SEMIGROUP APPROACH FOR PARTIAL DIFFERENTIAL EQUATIONS OF EVOLUTION

SEMIGROUP APPROACH FOR PARTIAL DIFFERENTIAL EQUATIONS OF EVOLUTION Istanbul Kemerburgaz University Istanbul Analysis Seminars 24 October 2014 Sabanc University Karaköy Communication Center

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u(x, t)

u(x, t) (i) Does it exist?

u(x, t) (i) Does it exist? (ii) Is it unique?

u(x, t) (i) Does it exist? (ii) Is it unique? (iii) Does it depend continuously on the initial data?

u(x, t) (i) Does it exist? (ii) Is it unique? (iii) Does it depend continuously on the initial data? Well-posed?

Solutions Classical solution: A solution of a "PDE of order k" which is at least k times continuously dierentiable.

Solutions Classical solution: A solution of a "PDE of order k" which is at least k times continuously dierentiable. Weak solution!!

Solutions Classical solution: A solution of a "PDE of order k" which is at least k times continuously dierentiable. Weak solution!! Sobolev space : Hilbert space which involves weak solutions H k (R).

Simple example du dt = f (u), u(0) = u 0 (1) Two Banach spaces : X and Y, with Y X.

Simple example du dt = f (u), u(0) = u 0 (1) Two Banach spaces : X and Y, with Y X. f : continuous from Y to X.

Simple example du dt = f (u), u(0) = u 0 (1) Two Banach spaces : X and Y, with Y X. f : continuous from Y to X. Assume that for all u 0, there exist a real number T > 0 and a unique solution u C([0, T ], Y ) satisfying equation (1).

Simple example du dt = f (u), u(0) = u 0 (1) Two Banach spaces : X and Y, with Y X. f : continuous from Y to X. Assume that for all u 0, there exist a real number T > 0 and a unique solution u C([0, T ], Y ) satisfying equation (1). du dt C([0, T ], X )).

Simple example du dt = f (u), u(0) = u 0 (1) Two Banach spaces : X and Y, with Y X. f : continuous from Y to X. Assume that for all u 0, there exist a real number T > 0 and a unique solution u C([0, T ], Y ) satisfying equation (1). du dt C([0, T ], X )). Assume that u 0 u is continuous (Y C([0, T ], Y ). locally well-posedness in Y. Very strong!!

Linear Homogeneous equation: u (t) + Au(t) = 0, 0 t T, u(0) = u 0

Linear u 0, solution u exists and is unique: Here, e ta = (ta) n n=0 n!. u(t) = e ta u 0

Linear Nonhomogeneous equation: u (t) + Au(t) = f (t), 0 t T, u(0) = u 0

Linear Lions, Yosida, Pazy... KATO!!! t u(t) = e ta u 0 + e (t s)a f (s)ds 0

Denitions Denition Let X be a Banach space. If the mapping T (t) : R + L(X ) satises the following properties, then it is a C 0 semigroup: (i) T (0) = I (ii) t, s 0, T (t + s) = T (t)t (s) (iii) lim t 0 T (t)x = x, x X.

Denition Denition Let {T (t)} t 0 be a C 0 semigroup. T (t)x x If the limit Ax := lim t 0 is dened, then the innitesimal t generator of T (t) is A. Domain of A is denoted by D(A): T (t)x x D(A) := {x X : lim t 0 t limit is in X.}

Cauchy Problem u (t) = Au(t), u(0) = u 0,

Cauchy Problem u (t) = Au(t), u(0) = u 0, (i) If a continuously dierentiable function u satises u(t) D(A) for all t 0 and satises the equation, then it is a classical solution,

Cauchy Problem u (t) = Au(t), u(0) = u 0, (i) If a continuously dierentiable function u satises u(t) D(A) for all t 0 and satises the equation, then it is a classical solution, (ii) If for a continuous function u, t u(s)ds D(A) and 0 t A u(s)ds = u(t) x, then it is a mild solution. 0

Idea u (t) + A(u, t)u = 0, u(0) = u 0 u(t) = e ta u 0 u(t) = T (t)u 0

Idea u (t) + A(u, t)u = 0, u(0) = u 0 u(t) = e ta u 0 u(t) = T (t)u 0 u (t) + A(ω, t)u = 0, u(0) = u 0 u(t) = T ω (t)u 0

Idea u (t) + A(u, t)u = 0, u(0) = u 0 u(t) = e ta u 0 u(t) = T (t)u 0 u (t) + A(ω, t)u = 0, u(0) = u 0 u(t) = T ω (t)u 0 ω Φ(ω) = T ω (t)u 0 = u Φ(ω) contraction mapping unique solution u (Banach xed point theorem)

Notation B(X ; Y ): Set of all bounded linear operators on X to Y G(X ): All negative generators of C 0 semigroups on X G(X, µ, β): A generates a C 0 semigroup {e ta } with e ta µe βt, 0 t <

Notation B(X ; Y ): Set of all bounded linear operators on X to Y G(X ): All negative generators of C 0 semigroups on X G(X, µ, β): A generates a C 0 semigroup {e ta } with e ta µe βt, 0 t < A G(X, 1, β) quasi-m-accretive

Heat equation u t κu xx = 0, u(x, 0) = u 0 (x), x R

Heat equation u t κu xx = 0, u(x, 0) = u 0 (x), x R Solution by classical method: u(x, t) = 1 4πκt e (x y)2 4κt u 0 (y)dy

Heat equation u t κu xx = 0, u(x, 0) = u 0 (x), x R Solution by classical method: By new notation, u(x, t) = 1 4πκt e (x y)2 4κt u 0 (y)dy u t + Au = 0, u(0) = u 0 where A = κd 2 and T (t)u x 0 = 1 (e x2 4κt 4πκt u 0 )(x).

Heat equation u t κu xx = 0, u(x, 0) = u 0 (x), x R Solution by classical method: By new notation, u(x, t) = 1 4πκt e (x y)2 4κt u 0 (y)dy u t + Au = 0, u(0) = u 0 where A = κd 2 and T (t)u x 0 = 1 (e x2 4κt 4πκt For all t, u(t) u 0 κd 2 x G(L 2, 1, 0). u 0 )(x).

General approach Evolution equation depending on time variable t and spatial variable x

General approach Evolution equation depending on time variable t and spatial variable x Ordinary dierential equation depending on time variable t in an innite dimensional Banach space

General approach Evolution equation depending on time variable t and spatial variable x Ordinary dierential equation depending on time variable t in an innite dimensional Banach space Choose an appropriate Banach space. Choose the corresponding operator A. Spatial derivatives and other restrictions are captured by either the Banach space or the operator.

Quasi-linear equations u t + A(u)u = f (u), t 0, u(0) = u 0. (2) X and Y Banach spaces Y X continuous and densely injected S : Y X topological isomorphism

Assumptions (A1) For given C > 0 and for all y Y satisfying y Y C, A(y) is quasi-m-accretive. In other words, A(y) generates a quasi-contractive C 0 semigroup {T (t)} t 0 in X satisfying T (t) e wt.

Assumptions (A1) For given C > 0 and for all y Y satisfying y Y C, A(y) is quasi-m-accretive. In other words, A(y) generates a quasi-contractive C 0 semigroup {T (t)} t 0 in X satisfying T (t) e wt. (A2) For all y Y, A(y) is a bounded linear operator from Y to X and (A(y) A(z))w X c 1 y z X w Y, y, z, w Y.

Assumptions (A3) SA(y)S 1 = A(y) + B(y), where B(y) L(X ) is uniformly bounded on {y Y : y Y M} and (B(y) B(z))w X c 2 y z Y w X, w X.

Assumptions (A3) SA(y)S 1 = A(y) + B(y), where B(y) L(X ) is uniformly bounded on {y Y : y Y M} and (B(y) B(z))w X c 2 y z Y w X, w X. (A4) The function f is bounded on bounded subsets {y Y : y Y M} of Y for each M > 0, and is Lipschitz in X and Y : and f (y) f (z) X c 3 y z X, y, z X f (y) f (z) Y c 4 y z Y, y, z Y. Here, c 1, c 2, c 3 and c 4 are constants depending only on max{ y Y, z Y }.

Conclusion 1 Theorem Assume (A1), (A2), (A3), (A4) hold. Given u 0 Y, there is a maximal T > 0, depending on u 0 and a unique solution u to (2) such that u = (u 0,.) C([0, T ), Y ) C 1 ([0, T ), X ). Moreover, the map u 0 u(u 0,.) is continuous from Y to C([0, T ), Y ) C 1 ([0, T ), X ). 1 T. Kato, Quasi-linear equations of evolution, with applications to partial dierential equations, in: Spectral theory and dierential equations, Lecture Notes in Math. 448, Springer-Verlag, Berlin, 1975, pp. 25 70.

Quasilinear wave equation u t + a(u)u x = b(u)u X = L 2 (R), Y = H 2 (R), S = 1 D 2 x A(u, t) = A(u) = a(u) x f (u, t) = b(u)u B(u) = [a (u)u xx + a (u)ux] S 2 1 2a (u)u x 2 S 1

Quasilinear wave equation u t + a(u)u x = b(u)u X = L 2 (R), Y = H 2 (R), S = 1 D 2 x A(u, t) = A(u) = a(u) x f (u, t) = b(u)u B(u) = [a (u)u xx + a (u)u 2 x] S 1 2a (u)u x 2 S 1 Alternatively, X = L 2 (R), Y = H s (R), s 2, S = (1 D 2 x ) s/2 or S = (1 D x ) s Gain regularity!!

Model equation u t + u x + 3 2 εuu x 3 8 ε2 u 2 u x + 3 16 ε3 u 3 u x + µ 12 (u xxx u xxt ) + 7ε 24 µ(uu xxx + 2u x u xx ) = 0, x R, t > 0. (3) Here, u(x, t) denotes free surface elevation, ε and µ represent the amplitude and shallowness parameters, respectively.

Quasilinear equation Rewrite equation (3) in the following way and dene the periodic Cauchy problem: where u t + A(u)u = f (u) x R, t > 0, (4) u(x, 0) = u 0 (x) x R, (5) u(x, t) = u(x + 1, t) x R, t > 0, (6) A(u) = (1 + 7 2 εu) x, (7) f (u) = (1 µ 12 2 x ) 1 x [2u + 5 2 εu2 1 8 ε2 u 3 + 3 64 ε3 u 4 7 48 εµu2 x]. (8)

Setting (i) X = L 2 [0, 1], (ii) Y = H s per, s > 3/2, (iii) S = Λ s, Λ = (1 2 x ) 1/2.

Assumption 1 If u H s per, s > 3/2, then the operator A(u) = (1 + u) x with the following domain, D(A) = {ω L 2 [0, 1] : (1 + u) x ω L 2 [0, 1]} L 2 [0, 1] is quasi-m-accretive.

Assumption 1 If (a) For all w D(A), there exists a real number β such that (Aw, w) X β w 2 X, (b) The range of A + λi is all of X for some (or all) λ > β, then we are done.

Assumption 2 For every ω H s per, s > 3/2, A(u) is bounded linear operator from H s to per L2 [0, 1] and (A(u) A(v))ω L 2 [0,1] c 1 u v L 2 [0,1] ω H s per.

Assumption 3 The operator B(u) = SA(u)S 1 A(u) = Λ s (u x + x )Λ s (u x + x ) = [Λ s, u x + x ]Λ s is bounded in L 2 [0, 1] for u H s per with s > 3/2.

Assumption 4 Let f (u) be the function (8). Then: (i) f is bounded on bounded subsets {u H s per of H s per for each M > 0. (ii) f (u) f (v) L 2 [0,1] c 3 u v L 2 [0,1]. (iii) f (u) f (v) H s per c 4 u v H s per, s > 3/2. : u H s per M}

Local well-posedness Given u 0 H s, s > 3 per, there exists a unique solution u to (3)-(6) 2 depending on u 0 as follows: u = u(u 0,.) C([0, T ), H s per) C 1 ([0, T ), L 2 [0, 1]). Moreover, the mapping u 0 H s per u(u 0,.) is continuous from H s to per C([0, T ), Hper) s C 1 ([0, T ), L 2 [0, 1]).

N. Duruk Mutluba³, Local well-posedness and wave breaking results for periodic solutions of a shallow water equation for waves of moderate amplitude, Nonlinear Anal. 97, 2014, 145-154. T. Kato, Quasi-linear equations of evolution, with applications to partial dierential equations, in:spectral theory and dierential equations, Lecture Notes in Math. 448, Springer-Verlag, Berlin, 1975, pp. 2570. A. Pazy, Semigroups of Linear Operators and Applications to Partial Dierential Equations, Applied Mathematical Sciences 44, Springer-Verlag, New York, U.S.A., 1983.