Blow-up at the Boundary for Degenerate Semilinear Parabolic Equations

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1 Blow-up at the Boundary for Degenerate Semilinear Parabolic Equations M. S. Floater Department of Mathematics Heriot-Watt University Riccarton Edinburgh EH14 7AS February 1989 Abstract. This paper concerns a superlinear parabolic equation, degenerate in the time derivative. It is shown that the solution may blow up in finite time. Moreover it is proved that for a large class of initial data blow-up occurs at the boundary of the domain when the nonlinearity is no worse than quadratic. Various estimates are obtained which determine the asymptotic behaviour near the blow-up. The mathematical analysis is then extended to equations with other degeneracies. 1. Introduction Consider the following semilinear problem xu t = u xx + u p in Ω (, ) (1.1) u(,t) = u(1,t) = t > (1.2) u(x,) = u (x) x Ω (1.3) where Ω is the unit interval (,1) and p > 1. The initial data u is assumed to be in C 1 ( Ω) with u () = u (1) =. Since the coefficient of u t is zero at x = we can regard (1.1) as degenerate. In this paper we investigate the effect of the degeneracy on the blow-up behaviour of the solution of (1.1) (1.3). It is shown that if u is large enough, sup x Ω u tends to infinity in finite time. More significantly, if 1 < p 2 and u is concave (this can be weakened to d dx (u (x)/x) ) u blows up at the boundary x =. Numerical evidence, see Floater (1988) and Stuart & Floater (1989), indicates that blow-up is away from the boundary when p > 2. The analytical results are generalised to x q u t = u xx + u p (1.4) for any q >. Blow-up at the boundary can occur when 1 < p q + 1. The motivation for studying (1.1) comes from the paper by Ockendon (1979). A model is derived for the dynamics of a fluid whose viscosity is temperature dependent in a channel. By making a reasonable approximation the equation xu t = u xx + e u 1

2 is found where u represents the temperature of the fluid, see Ockendon (1979) and Lacey (1984). The boundary x = corresponds to one boundary of the channel (the other boundary of the channel is disregarded). So that analytical results may be obtained we have approximated e u by u p. We begin in Section 2 by constructing a unique classical solution to (1.1) (1.3) in some time interval (,t ). Further, due to a continuation theorem, u can be continued for all t > or until blow-up. In Section 3 we show that if u is large enough, u blows up at some finite time T say. Now we call x Ω a blow-up point if there exists a sequence (x n,t n ) Ω (,T) such that t n T, x n x and u(x n,t n ) as n. The blow-up set S Ω is defined as the set of all blow-up points. Theorem 3.5 states that if 1 < p 2 and d dx (u (x)/x) then S = {}. This contrasts with a result by Friedman & McLeod (1985). Using the reflection principle they showed that for the solution of v t = v xx + v p with the same conditions (1.2) (1.3), the blow-up set S v is a compact subset of Ω. This implies that S v is bounded away from the boundary Ω. Under the assumptions of Theorem 3.5 the asymptotics of u as t approaches T are analysed in Section 4. It is demonstrated that and if γ < (p 1)/2, if γ > (p 1)/2 and u + u p, u γ (x,t)dt C γ for all t (,T) (1.5) u γ (x,t)dt as t T. (1.6) We also obtain information concerning the rate at which the maximum of u tends to the boundary. Letting s(t) be any point such that u(s(t),t) = sup x Ω u(x,t), we find C 1 (T t) 1/3 s(t) C 2 (T t) 1/(3+δ) for any δ >. (1.7) In the case p = 2, these estimates are in agreement with Lacey (1984). Using formal methods he conjectured that u(x,t) K 1 lnx x 2 as x and (assuming s(t) is unique) s(t) K 2 { (T t) ln(t t)} 1/3 as t T. Finally, in Section 5 the analytical results are extended to the more general equation x q u t = u xx + u p. Interestingly, the integral bounds (1.5), (1.6) are unchanged, i.e. independent of q. Meanwhile the bounds on s(t) in (1.7) extend to C 1 (T t) 1/(q+2) s(t) C 2 (T t) 1/(q+2+δ) for any δ > and are independent of p. We also make some remarks on the likeness of u t = u xx + up x q. 2

3 2. Existence and uniqueness of solution In this section it is demonstrated that a classical solution to (1.1) (1.3) exists. Uniqueness follows easily from the maximum principle. Our approach to existence is to define the new problem xu t = u xx + u p in (ǫ,1) (, ) (2.1) u(ǫ,t) = u(1,t) = t > (2.2) u(x,) = u (x) x (ǫ,1) (2.3) The initial data u : (ǫ,1) IR is simply the truncation of the initial data in (1.3). Even though the compatibility condition u (ǫ) = does not hold, it is well known that a C 2,1 solution u ǫ of (2.1) (2.3) exists on some time interval (,t ), see Friedman (1964). The solution u ǫ is continuous in (ǫ,1] [,t ) and [ǫ,1] (,t ). The idea in proving existence for (1.1) (1.3) is that lim ǫ u ǫ (x,t) = u(x,t) is a good candidate for the solution. The remainder of Section 2 is devoted to proving that u is indeed the classical solution in Ω (,t ) and can be continued in t until either blow-up or infinity. Lemma 2.1. Let 1 > ǫ 1 > ǫ 2 > and suppose u ǫ1 and u ǫ2 are solutions of (2.1) (2.3) on (,t ). Then u ǫ2 u ǫ1 for all x (ǫ 1,1), t (,t ). Proof. Define w = u ǫ2 u ǫ1 in (ǫ 1,1) (,t ). Then w t 1 x w xx = 1 x {f(u ǫ 2 ) f(u ǫ1 )} = f (k) x w for some k between u ǫ1 and u ǫ2. By the maximum principle, u ǫ1 and u ǫ2 are positive in their respective domains. In particular u ǫ2 > when x = ǫ 1. Therefore w > at x = ǫ 1. Further, w = at x = 1 and w = at t =. Hence by the maximum principle, w > in (ǫ 1,1) (,t ) as required. Due to the monotonicity of the u ǫ solutions, the pointwise limit of {u ǫ } ǫ> as ǫ would be well defined were it not that t depends on ǫ and {u ǫ } ǫ> may tend to infinity at some point (x,t). In this context it is interesting to study xu t =u p, u(x,) = sinπx. (2.4) The solution can be found explicitly as u(x,t) = β β x β /(T(x) t) β where T(x) = x/(p 1)u p 1 (x). For each x (,1), there is an associated blow-up time T(x). Now observe that since u (x) πx as x, T(x) as x when p < 2. Therefore there cannot be a solution of (2.4) in C((,1) (,t )) for any t >. We may conclude from this that for 1 < p < 2 the diffusion term u xx in (1.1) (1.3) plays a vital role in smoothing out the solution. The next lemma shows that diffusion really does ensure existence. 3

4 Lemma 2.2. There exists t and an a priori bound h C 2,1 ([,1] [,t ]) depending only on u and p such that ǫ > there exists a solution u ǫ of (2.1) (2.3) in (ǫ,1) (,t ) with u ǫ h. Proof. The idea in constructing h is to split the domain Ω into three parts (,δ), (δ, 1 δ), and (1 δ, 1). To show h is an upper solution for (1.1) (1.3) we use the diffusion term h xx in (,δ), (1 δ, 1) and the term xh t in (δ, 1 δ) to overcome the nonlinearity h p. Consider the following definitions: (i) Let ψ(x) = x(1 x)/2. (ii) Choose k s.t. k ψ(x) u (x). (iii) Choose δ (, 1 2 ) s.t. δ (2k ) (p 1)/p. (iv) Let k(t) solve k() = k and k = k p /4 p 1 δ. (v) Set h(x, t) = k(t)ψ(x). (vi) Define t by k(t ) = δ p/(p 1). We claim that h is an upper solution for (1.1) (1.3) in Ω (,t ). Substituting h into (1.1) gives J = xh t h xx h p = xk ψ + k k p ψ p. For x (,δ), t (,t ), J k k p ψ p k(1 k p 1 {δ(1 δ)/2} p ) k(1 k p 1 δ p ) since k p 1 δ p for t t. Exactly the same calculation holds for x (1 δ, 1). For x (δ, 1 δ), t (,t ), J xk ψ k p ψ p ψ(δk k p /4 p 1 ) = by definition of k. Finally let ǫ >. Since h is an upper solution for (1.1) (1.3) in (,1) (,t ) it is also an upper solution for (2.1) (2.3) in (ǫ,1) (,t ) (h is positive at x = ǫ). Hence the solution u ǫ of (2.1) (2.3) exists in (ǫ,1) (,t ) and u ǫ h, see Sattinger (1972). This ends the proof. Due to Lemmas 2.1 and 2.2 a function u can be constructed as u(x,t) = { limǫ u ǫ (x,t) for x (,1], t [,t ); for t [,t ). (2.5) We now show that u is a classical solution of (1.1) (1.3). 4

5 Theorem 2.3. Under the assumption that u C 2+α (,1), some α (,1), u as given in (2.5) is a classical solution of (1.1) (1.3) in Ω (,t ). Proof. It is required to prove that u is C 2,1 in (,1) (,t ) and continuous in [,1] [,t ). Choose a point (x 1,t 1 ) (,1) (,t ). Then select a domain Q = (a,b) (,t 2 ) s.t. < a < x 1 < b < 1 and < t 1 < t 2 < t. Now since u ǫ u pointwise in Q, u ǫ u in any L q (Q). Further, as u ǫ is uniformly bounded by h in Q, the L q estimate of LSU (1968) yields (u ǫ ) q + (u ǫ ) t q dxdt C Q where C is independent of ǫ. Now choose q so that q > 2/(1 α). Then Sobolev s embedding gives u ǫ C α (Q) C. Finally, an application of the C α estimate of LSU (1968) yields u ǫ C 2+α,1+α (Q) C where C depends on Q but not on ǫ. Now an appeal to the Ascoli-Arzela theorem shows u C 2+α,1+α (Q) with u C 2+α,1+α (Q) C, see for example Friedman (1964). This shows that u is C 2,1 at (x 1,t 1 ) and continuous at (x 1,). It is clear that u is continuous at the boundaries x =, x = 1 because of the estimate u ǫ h and the sandwich lemma. Having proved existence, it is straightforward to prove uniqueness and positivity. Proposition 2.4. The solution u defined by (2.5) is unique and positive. Proof. We already noted the fact that u ǫ in (ǫ,1) (,t ) and consequently, u in (,1) (,t ). An application of the (strong) maximum principle shows u > in (,1) (,t ) unless u. To show u is unique, let u and v be two solutions of (1.1) (1.3) in (,1) (,t ). Then, if w = v u, xw t = w xx + pθ p 1 w where θ = λ(x,t)u + (1 λ(x,t))v, for some λ(x,t) (,1). Further, w = at x =, x = 1 and w = at t =. Hence the maximum principle, modified to account for the degenerate coefficient of w t, implies w = in (,1) (,t ). See Friedman (1964). Thus far it has been proved that a unique classical solution exists in some time interval (,t ). To complete the existence theory we prove a continuation theorem. Theorem 2.5. Let T be the supremum over t for which there is a solution in (,t ). Then there is a solution u in (,T). If T < then u is unbounded in Ω (,T). Proof. Given any (x 1,t 1 ) Ω (,T) we can choose a solution v t (x,t) in Ω (,t ) with < t 1 < t < T. Then set u(x 1,t 1 ) = v t (x 1,t 1 ). This defines u : Ω (,T) IR. It is well defined by the uniqueness of v t and u is a solution in Ω (,T). Now suppose T < and u is bounded in Ω (,T). We need to show that u can be continued into a larger time interval (,T + t ). This is done by extending the a priori 5

6 bound h first. The only potential problem is that u x (,t) may grow large as t T. We must show u is well behaved at t = T. Suppose u M in Ω (,T) and define ψ(x) = Kx(1 x) where K = max{m p /2, sup(u (x)/x(1 x))}. x Ω Then ψ(x) u (x) and ψ = at x =, x = 1. Moreover xψ t ψ xx {xu t u xx } = K/2 u p K/2 M p. Therefore ψ is an upper solution of u in Ω (,T). In particular u is bounded by ψ(x) at t = T (and so u is indeed well behaved at (,T)). Now as in Lemma 2.2 we can construct an upper solution h(x,t) in Ω (T,T + t ) of (1.1) (1.2) with initial data ψ(x) at t = T. Finally set { ψ(x) for < t T, H(x,t) = h(x,t) for T t T + t. Since H is an a priori upper bound for any u ǫ in Ω (,T + t ), u ǫ can be continued into Ω (,T + t ). By repeating theorem 2.3 but replacing h by H and t by T + t we find u(x,t) = lim ǫ u ǫ (x,t) is a solution of (1.1) (1.3) in Ω (,T + t ). This contradicts the definition of T. To end this section an extra result is presented which will be required in Section 3. Lemma 2.6. The t derivative u t of the solution u is continuous at x = and x = 1. Proof. Since (u ǫ ) t u t as ǫ it is only necessary to find a uniform upper bound h(x,t) for (u ǫ ) t in Ω (,t ), any t < T. The construction of h follows similar lines to Lemma 2.2. Choose t (,T). Then u M in Ω (,t ). Define h as follows: (i) Let ψ = 1 2x(1 x). (ii) Choose k s.t. k xψ(x) u (x) + u p (x). (iii) Choose δ 1/pM p 1. (iv) Let k(t) = k exp(pm p 1 t/δ). (v) Set h(x, t) = k(t)ψ(x). Now we show h is an upper solution for (u ǫ ) t in Ω (,t ). Differentiating (1.1) w.r.t. t and substituting h for u gives J = xh t h xx pu p 1 h = xk ψ + k pu p 1 kψ. 6

7 For x (,δ), t (,t ), For x (δ, 1 δ), t (,t ), J k pu p 1 kψ k(1 pm p 1 δ(1 δ)/2) k(1 pm p 1 δ). J xk ψ pm p 1 kψ ψ(δk pm p 1 k) =. Thus h is an upper solution for u t and also for (u ǫ ) t, any ǫ, in Ω (,t ). Letting u ǫ u and using a similar procedure as in Theorem 2.3 it is shown that u t is bounded and continuous at the boundary of Ω. 3. The main theorem The main object of this section is to ascertain the blow-up set. Before doing this (in Theorem 3.5) we show that u tends to infinity in finite time provided that u is large enough. This is done by modifying the eigenfunction method due to Kaplan (1963). Consider the linear boundary value problem φ = λxφ in Ω, φ = in Ω. (3.1) There is a unique solution φ(x) = kx 1/2 J 1/3 (µ 1 x 3/2 ), λ = 9µ 2 1/4 (3.2) for which φ(x) > and xφ(x)dx = 1. The constant k is 1/ x3/2 J 1/3 (µ 1 x 3/2 )dx and µ 1 is the first positive zero of J 1/3, the Bessel function of the first kind of order 1/3, see Abramowitz & Stegun (1965). That φ solves (3.1) is evident from the Bessel equation x 2 J 1/3 + xj 1/3 + (x2 1/9)J 1/3 =. With φ and λ as defined in (3.2) a condition for blow-up is as follows. Proposition 3.1. Suppose u solves (1.1) (1.3) and Then u blows up in finite time. Proof. Set xφ(x)u (x)dx > λ 1/(p 1). U(t) = xφ(x)u(x, t)dx. 7

8 Then multiplying (1.1) by φ and integrating over x leads to φxu t = φu xx + Integration by parts and Jensen s inequality imply ( ) xφu = λ xφu + t λ λ xφu + Therefore U is a solution of the initial value problem U U( λ + U p 1 ), U() = φu p. φu p xφu p ( p xφu + xφu). xφ(x)u (x)dx. By hypothesis, U() > λ 1/(p 1) and therefore U tends to infinity in finite time. Hence u ceases to exist at some finite time T. By Theorem 2.5 there is a sequence (t n ) with t n T and sup x Ω u(x,t n ) as n. The main purpose of this paper is to determine the blow-up set S. This is not immediately apparent from equation (1.1). It turns out that a change of variables can be made transforming (1.1) into a more recognisable form. The transformation is really a composition of two distinct transformations. The first, u u/x has the effect of turning the Dirichlet boundary condition into a Neumann boundary condition. The second, x (2/3)x 3/2 is used to transform the operator xu t u xx into the well known v t v yy although terms in v y are also introduced. This latter substitution is used in the analysis of Airy s equation w = xw. Lemma 3.2. Suppose u is the solution of (1.1) (1.3) in Ω (,T). Define v(y,t) by Then v satisfies u(x,t) = xv(y,t), y = (2/3)x 3/2. (3.3) v t = v yy v y y + ay(2/3)(p 2) v p (3.4) v y (,t) = v(2/3,t) = (3.5) v(y,) = u (x)/x (3.6) where a = (3/2) (2/3)(p 2). Also v is C 2,1 in (,2/3) (,T), continuous in [,2/3] [,T) and v y is continuous at y =. Proof. The equation for v (3.4) follows from direct calculation. The only remaining property of v which is not obvious is v y (,t) =. By Lemma 2.6, we know u t is continuous at x = and so from (1.1), u xx is continuous at x = as well. Thus we can write u(x,t) = u x (,t)x + x s 8 u xx (r,t)drds.

9 Then and Therefore from (3.3), v(y,t) = u(x,t)/x = u x (,t) + 1 x x s u xx (r,t)drds x v(y,t) = 1 x s x 2 u xx (r,t)drds + 1 x u xx (s,t)ds x = 1 x x 2 su xx (s,t)ds. x v y = x 1/2 v x = x 5/2 su xx (s,t)ds. Now from (1.1), u xx (,t) =. Moreover since u p u p x(,t)x p as x and xu t = o(x) as x we have u xx = O(x p ) for 1 < p 2 and u xx = O(x 2 ) for p > 2. Therefore when 1 < p 2, v y x 5/2 x s(ks p )ds = Kxp 1/2 p + 2 = Cy 2(p 1/2)/3 as y. When p > 2, v y x 5/2 x s(ks 2 )ds = Kx3/2 4 = Cy as y. This ends the proof. Remark. This is precisely Note that in the case p = 2, equation (3.4) simplifies to v t = v yy v y y + v2. (3.7) v t = v + v 2 (3.8) where y is the radial coordinate except that the dimension is apparently 8/3! Even though the Laplacian operator is meaningless in 8/3 dimensions we would expect (3.7) and (3.8) to have similar properties. The main reason for considering v is that there is a maximum principle for v y if p 2. This is because the coefficient y (2/3)(p 2) in equation (3.4) is decreasing in y for p 2. For p > 2 the opposite is true. Lemma 3.3. Suppose 1 < p 2 and v y (y,) for y (,2/3). Then v y < for all y (,2/3), t (,T). Proof. Let w = v y and differentiate (3.4) with respect to y to obtain w t = w yy w y y 5 w 3 y 2 + a{2 3 (p 2)y(2/3)(p 7/2) v p + py (2/3)(p 2) v p 1 w} (3.9) 9

10 and since 1 < p 2 we have w t w yy 5 3 w y y bw where b is bounded above in (,2/3) (,t ), any t < T. At the boundaries, w = v y = at t = and since v, v y < at y = 2/3. Finally, since w at t =, the maximum principle shows w < in (,2/3) (,t ) and hence w < in (,2/3) (,T) as required. Now suppose u blows up at t = T. Then v = u/x x blows up at T also. A consequence of Lemma 3.3 is that y = is a blow-up point of v. The essence of proving Theorem 3.5 is showing that y = is the only blow-up point of u. First we need a lemma. The idea is due to Mueller & Weissler (1985). Lemma 3.4. Suppose 1 p < 2 and v y (y,) for y (,2/3). Suppose y (,2/3) is a blow-up point of v. Then if < y 1 < y, v(y 1,t) as t T. Proof. Choose y 1 (,y ) and let y 2 = (y 1 + y )/2. Since v y <, it is clear that limsup t T v(y 2,t) = (note that it is not necessarily true that limsup t T v(y,t) = ). We show that lim t T v(y 1,t) =. Observe that v t v yy = 5 3 v y y + ay(2/3)(p 2) v p 2 v y y since v y <. Define Ω to be the ball in IR 3 of radius 2/3, centre. Letting y = x and choosing τ (,T) we find v(y,t) w(x,t) where w t = w in Ω, w Ω = and w( x,τ) = v(y,τ). Denoting the Green s function for w by G, we have w(x,t) = G(x,z,t τ)w(z,τ)dz Ω and v(y,t) G(x,z,t τ)v( z,τ)dz Ω for any x s.t. y = x. Now choose a sequence (τ n ) (,T) such that τ n T and v(y 2,τ n ) as n. For any t (τ n,t) we have v(y 1,t) G(x 1,z,t τ n )v( z,τ n )dz Ω for any x 1 s.t. y 1 = x 1. Further, since v y < and G, v(y 1,t) v(y 2,τ n ) G(x 1,z,t τ n )dz. x y 2 But due to the asymptotics of G, x y 2 G(x 1,z,t τ n ) 1 2 once t τ n is small enough. Hence letting n shows v(y 1,t) as t T. Now comes the main result. The method is based on that of Friedman & McLeod (1985) with refinements. Recall that S represents the set of blow-up points of u. 1

11 Theorem 3.5. Suppose 1 < p 2 and that the solution u of (1.1) (1.3) blows up at t = T. If x (u (x)/x) for x (,1) then S = {}. Proof. Since v = u/x u, v blows up as t T. By Lemma 3.3, y = is a blow-up point of v. Clearly if y = is the only blow-up point of v then x = is the only blow-up point of u. Suppose y (,2/3) is a blow-up point of v. Choose y 1 s.t. < y 1 < y. Then, by Lemma 3.4, v(y 1,t) as t T. We show that this is not, in fact, the case. Choose λ (1,p) and define J = v y + ǫc(y)v λ where ǫ > is to be determined. The function c is defined as y if y y /3 c(y) = h(y) if y /3 y 2y /3 (y y) 3 if 2y /3 y y where h(y) is any positive function which completes the piecewise definition of c in such a way that c C 2 (,y ). It is important only that c is positive and has the appropriate behaviour near and y. Our task is to show J in (,y ) (t,t) for some t > via the maximum principle. From (3.4), J t J yy 5 3 J y y = 5 v y 3 y (p 2)ay(2/3)(p 7/2) + ay (2/3)(p 2) pv p 1 v y + ǫcλv λ 1 v t ǫ{c v λ + 2c λv λ 1 v y + cλ(λ 1)v λ 2 v 2 y + cλv λ 1 v yy } 5 3y {ǫc v λ + ǫcλv λ 1 v y } now noting p 2, Thus to show J t J yy 5 3 bj ǫc{ay (2/3)(p 2) (p λ)v p+λ 1 ( + c + 5 3y c 5 ) v λ 3y 2 c c 2ǫc λv 2λ 1 }. J y y bj it is sufficient to show ( a(p λ)v p+λ 1 /y (2/3)(2 p) 5 1 2ǫλc v 2λ 1 + 3y 2 5 c ) 3y c c v λ. (3.1) c Assuming v 1, setting ǫ a(p λ)/(4λy (2/3)(2 p) 1 max y (,y1 ){c (y)}) ensures a(p λ)v p+λ 1 /y (2/3)(2 p) 1 2ǫλc v 2λ 1 a(p λ)v 2λ 1 /2y (2/3)(2 p) 1. Further, by construction 5 3y 2 5 c 3y c c c M in (,y ) 11

12 for some bound M >. Therefore, since v(y 1,t) as t T we can choose t close enough to T so that (3.1) holds for y = y 1, t (t,t). Since v y <, (3.1) holds for any y (,y 1 ), t (t,t). Finally, to apply the maximum principle to J observe that J at y = and y = y 1 since c() = c(y 1 ) =. To ensure J at t = t, y (,y 1 ) note that as y, v y = O(y (2/3)(p 1/2) ), see Lemma 3.2. So for small enough y, v y Ky. Since c(y) and v(y,t) are bounded at t = t it follows that J provided ǫ is chosen small enough. Thus by the maximum principle v y + ǫc(y)v λ in (,y 1 ) (t,t) (3.11) as required. Now dividing (3.11) by v λ and integrating w.r.t. y over (,y 1 ) gives 1 v λ 1 (y 1,t) 1 v λ 1 (,t) Hence v(y 1,t) δ 1/(λ 1) which is a contradiction. y1 ǫ(λ 1) c(z)dz = δ. Blow-up at the boundary has been proved when 1 < p 2 and (d/dx)(u (x)/x). Particular examples of initial data u for which (d/dx)(u (x)/x) are λ sin πx and λx(1 x), any λ >. More generally, d2 u dx = d 2 dx ( u x ). 4. Asymptotics near the singularity In this section it is assumed that 1 < p 2, (d/dx)(u (x)/x) and u blows up at t = T. Firstly, upper and lower integral bounds for u are calculated. These bounds largely determine the shape of u as t T. Secondly, bounds on the the maximum of u are calculated. These determine the rate at which the maximum of u tends to. Let s(t) be any x(t) (,1) such that u(x(t),t) = sup x (,1) u(x,t). We will show that C 1 (T t) 1/3 s(t) C 2 (T t) 1/(3+δ) for any δ >. Integral bounds. The approach used to find the bounds is to find the corresponding bounds for v and then transfer them back to u via (3.3). To ease notation let α = (2/3)(2 p) so that α < 2/3 and let a = (3/2) (2/3)(p 2) as before. Theorem 4.1. Let u solve (1.1) (1.3) with 1 < p 2 and (d/dx)(u (x)/x) where T is the blow-up time. Suppose that u C 2 [,1] and u + u p in (,1). Then u γ (x,t)dx as t T provided γ > (p 1)/2. Proof. For this proof it is easier to treat v as a function of x rather than y. Let u(x,t) = xv(x, t). Then xv t = v xx + 2 v x x + xp 1 v p (4.1) 12

13 with v(x,) = u (x)/x and v(1,t) =. At x =, the analysis in Lemma 3.2 shows and since 1 < p 2, x v x = x 2 su xx (s,t)ds v x x 2 x s(ks p )ds = Kxp p + 2 as x. Also v x < in (,1) (,T) since v y <. Fix t (,T). Multiply equation (4.1) by v x and integrate w.r.t. x over (,x) to obtain x v x xv t = 1 x v 2 x 2 v2 x x + 2 x + x p 1 v p v x. Now by the maximum principle u t > and so v t = u t /x >. Further, since v x < we find x 1 2 v2 x 1 x p 1 (v p+1 ) x p + 1 = 1 [ x p 1 v p+1] x p p 1 p + 1 where v(t) denotes v(, t). Thus Integrating over (,x) gives x x p 2 v p+1 p 1 x p + 1 vp+1 x p 2 vp+1 x p 1 p + 1 v x Cx (p 1)/2 v (p+1)/2. v v 2Cx(p+1)/2 v (p+1)/2 p + 1. (4.2) Define x(t) to be the unique point in (,1) at which v( x(t),t) = v(t)/2. Then letting x = x in (4.2), x K v (p 1)/(p+1) (4.3) for some K >. Finally, u γ (x,t)dx = vγ 2 γ x γ v γ (x,t)dx x x γ dx γ+1 vγ x = 2 γ (γ + 1) Kγ+1 2 γ (γ + 1) vγ (γ+1)(p 1)/(p+1). 13

14 Therefore uγ (x,t)dx as t T provided γ (γ + 1)(p 1) p + 1 > which is equivalent to γ > (p 1)/2. To obtain the corresponding upper bound an improvement is made on the inequality (3.11) used to show 1-point blow-up. Lemma 4.2. Let u solve (1.1) (1.3) on (,1) (,T) with (d/dx)(u (x)/x) and 1 < p 2. For any λ (1,p) there exists ǫ > such that in (,1/3) (T/2,T). v y ǫy 1 α v λ (4.4) Proof. Recall the equation (3.4) for v. Fix λ (1,p) and set J = v y + ǫy 1 α v λ where ǫ is yet to be determined. Then J t J yy 5 3 J y y = 5 v y 3 y 2 + apvp 1 v y y α substituting v y = J ǫy 1 α v λ and using v 2 y, αavp y α+1 + ǫy1 α λv λ 1 v t ǫ{ α(1 α)y α 1 v λ + 2(1 α)y α λv λ 1 v y + y 1 α [λ(1 λ)v λ 2 vy 2 + λv λ 1 v yy ]} 5 ǫ 3 y {(1 α)y α v λ + y 1 α λv λ 1 v y } bj ǫy α 1 v λ ǫapy 1 2α v p+λ 1 αay α 1 v p + ǫaλy 1 2α v p+λ 1 + ǫα(1 α)y α 1 v λ + 2ǫ 2 (1 α)λy 1 2α v 2λ (1 α)ǫy α 1 v λ where b is bounded above in (,1/3) (T/2,t ), t < T, =bj ǫy 1 2α {a(p λ)v p+λ 1 2ǫ(1 α)λv 2λ 1 } ( } 5 y {αav α 1 p ǫ α + α(1 α) )v λ 3 =bj ǫy 1 2α v 2λ 1 {a(p λ)v p λ 2ǫ(1 α)λ} ( )} 8 ǫαy α 1 v {av λ p λ ǫ 3 α as long as ǫ is small enough. Note inf (,1/3) (T/2,T) v >. To apply the maximum principle to J observe J = at y =. Further, it is known that v is bounded at y = 1/3 for all t (T/2,T) by Theorem 3.5. Also v y is bounded 14

15 above by a negative constant for all t (T/2, T). To understand this let v be the solution (for all t > ) of v t = v yy + 5 v y (4.5) 3 y with conditions (3.5) and (3.6). A comparison of equations (4.5) and (3.4) shows that v is a lower solution for v. Then differentiating (4.5) w.r.t. y and comparing with (3.9) shows that v y is an upper solution for v y. Thus v y δ in {1/3} (T/2,T). So we can ensure J at y = 1/3 by choosing ǫ even smaller if necessary. Finally, to ensure J in (,1/3) {T/2} recall v y = O(y (2/3)(p 1/2) ) = O(y 1 α ) as y, see Theorem 3.5. So letting ǫ be small enough implies J in (,1/3) {T/2}. Hence the maximum principle applies to show J in (,1/3) (T/2,T) as required. By using (4.4) an upper bound for u can be computed. Theorem 4.3. Let u solve (1.1) (1.3) on (,1) (,T) where (d/dx)(u (x)/x) and 1 < p 2. Then, provided < γ < (p 1)/2, there exists a bound C γ such that u γ (x,t)dx C γ for all t (,T). Proof. Let λ (1,p). Then by Lemma 4.2 there is ǫ > with v y ǫy 1 α v λ in (,1/3) (T/2,T). Now dividing by v λ and integrating from to y yields v K λ y (2 α)/(λ 1). To transfer back to u, let v = u/x, y = (2/3)x 3/2 and remembering α = (2/3)(2 p) we find u C λ x (2+p λ)/(λ 1). Therefore u γ (x,t)dx C γ λ x γ(2+p λ)/(λ 1) dx C γ provided γ < λ p λ. (4.6) But (4.6) holds for any γ < (p 1)/2 if λ is chosen close enough to p. Bounds on s(t). It appears difficult to show that at any given t, the maximum of u over x in (,1) is attained at a unique point (there is no maximum principle for u x ). Despite this, we can obtain the following bounds. Let s 1 (t) (,1) be the infimum over s (,1) for which u(s,t) = sup x Ω u(x,t). Similarly, let s 2 (t) be the supremum over all such s. It will be shown that C 1 (T t) 1/3 s 1 (t) s 2 (t) C 2 (T t) 1/(3+δ) for any δ >. 15

16 Lemma 4.4. Suppose u + u p. There exists ǫ > such that xp 1 u t ǫu p for any x in (,1), t (t,t). Proof. There are difficulties in applying the maximum principle to both u(x,t) and v(y,t). Instead we work with v(x,t) = u(x,t)/x. Now set J = v t ǫv p. Then from equation (4.1), xj t J xx 2 J x x xp 1 pv p 1 J = ǫp(p 1)v p 2 v 2 x. As v x (,t) =, so J x (,t) =. Also J(1,t) = and J(x,t ) if ǫ is chosen small enough. Hence, by the maximum principle, J and consequently x p 1 u t ǫu p as required. Theorem 4.5. Suppose 1 < p 2, (d/dx)(u (x)/x) and u blows up at t = T. Under the additional assumption that u + u p, there exists C 1 > such that s 1 (t) C 1 (T t) 1/3 for all t (,T). Proof. Fix t (,T). Let x(t) be the least member of (,1) for which u x ( x(t),t) =. Clearly x(t) s 1 (t). Now multiply (1.1) by u x and integrate over (x, x) to obtain Therefore x x xu t u x = Integrating over (, x) shows x x (u 2 x) x /2 + x x u x Ku (p+1)/2 ( x,t) u p = u2 x 2 + up+1 ( x,t) u(x,t) p + 1 p + 1. for x (, x). x Cu (p 1)/2 ( x,t). (4.7) Now from Lemma 4.4, x p 1 u t ǫu p. Integrating w.r.t. t over (t,t) implies u Now combining (4.7) and (4.8) yields which completes the proof. Dx2/(p 1). (4.8) (T t) 1/(p 1) x(t) C 1 (T t) 1/3 To obtain an upper bound for s 2 (t) another lemma is needed. Lemma 4.6. Let ū(t) = u(s 2 (t),t). Then ū is Lipschitz continuous a.e. and satisfies s 2 ū ū p. Proof. It is straightforward to adapt the proof by Friedman & McLeod (1985) regarding u t = u xx + u p to the equation (1.1). 16

17 Theorem 4.7. Let u be the solution of (1.1)-(1.3) where 1 < p 2, (d/dx)(u (x)/x) and suppose u blows up at t = T. Then given δ > there exists C 2 > such that s 2 (t) C 2 (T t) 1/(3+δ). Proof. Let ū(t) = u(s 2 (t),t). Then from Lemma 4.6 and Theorem 4.5, Integrating over (t,t) implies ū ū p 1 s 2 1 s 1 Now from Lemma 4.2, for any λ (1,p), 1 C 1 (T t) 1/3. ū (p 1) K 1 (T t) 2/3. (4.9) v y ǫy 1 α v λ for some ǫ >. Substituting u = xv, y = (2/3)x 3/2, α = (2/3)(2 p) gives xu x u ǫ Now at x = s 2, u x = and therefore Eliminating ū from (4.9) and (4.1) yields Finally we can arrange that ( ) (2/3)(p 1/2) 2 x p λ+2 u λ. 3 ū K 2 s (p λ+2)/(λ 1) 2. (4.1) s 2 (t) C λ (T t) (2/3)/{(p 1)(p λ+2)}. 2(λ 1) 3(p 1)(p λ + 2) = δ by an appropriate choice of λ. 17

18 5. Generalisations To complete this paper the previous analysis is extended (with very little effort) to equation (1.4). Thus the new problem is x q u t = u xx + u p in Ω (, ) (5.1) u(,t) = u(1,t) = t > (5.2) u(x,) = u (x) x Ω (5.3) where p > 1 and q >. It is not necessary to go through much detail since the proofs are almost identical. Existence and uniqueness of solution can be demonstrated in exactly the same way as before. To obtain asymptotic results concerning u the key transformation is now Then u(x,t) = xv(y,t), y = 2 q + 2 x(q+2)/2. v t =v yy + q + 4 v y q + 2 y + ay2(p q 1)/(q+2) v p v(y,) =u (x)/x with a = {(q + 2)/2} 2(p q 1)/(q+2). Note the dimension is now 2(q + 3)/(q + 2) which is strictly between 2 and 3 for all q >, c.f. the case q = 1 when the dimension is 8/3. The behaviour of v near y = maybe different depending on p and q. In all cases v y =, v y D as y, or v y as y. Following Lemma 3.2, x v y = x q/2 v x = x q/2 2 su xx (s,t)ds. Now the behaviour of u xx is found from (5.1) where x q u t = O(x q+1 ) and u p = O(x p ) as x. Therefore when p q + 1, v y x q/2 2 x s(ks p )ds = Kxp q/2 p + 2 = Cy 2(p q/2)/(q+2). Thus v y if 1 < p 2, v y if p < q/2, and v y D if p = q/2 (v y is negative near y = since u xx is also). When p > q + 1, u xx Kx q+1 so that v y x q/2 2 x s(ks q+1 )ds = Kxq/2+1 q + 3 = Cy as y. Theorem 5.1. Suppose 1 < p q + 1 and that the solution of (5.1) (5.3) blows up at t = T. If x (u (x)/x) for x (,1) then S = {}. Proof. We show v blows up only at y =. By the maximum principle, v y <, see Lemma 3.3. Now suppose y (,2/(q + 2)) is a blow-up point of v. By an argument similar to Lemma 3.4, we can choose any y 1 (,y ) such that v(y 1,t) as t T. 18

19 Choose λ (1,p) and define J = v y + ǫc(y)v λ. As in Theorem 3.5, the maximum principle can be applied to J to deduce v y + ǫc(y)v λ in (,y 1 ) (t,t). This leads to a contradiction, see Theorem 3.5. Numerical evidence given by Stuart & Floater (1989) indicates that when p > q + 1, blow-up occurs away from the boundary. Integral bounds. The asymptotic bounds in Section 4 generalise in a straightforward manner. Letting α = 2(q + 1 p)/(q + 2), Theorem 4.1 now becomes Theorem 5.2. Theorem 5.2. Let u solve (5.1) (5.3) with 1 < p q +1 and (d/dx)(u (x)/x) where T is the blow-up time. Suppose that u C 2 [,1] and u + u p in (,1). Then u γ (x,t)dx as t T provided γ > (p 1)/2. By obtaining the bound v y ǫy 1 α v λ we can extend Theorem 4.3. Theorem 5.3. Let u solve (5.1) (5.3) on (,1) (,T) where (d/dx)(u (x)/x) and 1 < p q + 1. Then provided < γ < (p 1)/2, there exists a bound C γ such that u γ (x,t)dx C γ for all t (,T). Bounds on s(t). Let s 1 (t) (respectively s 2 (t)) be the infimum (respectively supremum) of s (,1) for which u(s,t) = sup x (,1) u(x,t). Lemma 4.4 holds as before when x is replaced by x q. The same transformation v(x,t) = u(x,t)/x is required. Note that v x (,t) = for any p > 1, q > despite v y (,t) being infinite if p > q/2. Using the inequality x p 1 u t ǫu p leads to Theorem 5.4. Theorem 5.4. Suppose 1 < p q + 1, (d/dx)(u (x)/x), and T is the blow-up time for u. Under the added assumption that u + u p, there exists C 1 > such that s 1 (t) C 1 (T t) 1/(q+2) for all t (,T). A simple extension of Lemma 4.6 is to show s q 2ū ū p. Then Theorem 4.7 can be generalised. Theorem 5.5. Let u be the solution of (5.1) (5.3) where 1 < p q+1, (d/dx)(u (x)/x) and suppose u blows up at t = T. Then given δ > there exists C 2 > such that s 2 (t) C 2 (T t) 1/(q+2+δ). As a final remark it is worth noting that the equation u t = u xx + up x q 19

20 has similar properties to (5.1) though there might not be a classical solution for large q and small p (especially p q 2). Letting u(x, t) = xv(x, t) yields v t = v xx + 2 v x x + xp q 1 v p. As for (5.1) the sign of the quantity p q 1 is critical in determining the blow-up set. By showing that v x < it follows that blow-up is at the boundary when p q + 1 and d dx (u (x)/x). Conversely we expect blow-up to be in (,1) when p > q + 1. Acknowledgement. I wish to thank Prof. J. B. McLeod for his generous support and encouragement. This work was financed by the Science and Engineering Research Council. 6. References M. Abramowitz & A. Stegun (1965) Handbook of mathematical functions, Dover Publications, New York. M. S. Floater (1988) Blow-up of solutions to nonlinear parabolic equations and systems, D. Phil. thesis, Oxford. A. Friedman (1964) Partial differential equations of parabolic type, Prentice-Hall, New Jersey. A. Friedman & J. B. McLeod (1985) Blow-up of positive solutions of semilinear heat equations, Ind. Univ. Math. J., 34, A. A. Lacey (1984) The form of blow-up for nonlinear parabolic equations, Proc. Roy. Soc. Edin., 98, O. A. Ladyženskaja, V. A. Solonnikov, & N. N. Ural ceva (1968) Linear and quasilinear equations of parabolic type, Amer. Math. Soc. Translations of Mathematical Monographs, Rhode Island. C. E. Mueller & F. B. Weissler (1985) Single point blow-up for a general semilinear heat equation, Ind. Univ. Math. J., 34, H. Ockendon (1979) Channel flow with temperature-dependent viscosity and internal viscous dissipation, J. Fluid Mech., 93, D. H. Sattinger (1972) Monotone methods in nonlinear elliptic and parabolic boundary value problems, Ind. Univ. Math. J., 21, A. Stuart & M. S. Floater (1989) Time-stepping and peak-tracking strategies for blow-up problems, in preparation. 2