ] =, ξ Γ, [µψ n] =, [ψ τ ] =, ξ S 3, ψ n =, ξ S. (2.8) Using the relation rotpe ψdξ = we write (2.1)-(2.5) as a nonlinear problem for u, q, h: Pe rotψdξ + [n Pe] ψds Γ 4

u t ν 2 uu + u q u T M (Lh) = f(x, t), u u =, ξ 1, t >, µ(h t Φ(h, u)) ψ(ξ, t)dξdt T + α 1 ProtPh rotψ(ξ, t)dξdt 1 3 µ 1 (L 1 u h) rotψ(ξ, t)dξdt 1 + Ψ(h, u) ψdsdt Γ = α 1 j(ξ, t) ψ(ξ, t)dξdt, 3 rotph =, ξ 2, h =, ξ 1 2 3, h n =, ξ S, [µh n] =, [h τ ] =, ξ S 3, (2.9) [µh n ] =, [h τ ] = ( AT An An 2 n ))[h n )] =, (T u (u, q) + [T M (Lh)])An =, ξ Γ, u(ξ, ) = v (ξ), ξ 1, h(ξ, ) = H (ξ), ξ i, i = 1, 2, 3, where Φ = L T L t L h + L T (u u ) L L h, Ψ = (u An )[µ]h τ. Theorem 2. Under the assumptions of Theorem 1, the problem (2.9) has a unique solution (u, q, h) satisfying the inequality u l+2,l/2+1 W 2 (Q 1 T ) + q W l,l/2 2 (Q 1 T ) + q l+1/2,l/2+1/4 W 2 (G T ) ( + h W l+1,l/2+1/2 2 (Q i T ) ( c + f W l,l/2 2 (Q 1 T ) + u W l+1 2 ( 1 ) H W l 2 ( i ) + j L 2 (,T ;W2 l( 3)) ) + j (l 1)/2 W 2 (,T ;W2 1(. 3)) (2.1) 5

3 Linear problems The proof of Theorem 2 is based on the analysis of some linear problems, namely, v t ν 2 v + p = f(ξ, t), v = f, ξ 1, t >, T (v, p)n = d(ξ, t), ξ Γ, v(ξ, ) = v (ξ), ξ 1, { roth(ξ) = k(x), h(ξ) =, ξ 1 2 3, [µh n] =, [h τ ] =, ξ Γ S 3, h n ξ S =, µh t ψ(ξ, t)dξdt + α 1 roth rotψ(ξ, t)dξdt 1 3 = g 1 (ξ, t) rotψ(y, t)dξdt 1 3 + µg 2 (y, t) ψ(ξ, t)dξdt, h(ξ, t) =, ξ 1 2 3, roth(ξ, t) =, ξ 2, [µh n ] =, [h τ ] =, ξ Γ, [µh n] =, [h τ ] =, ξ S 3, h n =, ξ S, h(ξ, ) = h (ξ), ξ 1 2 3, where ψ(ξ, t) is an arbitrary test vector field satisfying the conditions (2.8). Theorem 3. Assume that Γ W l+3/2 2, 3/2 < l < 2, and that the compatibility conditions f W l,l/2 2 (Q 1 T ), v W l+1 2 ( 1 ), f = F, f L 2 (, T ; W l+1 2 ( 1 )), F t W l/2 2 (, T ; L 2 ( 1 )), d W l+1/2,l/2+1/4 2 (G T ), (3.1) (3.2) (3.3) v (ξ) = f(ξ, ), (S(v )n) τ = d τ, ξ Γ are satisfied. Then the problem (3.1) has a unique solution v W 2+l,1+l/2 2 (Q 1 T ), p W l,l/2 1 (Q 1 T ) with p ξ Γ W l+1/2,l/2+1/4 2 (G T ) and v l+2,l/2+1 W 2 (Q 1 T ) + p W l,l/2 2 (Q 1 T ) + p W l+l/2,1/2+l/4 2 (Γ ) c( f L2 (,T ;W l+1 2 ( 1 )) + F t l/2 W 2 (,T ;L 2 ( 1 )) + d l+1/2,l/2+1/4 W 2 (G T ) + v W l+1 2 ( ) ). (3.4) 6

Let H l () be the space of vector fields ψ W2 l (), i = 1, 2, 3, satisfying (2.8); U(): finite dimensional set of vector fields satisfying (2.8) and, in addition, rotψ(ξ) =, ξ 1 2 3 ; H l (): the subset of Hl () whose elements satisfy the orthogonality condition µψ(ξ) u(ξ)dξ =, u U(). Theorem 4. Let Γ, S, S 3 W l+3/2 2, l (3/2, 2). For arbitrary k W r 2 ( i), r [, l] i = 1, 2, 3, such that k =, ξ 1 2 3, [k n] =, ξ Γ S 3, k n S = the problem (3.2) has a unique solution belonging to W r+1 2 ( i ), i = 1, 2, 3 and orthogonal to U(). The solution satisfies the inequality then Moreover, if h W 1+r 2 ( i ) c k W r 2 ( i ). (3.5) k = rotk(ξ), [K τ ] Γ S 3 =, K τ S =, (3.6) h L2 () c K L2 (). (3.7) Corollary 1. If h H r+1 (), then c 1 h W r+1 2 ( i ) roth W2 r( 1 3 ) c 2 h W r+1 2 ( i ). Corollary 2. If k W r,r/2 2 (Q i T ), i = 1, 2, 3, then h L 2(, T : W2 r+1 ( i )) W r/2 2 (, T ; W2 1( i)) and ( h L2 (,T :W r+1 2 ( i )) + h W r/2 2 (,T ;W2 1( i)) ) (3.8) Moreover, if (3.6) holds, then c k W r,r/2 2 ( i ). h W (r+1)/2 2 (,T ;L 2 ()) c K W (r+1)/2 2 (,T ;L 2 ()). (3.9) Theorem 5. Let l (3/2, 2). For arbitrary h, g 1, g 2 satisfying the conditions h W l 2( i ), i = 1, 2, 3, h (y) =, y 1 2 3, roth (y) =, y 2, [µh n] =, [h τ ] =, y Γ S 3, h n =, y S, 7 (3.1)

g 1 L 2 (, T ; W l 2( 1 3 )) W (l 1)/2 2 (, T ; W 1 2 ( 1 3 )), g 1 =, g 1 n =, y Γ S 3, g 2 W l 1,(l 1)/2 2 (Q i T ), i = 1, 2, 3, [µg 2 n] =, [g 2τ ] =, y Γ S 3, g 2 n =, y S the problem (3.3) has a unique solution h W 1+l,(1+l)/2 2 (Q i T ), satisfying the inequality (3.11) h W 1+l,(1+l)/2 2 (Q i T ) c( ( h W l 2 ( i ) + g 2 l 1,(l 1)/2 W 2 (Q i )) T + ( g 1 L2 (,T ;W2 l( i)),3 + g 1 W (l 1)/2 2 (,T ;W 1 2 ( i) )), (3.12) and Moreover, there exist e L 2 (, T ; W2 l( i)) W (l 1)/2 2 (, T ; W2 1( i)), i = 1, 2, 3, such that µ(h t g 2 ) = rote(ξ, t), ξ 1 2 3, roth = α(e + g 1 ). ξ 1 3, (3.13) e(ξ, t) =, ξ 2, [e τ ] =, ξ Γ S 3, e τ =, ξ S, ( e L2 (,T ;W l 2 ( i)) + e W (l 1)/2 2 (,T ;W 1 2 ( i)) ) c( ( h W l 2 ( i ) + g 2 l 1,(l 1)/2 W 2 (Q i )) T + ( g 1 L2 (,T ;W2 l( i)) + g 1 (l 1)/2 W 2 (,T ;W2 1( i) )).,2 (3.14) We describe briefly the (formal) construction of e. We decompose h, h, ψ, g 2 in the sums h = h + h, h = h + h, ψ = ψ + ψ, g 2 = g 2 + g 2, where h, h, ψ, g 2 U() and h, h, ψ, g 2 are orthogonal to U(). It is easily verified that in the case h = h, g 2 = g 2 the problem (3.3) reduces to µ(h t g 2) ψ dξdt =, 8

i.e., and in the case h = h, h t g 2 =, h t= = h, (3.15) g 2 = g 2 we have h = h [1]. Let E be the solution of the problem rote = µ(h t g 2), E =, ξ, E τ =, ξ S. (3.16) It is easily seen that T 1 3 ( E + α 1 roth g 1 ) rotψ (y, t)dξdt =, which implies E + α 1 roth g 1 = χ 1 (ξ, t), ξ 1 3, (3.17) in view of Theorem 4. We set e = E + χ 1 (ξ, t), ξ 1 3, b e = E + χ 2 (ξ, t) + C j (t)v k (ξ), ξ 2, k=1 (3.18) where χ 2 is a solution of the problem 2 χ 2 (ξ, t) =, ξ 2, χ 2 (ξ, t) = χ 1 (ξ, t), ξ Γ S 3, χ 2 (ξ, t) =, ξ S (3.19) and v k are the Dirichlet vector fields in 2, i.e., v k = φ(ξ, t), 2 φ k (ξ, t) =, ξ 2, φ k (ξ, t) = δ jk, ξ S j, φ k (ξ, t) =, ξ S, (3.2) S j are all the connected components of 2, except S, j, k = 1,..., b, b is the second Betti number of the domain 2. The coefficients C j (t) may be found from the additional normalization conditions S k e nds =, that gives = (E + χ 2 (ξ, t) + S k b C j (t)v k (ξ)) nds. (3.21) The estimate (3.14) follows from (3.21)and from the estimates of solutions of (3.15), (3.16), (3.19), (3.2). k=1 9

4 Nonlinear problem We go back to Sec. 2 and write our main nonlinear problem in the form u t ν 2 u + q = f(ξ, t) + l 1 (u, q), u = l 2 (u), ξ 1, t >, T (u, q)n = l 3 (u), ξ Γ, µh t ψ(y, t)dydt + α 1 roth rotψ(y, t)dydt 1 3 = l 4 (u, h) rotψ(y, t)dydt 1 3 + µl 5 (u, h) ψ(y, t)dydt, roth = l 6 (u, h), ξ 2, [µh n] =, [h τ ] = l 7 (u, h), ξ Γ, [µh n] =, [h τ ] =, ξ S 3, h n =, ξ S, u(ξ, ) = v (ξ), ξ 1, h(ξ, ) = H (ξ), ξ i, i = 1, 2, 3, (4.1) where l 1 = ν( 2 u 2 )u + ( u )q + u T M (Lh) f(x, t) + f(ξ, t), l 2 = ( u ) h = (I A T )h, l 3 = Π (Π S(u)n ΠS u (u)n)) + νn ((n S(u)n ) (n S u (u)n)) + n [n T M (Lh)n], l 4 = α 1 (roth ProtPh) + µ(l 1 u h) (n Ψ ), l 5 = Φ(u, h) + µ 1 rot(n Ψ ), l 6 = rot(i P)h, (4.2) l 7 = ( AT An An 2 n )[h n ], where n and Ψ are extensions of n, Ψ from Γ in 1, Πf = f n(n f), Π f = f n (n f). 1

We solve the problem (4.1) by successive approximations, according to the scheme u m+1,t ν 2 u m+1 + q m+1 = f(ξ, t) + l 1 (u m, q m ), u m+1 = l 2 (u m ), ξ 1, t >, T (u m+1, q m+1 )n = l 3 (u m, h m ), ξ Γ, µh m+1t ψ(y, t)dydt + α 1 roth m+1 rotψ(y, t)dydt 1 3 = l 4 (u m, h m ) rotψ(y, t)dydt 1 3 + µl 5 (u m, h m ) ψ(y, t)dydt, roth m+1 = l 6 (u m, h m ), ξ 2, [µh m+1 n] =, [h m+1,τ ] = l 7 (u m, h m ), ξ Γ, [µh m+1 n] =, [h m+1,τ ] =, ξ S 3, h m+1 n =, ξ S, u m+1 (ξ, ) = v (ξ), ξ 1, h m+1 (ξ, ) = H (ξ), ξ i, i = 1, 2, 3, (4.3) u =, q =, ρ =, h =. Using Theorems 3-5 and estimates of the nonlinear terms (they are omitted), we obtain where β >, Y m+1 (T ) c 1 T β Ym(t) j + c 2 N(T ), (4.4) j=1 Y m (T ) = u m l+2,l/2+1 W 2 (Q 1 T ) + q m l,l/2 W 2 (Q 1 T ) + q m l+1/2,l/2+1/4 W 2 (G T ) + h m l+1,l/2+1/2 W 2 (Q i ), T N(T ) = f l,l/2 W 2 (WT i ) + j L 2 (,T ;W2 l( 3)) + j (l 1)/2 W 2 (,T ;W2 1( 3)) + v W l+1 2 ( 1 ) + h W l 2 ( i ). It follows from (4.4) that in the case of small T Y m+1 (T ) 2c 2 N(T ). (4.5) 11

The proof of the convergence of the sequence (u m, q m, ρ m, h m ) is based on the estimate of the differences (u m+1 u m, q m+1 q m, ρ m+1 ρ m, h m+1 ρ m ), as in [2]. This yields the proof of Theorem 2. The reconstruction of e can be made in the same manner as in the linear case above, which completes the proof of Theorem 1. References 1. S.Mosconi, V.A.Solonnikov, On a problem of magnetohydrodynamics in a multiconnected domain, Nonlinear Anal., 74 No 2 (211), 462-478. 2. M.Padula, V.A.Solonnikov, On the free boundary problem of magnetohydrodynamics, Zapiski Nauchn. Semin. POMI 385 (21), 135-186. 3. G.Stroemer, About the equations of non-stationary magnetohydrodynamics, Nonlin. Anal. 52 (23), 1249-1273. 12