02.12.2012 Views

Quasilinear parabolic problems with nonlinear boundary conditions

Quasilinear parabolic problems with nonlinear boundary conditions

Quasilinear parabolic problems with nonlinear boundary conditions

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

Letting k(t) = e −t and ξ(t) = e −t S(t)h(0), t ≥ 0, we thus have<br />

Define now r ∈ L1, loc(R+) by means of<br />

Then (3.16) implies<br />

in particular<br />

Since k ∈ L1(R+) and<br />

v2(t) = ξ(t) + (k ∗ ξ)(t), t ≥ 0. (3.16)<br />

r(t) + (r ∗ k)(t) = k(t), t ≥ 0.<br />

ξ(t) = v2(t) − (r ∗ v2)(t), t ≥ 0,<br />

Aξ(t) = Av2(t) − (r ∗ Av2)(t), t ≥ 0.<br />

1 + ˆ k(λ) = 1 + 1<br />

λ+1 �= 0, Re λ ≥ 0,<br />

it follows by the halfline Paley-Wiener theorem (see [39, Chapter 2, Theorem 4.1, p. 45])<br />

that r ∈ L1(R+). So, letting b(t) = t κ−1 e −t , t ≥ 0, and Bκ = (b∗) −1 ∈ S(Lp(R+; X)),<br />

we see <strong>with</strong> the aid of Young’s inequality and Corollary 2.8.1 that<br />

r ∗ Av2 = r ∗ b ∗ BκAv2 = b ∗ r ∗ BκAv2 ∈ H κ p (R+; X).<br />

Therefore Aξ ∈ Hκ p (R+; X), and so, by Theorem 3.1.3, we get h(0) = v(0) = u(0) =<br />

f(0) ∈ DA(1 + κ/α − 1/pα, p). This proves condition (ii).<br />

To prove the converse direction, suppose that the <strong>conditions</strong> (i),(ii) and (iii) are<br />

fulfilled. Uniqueness is an immediate consequence of Theorem 3.1.1. As for existence,<br />

we build a solution u ∈ Z of (3.1) as follows. Put x = f(0) and y = ˙ f(0). If J = [0, T ],<br />

let u1 be the solution of<br />

w(t) + (a ∗ Aw)(t) = f(t) − x − ty, t ∈ J.<br />

This solution exists and lies in 0H α+κ<br />

p (J; X) ∩ Hκ p (J; DA), thanks to Theorem 3.1.1. By<br />

Theorem 3.1.2, we also have S(·)x, (1 ∗ S)(·)y ∈ Z, so that u := u1 + Sx + 1 ∗ Sy ∈ Z. It<br />

is easy to see that u solves (3.1). In case of J = R+, we extend v1, v2 ∈ Hα+κ p ([0, 1]; X)∩<br />

Lp([0, 1]; DA) defined by v1(t) = S(t)x, v2(t) = (1 ∗ S)(t), t ∈ [0, 1], to functions u1, u2 ∈<br />

Z and set f1 = f − (u1 + a ∗ Au1) − (u2 + a ∗ Au2). By construction, f1 ∈ 0H α+κ<br />

p (R+; X).<br />

So, due to Theorem 3.1.1, we can define u3 ∈ Z as the solution of v+a∗Av = f1, t ∈ R+.<br />

Clearly, u := u1 + u2 + u3 ∈ Z solves (3.1). Hence the proof is complete. �<br />

Remarks 3.1.3 (i) In Section 3.3 we will prove a corresponding result for the case of a<br />

compact time-interval J where κ ∈ (1/p, 1 + 1/p).<br />

(ii) In the case of a compact interval J, one can weaken the assumptions on both a<br />

and A. Due to the transformation property of (3.1) discussed at the end of Section 2.8,<br />

it suffices to assume that µ + A ∈ RS(X) <strong>with</strong> θa + φR µ+A < π for some µ ≥ 0. As for<br />

the kernel, the theorem is also true, if a is of the form a = b + dk ∗ b, where b is like a in<br />

the statement of Theorem 3.1.4 and k ∈ BVloc(R+) <strong>with</strong> k(0) = k(0+) = 0. This follows<br />

from a straightforward perturbation argument, cp. [63, Section 8.5].<br />

43

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!