Info

in C!o"(K+) as a —» oo, where A is as in the statement of Lemma 5.1. As a main consequence, du(-,u0) _ du(-,u0) __e±i dv(-,a) _

- V>i(0) + o(l)) = Au^ + o (uo^) , as wo —> oo. This finishes the proof of Theorem 5.1. □

We proceed now to the proof of Lemma 5.1.

Proof of Lemma 5.1. Fix R,h positive and choose Dr,h any bounded C2,a sub domain of containing {x : \x'\ < R, 0 < xn < h}, for instance a regularization of the latter domain which contains it. Consider the auxiliary problem,

By using a standard weak comparison argument it follows that problem (24) admits at most a unique positive solution. Such solution can be obtained by the method of sub and supersolutions by taking v = 0 as subsolution, v = 1 as a supersolution. Thus (24) admits a unique positive solution VR,h € C2'a(Dntfl) which satisfies,

Now observe that if v = v(z) G Clo'°(R+) is any nonnegative bounded solution to (21) (and so satisfying (22)), then

defines a subsolution to problem (24) which, as a consequence of weak comparison, satisfies v(z) < vRth{z) in DrHence,

0 < v(z) < vR>h(z) z G DRih, where strong comparison has been also employed.

For n G N set Dn := nDR<h = {nz : z G DR,h} while vn(z) designates the solution to problem (24) in Dn. Just for the same argument as the one given above we arrive at:

0 < v{z) < vn+k{z) < vn(z) <1 z G Dn, for every n, k G N. In fact, the restriction of vn+k to Dn defines a strict subsolution to (24) in Dn that can be strongly compared with vn.

Using the Lq and Schauder's estimates as has been already done we get the convergence,

Was this article helpful?

0 0

Post a comment