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Lemma 3. Let (HO), (H'l), (H2) and (H'3) be fulfilled, then there exist T and /5 E
(0,00) with inf u(t,.; j-L, 0) ;::: j -L/5 for all j -L E R+ and t ;::: T.
Proof. It follows from (H'3) that we can find a K E (0,00) with g(y) ::; Ky for all y E R+.
Let, denote the solution of
{
c( x )ot/(t, x) - div (k grad ,(t,' ))(x) = j-Lq( t)[1 - sup a] - K,( t, x)
(t,x) E (0,00) x M,
,(0,,) == 0.
We have u( t, x; j-L, J9) ~ ,( t, x) by comparison, since, ;::: 0, u(O,'; j-L, J9) ;::: ° and g(y) ::; Ky
for all y E R+, hence J-Lq(t)[1 - sup a]- K,(t, x) ::; j-LQ(t, x)[1 - a(x, ,(t, x ))]- g(!(t, x))
(net radiation flux deficit). For the same reason, , ~ 1j J follows provided that 1j J solves
.
j -L
K
1jJ(t) = liclloo [1 - supa]q(i) - inf e 1jJ(t),
1jJ(0) = 0.
S · b*'-"
d *._[1-supa]r 1 ()
(b*)d
ettmg
.- inf e an q.- Ilell oo Jo q s exp s s, we get
1jJ(t) ~ q*j-L[exp(b*([t]- i)) - exp( -b*t)]/[exp(b*) - 1]
for t E [1,00), hence choosing T = 1 + In(2)/b* and /5 = q*/2exp(2b*), e.g., we obtain
1jJ(t) ;::: j-LD for all i E [T,OO). Thus, u(t,X;j-L,O) ;::: fLD for all x E M, t E [T,OO) and
fL E R+.
Clearly, u(·,·; j-L, J9) ~ u(','; fL, 0) for J9 E C+(M), which yields the a priori bound
from below
This observation allows us to derive that the time-I-map II(fL, .) has at most one fixed
point for large fl.
Corollary. Let (HO), (H'l), (H2) and (H'3) be satisfied, then there exists a fL* E (0,00)
such that II(fL,.) has at most one fixed point for fL E [fL*, 00).
Proof. Choose Y E (y*, 00) (y* as in (H'3)) and K E (0,00) with g(y) ~ Ky for y E [y,oo).
Set E:= 2li;t with D > ° as in Lemma 2 and select y E [y,oo) with y l(02Ct)(x, y)1 ::; E
for y E [fj, 00). Finally, choose fL* E (0,00) with /l*D ~ y. We have for x E M,
I" E [1"*,00) and y E [j-LD, 00):
g'(y) + ILQ(t, x)(o2a)(x, y) ;::: g'(y) - fL IIQlloo l(02 a )(x,y)1 ~ e*g(y) - IlIIQlloo ~ ;:::
y
y
[e' 9(fLD) - fL IIQlloo E]/y ~ [e' KfLD - II I I Qlloo c' KII /(21IQll oo )]/y = p.e' KD / (2y) > °
Lemma 3. Let (HO), (H'l), (H2) and (H'3) be fulfilled, then there exist T and /5 E
(0,00) with inf u(t,.; j-L, 0) ;::: j -L/5 for all j -L E R+ and t ;::: T.
Proof. It follows from (H'3) that we can find a K E (0,00) with g(y) ::; Ky for all y E R+.
Let, denote the solution of
{
c( x )ot/(t, x) - div (k grad ,(t,' ))(x) = j-Lq( t)[1 - sup a] - K,( t, x)
(t,x) E (0,00) x M,
,(0,,) == 0.
We have u( t, x; j-L, J9) ~ ,( t, x) by comparison, since, ;::: 0, u(O,'; j-L, J9) ;::: ° and g(y) ::; Ky
for all y E R+, hence J-Lq(t)[1 - sup a]- K,(t, x) ::; j-LQ(t, x)[1 - a(x, ,(t, x ))]- g(!(t, x))
(net radiation flux deficit). For the same reason, , ~ 1j J follows provided that 1j J solves
.
j -L
K
1jJ(t) = liclloo [1 - supa]q(i) - inf e 1jJ(t),
1jJ(0) = 0.
S · b*'-"
d *._[1-supa]r 1 ()
(b*)d
ettmg
.- inf e an q.- Ilell oo Jo q s exp s s, we get
1jJ(t) ~ q*j-L[exp(b*([t]- i)) - exp( -b*t)]/[exp(b*) - 1]
for t E [1,00), hence choosing T = 1 + In(2)/b* and /5 = q*/2exp(2b*), e.g., we obtain
1jJ(t) ;::: j-LD for all i E [T,OO). Thus, u(t,X;j-L,O) ;::: fLD for all x E M, t E [T,OO) and
fL E R+.
Clearly, u(·,·; j-L, J9) ~ u(','; fL, 0) for J9 E C+(M), which yields the a priori bound
from below
This observation allows us to derive that the time-I-map II(fL, .) has at most one fixed
point for large fl.
Corollary. Let (HO), (H'l), (H2) and (H'3) be satisfied, then there exists a fL* E (0,00)
such that II(fL,.) has at most one fixed point for fL E [fL*, 00).
Proof. Choose Y E (y*, 00) (y* as in (H'3)) and K E (0,00) with g(y) ~ Ky for y E [y,oo).
Set E:= 2li;t with D > ° as in Lemma 2 and select y E [y,oo) with y l(02Ct)(x, y)1 ::; E
for y E [fj, 00). Finally, choose fL* E (0,00) with /l*D ~ y. We have for x E M,
I" E [1"*,00) and y E [j-LD, 00):
g'(y) + ILQ(t, x)(o2a)(x, y) ;::: g'(y) - fL IIQlloo l(02 a )(x,y)1 ~ e*g(y) - IlIIQlloo ~ ;:::
y
y
[e' 9(fLD) - fL IIQlloo E]/y ~ [e' KfLD - II I I Qlloo c' KII /(21IQll oo )]/y = p.e' KD / (2y) > °