A global solution curve for a class of free boundary value problems arising in plasma physics

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1 A global solution curve for a class of free boundary value problems arising in plasma physics Philip Korman epartment of Mathematical Sciences University of Cincinnati Cincinnati Ohio Abstract We study the existence and multiplicity of solutions and the global solution curve of the following free boundary value problem, arising in plasma physics, see R. Temam [8], and H. Berestycki and H. Brezis [3]: find a function u(x) and a constant b, satisfying u + g(x, u) = p(x) in u u = b,. Here R n, is a bounded domain, with a smooth boundary. This problem can be seen as a PE generalization of the periodic problem for one-dimensional pendulum-like equations. We use continuation techniques. Our approach is suitable for numerical computations. Key words: Free boundary value problem, global solution curve. AMS subject classification: 35J5, 78A30. Introduction We study the existence and multiplicity of solutions, and the global solution structure of the following free boundary value problem, arising in plasma physics, see R. Temam [8], H. Berestycki and H. Brezis [3], and also K.-C. Chang [6]. On a bounded domain R n, with a smooth boundary, we seek to find a function u(x) and a constant b, satisfying (.) u + g(x, u) = p(x) in u = b, u.

2 Here g(x, u) and p(x) are given functions, n denotes the unit normal vector on, pointing outside. More recently, this problem was studied in P. Amster, P.L. e Napoli, and M.C. Mariani [2]. As explained in [2], this problem can be seen as a PE generalization of the periodic problem for the one-dimensional pendulum-like equation: (.2) u + g(t, u) = p(t) u(0) = u(t) = b, u (0) = u (T). with p(t) being T-periodic, and g(t, u) is T-periodic in t. Indeed, writing the second boundary condition in (.2) as T 0 u dt = 0, we see that a similar condition for (.) is u dx = 0, which by the divergence theorem is equivalent to the second boundary condition in (.). There exists an enormous literature for the periodic problem (.2), see e.g., the review paper of J. Mawhin [4], and [4], [5], [8], [9], [0], [5], [6], [7]. By constrast, not much is known for the more general problem (.). We use continuation techniques to study the solution curves for the problem (.), similarly to our approach to the pendulum-like equations [0]. One can think of the problem (.) as being at resonance, i.e., some conditions on p(x) are necessary for its solvability. Indeed, decompose p(x) = µ 0 +θ(x), p(x) dx, and with µ 0 = (.), we get (.3) θ(x) dx = 0. Integrating the equation in g(x, u(x)) dx = µ 0, where is the volume of. Assume that the following limits exist: g(x, ± ) = lim g(x, u), uniformly in x, with g(x, ± ) L (), u ± and suppose that (.4) g(x, ) < g(x, u) < g(x, ) for all x. Then, from (.3) and (.4), we obtain g(x, ) dx < µ 0 < (.5) g(x, ) dx to be a necessary condition for existence of solutions. Similarly to the classical paper of E.M. Landesman and A.C. Lazer [3], we give conditions under which the condition (.5) is sufficient for existence. In order to use continuation techniques, we will embed the problem (.) into a family of problems (.6) u + kg(x, u) = p(x) in u = b, u, 2

3 with 0 k. At k =, one recovers the original problem (.), while at k = 0 the equation is linear. It turns out that at k = 0, the problem has infinitely many solutions, and one may fix a unique solution, by additionally u(x) dx ξ. One then performs continuation prescribing any value of in k on curves of fixed average. Namely, for any ξ R, one solves the following problem: find (u, b, µ), as a function of k, solving (.7) u + kg(x, u) = µ + θ(x) in u = b, u u(x) dx = ξ. To prove the solvability of (.7), we set up the appropriate function spaces, and show that the corresponding differential operator is one-to-one, and onto, so that the Implicit Function Theorem applies. Onto is the harder part. Once the continuation process is completed, and one has a solution of the problem (.7) at k =, it remains to show that one can select a value of ξ, at which µ = µ 0, thus giving us a solution of the original problem (.). This part is also accomplished by continuation. In addition to a result of Landesman-Lazer type, we obtain an existence result of.g. de Figueiredo and W.-M. Ni [7] type, which does not require that limits at infinity exist. We also give a condition for the existence of at least two solutions for the problem (.). We show that ξ, the average value of the solution, is a global parameter, and then we study the curve µ = µ(ξ ), yielding a multiplicity result. The continuation approach of this paper is similar to our paper on the pendulum-like equations, see [0]. In addition to its conceptual clarity, this approach is suitable for numerical computation of all solutions of (.). Every solution can be obtained by two continuations, first in k, 0 k, and then in ξ. These solutions curves are easy to compute numerically e.g., by the predictor-corrector method, since we prove that no turning points (or other singularities) are encountered. We had performed similar numerical computations in []. 2 Continuation of solutions of any fixed average We begin with the following linear problem (2.) u = b, u = θ(x) in u u(x) dx = ξ. 3

4 Lemma 2. Given any θ(x) L 2 (), with θ(x) dx = 0, and any ξ R, we can find a unique pair (u, b) W 2,2 () W,2 0 () R, solving (2.). Proof: Let v(x) W 2,2 () W,2 0 () be the solution of v = θ(x) in, v = 0 on. By the divergence theorem, v n ds = 0. Then u = v + b will give us solution of (2.), if we select the constant b, so that (v + b) dx = ξ. Uniqueness follows from the fact that any harmonic in function, satisfying ds = 0, is a constant. u n Corollary Assume in addition that θ(x) L (), and ξ = 0. Then there is a constant c, such that u L ( ) c. Proof: Using the W 2,p estimates and the Sobolev embedding theorem, we conclude an estimate for v L ( ), and then for b, giving us an estimate of u = v + b. The following Poincaré s inequality is well known. By c 0 we denote the best (largest) constant. Lemma 2.2 Assume that u(x) W,2 0 () is any function satisfying u(x) dx = 0. Then there is a constant c 0, depending only on, such that u 2 dx c 0 u 2 dx. We consider next the following linear problem: given the function a(x), find a triple (w(x), µ, b) W 2,2 () R R, satisfying (2.2) w + a(x)w = µ, in w w = b, w(x) dx = 0. Lemma 2.3 Assume that a(x) < c 0, for all x. Then the only solution of the problem (2.2) is w(x) 0, µ = 0, b = 0. 4

5 Proof: Multiply the equation in (2.2) by w, and integrate over. Since w w dx = w 2 w dx + b n ds = w 2 dx, we have by Poincaré s inequality c 0 w 2 dx > a(x)w 2 dx = w w dx = w 2 dx c 0 from which we conclude that w(x) 0, and then µ = 0, and b = 0. Theorem 2. Consider the problem w 2 dx, (2.3) u + kg(x, u) = µ + θ(x) in u = b, u u(x) dx = ξ, where θ(x) L 2 () is a given function. The function u(x) and the constants µ and b are unknown. Assume that g(x, u) C ( R) satisfies (2.4) g u (x, u) M, for all x, and u R, and (2.5) M < min(c 0, λ 2 ). Then for any 0 k, and ξ R, there exists a unique solution (u(x), b, µ) W 2,2 () R R, solving (2.3). Proof: We perform continuation in k. When k = 0, we take µ = 0, and obtain the unique solution (u(x), 0, b) by Lemma 2.. We consider two cases. Case I: ξ = 0. We shall recast the problem (2.3) in the operator form, after we define the appropriate spaces. We shall denote by W 2,2 () W,2 0 () + R the functions of the form v(x) + b, with v(x) W 2,2 () W,2 0 (), and b R. efine { } H 2 = u(x) W 2,2 () W,2 u 0 () + R u(x) dx = 0,. One checks that H 2 is a Banach space. Consider the map F(u, µ, k) = u + kg(x, u) µ : H 2 R R L 2 (). 5

6 Then the problem (2.3) can be written in the operator form (2.6) F(u, µ, k) = θ(x). Observe that the unknown constant b is now placed in the definition of the space H 2. We wish to apply the Implicit Function Theorem to perform continuation in k. For that we need to show that the linearized operator is one-to-one and onto. (2.7) F (u,µ) (u, µ, k)(w,µ ) = w + kg u (x, u)w µ It is one-to-one, because the only solution of the problem w + kg u (x, u)w µ = 0, in w w = b, w(x) dx = 0 is w(x) 0, µ = 0, in view of Lemma 2.3. Turning to the onto part, we need to show that given any θ(x) L 2 (), one can find w H 2, and µ R, solving (2.8) We consider two further cases. w + kg u (x, u)w = µ + θ(x), in w w = b, w(x) dx = 0. Case. The operator L : W 2,2 () W,2 0 () L 2 (), defined by Lw w + kg u (x, u)w, subject to w = 0 on is invertible. Let w = L (), i.e., w satisfies (2.9) w + kg u (x, u)w =, in, w = 0 on. We also consider z = L (kg u (x, u)), i.e., z satisfies (2.0) z + kg u (x, u)z = 0, in, z = on. We shall build the solution of (2.8), by using w and z. Multiplying (2.9) by z, subtracting the equation (2.0) multiplied by w, and integrating (2.) w z dx = n ds. 6

7 Sub-case i. The integrals in (2.) are both non-zero. We now construct the solution of (2.8) in the form w = µ (w + bz) + L (θ(x)) + b z, with the constants µ, b and b to be selected. Clearly, w satisfies the equation in (2.8), while w = b + b. Since z dx 0, we can select b and b, such that (w + bz) dx = (L (θ) + b z) dx = 0. It follows that w(x) dx = 0, for any µ. The function W w + bz satisfies W + kg u (x, u)w =, in W = b, W(x) dx = 0. By Lemma 2.3, it follows that W n ds 0. Hence, we can select µ, so that w. Sub-case ii. The integrals in (2.) are both zero. Since z(x) dx = 0, it follows by Lemma 2.3 that (2.2) z n ds 0. Similarly, since w n (2.3) ds = 0, we conclude by Lemma 2.3 that w (x) dx 0. We now construct the solution of (2.8) in the form w = µ w + L (θ(x)) + bz. Clearly, w satisfies the equation in (2.8), while w = b. By (2.3) we can choose µ, so that w(x) dx = 0, for all b, and then we choose b, so that ds = 0, by using (2.2). w n Case 2. The operator L is not invertible. 7

8 Since we have assumed kg u (x, u) < λ 2, it follows that the null space of L is one dimensional, spanned by some w, with w(x) > 0 on. By the Fredholm alternative, given any f(x) L 2 (), L (f(x)) is defined, i.e., the problem w + kg u (x, u)w = f(x), in, w = 0 on is solvable, if and only if f(x) w(x) dx = 0. We can choose a constant s, so that (s + θ(x)) w(x) dx = 0, which implies that L (s + θ(x)) is defined. Similarly, we select a constant t, so that L ( kg u (x, u) + t) is defined, and we set z = + L ( kg u (x, u) + t), i.e., z satisfies (2.4) z + kg u (x, u)z = t, in, z = on. Sub-case i. z dx 0. We construct the solution of (2.8) in the form w = L (s + θ(x)) + b z + a( w + b 2 z), with s fixed above, and the constants a, b and b 2 to be selected. We choose b and b 2, so that [ ] L (s + θ(x)) + b z dx = ( w + b 2 z) dx = 0. Then w(x) dx = 0, for any constant a. The function W(x) = w + b 2z satisfies W + kg u (x, u)w = b 2 t, in W = b 2, W(x) dx = 0. By Lemma 2.3, it follows that W n ds 0. We select a so that w n ds = 0. Then w satisfies (2.8). (Observe that Lw = s + (b + ab 2 )t + θ(x) µ + θ(x), w = b + ab 2 b.) Sub-case ii. z dx = 0. By Lemma 2.3, and (2.4), it follows that z n ds 0. Let w(x) = L (s + θ(x)) + a w + bz. Choose the constant a, so that [ ] L (s + θ(x)) + a w dx = 0. 8

9 Then, w(x) dx = 0, for any b. By Lemma 2.3, n z ds 0. Then choosing b, so that w n ds = 0, we conclude that w(x) satisfies (2.8). (Here Lw = s + bt + θ(x) µ + θ(x), w = b.) Case II: ξ 0. We reduce it to the case ξ = 0, by letting v = u ξ. Then v(x) satisfies (2.5) v + kg(x, v + ξ ) = µ + θ(x) in v v = b ξ, v(x) dx = 0. We perform the continuation in k exactly the same way as before, since the bound on g v (x, v + ξ ) remains the same. We conclude that each solution of (2.3) can be continued locally in k. To show that solutions can be continued for all 0 k, we need an a priori bound on (u, b, µ)(k). Write u(x) = ξ + U(x), with U(x) dx = 0. We have (2.6) U + kg(x, ξ + U) = µ + θ(x) in U = b ξ, U. By our conditions (2.4) and (2.5), we can find positive constants c and c 2, with c < c 0, such that (2.7) g(x, ξ + U) c ξ + U + c 2, for all x, and U R. Multiplying (2.6) by U, and using the Lemma 2.2, we conclude a bound on U L 2 () (keep in mind that ξ is arbitrary, but fixed). Using this bound in (2.6), and the estimate (2.7), we obtain a bound on U W 2,2 (). We conclude that (2.8) u W 2,2 () c 3 ξ + c 4 c 5. Integrating the equation (2.6) over, using (2.7) and the bound on U L 2 (), we obtain a bound on µ. Writing u = b + V, and proceeding similarly, we conclude an L 2 bound on V, which implies a bound on b, since u L 2 () is bounded. This theorem implies that for any ξ we have a curve of solutions (u, b, µ)(k) W 2,2 () R R solving (2.3), and at k =, we have a solution of (2.9) u + g(x, u) = µ + θ(x) in u = b, n u ds = 0 u(x) dx = ξ. 9

10 Recall that the original problem (.), after we decomposed p(x) = µ 0 +θ(x), is (2.20) u + g(x, u) = µ 0 + θ in u = b, u. Hence, it remains to show that one can choose ξ, so that µ = µ 0 in (2.9). The corresponding u(x) (from (2.9)) is then a solution of (2.20). Remark If the domain is an interval in one dimension, then c 0 = λ 2. Indeed, suppose = ( L, L). Represent u(x) W,2 0 ( L, L) by its Fourier series u(x) = a 0 + Σ n=a n cos nπ L x + b n sin nπ L x. If L L u(x) dx = 0, then a 0 = 0, and hence L L u 2 (x) dx π 2 L 2 L L u 2 (x) dx = 0, i.e., c 0 = π2 L 2. The irichlet eigenvalues are λ n = n2 π 2 (2L) 2, so that c 0 = λ 2. 3 Continuation in ξ We have just seen that for each ξ the problem (3.) u + g(x, u) = µ + θ(x) in u = b, u has a solution (u, b, µ) W 2,2 () R R, with the average value of u(x) equal to ξ. We now show that all of these solutions lie on a unique continuous solution curve, which is globally parameterized by ξ. Theorem 3. Assume that the conditions of the Theorem 2. hold. Then any solution of (3.) can be continued in ξ, for all < ξ <, giving us a curve of solutions (u, b, µ)(ξ ) W 2,2 () R R. All solutions of the problem (3.) lie on this continuous solution curve. Moreover, for each ξ there exists a unique solution (u, b, µ) of (3.), i.e., ξ is a global parameter. Proof: We use the Implicit Function Theorem to show that any solution of (3.) can be continued in ξ. The proof is essentially the same, as the 0

11 one above for continuation in k. We let u = ξ + v, with v(x) dx = 0. efining the map F(v, µ, ξ ) = v + g(x, v + ξ ) µ : H 2 R R L 2 (), we recast the problem (3.) in the operator form (3.2) F(v, µ, ξ ) = θ(x). Similarly to the Theorem 2., we apply the Implicit Function Theorem to perform local continuation in ξ, by showing that the linearized operator F (v,µ) (u, µ, ξ )(w, µ ) = w + g u (x, v + ξ )w µ is one-to-one and onto. By the estimate (2.8), v W 2,2 () does not grow more than linearly in ξ, so that we can continue the solutions for all < ξ <. Turning to the uniqueness, assume that there are two different triples of solutions (u (x), b, µ ) and (u 2 (x), b 2, µ 2 ), with u (x) and u 2 (x) having the same average value ξ 0. By the Theorem 2., we can continue both triples for decreasing k on two curves of solutions, with fixed average ξ. 0 These curves do not intersect, since at all points the Implicit Function Theorem applies. At k = 0, we obtain a contradiction with the uniqueness part of Lemma Existence and multiplicity of solutions We now discuss which values of µ are covered in (3.), as we continue solutions in ξ (, ), which translates into existence results for our problem (.). In this section we shall assume g(x, u) to be bounded, which will provide us with uniform in ξ estimates of solutions in (3.). We have the following result of Landesman-Lazer type. Theorem 4. Assume that the function g(x, u) L ( R) satisfies the conditions (.4), (2.4) and (2.5). Assume also that θ(x) L (). Then the condition (.5) is necessary and sufficient for the existence of solution for the problem (.). Proof: We proved above that the condition (.5) is necessary for solvability, we now prove its sufficiency. By the Theorem 2., for any ξ R, there

12 exists a triple (u, b, µ) = (u, b, µ)(ξ ) solving the problem (.), with the average of u(x) equal to ξ. As before, we write u = ξ +U, with U = U(ξ ) satisfying U dx = 0. By the Theorem 3., µ = µ(ξ ) is continuous in ξ, and integrating the equation (3.), we express (4.) µ = g (x, ξ + U(ξ )) dx. We need to show that we can choose ξ = ξ, 0 so that µ(ξ) 0 = µ 0. (Recall that in (.) we set p(x) = µ 0 + θ(x).) This will follow, once we prove the existence of the limits (4.2) lim g (x, ξ + U(ξ )) dx = g (x, ) dx, ξ (4.3) lim ξ g (x, ξ + U(ξ )) dx = g (x, ) dx, since µ 0 is assumed to lie in between these limits. In view of (4.), U = U(ξ ) satisfies U = g (x, ξ + U) + g (x, ξ + U) dx + θ(x), in U U = b, U(x) dx = 0, with b = b ξ, a new unknown constant. Since g(x, u) is bounded, by Corollary to Lemma 2., there is a constant c, so that U(ξ ) L ( ) c, uniformly in ξ R, from which the limits (4.2) and (4.3) follow. The following result is similar to the one in.g. de Figueiredo and W.-M. Ni [7], who considered a irichlet problem. Theorem 4.2 Consider the problem (4.4) u + g(x, u) = θ(x) in u = b, n u ds = 0, where θ(x) L () is a given function, satisfying θ(x) dx = 0. Assume that the function g(x, u) L ( R) satisfies the conditions (2.4) and (2.5), and in addition assume there is a constant G > 0 such that (4.5) ug(x, u) > 0, for all u > G, and x. Then the problem (4.4) has a solution (u, b) W 2,p () R, for any p >. 2

13 Proof: The proof is similar to the one above. Here we have µ 0 = 0. Using (4.5), we see from (4.) that µ > 0 (< 0) for ξ large and positive (negative). By continuity, we have µ(ξ 0) = 0 for some ξ0. We also have the following multiplicity result. Theorem 4.3 Consider again the problem (4.4), and assume that the conditions of the Theorem 4.2 hold. Assume, in addition, that the limits g(x, ± ) = lim u ± g(x, u) exist, uniformly in x, and (4.6) g(x, ± ) 0. Then there exists an interval (µ, µ + ), with µ < 0 and µ + > 0, so that for µ (µ, µ + ) \ {0} the problem (4.4) has at least two solutions (u, b ) and (u 2, b 2 ). Proof: By the Theorem 2., for any ξ R, there exists a triple (u, b, µ) = (u, b, µ)(ξ ) solving the problem (4.4), with average of u(x) equal to ξ. As in Theorem 4., we continue this solution in ξ, paying particular attention to the curve µ = µ(ξ ). From the formula (4.) and the condition (4.5) it follows that µ(ξ ) is positive (negative) for ξ large and positive (negative). The condition (4.6) implies that lim ξ µ(ξ ) = 0, i.e., both sides of the curve µ = µ(ξ ) tend to zero. Hence, we can find some intervals (µ, 0) and (0, µ + ), which are covered at least twice by the function µ = µ(ξ ). References [] A. Ambrosetti and G. Prodi, On the inversion of some differentiable mappings with singularities between Banach spaces, Ann. Mat. Pura Appl. 93 (4), (972). [2] P. Amster, P.L. e Napoli, and M.C. Mariani, Existence of solutions to N-dimensional pendulum-like equations, Electron. J. ifferential Equations, No. 25, 8 pp. (electronic) (2004). [3] H. Berestycki and H. Brezis, On a free boundary problem arising in plasma physics, Nonlinear Anal. 4, no. 3, (980). [4] A. Castro, Periodic solutions of the forced pendulum equation, ifferential equations (Proc. Eighth Fall Conf., Oklahoma State Univ., Stillwater, Okla., 979), pp , Academic Press, New York-London- Toronto, Ont.,

14 [5] J. Cepicka, P. rabek and J. Jensikova, On the stability of periodic solutions of the damped pendulum equation, J. Math. Anal. Appl. 209, (997). [6] K.-C. Chang, Remarks on free boundary problems for the flux equations in plasma physics, Comm. Partial ifferential Equations 5, no. 7, (980). [7].G. de Figueiredo and W.-M. Ni, Perturbations of second order linear elliptic problems by nonlinearities without Landesman-Lazer condition, Nonlinear Anal. 3 no. 5, (979). [8] G. Fournier and J. Mawhin, On periodic solutions of forced pendulumlike equations, J. ifferential Equations 60, no. 3, (985). [9] G. Katriel, Periodic solutions of the forced pendulum: exchange of stability and bifurcations, J. ifferential Equations 82, no., -50 (2002). [0] P. Korman, A global solution curve for a class of periodic problems, including the pendulum equation, Z. Angew. Math. Phys, (ZAMP) 58, (2007). [] P. Korman, Global solution curves for boundary value problems, with linear part at resonance, Nonlinear Anal. 7, no. 7-8, (2009). [2] P. Korman, Curves of equiharmonic solutions, and ranges of nonlinear equations, Adv. ifferential Equations 4, no. 9-0, (2009). [3] E.M. Landesman and A.C. Lazer, Nonlinear perturbations of linear elliptic boundary value problems at resonance, J. Math. Mech. 9, (970). [4] J. Mawhin, Seventy-five years of global analysis around the forced pendulum equation, In: R. P. Agarwal and F. Neuman and J. Vosmansky (eds.): Proceedings of Equadiff 9, Masaryk University, Brno, pp (997). [5] R. Ortega, Stability and index of periodic solutions of an equation of uffing type, Boll. Un. Mat. Ital. B (7), no. 3, (989). [6] R. Ortega, Topological degree and stability of periodic solutions for certain differential equations, J. London Math. Soc. (2) 42, no. 3, (990). [7] G. Tarantello, On the number of solutions of the forced pendulum equations, J. ifferential Equations 80, (989). [8] R. Temam, A nonlinear eigenvalue problem: Equilibrium shape of a confined plasma, Arch. Rat. Mech. Anal. 60, no., 5-73 (975/76). 4