Sixth Mississippi State Conference on Differential Equations and Computational Simulations, Electronic Journal of Differential Equations, Conference 15 (2007), pp. 51–65. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp)

EXISTENCE AND NON-EXISTENCE RESULTS FOR A NONLINEAR HEAT EQUATION

CANAN CELIK

Abstract. In this study, we consider the nonlinear heat equation

u_{t}(x, t) = Δu(x, t) + u(x, t)^{p} in Ω × (0, T),

Bu(x, t) = 0 on ∂Ω × (0, T),
u(x, 0) = u_{0}(x) in Ω,

with Dirichlet and mixed boundary conditions, where Ω ⊂ ℝ^{n} is a smooth bounded domain and p = 1 + 2/n is the critical exponent. For an initial condition u_{0} ∈ L^{1}, we prove the non-existence of local solution in L^{1} for the mixed boundary condition. Our proof is based on comparison principle for Dirichlet and mixed boundary value problems. We also establish the global existence in L^{1+ϵ} to the Dirichlet problem, for any fixed ϵ > 0 with ∥u_{0}∥_{1+ϵ} sufficiently small.

1. Introduction

In this paper we study the existence and non-existence results for the initial and boundary value problems of the form

u_{t}(x, t) = Δu(x, t) + |u(x, t)|^{p−1}u(x, t) in Ω × (0, T),
Bu(x, t) = 0 on ∂Ω × (0, T),
u(x, 0) = u_{0}(x) in Ω,

(1.1)

where Ω ⊂ ℝ^{n} is a smooth bounded domain, B denotes the corresponding boundary condition and p is the critical exponent, which will be specified later.

Throughout this paper we use the following definition for the solution.

Definition 1.1. A function u = u(x, t) is a mild solution of initial and boundary value problem (1.1) in Ω̄ × [0, T], T being a positive number, if and only if u ∈ C([0, T], L^{q}(Ω)) satisfies the integral equation

u(x, t) = ∫_{Ω}K(x, y, t)u_{0}(y)dy +∫_{0}^{t}∫_{Ω}K(x, y, t − s)|u(y, s)|^{p−1}u(y, s) dy ds, (1.2)

2000 Mathematics Subject Classification. 35K55, 35K05, 35K57, 35B33. Key words and phrases. Nonlinear heat equation; mixed boundary condition; global existence; critical exponent. \begin{matrix}c \\ ◯ \end{matrix}2007 Texas State University - San Marcos. Published February 28, 2007.

where u_{0} ∈ L^{q}(Ω) and K denotes the heat kernel for the linear heat equation with Dirichlet boundary condition.

It is well known that if the initial condition u_{0} ∈ L^{∞}(Ω), there exists a unique solution of (1.1) on [0, T_{max}). However the question of local existence and uniqueness was interesting when u_{0} ∉ L^{∞}(Ω); i.e., u_{0} ∈ L^{q}(Ω) where 1 ≤ q < ∞.

This type of initial and boundary value problem was studied by many authors. First Fujita [4, 5] studied this problem for classical solutions and established the following results for the Cauchy problem

\begin{matrix}u_{t}(x, t) = Δu(x, t) + u^{p}(x, t) in ℝ^{n} × (0, T), \\ u(x, t) = u_{0}(x) on ℝ^{n}, \end{matrix}  (1.3)

where u_{0}(x) ≥ 0.

(i) If n(p − 1)/2 < 1, no non-negative global solution exists for any non-trivial initial data u_{0} ∈ L^{1}. That is, every positive solution to this initial value problem blows up in a finite time.

(ii) If n(p − 1)/2 > 1, global solution do exist for small initial data. To be precise, for any k > 0, δ can be chosen such that problem (1.3) has a global solution whenever 0 ≤ u_{0}(x) ≤ δe^{−k|x|^{2}}.

Later Hayakawa [8] showed that the critical case for the Cauchy problem (1.3), that is n(p − 1)/2 = 1, also belongs to blow up case for n = 1, 2. After Hayakawa the same result was proved by Kobayashi, Sirao and Tanaka [9] for general n. Also Weissler [12] obtained that for n(p − 1)/2 ≤ 1 with 1 ≤ p < ∞, non-negative L^{p} solutions to the Cauchy problem blow up in L^{p} norm in finite time. However in the case n(p − 1)/2 > 1, he obtained sufficient conditions for the initial data u_{0} which guarantee the existence of global solution.

The initial and boundary value problem was considered by Weissler [13, 14] who obtained some important local existence and uniqueness results in C([0, T], L^{q}(Ω)). In [13], he considered the case q > n(p − 1)/2 and q > p and proved the local existence and uniqueness in C([0, T], L^{q}(Ω)) for which the same argument works for q > n(p − 1)/2 and q = p. Later local existence in C([0, T], L^{q}(Ω)) with q ≥ n(p − 1)/2 and q > 1 or q > n(p − 1)/2 and q ≥ 1 was proved in [14]. In the same study he also proved a uniqueness result in a smaller class which was improved in [2]. If we notice, while the applications of the abstract results in [13] and [14] are given on a bounded domain, they work equally well on all of ℝ^{n}.

The first non-uniqueness result for this problem was given by Haraux-Weissler in [7]. Further non-uniqueness results which were motivated by [7] came in [1] and [10]. A unique solution to the initial and boundary value problem (1.1) in L^{p_{1}}(0, T; L^{p_{2}}) was constructed by Giga [6]. In his work, the main relation between p_{1} and p_{2} was \frac{1}{p_{2}} = \frac{n}{2}(\frac{1}{r} − \frac{1}{p_{1}} ), p_{1} > r, provided that the initial data u_{0} ∈ L^{r} with r = n(p − 1)/2 > 1.

As we stated above, it is well known that if u_{0} ∈ L^{∞}(Ω), there exists a unique solution defined on a maximal interval [0,T_{max}) and Brezis and Cazenave [2] considered the case if u_{0} ∉ L^{∞}(Ω); i.e., u_{0} ∈ L^{q}(Ω) for some 1 ≤ q < ∞. And they studied two cases.

(i) If q > n(p − 1)/2 (resp. q = n(p − 1)/2) and q ≥ 1 (resp. q > 1) with n ≥ 1, they obtained existence and uniqueness of a local solution in C([0, T], L^{q}(Ω)) which is a classical solution of the initial boundary value problem on Ω × (0, T).

(ii) If q < n(p − 1)/2, they showed that there exists no local solution in any reasonable sense for some initial data u_{0} ∈ L^{q}(Ω).

Remark 1.2. The quantity q = n(p − 1)/2 plays a critical role for this initial and boundary value problem. To see that q = n(p − 1)/2 is the critical exponent, we observe the following dilation argument. If v is the solution of

v_{t} = v_{xx} + v^{p},
v(x, 0) = g(x),

then u(x, t) = k^{s}v(kx, k^{2}t) is the solution of

u_{t} = u_{xx} + u^{p},
u(x, 0) = k^{s}g(kx).

By using the equation u_{t} = u_{xx} + u^{p} with u(x, t) = k^{s}v(kx, k^{2}t), we get

k^{s+2}v_{t} = k^{s+2}v_{xx} + k^{sp}v^{p}

which gives the condition s = 2/(p − 1). Also note that,

∥u(0)∥_{q}^{q} = ∫ u^{q}(x, 0)dx = ∫ k^{sq−n}v^{q}(y, 0)dy = ∥v(0)∥_{q}^{q}

when s = n/q. So, combining these two conditions for s, we have q = n(p − 1)/2, which is the critical exponent.

The question of existence and uniqueness of a local solution for the doubly critical case which is q = n(p − 1)/2 and q = 1, which was proposed by Brezis and Cazenave [2], was wide open even for n = 1,(p = 3). In [3], we proved that for u_{0} = ∑_{j=1}^{∞} \frac{1}{j^{2}} e^{j^{24}}e^{−(e^{2j^{24}} )x^{2}}, Dirichlet problem has no non-negative local solution in L^{1}(−1, 1). Moreover we generalized the non-existence result of local solution for n-dimensional case with the critical exponent p = 1 + \frac{2}{n}. More general nonlinearity was also considered for Dirichlet boundary value problem.

This paper is organized as follows: In Section 2, we summarize the non-existence results of local solution for both cauchy and the Dirichlet problems for doubly critical case q = n(p − 1)/2 and q = 1, first for one dimensional case and then n-dimensional case.

In Section 3, we consider the same problem with mixed boundary conditions; that means having u and the normal derivative of u in the boundary condition. We prove the non-existence of local solution for the mixed boundary conditions by using the comparison argument for the kernels with Dirichlet and mixed boundary conditions.

In Section 4, we give the sufficient conditions for q and the initial data u_{0} to guarantee the existence of the global solution to the Dirichlet problem. In particular, we prove the global existence in L^{1+ϵ} with ∥u_{0}∥_{1+ϵ} sufficiently small and ϵ > 0, for the critical exponent p = 3.

2. Non-existence of a Local Solution

In the following two sections, we summarize the non-existence of local solution for both cauchy and Dirichlet problems.

2.1. Non-existence of Local Solutions for the Cauchy Problem. To motivate the proof of the non-existence of local L^{1} solution for the Dirichlet problem, we first prove the non-existence of local L^{1} solution for the Cauchy problem for the one dimensional case,

\begin{matrix}u_{t}(x, t) = u_{xx}(x, t) + u^{3}(x, t) in ℝ × (0, T), \\ u(x, t) = u_{0}(x) on ℝ, \end{matrix}  (2.1)

for some u_{0} ∈ L^{1}(ℝ). First we choose a particular initial data u_{0}(x) = ke^{−k^{2}x^{2}} ∈ L^{1}(ℝ) and observe the following result. After this observation, we prove in Theorem 2.1. that the Cauchy problem has no local L^{1} solution for initial data of the form u_{0} = ∑_{k} c_{k}ke^{−k^{2}x^{2}} where c_{k} ≥ 0 will be determined later.

We denote the solution of the Cauchy problem (2.1) by u_{c}. For the Cauchy problem (2.1) with the initial data u_{0}(x) = ke^{−k^{2}x^{2}} ∈ L^{1}(ℝ), using that

∫_{ℝ} \frac{e^{− \frac{(x−y)^{2}}{4t}}}{\sqrt{4πt}} e^{−ay^{2}}dy = \frac{e^{\frac{−ax^{2}}{1+4at}}}{\sqrt{1 + 4at}},

the linear solution is

u_{c}^{L}(x, t) = \frac{k}{\sqrt{1 + 4k^{2}t}}e^{\frac{−k^{2}x^{2}}{1+4k^{2}t}} .

So the solution to the Cauchy problem (2.1) is

u_{c}(x, t) = u_{c}^{L}(x, t) + ∫_{0}^{t} ∫_{ℝ} K_{c}(x, y, t − s)(u_{c}(y, s))^{3} dy ds,

where the Cauchy kernel is

K_{c}(x, y, t) = \frac{e^{− \frac{|x−y|^{2}}{4t}}}{\sqrt{4πt}} .

Note that

(u_{c}^{L}(y, s))^{3} = \frac{k^{3}}{(1 + 4k^{2}s)^{3/2}} e^{\frac{−3k^{2}y^{2}}{1+4k^{2}s}}

and ∫_{ℝ} K_{c}(x, y, t − s)dx = 1. Since u_{c}^{L}(x, t) > 0 and u_{c}(x, t) > u_{c}^{L}(x, t), we have

u_{c}(x, t) > ∫_{0}^{t} ∫_{ℝ} K_{c}(x, y, t − s)(u_{c}^{L}(y, s))^{3} dy ds.

Then for any δ > 0,

∥u_{c}∥_{L^{1}(ℝ×(0,δ))} = ∫_{0}^{δ} ∫_{ℝ} u_{c}(x, t) dx dt

> ∫_{0}^{δ} ∫_{ℝ} ∫_{0}^{t} ∫_{ℝ} K_{c}(x, y, t − s)(u_{c}^{L}(y, s))^{3} dy ds dx dt

= ∫_{0}^{δ} ∫_{0}^{t} ∫_{ℝ}(u_{c}^{L}(y, s))^{3} ∫_{ℝ} K_{c}(x, y, t − s)dx dy ds dt

= ∫_{0}^{δ} ∫_{0}^{t} ∫_{ℝ} \frac{k^{3}e^{\frac{−3k^{2}y^{2}}{1+4k^{2}s}}}{(1 + 4k^{2}s)^{3/2}} dy ds dt

= \sqrt{\frac{π}{3}} ∫_{0}^{δ} ∫_{0}^{t} \frac{k^{2}}{(1 + 4k^{2}s)} ds dt = \sqrt{\frac{π}{3}} ∫_{0}^{δ} \frac{1}{4} ln(1 + 4k^{2}t)dt.

We can use the above estimate to establish the following theorem for the Cauchy problem.

Theorem 2.1. For some initial data u_{0} ∈ L^{1}(ℝ), u_{0} ≥ 0, the Cauchy problem (2.1) has no non-negative local mild solution in C([0, δ], L^{1}(ℝ)) for all δ > 0.

Proof. Let u_{0} = ∑_{k} c_{k}ke^{−k^{2}x^{2}} where c_{k} ≥ 0 will be determined later. The linear solution to the Cauchy problem (2.1) is

u_{c}^{L}(x, t) = ∑_{k} \frac{c_{k}k}{\sqrt{1 + 4k^{2}t}}e^{\frac{−k^{2}x^{2}}{1+4k^{2}t}} .

Note that

(u_{c}^{L}(y, s))^{3} ≥ ∑_{k} \frac{c_{k}^{3}k^{3}}{(1 + 4k^{2}s)^{3/2}} e^{\frac{−3k^{2}y^{2}}{1+4k^{2}s}} .

Since the solution of the problem satisfies

u_{c}(x, t) > ∫_{0}^{t} ∫_{ℝ} K_{c}(x, y, t − s)(u_{c}^{L}(y, s))^{3} dy ds,

we have that for any δ > 0,

∥u_{c}∥_{L^{1}(ℝ×(0,δ))} = ∫_{0}^{δ} ∫_{ℝ} u_{c}(x, t) dx dt

> ∫_{0}^{δ} ∫_{ℝ} ∫_{0}^{t} ∫_{ℝ} K_{c}(x, y, t − s)(u_{c}^{L}(y, s))^{3} dy ds dx dt

= ∫_{0}^{δ} ∫_{0}^{t} ∫_{ℝ}(u_{c}^{L}(y, s))^{3} ∫_{ℝ} K_{c}(x, y, t − s)dx dy ds dt

≥ ∫_{0}^{δ} ∫_{0}^{t} ∫_{ℝ} ∑_{k} \frac{c_{k}^{3}k^{3}}{(1 + 4k^{2}s)^{3/2}} e^{\frac{−3k^{2}y^{2}}{1+4k^{2}s}} dy ds dt

= ∑_{k} C ∫_{0}^{δ} ∫_{0}^{t} \frac{c_{k}^{3}k^{2}}{(1 + 4k^{2}s)} ds dt

= ∑_{k} C ∫_{0}^{δ} \frac{1}{4}c_{k}^{3} ln(1 + 4k^{2}t)dt

≥ C ∑_{k} ∫_{δ/2}^{δ} c_{k}^{3} ln(1 + 4k^{2}t)dt

≥ C ∑_{k} δc_{k}^{3} ln(1 + 2k^{2}δ).

Now we choose c_{k}’s such that (i) ∑_{k} c_{k}^{3}δ ln(1 + 2k^{2}δ) = ∞; and

(ii) ∑_{k} c_{k} < ∞ so that u_{0} ∈ L^{1}.

We will find a sequence {k_{j}}_{j=1}^{∞} such that c_{k} ≠ 0 only when k = k_{j}. For k big enough, we have

c_{k}^{3}δ ln(1 + 2k^{2}δ) ≥ c_{k}^{3}\sqrt{ln k}.

So if we set c_{k}^{3}\sqrt{ln k} = (ln k)^{1/4} for some k = k_{j} → ∞ as j → ∞; that is, if we choose c_{k} = \frac{1}{(ln k)^{1/12}} , for k = k_{j}, j = 1, 2, ⋯, condition (i) will be satisfied for all δ > 0.

For (ii), we set \frac{1}{(ln k)^{1/12}} = \frac{1}{j^{2}} which implies that k = e^{j^{24}} = k_{j}, j = 1, 2, · · ·. Then for

u_{0}(x) = ∑_{j=1}^{∞} \frac{1}{j^{2}} e^{j^{24}}e^{−(e^{2j^{24}} )x^{2}},

the Cauchy problem (2.1) has no local solution in L^{1}. □

2.2. Non-existence of Local Solutions for the Dirichlet Problem. In this section we will prove that there is no local L^{1} solution to the one dimensional Dirichlet problem

\begin{matrix}u_{t}(x, t) = u_{xx}(x, t) + u^{3}(x, t) in (−1, 1) × (0, T) \\ u(±1, t) = 0, \\ u(x, 0) = u_{0}(x) in (−1, 1), \end{matrix}  (2.2)

for some initial data u_{0} ∈ L^{1}. First by using the lower solution argument with a cutoff function we prove some estimates for a family of particular initial data and then we will construct the initial data u_{0} ∈ L^{1}(−1, 1) for which there is no local L^{1} solution to the Dirichlet problem (2.2).

Throughout this section we denote the solution of the Dirichlet problem (2.2) by u_{d}. Let u_{0}(x) = ke^{−k^{2}x^{2}}, then by Fourier series, the linear solution to the problem (2.2) is

u_{d}^{L}(x, t) = ∫_{−1}^{1} K_{d}(x, y, t)ke^{−k^{2}y^{2}}dy

where

K_{d}(x, y, t) = ∑_{j=1}^{∞} e^{−j^{2}π^{2}t} sin jπ(\frac{1 + y}{2} ) sin jπ(\frac{1 + x}{2})

is the Dirichlet kernel. Then the solution of (2.2) is

u_{d}(x, t) = u_{d}^{L}(x, t) + ∫_{0}^{t} ∫_{−1}^{1} K_{d}(x, y, t − s)u_{d}^{3}(y, s) dy ds.

The difficulty for the Dirichlet problem is that we do not have a good explicit formula for the solution in a bounded domain as we have for the Cauchy problem. In the proof for the Cauchy problem, ∫_{R} K_{c}(x, y, t − s)dx = 1 is used to simplify the argument. For the Dirichlet problem we do not have that but we know that ∫_{Ω} K_{d}(x, y, t)dx decays exponentially for t > 0.

The idea behind the proof of the non-existence of local solution for the Dirichlet problem is to use the non-existence of local solution for the Cauchy problem (which was proved in the previous section). To do this, we first construct a cutoff function and estimate the Dirichlet kernel K_{d} in terms of the Cauchy kernel K_{c}. Then using this estimate we find a lower bound for the linear solution of the Dirichlet problem u_{d}^{L} in terms of the linear solution of the Cauchy problem u_{c}^{L}. Combining the estimates for K_{d} and u_{d}^{L}, we can estimate ∥u_{d}∥_{L^{1}} and obtain the non-existence result of local solution for the Dirichlet problem.

In the first lemma below, we estimate the Dirichlet kernel K_{d} in terms of the Cauchy kernel K_{c} by constructing a cutoff function in a concentrated domain.

Lemma 2.2. For |x − y| ≤ \frac{1}{\sqrt{2}} and |x| ≤ \frac{1}{4},

K_{d}(x, y, t) ≥ e^{−12t}(\frac{1}{1 + (x − y)^{2}} − \frac{1}{1 + \frac{1}{2}} )K_{c}(x, y, t).

Proof. The crux of the proof of this lemma is to construct a cutoff function g so that the function

w(x, y, t) = gK_{c}(x, y, t) − K_{d}(x, y, t)

will satisfy the following conditions:

(i) L(w) ≤ 0 where L(w) = w_{t} − w_{xx},

(ii) w ≤ 0 on the boundary |x − y| = \frac{1}{\sqrt{2}} with |x| ≤ \frac{1}{4} and at t = 0.

Since L(K_{d}) = 0, we want w to satisfy

L(w) = L(gK_{c} − K_{d}) = L(gK_{c}) = L(g)K_{c} − 2∇K_{c}.∇g ≤ 0.

Note that K_{c}(x, y, t) = ^{\frac{e^{− \frac{|x−y|^{2}}{4t}}}{\sqrt{4πt}}} implies

∇K_{c} = \frac{−2(x − y)}{4t} \frac{e^{− \frac{|x−y|^{2}}{4t}}}{\sqrt{4πt}} = \frac{−(x − y)}{2t}K_{c}.

So g must satisfy

L(w) = L(g)K_{c} + \frac{(x − y)}{t}K_{c} · ∇g ≤ 0;

that is,

g_{t} − g_{xx} + \frac{(x − y)}{t}g_{x} ≤ 0

in the one-dimensional case. To satisfy this inequality and to have zero on the boundary (i.e., when y = x ± \frac{1}{\sqrt{2}}), we choose the cutoff function as

g(x, y, t) = e^{−αt}(\frac{1}{1 + (x − y)^{2}} − \frac{1}{1 + \frac{1}{2}} )

where α will be determined later. So we want

g_{t} − g_{xx} + \frac{(x − y)}{t}g_{x}

= −α(\frac{1}{1 + (x − y)^{2}} − \frac{1}{1 + \frac{1}{2}} ) − (\frac{6(x − y)^{2} − 2}{(1 + (x − y)^{2})^{3}} ) − \frac{2(x − y)^{2}}{t(1 − (x − y)^{2})^{2}} ≤ 0.

If |x − y| ≥ \frac{1}{\sqrt{3}}, then g_{t} − g_{xx} + \frac{(x−y)}{t}g_{x} ≤ 0 for any α ≥ 0. If |x − y| < \frac{1}{\sqrt{3}}, we need to find α so that this expression will be negative. So, if we choose

α ≥ \frac{\frac{2−6(x−y)^{2}}{(1+(x−y)^{2})^{3}}}{(\frac{1}{1+(x−y)^{2}} − \frac{1}{1+ \frac{1}{2}} )};

i.e., if α ≥ 12, then we have

g_{t} − g_{xx} + \frac{(x − y)}{t}g_{x} ≤ 0.

Set α = 12. Then the cutoff function is

g(x, y, t) = e^{−12t}(\frac{1}{1 + (x − y)^{2}} − \frac{1}{1 + \frac{1}{2}} ).

So L(w) ≤ 0 and (i) is satisfied.

Now we show that condition (ii) will be satisfied by w with this cutoff function; i.e., we will show that w ≤ 0 on the boundary and at t = 0. Since g(x, x± \frac{1}{\sqrt{2}}, t) = 0, w ≤ 0 on the boundary |x − y| = \frac{1}{\sqrt{2}} and |x| ≤ \frac{1}{4}. To show that w ≤ 0 at t = 0, it is sufficient to show that

lim_{t→0} ∫_{x− \frac{1}{\sqrt{2}}}^{x+ \frac{1}{\sqrt{2}}} w(x, y, t)h(y)dy ≤ 0 for every h ≥ 0.

Since

lim_{t→0} ∫_{x− \frac{1}{\sqrt{2}}}^{x+ \frac{1}{\sqrt{2}}} w(x, y, t)h(y)dy = ∫_{x− \frac{1}{\sqrt{2}}}^{x+ \frac{1}{\sqrt{2}}} (g(x, y, t)K_{c}(x, y, t) − K_{d}(x, y, t))h(y)dy
≤ Ch(x) − h(x)

where g ≤ C ≤ 1, we have w ≤ 0 at t = 0. Hence, by the maximum principle, (i) and (ii) imply that w ≤ 0 inside the domain, which implies

K_{d}(x, y, t) ≥ e^{−12t}(\frac{1}{1 + (x − y)^{2}} − \frac{1}{1 + \frac{1}{2}} )K_{c}(x, y, t)

when |x − y| ≤ \frac{1}{\sqrt{2}} and |x| ≤ \frac{1}{4}. □

In the following lemma we use the estimate for K_{d} to get an estimate for u_{d}^{L} which is the linear solution of the Dirichlet problem.

Lemma 2.3. For |x| ≤ \frac{1}{4}, u_{d}^{L}(x, t) ≥ c_{1}e^{−12t}u_{c}^{L}(x, t) with u_{d}(x, 0) = ke^{−k^{2}x^{2}}, where c_{1} is a uniform constant independent of k.

Proof. To show this estimate for the linear solution of the Dirichlet problem, the main idea is to use the crucial estimate for the Dirichlet kernel in terms of the Cauchy kernel that we proved in Lemma 2.2. The linear solution for the Dirichlet problem is

u_{d}^{L}(x, t) = ∫_{−1}^{1} K_{d}(x, y, t)ke^{−k^{2}y^{2}}dy.

Using the estimate of Lemma 2.2. for K_{d}(x, y, t), we have

u_{d}^{L}(x, t) = ∫_{−1}^{1} K_{d}(x, y, t)ke^{−k^{2}y^{2}}dy

≥ ∫_{x− \frac{1}{\sqrt{2}}}^{x+ \frac{1}{\sqrt{2}}} K_{d}(x, y, t)ke^{−k^{2}y^{2}}dy (since |x − y| ≤ \frac{1}{\sqrt{2}})

≥ ∫_{x− \frac{1}{\sqrt{2}}}^{x+ \frac{1}{\sqrt{2}}} e^{−12t}(\frac{1}{1 + (x − y)^{2}} − \frac{1}{1 + \frac{1}{2}} )K_{c}(x, y, t)ke^{−k^{2}y^{2}}dy

≥ ∫− \frac{1}{4}^{\frac{1}{4}} ce^{−12t}K_{c}(x, y, t)ke^{−k^{2}y^{2}}dy

(since |x| ≤ \frac{1}{4} implies (\frac{1}{1 + (x − y)^{2}} − \frac{1}{1 + \frac{1}{2}} ) ≥ c)

= ce^{−12t} ∫− \frac{1}{4}^{\frac{1}{4}} \frac{e^{− \frac{(x−y)^{2}}{4t}}}{\sqrt{4πt}} ke^{−k^{2}y^{2}}dy

= ce^{−12t} \frac{e^{\frac{−x^{2}}{4t}}}{\sqrt{4πt}} ∫ \frac{k}{4} ; \frac{−k}{4} e^{\frac{xs}{2kt} −( \frac{1+4k^{2}t}{4k^{2}t})s^{2}}ds (where s = ky)

= ce^{−12t} \frac{e^{\frac{−x^{2}}{4t} + \frac{k^{2}x^{2}}{4k^{2}t(1+4k^{2}t)}}}{\sqrt{4πt}} ∫ \frac{k}{4} ; \frac{−k}{4} e^{−( \frac{1+4k^{2}t}{4k^{2}t})(s− \frac{kx}{1+4k^{2}t})^{2}}ds

= ce^{−12t} \frac{e^{\frac{−k^{2}x^{2}}{1+4k^{2}t}}}{\sqrt{4πt}} \begin{matrix}√ & \overline{4k^{2}t} \\ √ & \overline{1 + 4k^{2}t} \end{matrix} ∫_{u_{1}}^{u_{2}} e^{−u^{2}} du

where u = \sqrt{\frac{1+4k^{2}t}{4k^{2}t}}(s− \frac{kx}{1+4k^{2}t}), u_{1} = \sqrt{\frac{1+4k^{2}t}{4k^{2}t}}(−\frac{k}{4} − \frac{kx}{1+4k^{2}t}) and u_{2} = \sqrt{\frac{1+4k^{2}t}{4k^{2}t}}(\frac{k}{4} − \frac{kx}{1+4k^{2}t}). The above expression is greater than or equal to

\begin{cases}ce^{−12t}^{\frac{ke^{\frac{−k^{2}x^{2}}{1+4k^{2}t}}}{\sqrt{π}\sqrt{1+4k^{2}t}}} ∫_{0}^{u_{2}} e^{−u^{2}}du & if − \frac{1}{4} ≤ x < 0, \\ ce^{−12t}^{\frac{ke^{\frac{−k^{2}x^{2}}{1+4k^{2}t}}}{\sqrt{π}\sqrt{1+4k^{2}t}}} ∫_{u_{1}}^{0} e^{−u^{2}}du & if 0 ≤ x ≤ \frac{1}{4} \end{cases}

≥ \frac{c}{\sqrt{π}} e^{−12t}u_{c}^{L}(x, t)c′

≥ c_{1}e^{−12t}u_{c}^{L}(x, t),

where c_{1} = c′ \frac{c}{\sqrt{π}}, c′ = min(c_{1}′, c_{2}′), ∫_{u_{1}}^{0} e^{−u^{2}}du ≥ c_{1}′, and ∫_{0}^{u_{2}} e^{−u^{2}}du ≥ c_{2}′. □

Now using the estimates in Lemmas 2.2 and 2.3, we prove that for some initial data u_{0} ∈ L^{1} there is no local solution to the Dirichlet problem (2.2) in L^{1}. Lemma 2.4. For any fixed δ > 0, we have ∥u_{d}∥_{L^{1}((−1,1)×(0,δ))} = ∞ with u_{0} = ∑_{k} c_{k}ke^{−k^{2}x^{2}} where the c_{k}’s will be determined later. Proof. For the solution of the Dirichlet problem, for any fixed δ > 0,

∥u_{d}∥_{L^{1}((−1,1)×(0,δ))} = ∫_{0}^{δ} ∫_{−1}^{1} u_{d}(x, t) dx dt

> ∫_{0}^{δ} ∫_{−1}^{1} ∫_{0}^{t} ∫_{−1}^{1} K_{d}(x, y, t − s)(u_{d}^{L}(y, s))^{3} dy ds dx dt

> ∫_{0}^{δ} ∫− \frac{1}{4}^{\frac{1}{4}} ∫_{0}^{t} ∫− \frac{1}{4}^{\frac{1}{4}} K_{d}(x, y, t − s)(u_{d}^{L}(y, s))^{3} dy ds dx dt

= ∫_{0}^{δ} ∫_{0}^{t} ∫− \frac{1}{4}^{\frac{1}{4}} (u_{d}^{L}(y, s))^{3} ∫− \frac{1}{4}^{\frac{1}{4}} K_{d}(x, y, t − s) dx dy ds dt.

Then using Lemma 2.2, we get

∫− \frac{1}{4}^{\frac{1}{4}} K_{d}(x, y, t − s)dx ≥ ∫ \frac{1}{4} ; \frac{−1}{4} ce^{−12(t−s)}K_{c}(x, y, t − s)dx

≥ ce^{−12(t−s)} ∫ \frac{1}{4} ; \frac{−1}{4} \frac{e^{− \frac{(x−y)^{2}}{4(t−s)}}}{\sqrt{4π(t − s)}}dx

= \frac{ce^{−12(t−s)}}{\sqrt{π}} ∫ \frac{\frac{1}{4} −y}{\sqrt{4(t−s)}} ; \frac{− \frac{1}{4} −y}{\sqrt{4(t−s)}} e^{−u^{2}}du (where u = \frac{x−y}{\sqrt{4(t−s)}})

≥ \frac{ce^{−12(t−s)}}{\sqrt{π}} ∫0^{\frac{\frac{1}{4} −y}{\sqrt{4(t−s)}}} e^{−u^{2}}du

≥ Ce^{−12(t−s)}

for some constant C. Using Lemma 2.3 for u_{d}^{L} and the fact that

(u_{c}^{L}(y, s))^{3} ≥ ∑_{k} \frac{c_{k}^{3}k^{3}e^{− \frac{3k^{2}y^{2}}{1+4k^{2}s}}}{(1 + 4k^{2}s)^{3/2}} ,

we have

∥u_{d}∥_{L^{1}((−1,1)×(0,δ))} ≥ ∫_{0}^{δ} ∫_{0}^{t} ∫− \frac{1}{4}^{\frac{1}{4}} (u_{d}^{L}(y, s))^{3} ∫− \frac{1}{4}^{\frac{1}{4}} K_{d}(x, y, t − s)dx dy ds dt

≥ ∫_{0}^{δ} ∫_{0}^{t} ∫− \frac{1}{4}^{\frac{1}{4}} (u_{d}^{L}(y, s))^{3}(Ce^{−12(t−s)}) dy ds dt

≥ ∫_{0}^{δ} ∫_{0}^{t} ∫ \frac{1}{4} ; \frac{−1}{4} Ce^{−12(t−s)}(c_{1}e^{−12s}u_{c}^{L}(y, s))^{3} dy ds dt

= ∫_{0}^{δ} ∫_{0}^{t} Cc_{1}e^{−12t−24s} ∫ \frac{1}{4} ; \frac{−1}{4} (u_{c}^{L}(y, s))^{3} dy ds dt

≥ ∫_{0}^{δ} ∫_{0}^{t} Cc_{1}e^{−12t−24s} ∫ \frac{1}{4} ; \frac{−1}{4} ∑_{k} \frac{c_{k}^{3}k^{3}e^{\frac{−3k^{2}y^{2}}{1+4k^{2}s}}}{(1 + 4k^{2}s)^{3/2}} dy ds dt

≥ ∑_{k} C′ ∫_{0}^{δ} ∫_{0}^{t} \frac{c_{k}^{3}k^{2}}{1 + 4k^{2}s} ds dt

= ∑_{k} C′ ∫_{0}^{δ} \frac{1}{4}c_{k}^{3} ln(1 + 4k^{2}t)dt,

for some constant C′. □

Now as we discussed for the Cauchy problem in previous section, we choose c_{k} = \frac{1}{(ln k)^{1/12}} which will imply that k = e^{j^{24}} = k_{j}, for j = 1, 2, · · ·. So for

u_{0}(x) = ∑_{j=1}^{∞} \frac{1}{j^{2}} e^{j^{24}}e^{−(e^{2j^{24}} )x^{2}},

using Lemmas 2.2, 2.3 and 2.4 we can prove the following theorem. Theorem 2.5. For u_{0} = ∑_{j=1}^{∞} \frac{1}{j^{2}} e^{j^{24}}e^{−(e^{2j^{24}} )x^{2}}, the Dirichlet problem (2.2) has no non-negative local mild solution in L^{1}(−1, 1).

We have also generalized this result to n-dimensional Dirichlet problem which is

u_{t}(x, t) = Δu(x, t) + |u(x, t)|^{\frac{2}{n}} u(x, t) in Ω × (0, T),

u(x, t) = 0 on ∂Ω × (0, T),
u(x, 0) = u_{0}(x) in Ω,

(2.3)

where Ω = (−1, 1) × ⋯ × (−1, 1) ⊂ ℝ^{n} and obtained the following result.

Theorem 2.6. For u_{0}(x) = ∑_{j=1}^{∞} \frac{1}{j^{2}} e^{nj^{\frac{16+8n}{n}}} e^{−(e^{2j \frac{16+8n}{n}} )|x|^{2}} ∈ L^{1} , the n - dimensional Dirichlet problem has no non-negative local mild solution in L^{1}((−1, 1) × ⋯ × (−1, 1)).

We have also considered the general nonlinearity for Dirichlet boundary value problem in [3] and proved the following result.

Corollary 2.7. The Dirichlet problem u_{t}(x, t) = Δu(x, t) + f(u) in Ω × (0, T),

u(x, t) = 0 on ∂Ω × (0, T),
u(x, 0) = u_{0}(x) in Ω,

(2.4)

where Ω ⊂ ℝ^{n} is any smooth bounded domain and f(u) ≥ |u|^{\frac{2}{n} +1}, has no nonnegative local mild solution for some u_{0} ≥ 0, in L^{1}.

3. Non-existence of Local Solutions for the Mixed Boundary Condition

In this section, we study the same problem with the mixed boundary conditions; i.e., considering the boundary conditions in terms of u and the normal derivative of u, which is denoted by \frac{∂u}{∂n}. Let ũ be the solution of the one-dimensional mixed boundary value problem,

\begin{matrix}ũ_{t}(x, t) = ũ_{xx}(x, t) + ũ^{3}(x, t) in (−1, 1) × (0, T), \\ \frac{∂ũ}{∂n}(x, t) + βũ(x, t) = 0 x = ±1 in (0, T), \\ ũ(x, 0) = u_{0} in (−1, 1) \end{matrix}  (3.1)

where β ≥ 0 is a constant.

First we prove the non-existence of local solution for the one-dimensional mixed boundary value problem (3.1) and then we generalize this result to the n - dimensional case. The main idea of the proof is to use the comparison principle for the kernels of heat semigroups with Dirichlet and mixed boundary conditions.

Recall that one-dimensional Dirichlet problem is

\begin{matrix}u_{t}(x, t) = u_{xx}(x, t) + u^{3}(x, t) in (−1, 1) × (0, T), \\ u(±1, t) = 0, \\ u(x, 0) = u_{0}(x) in (−1, 1). \end{matrix}  (3.2)

where u_{0}(x) = ∑_{j=1}^{∞} \frac{1}{j^{2}} e^{j^{24}}e^{−(e^{2j^{24}} )x^{2}} for which there is no local solution.

Let K(t) and K̃(t) be the heat kernels on (−1, 1) with Dirichlet and mixed boundary conditions respectively. Now we claim that K̃(t) ≥ K(t) on (−1, 1). In fact, let u_{0} be any positive initial data, and let u and ũ be the solution to linear heat equation with Dirichlet and mixed boundary conditions respectively. Hence,

\begin{matrix}(ũ − u)_{t} = (ũ − u)_{xx} in (−1, 1) × (0, T), \\ \frac{∂(ũ − u)}{∂n} + β(ũ − u) = −\frac{∂u}{∂n} ≥ 0, x = ±1, t ∈ (0, T), \\ (ũ − u)(x, 0) = 0 x ∈ (−1, 1), \end{matrix}  (3.3)

and by the maximum principle, we have ũ ≥ u; i.e., K̃(t) ≥ K(t) on (−1, 1).

Now suppose for the same initial data u_{0}(x) = ∑_{j=1}^{∞} \frac{1}{j^{2}} e^{j^{24}}e^{−(e^{2j^{24}} )x^{2}} problem (3.1) has a solution ũ ∈ C([0.T], L^{1}(−1, 1)), then

ũ(t) = K̃(t)u_{0} + H̃(ũ)(t) ≥ K(t)u_{0} + H(ũ)(t),

where H(u)(t) = ∫_{0}^{t} K(t − s)u^{3}(s)ds and H̃(u)(t) = ∫_{0}^{t} K̃(t − s)u^{3}(s)ds. Let ũ(t) = u_{1}(t), and define by iteration,

u_{k+1}(t) = K(t)u_{0} + H(u_{k})(t).

It follows by induction that

K(t)u_{0} ≤ u_{k+1}(t) ≤ u_{k}(t).

Thus, the u_{k}(t) converge v(t) (by monotone convergence theorem) which must be a solution of the integral equation v(t) = K(t)u_{0} + H(v)(t), which is a contradiction that proves the following theorem for the one dimensional mixed boundary value problem.

Theorem 3.1. For u_{0}(x) = ∑_{j=1}^{∞} \frac{1}{j^{2}} e^{j^{24}}e^{−(e^{2j^{24}} )x^{2}} there is no local L^{1} solution to the one-dimensional mixed boundary value problem (3.1).

Since we can generalize the non-existence result of local solution to the ndimensional Dirichlet problem, we can also obtain the non-existence of local solution for the n-dimensional mixed boundary value problem by the similar comparison argument as above.

Theorem 3.2. For the n-dimensional mixed boundary value problem,

ũ_{t}(x, t) = Δũ(x, t) + |ũ(x, t)|^{\frac{2}{n}} ũ(x, t) in (0, T) × Ω,

\frac{∂ũ}{∂n} + βũ = 0 on (0, T) × ∂Ω,

ũ(x, 0) = u_{0}(x) in Ω,

(3.4)

where Ω ⊂ ℝ^{n} and u_{0}(x) = ∑_{j=1}^{∞} \frac{1}{j^{2}} e^{nj^{\frac{16+8n}{n}}} e^{−(e^{2j \frac{16+8n}{n}} )|x|^{2}} ∈ L^{1}, there is no nonnegative local solution in L^{1}.

In the previous sections we give the non-existence results of local solution for some initial data u_{0} ∈ L^{1} for the Dirichlet and mixed boundary value problems. However in the next section, we establish the existence of global solution for some u_{0} ∈ L^{1+ϵ} sufficiently small.

4. Existence of a Global Solution for Small Initial Data

In this section, we give sufficient conditions on q and u_{0} for the existence of global solution to the Dirichlet problem

\begin{matrix}u_{t}(x, t) = u_{xx}(x, t) + u^{3}(x, t) in (−1, 1) × (0, T) \\ u(±1, t) = 0 in (0, T) \\ u(x, 0) = u_{0} in (−1, 1). \end{matrix}  (4.1)

where u_{0} ∈ L^{q}(−1, 1). Mainly we will prove that if the initial condition u_{0} ∈ L^{1+ϵ} for any fixed ϵ > 0, then the Dirichlet problem (4.1) has a global solution in L^{1+ϵ} for ∥u_{0}∥_{1+ϵ} sufficiently small.

For notational simplicity, we will use the following form of the solution throughout this section;

u(t) = e^{tΔ}u_{0} + ∫_{0}^{t} e^{(t−s)Δ}u^{3}(s)ds

where e^{tΔ}u_{0} denotes the linear solution of problem (4.1). Before proving the global existence result, we give some interpolation inequalities for the linear solution of (4.1).

Remark 4.1. Let 1 < p < 2. Then it is easy to see that for all t ≥ 0, e^{tΔ} : L^{1} → L^{1} with norm M_{1} ≤ 1; i.e.,

∥e^{tΔ}ϕ∥_{1} ≤ ∥ϕ∥_{1},

and e^{tΔ} : L^{2} → L^{2} with norm M_{2} = e^{tγ_{1}} where γ_{1} is the first (negative) eigenvalue of the Laplacian with vanishing Dirichlet boundary condition; i.e.,

∥e^{tΔ}ϕ∥_{2} ≤ e^{tγ_{1}}∥ϕ∥_{2}.

Now we recall the following interpolation theorem.

Theorem 4.2 (Nirenberg [11]). Let T (t) be a continuous linear mapping of L^{p} into L^{p} with norm M_{1} and L^{q} into L^{q} with norm M_{2}. Then T (t) is a continuous mapping of L^{r} into L^{r} with the norm M ≤ M_{1}^{λ}M_{2}^{1−λ} where 1 ≤ p ≤ r ≤ q ≤ ∞ and \frac{1}{r} = \frac{λ}{p} + \frac{1−λ}{q}.

By this theorem, we conclude that e^{tΔ} : L^{p} → L^{p} with the norm M ≤ M_{1}^{λ}M_{2}^{1−λ} with λ = \frac{2−p}{p}. Hence

∥e^{tΔ}ϕ∥_{p} ≤ e^{tγ_{1} ^{\frac{2p−2}{p}}} ∥ϕ∥_{p} for 1 < p < 2.

For p ≥ 2, we can get a similar estimate by the following interpolation argument. e^{tΔ} : L^{2} → L^{2} with norm M_{1} = e^{tγ_{1}} as above. Note that for all t ≥ 0, e^{tΔ} : L^{∞} → L^{∞} with norm M_{2} ≤ 1 by the maximum principle; i.e.,

∥e^{tΔ}ϕ∥_{∞} ≤ ∥ϕ∥_{∞}.

Then again by the interpolation theorem [11] of Nirenberg, we have e^{tΔ} : L^{p} → L^{p} with the norm M ≤ M_{1}^{λ}M_{2}^{1−λ} with λ = \frac{2}{p}; i.e.,

∥e^{tΔ}ϕ∥_{p} ≤ e^{tγ_{1} ^{\frac{2}{p}}} ∥ϕ∥_{p} ∀p ∈ [2, ∞).

First by the following lemma, we get the estimate on ∥u(s)∥_{∞} in terms of ∥u_{0}∥_{1+ϵ}, which is a crucial estimate to prove the global existence theorem.

Lemma 4.3. ∥u(s)∥_{∞} ≤ G(M, T)∥u_{0}∥_{1+ϵ}s^{− \frac{1}{2(1+ϵ)}} , where G(M, T) : R^{+} × R^{+} → R^{+} is an increasing function of M and T respectively when ∥u(s)∥_{1+ϵ} ≤ M + 1 on [0, T].

For the proof of the above lemma, we use the following local existence theorem by Brezis and Cazenave (Theorem 1 in [2]).

Theorem 4.4. Assume q > n(p − 1)/2 (resp. q = n(p − 1)/2) and q ≥ 1 (resp. q > 1), n ≥ 1. Given any u_{0} ∈ L^{q}, there exist a time T = T (u_{0}) > 0 and a unique function u ∈ C([0, T], L^{q}) with u(0) = u_{0}, which is a classical solution of (4.1). Moreover, we have smoothing effect and continuous dependence; namely,

∥u − v∥_{L^{q}} + t^{n/2q}∥u − v∥_{L^{∞}} ≤ C∥u_{0} − v_{0}∥_{L^{q}}.

for all t ∈ (0, T] where T = min(T (u_{0}), T (v_{0})) and C can be estimated in terms of ∥u_{0}∥_{L^{q}} and ∥v_{0}∥_{L^{q}}.

Proof of Lemma 4.3. By replacing q = 1 + ϵ and p = 3 in this theorem, we get an L^{∞} estimate for the solution u of problem (4.1),

∥u(s)∥_{∞} ≤ G(M, T)∥u_{0}∥_{1+ϵ}s^{− \frac{1}{2(1+ϵ)}}

where G(M, T) = e^{c_{1}(M+1)^{\frac{6+6ϵ}{2+3ϵ}} T ^{\frac{3ϵ}{2+3ϵ}}} and c_{1} = c_{1}(ϵ). □

Next we prove the following global existence theorem.

Theorem 4.5. For any fixed ϵ > 0, there is a δ > 0 such that ∥u_{0}∥_{1+ϵ} ≤ δ implies that u(t) globally exists and

∥u(t)∥_{1+ϵ} ≤ 2δ for 0 ≤ t ≤ ∞.

Proof. Let us choose δ > 0 such that \frac{1+ϵ}{ϵ}(G(2δ, 1)δ)^{2} < −γ and e^{^{\frac{1+ϵ}{ϵ}} (G(2δ,1)δ)^{2}} < 2. By the local existence theorem that is stated in Lemma 4.3, there exists T_{1} > 0 such that u(t) exists on [0, T_{1}] and ∥u(t)∥_{1+ϵ} ≤ 2δ for t ∈ [0, T_{1}]. Now we will prove the following claims to prove the global existence result. Claim 1. ∥u(t)∥_{1+ϵ} < 2δ for 0 ≤ t ≤ 1. Claim 2. ∥u(1)∥_{1+ϵ} ≤ δ. Proof of Claim 1. Suppose Claim 1 is not true. Then set

T̃ = min{t > 0|∥u(t)∥_{1+ϵ} ≥ 2δ}.

We have ∥u(T̃)∥_{1+ϵ} = 2δ and ∥u(t)∥_{1+ϵ} < 2δ for all 0 ≤ t < T̃ with 0 < T̃ ≤ 1. Consequently, for any t ∈ [0, T̃], the Remark 4.1 in this section indicates that there is a γ < 0 (γ = γ_{1} \frac{2ϵ}{1+ϵ}) such that

∥u(t)∥_{1+ϵ} ≤ e^{γt}∥u_{0}∥_{1+ϵ} + ∫_{0}^{t} ∥e^{(t−s)Δ}u^{3}(s)∥_{1+ϵ}ds
≤ e^{γt}∥u_{0}∥_{1+ϵ} + ∫_{0}^{t} e^{γ(t−s)}∥u(s)∥_{1+ϵ}∥u(s)∥_{∞}^{2}ds
≤ e^{γt}∥u_{0}∥_{1+ϵ} + ∫_{0}^{t} e^{γ(t−s)}∥u(s)∥_{1+ϵ}Bs^{− \frac{1}{1+ϵ}} ds.

where B = (G(2δ, T̃))^{2}∥u_{0}∥_{1+ϵ}^{2} for notational simplicity. Then

e^{−γt}∥u(t)∥_{1+ϵ} ≤ ∥u_{0}∥_{1+ϵ} + B ∫_{0}^{t} e^{−γs}∥u(s)∥_{1+ϵ}s^{− \frac{1}{1+ϵ}} ds := H(t),

where, we denote the right-hand side of the above inequality by H(t). Then

H′(t) = Be^{−γt}∥u(t)∥_{1+ϵ}t^{− \frac{1}{1+ϵ}} ≤ BH(t)t^{− \frac{1}{1+ϵ}} .

Hence \frac{H′(t)}{H(t)} ≤ Bt^{− \frac{1}{1+ϵ}}. Integrating from 0 to t yields

ln \frac{H(t)}{H(0)} ≤ \frac{1 + ϵ}{ϵ}Bt^{\frac{ϵ}{1+ϵ}} and H(t) ≤ H(0)e^{\frac{1+ϵ}{ϵ} Bt^{\frac{ϵ}{1+ϵ}}} ,

which implies

∥u(t)∥_{1+ϵ} ≤ e^{γt}H(t) ≤ ∥u_{0}∥_{1+ϵ}e^{γt}e^{\frac{1+ϵ}{ϵ} Bt^{\frac{ϵ}{1+ϵ}}} .  (4.2)

Note that for t ∈ [0, T̃] ⊂ [0, 1],

e^{γt}e^{\frac{1+ϵ}{ϵ} Bt^{\frac{ϵ}{1+ϵ}}} ≤ e^{\frac{1+ϵ}{ϵ} B} ≤ e^{\frac{1+ϵ}{ϵ} (G(2δ,1))^{2}δ^{2}} < 2,

by the choice of δ. Then by (4.2), ∥u(T̃)∥_{1+ϵ} < 2δ which is contrary to the fact that ∥u(T̃)∥_{1+ϵ} = 2δ. So the proof of Claim 1 is completed. Proof of Claim 2. After proving Claim 1, Claim 2 follows immediately by the inequality (4.2) since

∥u(1)∥_{1+ϵ} ≤ δe^{γ+ \frac{1+ϵ}{ϵ} B} ≤ δ

by the choice of δ.

Finally, the existence of a global solution will follow by a simple induction argument. By Claim 1 and 2, we can easily conclude that ∥u(j)∥_{1+ϵ} ≤ δ for each j = 0, 1, . . .. Using u(j) as new initial data, we can solve the problem for t ∈ [j, j + 1], and we get ∥u(t)∥_{1+ϵ} ≤ 2δ for all t ∈ [j, j + 1] by Claim 2. □

As a conclusion, we can state the generalized global existence result as follows.

Corollary 4.6. For q > 1, problem (4.1) has a global solution in L^{q} for all u_{0} ∈ L^{q} with ∥u_{0}∥_{q} sufficiently small.

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Canan Celik Departmane to Mathematics, TOBB Economics and Technology University, Sögütözü cad. No. 43. 06560 Sögütözü, Ankara, Turkey

E-mail address: canan.celik@etu.edu.tr