Sub–Critical Regime Sample Clauses

Sub–Critical Regime. 4.1.1 Vanishing and boundedness of the solution

Sub–Critical Regime. We adopt a probabilistic approach in showing existence of a continuation of the solution. To this end, consider (5.0.3) and note that by Theorem 4.1.1 we have that the initial data, v0 is bounded. In the following, we will use the flow line formulation to show that there exists a unique solution to (5.0.3), but the solution may not attain the initial data of (5.0.3). Indeed, we prove the attainment of the initial data by using a suitable ansatz for the solution v of (5.0.3). Let z(l, t) = 1 t2α sinh 2g(lt−α) and write w(z, t) = v(l, t). Then, by standard calculations it follows that w solves = t4α + 4z2 2 + α ∂w √ ∂2w + 4z t4α−1 4α 2z ∂w (5.1.1) 2 √ ∂z √ t4α + 4z2 w(z, 0) = w0(z) where w0(z) = v0(2 |z|) ∈ L∞(R). We note that this equation is the same as equation (3.3.1), but with (1 − t) replaced by t and the sign of the coefficient of w reversed. By the same asymptotic analysis process as in Section 4.1.2, the important equation near the singularity is t4α−1 ∂t = αt4α + 4z2 w˜ whose solution, up to some function A : R → R, is given by w˜(z, t) = A(z)(t4α + 4z2)1/4. This provides us with a guess as to what the solution looks like near the singularity. We now write w(z, t) = (t4α + 4z2)1/4ϕ(z, t) and so ϕ solves = t4α + 4z2 2 + ∂ϕ √ ∂2ϕ 2t4α 4α 6z ∂ϕ 2 3/2 ϕ + √ w(z, 0) (t + 4z ) t4α + 4z2 ∂z

Sub–Critical Regime. Consider (6.2.1) with α < 1 . We aim to prove the analogue of Theorem 4.1.1 and Theorem 4.1.3 for this equation. Indeed, we have the following.