Appearance of Anomalous Singularities in. a Semilinear Parabolic Equation. (Tokyo Institute of Technology) with Shota Sato (Tohoku University)

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1 Appearance of Anomalous Singularities in a Semilinear Parabolic Equation Eiji Yanagida (Tokyo Institute of Technology) with Shota Sato (Tohoku University) 4th Euro-Japanese Workshop on Blow-up, September 6-10, 2010, Lorentz Center, Leiden University, Netherlands.

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3 We study singular solutions of u t = u + u p in R N. 1. Solution with a (moving) singularity 2. Appearance of an anomalous singularity: radial case 3. Appearance of an anomalous singularity: non-radial case 4. More general solutions

4 1. Singular solution We consider singular solutions of the Fujita equation (F) u t = u + u p, x R N, where p > 1. For N 3 and p > state N, (F) has a singular steady N 2 where ξ 0 R N is arbitrary and u = L x ξ 0 m, m := 2 p 1, L := { m(n m 2) } 1 p 1. The singularity of u = L x ξ 0 m persists for all t > 0, but it does not move in time.

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6 We define a solution with a moving singularity as follows. Definition 1. u(x, t) is a solution of (F) with a moving singularity at ξ(t) R N if the following conditions hold for some T (0, ]: (i) u, u p C([0, T ); L 1 loc (RN )) satisfy (F) in the distribution sense. (ii) u(x, t) is defined on R N \ {ξ(t)} [0, T ), and is C 2 with respect to x and C 1 with respect to t. (iii) u(x, t) as x ξ(t) for every t [0, T ).

7 Consider the initial value problem (P) { ut = u + u p, x R N \ {ξ(t)}, t > 0, u(x, 0) = u 0 (x) 0, x R N \ {ξ(0)}, where ξ(t) : [0, ) R N is prescribed. [Assumptions] (A1) (A2) N 3 and N N 2 < p < p := N + 2 N 1 N N 1. ξ(t) is sufficiently smooth. (A3) u 0 (x) is nonnegative and continuous in x R N \ ξ(0), and is uniformly bounded for x ξ(0) 1. (A4) u 0 (x) = Lr m + o(r m ) as r = x ξ(0) 0.

8 Known facts (Sato-Y, JDE 2009, DCDS 2010): (i) (Time-local existence) For some time interval [0, T ), there exists a solution u of (P) with a singularity at ξ(t) such that u(x, t) = Lr m + o(r m ) as r = x ξ(t) 0 for all t [0, T ). (ii) (Uniqueness) If u 1 and u 2 are two solutions of (P) such that u 1 (x, t) u 2 (x, t) = o(r λ 2 ) as r = x ξ(t) 0, then u 1 u 2. (iii) (Global existence) For some ξ(t) and u 0 (x), the solution exists for all t (0, ) and is asymptotically radially symmetric as t.

9 Why N N 2 < p < p? Assume that a solution u(x, t) with a singularity at ξ(t) is close to the singular steady state u = L x ξ(t) m, and formally expand the solution u(x, t) as follows: where u(x, t) = Lr m + [m] i=1 b i (ω, t)r m+i + v(y, t), m = 2 1, y = x ξ(t), r = y, ω = p 1 y y SN 1. Substitute this expansion into the equation and equate each power of r to obtain a system of equations for b i (ω, t).

10 These equations are solvable and the remainder term v(y, t) must satisfy v t = v + ξ t v + plp 1 y 2 v + o( y 2 ). This equation is well-posed if and only if 0 < pl p 1 < (N 2)2. 4 These inequalities hold if N > 2 and N N 2 < p < p = N + 2 N 1 N N 1. L := n 2 p 1 N 2 p 1 2 o 1 p 1.

11 Question If a solution is not global in time, what happens at the maximal existence time? Possibilities: Blow-up at x ξ(t). Appearance of an anomalous singularity: At some t = T <, the leading term of u at ξ(t) becomes different from Lr m : u(x, t) L x ξ(t) m for t (0, T ), u(x, t) L x ξ(t) m as t T.

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13 2. Appearance of anomalous singularities: radial case We consider a radially symmetric backward self-similar solution of (F) of the form u(x, t) = (T t) 1/(p 1) ϕ(r), r = (T t) 1/2 x, where ϕ satisfies ϕ rr + N 1 ϕ r r r 2 ϕ r m 2 ϕ + ϕp = 0, r > 0. With the condition lim r rm ϕ(r) = α > 0, we can show that there exists a unique solution ϕ = ϕ α (r). We note that ϕ α Lr m for α = L.

14 Lemma. Let N 3 and N N 2 < p < p. (i) For every α (0, ), ϕ α (r) is positive and strictly decreasing in r > 0. Moreover, ϕ α (r) = Lr m + o(r m ) as r 0. (ii) There exists α > L such that ϕ α (r) is strictly increasing in α (0, α ) for every r > 0. Proof. For (i), we use a (radial version) of the Pohozaev identity introduced by Giga-Kohn (1985). For (ii), we use the linearized equation at ϕ = Lr m and apply the Sturm Oscillation Theory.

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16 By this lemma, we obtain the following result. Theorem 1. self-similar solution Let N 3 and N N 2 < p < p. Then the backward u(x, t) = (T t) m/2 ϕ α ((T t) 1/2 x ) has the following properties: (i) The solution exists for t (0, T ) and has a singularity at the origin: (ii) The solution satisfies u(x, t) = L x m + o( x m ) as x 0. lim u(x, t) = t T α x m.

17 Proof. Since ϕ α (r) = Lr m + o(r m ) as r 0, αr m + o(r m ) as r, we have u(x, t) = (T t) m/2 ( ϕ α (T t) 1/2 x ) (T t) m/2 L { (T t) 1/2 x } m as x 0, = L x m whereas u(x, t) = (T t) m/2 ( ϕ α (T t) 1/2 x ) (T t) 1/(p 1) α { (T t) 1/2 x } m as t T, = α x m.

18 Remark. In fact, the backward self-similar solution can be expanded as follows: as x 0. u(x, t) = (T t) m/2 ϕ α ((T t) 1/2 x ) = L x m + C(α)(T t) (λ 2 m)/2 x λ 1 + h.o.t.

19 3. Appearance of anomalous singularities: non-radial case We consider a backward self-similar solution of the form u(x, t) = (T t) m/2 ϕ(z), z = (T t) 1/2 x a, where a 0 is fixed and (BSS) z ϕ z + a 2 z ϕ m 2 ϕ + ϕp = 0, z R N \ {0}.

20 N Theorem 2. Let N 3 and N 2 < p < p. Then for any α (0, α ) and ε > 0, there exists a constant δ > 0 such that if a < δ, (BSS) has a positive solution with the following properties: (i) ϕ(z) = L z m + o( z m ) as z 0. (ii) (α ε) z m < ϕ(z) < (α + ε) z m for large z. This theorem implies that the self-similar solution u(x, t) = (T t) m/2 ϕ((t t) 1/2 x a) has a moving singularity at x = ξ(t) := (T t) 1/2 a anomalous at t = T. which becomes

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22 [Outline of the proof of Theorem 2] First we consider the formal expansion of a solution ϕ(z) of (BSS) z ϕ z + a 2 z ϕ m 2 ϕ + ϕp = 0, z R N \ {0}. We expand ϕ(z) as where ϕ(z) = ϕ α (r) + [m] i=1 b i (ω)r m+i + v(z), r = z, ω = z r SN 1. Substituting this expansion into the equation and equating each power of r, {b i (ω)} are determined uniquely.

23 The remainder term v(z) must satisfy v z + a 2 v m 2 v + plp 1 z 2 v + o( z 2 ) = 0 in R N \ {0}, This equation has a solution v(z) if 0 < pl p 1 < (N 2)2. 4 These inequalities are satisfied if N N 2 < p < p

24 Taking into account of the formal analysis, we can show the existence of a solution as follows: Step 1 : Construct suitable comparison functions on R N \ {0} by modifying ϕ α. Step 2 : Construct a sequence of approximate solutions on annular domains D k := {z R N : 1 k < z < k}, k = 2, 3,... and find a convergent subsequence. Step 3 : Show that the limiting function is a desired solution.

25 4. More general solutions In Theorem 3, the singular point ξ(t) := (T t) 1/2 a moves along a straight line. We can extend the result to more general motion of a singular point. Theorem 3. Let N 3 and N N 2 < p < p. If ξ(t) satisfies ξ (j) (t) β(t + 1) j+ 1 2, j = 1,..., j0, for sufficiently small β > 0 and some j 0, there exists a solution with a singularity at ξ(t) that becomes anomalous at t = T, where T is arbitrarily prescribed. Proof. We modify the proof of Theorem 2.

26 Remark. In particular, Theorem 3 implies that the singular point ξ(t) can trace any prescribed smooth curve, though the speed becomes lower and lower with the rate ξ t = O(t 1/2 ).

27 Thank you for your attention!