A bifurcation for a generalized Burger's equation in dimension one

We consider a generalized Burger's equation (dtu = dxxu - udxu + up - {\lambda}u)in a subdomain of R, under various boundary conditions. First, using some phase plane arguments, we study the existence of stationary solutions under Dirichlet or Neumann boundary conditions and prove a bifurcation depending on the parameters. Then, we compare positive solutions of the parabolic equation with appropriate stationary solutions to prove that global existence can occur for the Dirichlet, the Neumann or the dissipative dynamical boundary conditions {\sigma}dtu+d{\nu}u = 0. Finally, for many boundary conditions, global existence and blow up phenomena for solutions of the nonlinear parabolic problem in an unbounded domain are investigated by using some standard super-solutions and some weighted L1-norms.


Introduction
Let Ω be a domain of the real line R, not necessarily bounded. Let p be a real number with p > 1, λ ∈ R and ϕ a non-negative continuous function in Ω. Consider the following nonlinear parabolic problem   where B(u) = 0 stands for the Dirichlet boundary conditions (u = 0), the Neumann boundary conditions (∂ ν u = 0) or the dynamical boundary conditions (σ∂ t u + ∂ ν u = 0 with σ a non-negative smooth function). In the first section, we study the stationary equation stemming from Problem 1. We aim to prove the existence of positive and signchanging solutions using phase plane arguments and dealing with the first order system We prove a bifurcation in the phase plane of this system, depending on the parameters λ and p, which influences the resolution of Equation 2 under the Dirichlet, Neumann and mixed boundary conditions. Then in a second section, using the comparison method from [2], we deduce from the solutions of the stationary Equation 2 some regular super-solutions for the Problem 1. Dealing with these super-solutions and with the blow-up results from [4], we investigate global existence and blow-up phenomena for the Problem 1 for different values of λ and p, and for the Dirichlet, the Neumann and the dynamical boundary conditions. We also examine both phenomena in unbounded domains: we obtain global existence results with the comparison method and using some well-known super-solutions (we mean explicit functions) for the Dirichlet, the Neumann and the dynamical boundary conditions. The blowing-up concerns the regular solutions of Problem 1 satisfying some growth order at infinity and some boundary conditions such that • ∂ ν u = 0 (Neumann b.c.), • ∂ ν u = g(u) with g a polynomial of degree 2 (nonlinear b.c.).
We use some weighted L 1 −norms: our technique is to prove the blowing-up of the solution by proving the blowing-up of appropriate L 1 −norms.
Before starting, let us define the kind of solution we look for: Definition 1.1. A function u is called a solution (or regular solution) of Equation 2 in Ω if u is of class C 2 (Ω) and satisfies the equation in the classical sense. A function u is called a solution (or regular solution) of Problem 1 in Ω if u is of class C(Ω × [0, T )) ∩ C 2,1 (Ω × (0, T )), where denotes T its maximal existence time, and satisfies the equations in the classical sense.

Stationary equation
In this section, we study the existence of positive and sign-changing solutions of Equation 2 using a phase plane method. For the theory of phase planes (nature of equilibrium, regularity, behaviour and uniqueness of trajectory), we refer to H.Amann's book [1]. Unless otherwise stated, we suppose p ∈ (1, ∞). First, we can note that System 3 has three equilibrium points if λ > 0: (0, 0), (λ p−1 ≥ 0 with degeneracy when equality occurs). If λ ≤ 0, then (0, 0) is the only equilibrium point of System 3. We will prove later that (0, 0) is a center.

Case λ > 0
Let λ be a positive real number and p > 1. We want to study the phase plane of the System 3. First we prove a lemma on the symmetry of the trajectories: Thus, we can reduce our phase plane analysis to the half plane R + × R. In order to draw the phase plane of the System 3, we write the ordinate v as a function depending on the abscissa u: v = f (u). We do not know the function f , but we can deduce its variations and convexity using the equations 3. For the variations, we have Then we have dv du > 0 in the sets {u > 0, v > 0, v > |u| p−1 − λ} and {u > 0, v < 0, v < |u| p−1 − λ}. On the other hand, dv du < 0 in the sets {u > 0, v < 0, v > |u| p−1 − λ} and {u > 0, v > 0, v < |u| p−1 − λ}. Next, we compute the convexity of the function f and we obtain We Proof. Let v 0 > 0 and consider (u, v) the solution of the System 3 with initial data (u(0), v(0)) = (0, v 0 ). The calculus of the variations (see Equation 4) ensures that (u(t), v(t)) ∈ A for small t > 0. We prove that there exist 0 < τ < ∞ such that v(τ ) = |u(τ )| p−1 − λ. It means that (u, v) is bounded in A. Since (u, v) belongs to A, we have Then dv du ≥ 0 in A implies v > v 0 as long as (u, v) ∈ A, and we obtain dv du ≤ u 1 + λ v 0 .

Integration gives
If p > 3, the intersection {v = |u| p−1 −λ}∩{v = 1 2 1+ λ v0 u 2 +v 0 } is non-empty for all v 0 > 0. If p = 3, we need to choose v 0 sufficiently big such that Then, the trajectory (u, v) belongs to the compact and, using dv du ≥ 0, we know that there exist 0 < τ < ∞ such that v(τ ) = |u(τ )| p−1 − λ. This argument proves that each solution of the System 3 with initial data (u(0), v(0)) = (0, v 0 ) is bounded in A if v 0 is big enough. Thanks to uniqueness of solution, it also proves the result for all the solutions initiated in A. Proof. Thanks to the symmetry (see Lemma 2.1), we just need to prove that for some initial data belonging to {0} × (0, ∞), there exists a trajectory which attains a point belonging to {0} × (−∞, 0). First, consider a trajectory (u, v) initiated at (0, v 1 ) with v 1 > 0. According to hypothesis 6, we know that (u, v) is bounded, and using its variations and its convexity (Equations 4 and 5), we can deduce that (u, v) attains the x−axis at a point (u 1 , 0) with u 1 > λ 1 p−1 (see Figure 1). Then, using the reverse system and one of its trajectories initiated at (0, v 2 ) with v 2 < −λ) (trajectories of reverse system and of System 3 have same support) , one can note that for u 0 > λ 1 p−1 , there exist a trajectory (w, z) with w(0) = u 0 and z(0) = 0 (see Figure 1). Finally, let us consider the trajectory (a, b) of System 3 containing the point (u 2 , 0). Thanks to the uniqueness of the solutions, and using the information on the variations and the convexity, we deduce that there exist two real numbers s < t such that a(s) = a(t) = 0, b(s) = v 0 and b(t) = v 3 (see Figure 1). Thus, the trajectory (a, b) is the periodic trajectory we look for. Now, analysing the phase plane of the System 3, we deduce the following results concerning the equation 1.
Theorem 2.4. Assume hypothesis 6 and λ > 0. For some α > 0 and for each boundary conditions there exists a unique positive solution of the Equation 1 Proof. We use the phase plane of System 3, see For the Neumann solution, consider 0 < µ 0 < λ 1 p−1 and the trajectory (µ, ν) of System 3 initiated at (µ 0 , 0). Since (µ, ν) can not cross the trajectory (u, v) ( see Figure 1), it must cross the x−axis at (µ 1 , 0) with λ 1 p−1 < µ 1 < u 1 . Thus, the abscissa of this trajectory is the Neumann solution we look for. Uniqueness of solution comes from standard ODE's theorems applied to the System 3. Finally, the length (2α) of the existence interval is governed by the time needed by the trajectory to go from its initial data to its "final data". Theorem 2.5. Assume hypothesis 6 and λ > 0. For some α > 0, there exists a periodic sign-changing solution of the Equation 1 Proof. We just need to choose one the periodic trajectories of the System 3 built in Lemma 2.3.

Remark 1.
Using the periodic solutions in the previous theorem, we can build four sign-changing solutions satisfying the four boundary conditions: Dirichlet, Neumann, mixed−1 and mixed−2 (see Theorem 2.4). Now, suppose that hypothesis 6 is not achieved. Then, we do not know if the solutions are bounded in {v > |u| p−1 − λ}: we will see in §2.3 that unbounded solutions appear. But in {v < |u| p−1 − λ}, the behaviour of the trajectories do not change.
Theorem 2.6. Let λ > 0. For some α > 0, there exists a unique positive solution of the Equation 1 with the mixed boundary conditions u (−α) = u(α) = 0. In addition, if then there exists a unique positive solution of the equation 1 under the Neumann boundary conditions.
Without hypothesis 6, we can not construct positive solutions anymore for the Dirichlet, Neumann or mixed−1 boundary conditions. If we do not impose the positivity, we obtain this result: Theorem 2.7. Let λ > 0. For some α > 0 and for each boundary conditions there exists a solution of the Equation 1 u − uu + u|u| p−1 − λu = 0 in (−α, α).
Proof. As we mentioned before, we consider the part {v < |u| p−1 − λ} of the phase plane of the System 3 (see Figure 1). For the Neumann solution, we consider the trajectory (a, b) between (u 2 , 0) and (−u 2 , 0). For the mixed−1 solution, we can also consider the trajectory (a, b), but only between (0, v 3 ) and (−u 2 , 0). Remark 2. The Neumann solution built above is sign changing, whereas the mixed−1 solution is negative.
Remark 3. In the general case, we can not build any solution with the Dirichlet boundary conditions using our phase plane method. Indeed, we will give a criterion in Theorem 2.18 concerning nonexistence of the Dirichlet solution.
Concerning the solutions in infinite interval, we can state: • a sign-changing solution w in R satisfying w (−∞) = w (∞) = 0 (Neumann).

Case λ ≤ 0
First note that the System 3 has only one equilibrium point (0, 0). As in the previous case, we can reduce our phase plane analysis to the half-plane R + × R since of Lemma 2.1. Again, we obtain some information on the variations of the trajectories of the System 3 using Equation 4. We have dv du = 0 along the curves In addition, thanks to Equation 5, we know that In this last part of the plane, we use the following lemma, similar to Lemma 2.2: Lemma 2.9. Let λ ≤ 0 and (u, v) be a trajectory of the System 3 with initial Proof. The calculus of the variations (see Equation 4) ensures that (u(t), v(t)) ∈ A for small t > 0. We prove that there exist 0 < τ < ∞ such that v(τ ) = |u(τ )| p−1 − λ. It means that (u, v) is bounded in A. Since (u, v) belongs to A and thanks to λ ≤ 0, we have Then, integration between 0 and u gives v ≤ 1 2 u 2 + v 0 .
Proof. As in Lemma 2.3, we can build periodic trajectories of 3 using the symmetry (Lemma 2.1).

Unbounded solutions
In the above paragraphs, we proved that all the trajectories of the System 3 are bounded for p ≥ 3, but if 1 < p < 3 we do not have a general answer: for example, we obtain some bounded trajectories when λ ≤ 0 (see Lemma 2.9), but with our method, we do not have (yet) any result when λ > 0. In this paragraph, we show that there exists unbounded trajectories for every λ ∈ R and for all p ∈ (1, 3). We start with a trajectory (u, v) with an initial data (0, v 0 ). Lemma 2.14. Let p ∈ (1, 3) and λ ∈ R. Suppose that v 0 > 2 max{−λ, 0} + 2 · (8) Then the trajectory (u, v) is not bounded.
Proof. We will show that under hypothesis 9, the trajectory (u, v) always lies above the curve v = 2u p−1 + 2 max{−λ, 0} . Thus, using dv du ≥ 0 (Equation 4), we obtain that (u, v) is not bounded. Ab absurdo, suppose that there exists and On the other hand, Equation 4 gives and thanks to condition 11, we obtain Then v(u) ≥ u 2 4 + v 0 . Hence, for u = u 1 , we have: and by definition 10 of u 1 , we have Hypothesis 9 implies Meanwhile, if we study both cases u 1 < 8 1 3−p and u 1 > 8 1 3−p , we remark that Equations 13 and 14 are not compatible. Thus, the trajectory (u, v) can not attain the curve v = 2u p−1 + 2 max{−λ, 0} .
Concerning Equation 1, we obtain the following results: Proof. The existence comes from the previous lemma. We just need to prove that the length of the existence interval is finite. Ab absurdo, suppose that there exists a positive and unbounded solution u of the Equation 1 in [0, ∞).
and for all t ∈ [0, ∞). Thanks to the choice of b, we have Because the solution u corresponds to a trajectory of the System 3 located in R × R + , we have ∂ t w > 0. Thus, w is super-solution of the following problem   By the comparison principle from [2], w ≥ v where v is the solution of the previous problem. But, according to [4], the solution v blows up in finite time.    With some assumption on the parameter λ, we can also build a trajectory (u, v) with an initial data (u 0 , 0) belonging to the abscissa axis.
Proof. We use the same method as in Lemma 2.14: we prove that, under hypotheses 15 and 16, the trajectory (u, v) always lies above the curve {v = βu p−1 − λ}. Ab absurdo, suppose that there exist x * > 0 such that u(x * ) = u 1 and Equation 4 gives Integration between u 0 and u 1 leads to definition 17 gives and we obtain Since of u 0 < u 1 , Equations 15 and 16 imply Hence, Equation 19 is a contradiction.
Concerning Equation 1, and reasoning as in Theorem 2.15, we obtain the following result.

Limiting case p = 1
In this paragraph, we study the case where the exponent p attains the limit 1. Then, Equation 1 becomes and the System 3 is written .
For λ = 1, (0, 0) is the only equilibrium point, while for λ = 1 the axis {v = 0} is a continuum of equilibria. We begin with the case λ = 1. Here, we have where c depends on the initial data. Thus, the phase plane is easily drawn, see Figure 3. Now, suppose λ = 1. One can compute the explicit trajectory Then, using the following equations we can draw the phase plane of the System 20, see Figure 3.

Bifurcation
According to the previous paragraphs, we can state that there exists a bifurcation of the phase plane of the System 3. First, we note that, for a fixed exponent p, the value of λ influences the phase plane of the System 3: for λ > 0, the System 3 admits three equilibrium points (a saddle point, an attractive equilibrium and a repulsive equilibrium). The distance between these equilibria goes to 0 when λ → 0, and for λ = 0, they collapse and generate a unique center, which persists for all negative λ (see Figure 4). Now, for a fixed λ, the value of the exponent p has an important role. With λ, the value of p governs the type of the equilibrium points (node, improper node, vortex). The exponent p also establishes if all the trajectories of the System 3 are bounded (p ≥ 3) or if there exists unbounded trajectories (1 ≤ p < 3). Moreover, when p attains the limit 1, the critical value of λ changes from 0 (if p > 1) to 1 (for p = 1). The case λ = 1 is special because when p → 1, the three equilibria of the System 3 (a saddle point, an attractive vortex and a repulsive vortex) generate a continuum of equilibria when p attains the limit 1 (see Figure  5).

Parabolic problem
In this section, we study the parabolic Problem 1 for many boundary conditions. First, we use the results concerning the stationary Equation 1 when the domain Ω is bounded. Then, we consider the case of unbounded domains: we investigate global existence using the comparison method, and blow-up phenomenon thanks to a L 1 −norm technique.

Comparison
We begin with the Dirichlet problem   where α > 0, p > 1, λ ∈ R and ϕ ∈ C 0 ([−α, α]) is non-negative. Thanks to the comparison principle [2] and with the results of the previous sections, we have: of Problem 21 if the initial data ϕ Using the comparison principle from [2], we prove that there exists a solution u of 21 satisfying 0 ≤ u ≤ β for all (x, t) ∈ [−α, α] × (0, ∞). Thus, u is a global positive solution. If 1 < p < 3 and λ > 0, then we just need to choose a positive solution β given in Theorem 2.6 (even if β(±α) > 0). For λ ≤ 0, we consider the Dirichlet solution given in Theorem 2.10. Now, we replace the Dirichlet boundary conditions by the dynamical boundary conditions. Consider the following problem   are non-negative. We obtain two results, depending on the sign of λ.
Proof. For the first statement, we just need to note that the constant function λ 1 p−1 is a super-solution of 22 when 0 ≤ ϕ ≤ λ 1 p−1 . For the second statement, we consider two cases: when p ≥ 3, we consider a positive solution w of 1 under the Neumann boundary conditions, see Theorem 2.4. Choosing 0 ≤ ϕ ≤ w, we obtain a super-solution of 22. If 1 < p < 3, we consider a trajectory (µ, γ) of 3 with 0 < µ(0) < λ 1 p−1 and γ(0) = 0. According to Equation 4, for a small x * > 0, we have γ(−x) < 0 and γ(x) > 0 for all x ∈ (0, x * ). Thus, µ satisfies Then, using these super-solutions and the comparison principle from [2], we prove first and second assertions. For the third statement, consider c > 0 such that The comparison principle from [2] implies that u > c, where u denote the solution of 22 with the initial data ϕ. Hence, there exists d > 0 such that Then, blow-up results from [4] imply the blowing-up in finite time of u.
Thanks to the blow-up results from [4], we know that u blows up in finite time.
Remark 4. The Neumann boundary conditions are included here, with the special case σ ≡ 0.

Global existence in unbounded domains
We study the Problem 1 under the Dirichlet, the Neumann and the dynamical boundary conditions when Ω is an unbounded domain. Using some explicit super-solutions, we look for global existence in the three types of unbounded domains: (−∞, 0), (0, ∞) and R. We begin with the case λ > 0: Thus, v is super-solution of 1 for the four boundary conditions above, and we conclude with the comparison principle [2].
If λ ≤ 0, we must add some restrictions, and we obtain the following results. Proof. We deal with the comparison principle [2] and the explicit function v(x, t) = Ae αx+(t+t0) 2 defined in R + × R + . Computing the partial derivatives, we have Thanks to p ≤ 2 and with αx Since v ≥ 0, the case of the Dirichlet boundary conditions is trivial. Choosing t 0 ≥ α 2σ , the case of the dynamical boundary conditions is verified thanks to Finally, we have a super-solution choosing A ≥ sup ϕ.
Remark 5. In the previous proof, one can see that the dynamical boundary conditions are satisfied for a more general coefficient σ verifying .
And replacing the function v by w(x, t) = Ae αx+(t+t0) n , we can consider smaller Proof. As in the previous theorem, we consider v(x, t) = Ae αx+(t+t0) 2 . Thanks to p = 2 and with some appropriate constants A and α, we have in Ω.
The case Ω = R (no boundary) and the case of Dirichlet boundary conditions are trivial. For Ω = (−∞, 0), we have ∂ ν v = ∂ x v = αv > 0 on the boundary. Thus, the Neumann boundary conditions and the dynamical boundary conditions with σ ≥ 0 are verified.
Theorem 3.6. Assume Ω = (−∞, 0), p > 3 and ϕ ∈ C(Ω). Then the Problem 1 admits a global positive solution if the initial data ϕ is sufficiently small and when B(u) = 0 stands for the Dirichlet, the Neumann or the dynamical boundary conditions with σ > 0 constant.
By definition of γ and p > 3, we have −2γ The case of the Dirichlet boundary conditions is clear because v ≥ 0. For the dynamical boundary conditions and the Neumann boundary conditions (σ ≡ 0), we use the definition of y and we have Thus, v is a super-solution of the Problem 1 as soon as ϕ ≤ v(·, 0) in Ω.

Blow up in unbounded domains
Here, using some weighted L 1 −norms, we examine blow-up phenomena for some solutions of Problem 1 in unbounded domains satisfying the Neumann, the Robin, and some nonlinear boundary conditions and this growth order at infinity: for all a > 0 and for all t > 0 lim |x|→∞ u(x, t)e −a|x| = 0 and lim Unless otherwise stated, we always suppose Ω = (0, ∞). We begin with a lemma which gives a criterion for the blowing-up of the solution.
Lemma 3.7. Let u be a solution of Problem 1 satisfying the condition 23. If there exists α > 0 such that blows-up in finite time, then u also blows-up in finite time.
Proof. Consider α > 0 such that N α blows-up in finite time. Using the following inequality and because N α blows up, we can deduce the blowing up in finite time of the function u(x, t)e − α 2 x . Then, thanks to the growth order condition 23, the solution u must blow up too.
We also need this technical lemma.
Proof. Let v be the positive solution of the following problem Then, the comparison principle and the maximum principle from [2] imply for all x ∈ [0, 1] and t > 0, see Lemma 2.1 in [3]. Thus, for all τ > 0, we obtain Proof. We aim to prove the existence of α > 0 and β > 0 such that Derivating the function N α , we obtain Using the growth order condition 23 and integrating by parts, we obtain Thanks to Lemma 3.8, and considering u from a time τ > 0, we can assume that ≥ 0. Then, the Neumann boundary conditions imply Shrinking α, we can suppose α ≤ −2λ and α ≤ 1.

Hölder inequality
Finally, we prove the blowing-up of N α in finite time. Integrating the differential inequality N α (t) ≥ βN p α (t) between 0 and t > 0, we obtain Since of −1 p−1 < 0, the right hand side term blows up at t = (p−1)β > 0. We conclude with Lemma 3.7. Proof. We follow the proof of Theorem 3.9. We just change the choice of α: let α > 0 such that α ≤ δ, and use the following minoration in Equation 24: Then, we return to Equation 25.
When λ = 0, the choice of α is too strict. Meanwhile, we obtain some blow-up results imposing some restrictions on the exponent p and on the initial data.
Let β ∈ (0, 1) and put it into the previous equation: If u ≤ 2α, we have α 2 − αu/2 ≥ 0, whereas u > 2α implies Thanks to 1 < p ≤ 3, Equation 26 is achievied by choosing α > 0 sufficiently small and β ∈ (0, 1) depending on p. Thus, we obtain Then, we can suppose that u(0, t) > c > 0 for all t > 0 (see Lemma 3.8), and with α < c/2 we have −αu(0, t) + u 2 (0,t) As in the proof of Theorem 3.9, we use Hölder inequality and we are led to N α ≥ δN p α with δ > 0 depending on α, β and p. Hence, N α blows-up in finite time, so does the solution u, see Lemma 3.7. Proof. The proof is similar to the previous one. Go back to Equation 26: since p > 3, we must choose α such that Under this condition, N α satisfies the differential inequality Because α can not be too small, we must use the assumption ϕ(0) > 2 Thus, with β very close to 1 and with α = 2 We conclude with Hölder inequality and the blowing up of N α .
Finally, if Ω = (−∞, 0), we must change the weight in N α and we obtain this results concerning the nonlinear boundary conditions. Proof. As in the case of Ω = (0, ∞), we use a weighted L 1 −norm: We compute N α (t) = 0 −∞ ∂ t u(x, t)e αx dx, and using the equations of Problem 1, integration by parts leads to Thanks to ∂ ν u(0, t) = ∂ x u(0, t) in (−∞, 0), choosing α = min{2c, d}, we obtain Hölder inequality leads to the differential equation N α (t) ≥ γN p α (t) with γ > 0. Hence N α and the solution u blow up in finite time.

Introduction
Let Ω be a domain of the real line R, not necessarily bounded. Let p be a real number with p > 1, λ ∈ R and ϕ a non-negative continuous function in Ω.

Consider the following nonlinear parabolic problem
where B(u) = 0 stands for the Dirichlet boundary conditions (u = 0), the Neumann boundary conditions (∂ ν u = 0) or the dynamical boundary conditions (σ∂ t u+∂ ν u = 0 with σ a non-negative smooth function). For the local existence of the positive solutions of this problem, we refer to von Below and Mailly's results [6] and references therein, [2], [4] and [7] . In the first section, we study the stationary equation stemming from Problem (1). We aim to prove the existence of positive and sign-changing solutions using phase plane arguments and dealing with the first order system We prove a bifurcation in the phase plane of this system, depending on the parameters λ and p, which influences the resolution of Equation (2) under the Dirichlet, the Neumann and the mixed boundary conditions. Then in a second section, using the comparison method from [3], we deduce from the solutions of the stationary Equation (2) some regular super-solutions for the Problem (1). Dealing with these super-solutions and with the blow-up results from [6], we investigate global existence and blow-up phenomena for the Problem (1) for different values of λ and p, and for the Dirichlet, the Neumann and the dynamical boundary conditions. We also examine both phenomena in unbounded domains: we obtain global existence results with the comparison method and using some well-known super-solutions (we mean explicit functions) for the Dirichlet, the Neumann and the dynamical boundary conditions. The blowing-up concerns the regular solutions of Problem (1) satisfying some growth order at infinity and some boundary conditions such that • ∂ ν u = g(u) with g a polynomial of degree 2 (nonlinear b.c.).
We use some weighted L 1 −norms: our technique is to prove the blowing-up of the solution by proving the blowing-up of appropriate L 1 −norms.
Before starting, let us define the kind of solution we look for: Definition 1.1. A function u is called a solution (or regular solution) of Equation (2) in Ω if u is of class C 2 (Ω) and satisfies the equation in the classical sense.
A function u is called a solution (or regular solution) of Problem (1) in Ω if u is of class C(Ω × [0, T )) ∩ C 2,1 (Ω × (0, T )) and satisfies the equations of Problem (1) in the classical sense in Ω × [0, T ) where T ∈ (0, ∞] denotes the maximal existence time of the solution u.

Stationary equation
In this section, we study the existence of positive and sign-changing solutions of Equation (2) using a phase plane method. Unless otherwise stated, we suppose p ∈ (1, ∞). For the theory of phase planes (nature of equilibrium, regularity, behaviour and uniqueness of trajectories), we refer to H.Amann's book [1]. Here we consider the system (u . We will prove later that (0, 0) is a center.

Case λ > 0
Let λ be a positive real number and p > 1. We want to study the phase plane of the System (3). First we prove a lemma on the symmetry of the trajectories: for all x ∈ (−a, a).
A simple calculus of the derivatives implies Then (w, z) is also a trajectory of the System (3), and it is symmetric to (u, v) with respect to the ordinates axis.
Thus, we can reduce our phase plane analysis to the half plane R + × R. In order to draw the phase plane of the System (3), we write the ordinate v as a function depending on the abscissa u: v = f (u). We do not know the function f , but we can deduce its variations and convexity using the equations (3). For the variations, when v = 0, we have Then we have dv du > 0 in the sets {u > 0, v > 0, v > |u| p−1 − λ} and {u > 0, v < 0, v < |u| p−1 − λ}. On the other hand, dv du < 0 in the sets {u > 0, v < 0, v > |u| p−1 − λ} and {u > 0, v > 0, v < |u| p−1 − λ}. Next, we compute the convexity of the function f and we obtain We the convexity is sign-changing in {u > 0, v > |u| p−1 − λ}. These arguments are sufficient to know the profile of the trajectories in the half plane {v < |u| p−1 −λ}.

We do not need to know how the trajectories behave in {u
Proof. Let v 0 > 0 and consider (u, v) the solution of the System (3) with initial data (u(0), v(0)) = (0, v 0 ). The calculus of the variations (see Equation (4)) ensures that (u(t), v(t)) ∈ A for small t > 0. We prove that there exist 0 Then dv du ≥ 0 in A implies v > v 0 as long as (u, v) ∈ A, and we obtain If p > 3, the intersection {v = |u| p−1 −λ}∩{v = 1 2 1+ λ v0 u 2 +v 0 } is non-empty for all v 0 > 0. If p = 3, we need to choose v 0 sufficiently big such that Then, the trajectory (u, v) belongs to the compact and, using dv du ≥ 0, we know that there exist 0 < τ < ∞ such that v(τ ) = |u(τ )| p−1 − λ. This argument proves that each solution of the System (3) with initial data (u(0), v(0)) = (0, v 0 ) is bounded in A if v 0 is big enough. Thanks to uniqueness of solution, it also proves the result for all the solutions initiated in A.
Then, we complete this phase plane analysis by proving the existence of periodic trajectories.  Proof. Thanks to the symmetry (see Lemma 2.1), we just need to prove that for some initial data belonging to {0} × (0, ∞), there exists a trajectory which attains a point belonging to {0} × (−∞, 0). First, consider a trajectory (u, v) initiated at (0, v 1 ) with v 1 > 0. According to hypothesis (6), we know that (u, v) is bounded, and using its variations and its convexity (Equations (4) and (5)), we can deduce that (u, v) attains the x−axis at a point (u 1 , 0) with u 1 > λ 1 p−1 (see Figure 1). Then, using the reverse system and one of its trajectories initiated at (0, v 2 ) with v 2 < −λ) (trajectories of reverse system and of System (3) have same support), one can note that for u 0 > λ 1 p−1 , there exists a trajectory (w, z) of (3) with w(0) = u 0 and z(0) = 0 (see Figure 1). Finally, let us consider the trajectory (a, b) of System (3) containing the point (u 2 , 0), where u 2 > max{u 0 , u 1 }. Thanks to the uniqueness of the solutions, and using the information on the variations and the convexity, we deduce that there exist two real numbers s < t such that a(s) = a(t) = 0, b(s) = v 0 and b(t) = v 3 (see Figure 1). Thus, the trajectory (a, b) is the periodic trajectory we look for. Now, analysing the phase plane of the System (3), we deduce the following results concerning the Equation (2). For the Neumann solution, consider 0 < µ 0 < λ 1 p−1 and the trajectory (µ, ν) of System (3) initiated at (µ 0 , 0). Since (µ, ν) can not cross the trajectory (u, v) ( see Figure 1), it must cross the x−axis at (µ 1 , 0) with λ 1 p−1 < µ 1 < u 1 . Thus, the abscissa of this trajectory is the Neumann solution we look for. Finally, the length (2α) of the existence interval is governed by the time needed by the trajectory to go from its initial data to its "final data".
with the mixed boundary conditions u (−α) = u(α) = 0. In addition, if then there exists a positive solution of the Equation (2) under the Neumann boundary conditions.
Proof. The first part of the statement comes from Theorem 2.4, the solution with mixed−2 boundary conditions is located in {v < |u| p−1 − λ}. The other part stems from Equation (7): in this case, the equilibrium (λ Without hypothesis (6), we can not construct positive solutions anymore for the Dirichlet, Neumann or mixed−1 boundary conditions. If we do not impose the positivity, we obtain this result: there exists a solution of the Equation (2) u − uu + u|u| p−1 − λu = 0 in (−α, α) for some α > 0.
Proof. As we mentioned before, we consider the part {v < |u| p−1 − λ} of the phase plane of the System (3) (see Figure 1). For the Neumann solution, we consider the trajectory (a, b) between (u 2 , 0) and (−u 2 , 0). For the mixed−1 solution, we can also consider the trajectory (a, b), but only between (0, v 3 ) and (−u 2 , 0).

Remark 2.
The Neumann solution built above is sign changing, whereas the mixed−1 solution is negative.

Remark 3.
In the general case, we can not build any solution with the Dirichlet boundary conditions using our phase plane method. Indeed, we will give a criterion in Theorem 2.18 concerning nonexistence of the Dirichlet solution.

Case λ ≤ 0
First note that the System (3) has only one equilibrium point (0, 0). As in the previous case, we can reduce our phase plane analysis to the half-plane R + × R since Lemma 2. Then, we have dv In addition, thanks to Equation (5), we know In this last part of the plane, we use the following lemma, similar to Lemma 2.2: Lemma 2.9. Let λ ≤ 0 and (u, v) be a trajectory of the System (3) with initial data (0, v 0 ).
To conclude this section, let us give this result concerning the periodic solutions: Theorem 2.13. Let λ ≤ 0. For some α > 0, there exists a sign-changing periodic solution of the Equation (2) u − uu + u|u| p−1 − λu = 0 in R.
Proof. As in Lemma 2.3, we can build periodic trajectories of (3) using the symmetry (Lemma 2.1).

Unbounded solutions
In the above paragraphs, we proved that all the trajectories of the System (3) are bounded for p ≥ 3, but if 1 < p < 3 we do not have a general answer: for example, we obtain some bounded trajectories when λ ≤ 0 (see Lemma 2.9), but with our method, we do not have (yet) any result when λ > 0. In this paragraph, we show that there exists unbounded trajectories for every λ ∈ R and for all p ∈ (1, 3). We start with a trajectory (u, v) with an initial data (0, v 0 ). Lemma 2.14. Let p ∈ (1, 3) and λ ∈ R. Suppose that v 0 > 2 max{−λ, 0} + 2 · 8 Then the trajectory (u, v) is not bounded.
Concerning Equation (2), we obtain the following results: Because the solution u corresponds to a trajectory of the System (3) located in R×R + , we have ∂ t w = ∂ x u > 0. Thus, w is super-solution of the following problem   By the comparison principle from [3], w ≥ v where v is the solution of the previous problem. But, according to With some assumption on the parameter λ, we can also build an unbounded trajectory (u, v) with an initial data (u 0 , 0) belonging to the abscissa axis.
Proof. We use the same method as in Lemma 2.14: we prove that, under hypotheses (15) and (16), the trajectory (u, v) always lies above the curve {v = βu p−1 − λ}. Ab absurdo, suppose that there exist x * > 0 such that and Equation (4) gives Integration between u 0 and u 1 leads to definition (17) gives and we obtain Since u 0 < u 1 , Equations (15) and (16) imply Hence, Equation (19) is a contradiction.
Concerning Equation (2), and reasoning as in Theorem 2.15, we obtain the following result. In addition, if p ∈ (2, 3), then α is finite.

Limiting case p = 1
In this paragraph, we study the case where the exponent p attains the limit 1. Then, Equation (2) becomes

and the System (3) is written
.
For λ = 1, (0, 0) is the only equilibrium point, while for λ = 1 the axis {v = 0} is a continuum of equilibria. We begin with the case λ = 1. Here, we have where c depends on the initial data. Thus, the phase plane is easily drawn, see Figure 3. Now, suppose λ = 1. One can compute the explicit trajectory Then, using the following equations we can draw the phase plane of the System (20), see Figure 3.

Bifurcation
According to the previous paragraphs, we can state that there exists a bifurcation of the phase plane of the System (3). First, we note that, for a fixed exponent p, the value of λ influences the phase plane of the System (3): for λ > 0, the System (3) admits three equilibrium points (a saddle point, an attractive equilibrium and a repulsive equilibrium). The distance between these equilibria goes to 0 when λ → 0, and for λ = 0, they collapse and generate a unique center, which persists for all negative λ (see Figure 4). Now, for a fixed λ, the value of the exponent p has an important role. With λ, the value of p governs the type of the equilibrium points (node, improper node, vortex). The exponent p also establishes if all the trajectories of the System (3) are bounded (p ≥ 3) or if there exists unbounded trajectories (1 ≤ p < 3). Moreover, when p attains the limit 1, the critical value of λ changes from 0 (if p > 1) to 1 (for p = 1). The case λ = 1 is special because when p → 1, the three equilibria of the System (3) (a saddle point, an attractive vortex and a repulsive vortex) generate a continuum of equilibria when p attains the limit 1 (see Figure 5).

Parabolic problem
In this section, we study the positive solutions of the parabolic Problem (1) for many boundary conditions. First, we use the results concerning the stationary Equation (2) when the domain Ω is bounded. Then, we consider the case of unbounded domains: we investigate global existence using the comparison method, and blow-up phenomenon thanks to a L 1 −norm technique.
The comparison principle from [3] implies that u > c, where u denote the solution of (22) with the initial data ϕ. Hence, there exists d > 0 such that and for all t > 0.
Then, blow-up results from [6] imply the blowing-up in finite time of u.
Thanks to the blow-up results from [6] and [8], we know that u blows up in finite time.
Remark 4. The Neumann boundary conditions are included here, with the special case σ ≡ 0.

Global existence in unbounded domains
We study the Problem (1) under the Dirichlet, the Neumann and the dynamical boundary conditions when Ω is an unbounded domain. Using some explicit super-solutions, we look for global existence in the three types of unbounded domains: (−∞, 0), (0, ∞) and R. We begin with the case λ > 0: On the boundary, we have: The choice of ϕ implies ϕ ≤ v(·, 0) in Ω. Thus, v is super-solution of (1) for the four boundary conditions above, and we conclude with the comparison method from [3].
If λ ≤ 0, we must add some restrictions, and we obtain the following results.
Proof. We deal with the comparison principle [3] and the explicit function v(x, t) = Ae αx+(t+t0) 2 defined in R + × R + . Computing the partial derivatives, we have Thanks to p ≤ 2 and with αx Since v ≥ 0, the case of the Dirichlet boundary conditions is trivial. Choosing t 0 ≥ α 2σ , the case of the dynamical boundary conditions is verified thanks to Finally, choosing A ≥ sup Ω ϕ, v is a super-solution of Problem (1) under the above boundary conditions. Thus, using the comparison method from [3], we prove that there exist a positive solution of Problem (1) bounded by v, and then, this solution must be global.

Remark 5.
In the previous proof, one can see that the dynamical boundary conditions are satisfied for a more general coefficient σ verifying .
Proof. As in the previous theorem, we consider v(x, t) = Ae αx+(t+t0) 2 . Thanks to p = 2 and with some appropriate constants A and α, we have in Ω.
The defined in R − × R + with A > 0, γ = 1 p−1 and y = −2σγ. A simple calculation leads to By definition of γ and p > 3, we have −2γ + 1 > 0. Since v p−1 ≤ A p−1 , and because −(x + y) > 0 for all x ∈ Ω, we obtain ∂ t v − ∂ 2 x v + v∂ x v − v p ≥ 0 by choosing A small enough. The case of the Dirichlet boundary conditions is clear because v ≥ 0. For the dynamical boundary conditions and the Neumann boundary conditions (σ ≡ 0), we use the definition of y and we have Thus, v is a super-solution of the Problem (1) as soon as we choose 0 ≤ ϕ ≤ v(·, 0) in Ω.

Blow up in unbounded domains
Here, using some weighted L 1 −norms, we examine blow-up phenomena for some solutions of Problem (1) in unbounded domains satisfying the Neumann, the Robin, and some nonlinear boundary conditions. We only consider regular solutions satisfying this standard growth order condition at infinity: for all a > 0 and for all t > 0 lim |x|→∞ u(x, t)e −a|x| = 0 and lim |x|→∞ |∂ x u(x, t)|e −a|x| = 0.
Unless otherwise stated, we always suppose Ω = (0, ∞). We begin with a lemma which gives a criterion for the blowing-up of the solution. Proof. Consider α > 0 such that N α blows-up in finite time. Using the following inequality and because N α blows up, we can deduce the blowing up in finite time of the function u(x, t)e − α 2 x . Then, thanks to the growth order condition (23), the solution u must blow up too.
Then, the comparison principle and the maximum principle from [3] imply ∂ t v(x, t) ≥ 0 and v(x, t) > 0.
Hence N α blows-up, and the solution u blows up too, see Lemma 3.7.
Finally, if Ω = (−∞, 0), we must change the weight in N α and we obtain this results concerning the nonlinear boundary conditions. Theorem 3.12. Let λ ≤ 0 and p ≥ 2. Then the Problem (1) admits no global positive solution when B(u) = 0 stands for the nonlinear boundary conditions ∂ ν u = g(u), where g is a function such that there exists c > 0 and d > 0 satisfying g(η) ≥ cη 2 + dη.