Applications of Semigroups to Nonlinear IVP’s
1.The Abstract IVP
Consider the following nonlinear initial value problem
u v ÝtÞ + AuÝtÞ = FÝuÝtÞÞ
0 < t < T,
uÝ0Þ = u 0
Ý1.1Þ
where ?A : D A í H generates a C 0 ? s/g of contractions on H. Of course this includes the
special case that the semigroup generated by ?A is analytic. A strong solution of (1) on [0,T]
is a function uÝtÞ 5 C 0 Ýß0, TÞ : HÞ V C 1 ÝÝ0, TÞ : HÞ which solves the equation and we will
define a funtion uÝtÞ to be a mild solution of (1) if uÝtÞ 5 C 0 Ýß0, TÞ : HÞ satisfies
t
uÝtÞ = SÝtÞu 0 + X SÝt ? sÞFÝuÝsÞÞds
0
0 < t < T.
Ý1.2Þ
The simplest existence proofs for problems like this make the assumption that F : V í H
is locally Lipschitz; i.e., V denotes a closed subspace of H (V = H is allowed) and for some
R > 0, there exists C R > 0 such that
||FÝuÞ ? FÝvÞ|| H ² C R ||u ? v|| V
-u, v 5 B R Ý0Þ Ð V
Ý1.3Þ
For some nonlinearities it will suffice to take V=H, while for others it will be necessary to
choose V to be an appropriate proper closed subspace of H. In these cases we will
suppose that S(t) maps H into V Ð D K with ||SÝtÞx|| V ² C V ||x|| H , and, for convenience we will
assume C V = 1.
To show that (1) has a mild solution under the assumption (3), let
t
®ÝuÞ = X SÝt ? sÞFÝuÝsÞÞds
0
and
vÝtÞ = SÝtÞu 0 ,
also
R = 2||u 0 || V
and
K R = RC R + ||FÝ0Þ|| H .
Then
||FÝuÞ|| H ? ||FÝ0Þ|| H ² ||FÝuÞ ? FÝ0Þ|| H ² C R ||u|| V ² RC R
and
||FÝuÞ|| H ² K R
-u 5 B R Ý0Þ Ð V.
This bound on ||FÝuÞ|| H implies
||®ÝuÝtÞÞ|| V ² T max 0²s²t²T ||SÝt ? sÞFÝuÝsÞÞ|| V ² TK R
if uÝtÞ 5 B R Ý0Þ for 0 ² t ² T. Now if we let
1
MR =
u 5 CÝß0, Tà: HÞ : ||uÝtÞ|| V ² R, 0 ² t ² T
Then for u 5 M R and 0 < T < R/Ý2K R Þ we have
||®ÝuÝtÞÞ|| V ² TK R < R/2 = ||u 0 || V
i.e.,
® : MR í MR
0²t²T<
for
R
2K R
In addition, for 0 ² t ² T,
||®ÝuÝtÞÞ ? ®ÝwÝtÞÞ|| H ² C R t||uÝtÞ ? wÝtÞ|| V
-u, v 5 M R
hence, for t < 1/C R , ® is a strict contraction on M R . Now let
T 0 = minß1/C R , R/2K R à
Then for u 5 M R and 0 ² t ² T 0 ,
||vÝtÞ + ®ÝuÝtÞÞ|| V ² ||u 0 || V + ||®ÝuÝtÞÞ|| V ² 2||u 0 || V = R
and it follows that M R 8 u $ v + ®ÝuÞ 5 M R is a strict contraction. Then there is a unique
fixed point, û 5 M R such that
ûÝtÞ = vÝtÞ + ®ÝûÝtÞÞ,
0 ² t ² T0
i.e., û is a mild solution of the IVP. In order to prove that û is, in fact, a strong solution to the
IVP, additional hypotheses on A or on F are needed. For example, if A generates an
analytic semigroup, then û would have the additional smoothness required of a strong
solution. Also if additional smoothness on F were assumed, we may be able to show the
mild solution is strong.
Since the solution has only been shown to exist for 0 ² t ² T 0 , it is referred to as a local
solution. In an effort to extend the solution to larger time, suppose we use u 1 = ûÝT 0 Þas the
initial condition for a new IVP and follow the same procedure to obtain a new mild solution
on an interval ßT 0 , T 1 à for some T 1 > T 0 . Repeating this procedure N times leads to solutions
on ß0, T 0 àWßT 0 , T 1 àWßT 1 , T 2 àW... W ßT N?1 , T N à= ß0, T N à. In general, the length |ßT j , T j+1 à| tends to
zero with increasing j due to the fact that R, C R , K R grow as T increases. However, if it is
known, say from some a-priori estimate of the solution, that any solution of the IVP must
satisfy ||uÝtÞ|| V ² C for 0 ² t ² T, then we may take R = maxß2||u 0 || H , Cà in the procedure just
described. Then we can divide [0,T] into subintervals ßT j , T j+1 à of uniform length and in this
way, obtain a solution for the interval [0,T]; i.e., a uniform bound on solutions implies a
global solution.
The nonlinear operator DßuÝtÞà= vÝtÞ + ®ßuÝtÞà:H í H may be interpreted as the
continuous flow on H associated with the IVP.
2
2. A Nonlinear Diffusion Equation on R n
Consider the problem
/ t uÝx, tÞ = 4 2 uÝx, tÞ + fÝuÝx, tÞÞ
uÝx, 0Þ = u 0 ÝxÞ
x 5 Rn, t > 0
x 5 Rn.
Ý2.1Þ
In this problem we take, instead of a Hilbert space H, the Banach space of functions which
are defined and continuous on R n and have a finite max. This linear space of functions
X = C b ÝR n Þ is a Banach space for the sup norm. We assume also that the nonlinearity,
f : R í R satisfies,
|fÝuÞ ? fÝvÞ| ² C R |u ? v|
- |u|, |v| ² R
Ý2.2Þ
Note that fÝuÞ = u 2 satisfies condition (2.2) for C R = 2R. Then (2.2) implies that
FÝuÞ = fÝuÝx, tÞÞ satisfies the condition (1.3) with H = V = X, and, since the composition of
continuous functions is continuous, that FÝuÞ = fÝuÝx, tÞ maps X to itself.
Since the operator A = ?4 2 on D A = áu 5 X : Au 5 Xâ = C 2 ÝR n Þ V C b ÝR n Þ can be
shown to generate a C 0 semigroup of contractions on X, it follows from the result of the
previous section that the initial value problem has a unique mild solution, û(x,t)
which satisfies,
t
ûÝtÞ = SÝtÞu 0 + X SÝt ? sÞFÝûÝsÞÞds
0 < t < T0
0
i.e.,
t
ûÝx, tÞ = X n KÝx ? y, tÞu 0 ÝyÞdy. + X X n KÝx ? y, t ? sÞfÝûÝy, sÞdyds.
R
0 R
Ý2.3Þ
where
KÝx, tÞ = 1/ 4^t e ?x
2 /4t
,
t > 0.
Since the semigroup generated by ?A = 4 2 is, in fact, analytic, we can show that the mild
solution to the IVP is actually a strong solution. This follows from the fact that when the
semigroup is analytic, the abstract IVP has a strong solution when the inhomogeneous term
fÝtÞ is only Lipschitz continuous in t. The condition (2.2) is sufficient to imply that
fÝtÞ = fÝuÝx, tÞÞ is Lipschitz in t for any uÝx, tÞ 5 X.
In addition, for this problem it is possible to use monotonicity methods to establish
uniform bounds on the solution under appropriate conditions on f. When f is such that such
bounds can be established, the solution can be shown to be global in t.
3. An IBVP in 1-dimension
Consider the problem
/ t uÝx, tÞ ? / xx uÝx, tÞ = fÝuÝx, tÞÞ
uÝx, 0Þ = u 0 ÝxÞ
uÝ0, tÞ = uÝ1, tÞ = 0
0 < x < 1, t > 0
0 < x < 1,
t > 0,
3
where we suppose f 5 C 1 ÝRÞ.
Let H = L 2 Ý0, 1Þ and V = H 10 Ý0, 1Þ. Then we can show that
V Ð C 0,J Ý0, 1Þ
for
0 < J ² 1/2.
i.e., for u 5 V, and 0 ² x, y ² 1,
x
|uÝxÞ ? uÝyÞ| = X y u v ÝsÞds ²
1/2
x
X y 1 2 ds
1
² |x ? y| 1/2 X u v ÝsÞ 2 ds
0
1/2
x
X y u v ÝsÞ 2 ds
1/2
² ||u|| V |x ? y| 1/2
Then it follows that for 0 ² x ² 1, |uÝxÞ| ² ||u|| V ; i.e., ||u|| K ² ||u|| V . In particular then for u 5 V,
fÝuÞ 5 H so F = fÝuÞ maps V to H. Now, for u, v 5 B R Ý0Þ Ð V,
1
||fÝuÞ ? fÝvÞ|| 2H = X 0 |fÝuÝxÞÞ ? fÝvÝxÞÞ| 2 dx
1
² Ýmax |s|²R |f v ÝsÞ|Þ 2 X |uÝxÞ ? vÝxÞ| 2 dx
0
² C R ||u ? v|| 2H ² C R ||u ? v|| 2V
and we see that f : V ë H is locally Lipschitz. It follows from the results of section 1 that
the abstract IVP has a unique mild solution, û 5 CÝß0, Tà: HÞ for T > 0, sufficiently small.
However, since the semigroup generated by ?A is, in fact, an analytic semigroup, the
Lipschitz smoothness of f is sufficient to imply that the mild solution is actually strong.
Note that we used that V Ð C 0,1/2 Ýß0, 1àÞ Ð H in order to assert that fÝuÞ 5 H for
u 5 V and that
u, v 5 B R Ý0Þ Ð V implies ||u|| K ² R, and ||v|| K ² R
which leads then to the result, |fÝuÞ ? fÝvÞ| ² max |s|²R |f v ÝsÞ| |u ? v|. i.e., this is a case where we
have to take V to be an appropriate closed subspace of H in order to get the behavior we
need for f.
4. A Semilinear IBVP on R 1
Consider the semilinear problem
/ t uÝx, tÞ ? / xx uÝx, tÞ + uÝx, tÞ/ x uÝx, tÞ = fÝuÝx, tÞÞ
uÝx, 0Þ = u 0 ÝxÞ
uÝ0, tÞ = uÝ1, tÞ = 0
0 < x < 1, t > 0
0 < x < 1,
Ý4.1Þ
t > 0,
where we suppose f 5 C 1 ÝRÞ. Let
4
FÝuÞ = fÝuÞ ? u/ x u
H = L 2 Ý0, 1Þ
f:VëH
Then
and
V = H 10 Ý0, 1Þ Ð C 0,1/2 Ýß0, 1àÞ
||u/ x u|| H ² ||u|| K ||/ x u|| H ² ||u|| 2V
so we have
F : V ë H. Moreover, for all u, v 5 B R Ý0Þ Ð V,
||u/ x u ? v/ x v|| H ² ||uÝ/ x u ? / x vÞ|| H + ||Ýu ? vÞ/ x v|| H
² ||u|| K ||u ? v|| V + ||u ? v|| K ||v|| V
² Ý||u|| V + ||v|| V Þ||u ? v|| V ² 2R||u ? v|| V
and this implies F is locally Lipschitz on V. It follows then that the abstract IVP has a
unique mild solution which can again be seen to be a strong solution due to the fact that ?A
generates an analytic semigroup on H. The strong solution is only local in t unless some
a-priori bound on the solution can be established.
5. A Semilinear IBVP on R n , n=2,3
The previous two examples were set in one space dimension where it happens that
V Ð C 0,J Ý0, 1Þ for 0 < J ² 1/2. For n ³ 2, the Sobolev embedding theorem changes the
situation and we have to deal more carefully with the function spaces in order to get the
Lipschitz behavior for the nonlinearity.
For U a bounded open set in R n , n ³ 2 and for J ³ 0, define
H J ÝUÞ =
u 5 H 0 ÝUÞ : > j³1 |V j | 2J |Ýu, j j Þ H | 2 < K
where áj j â j³1 denote the orthonormal family of eigenfunctions for A = ?4 2 on
V = H 10 ÝUÞ; i.e.,
H = H 0 ÝUÞ 8 u = > j³1 Ýu, j j Þ H j j
H1 = DA =
||u|| 2H = > j³1 |Ýu, j j Þ H | 2
u 5 H : Au = > j³1 V j Ýu, j j Þ H j j 5 H
i.e., u 5 D A iff ||Au|| 2H = > j³1 |V j | 2 |Ýu, j j Þ H | 2 < K
for u 5 H J ,
A J u = > j³1 V Jj Ýu, j j Þ H j j
0 ² J ² 1,
5
||u|| 2J = ||A J u|| 2H = > j³1 |V j | 2J |Ýu, j j Þ H | 2
This defines a sequence of linear spaces,
D A = H 1 Ð H J Ð H 0 = H 0 ÝUÞ,
0 < J < 1.
Evidently, H J is a Hilbert space for
Ýu, vÞ J = Ýu, vÞ H + ÝA J/2 u, A J/2 vÞ H = > j³1 Ý1 + |V j | 2J Þ|u j v j |
i.e.,
||u|| 2J = ||u|| 2H + ||A J u|| 2H
And since this can be seen to be the graph norm on D A , it follows from the closed graph
theorem that H J is a Banach space for this norm. Of course the norm then supports this
inner product and H J becomes a Hilbert space. In particular, H 1/2 = H 10 ÝUÞ.
Embedding Results
We state now some results regarding the embedding of the H J spaces.
If
H 10 ÝUÞ V H 2 ÝUÞ Ð D A = H 1 Ð H J Ð H 0 = H 0 ÝUÞ,
0 < J < 1.
then we can show that
H J is continuously embedded in W p,q ÝUÞ if
2J > p
2J ? n/2 > p ? n/q
H J is continuously embedded in C m ÝŪÞ if 2J ? n/2 > m
Now consider
x 5 U Ð Rn, t > 0
/ t uÝx, tÞ ? 4 2 uÝx, tÞ = fÝuÝx, tÞÞ
x5U
uÝx, 0Þ = u 0 ÝxÞ
uÝx, tÞ = 0
x 5 @, t > 0,
where f 5 C 1 ÝRÞ. Then FÝuÞ = fÝuÝx, tÞÞ : H J í H provided H J Ð C 0 ÝUÞ; i.e., for J > n/4.
In addition, F is locally Lipschitz if
u, v 5 B R Ý0Þ Ð H J implies ||u|| K , ||v|| K ² R
Again, we need H J continuously embedded in C 0 ÝŪÞ which means that J > n/4. It follows
that for u 0 5 H J with J > n/4 there is a unique mild solution for the IBVP, ûÝtÞ 5 CÝß0, Tà: HÞ
for sufficiently small T > 0. Since the semigroup generated by ? A is analytic here, the
solution is actually a strong solution belonging to C 0 Ýß0, Tà: HÞ VC 1 ÝÝ0, TÞ : HÞ . Note that
for n ³ 2 it is not sufficient to choose H 1/2 = H 10 ÝUÞ as the closed subspace of H which leads
6
to Lipschitz behavior for F.
Now let us consider the IBVP in the more difficult case where n = 3 and the
nonlinearity FÝuÞ = fÝuÝx, tÞÞ is given by
3
fÝuÞ = > i=1 uÝ/u//x i Þ
This nonlinearity is more difficult to deal with than the previous f 5 C 1 ÝRÞ and we need
some lemmas before trying to prove existence of the solution to the IBVP.
Lemma 1 There exists a constant C > 0, such that for all u 5 H 1 = D A ,
Lemma
-x, y 5 R 3
|uÝxÞ ? uÝyÞ| ² C ||Au|| H |x ? y| 1/2
R3
Proof- For j 5 C K0 ÝUÞ we have the classical representation for a solution of Poisson’s
equation in terms of a fundamental solution, (cf sec 2.2.1 in the Evans text)
4 2 jÝyÞ
dy
U |x ? y|
for C an appropriate constant. Applying the C-S inequality to this expression leads to
jÝxÞ = C X
1 ? 1
|x ? y|
|z ? y|
|jÝxÞ ? jÝzÞ| 2 ² C 2 XU 4 2 jÝyÞ
² C 2 X |4 2 jÝyÞ| 2 dy 6 X
U
But
XU
1 ? 1
|x ? y|
|z ? y|
2
U
dy
2
1 ? 1
|x ? y|
|z ? y|
2
dy
dy ² C U |x ? z|
for C U > 0 depending only on U. Then it follows that
|jÝxÞ ? jÝzÞ| ² C||Aj|| H |x ? z| 1/2
Since C K0 ÝUÞ is dense in D A = H 1 Ð C 0 ÝŪÞ, we can approximate any u 5 D A by
áj n â Ð C K0 ÝUÞ and pass to the limit to get the result.n
Lemma 2 There exists a constant C > 0, such that for all u 5 H 1 = D A ,
Lemma
||u|| 4K ² C ||Au|| 3H ||u|| H
Proof- The embedding results imply D A = H 1 Ð C 0 ÝŪÞ and, assuming the boundary @ is
smooth, we have that u| @ = 0, since H 10 ÝUÞ V H 2 ÝUÞ is dense in D A = H 1 . Now if u is
identically zero, the result is trivial so suppose ||u|| K = ess ? sup U |uÝxÞ| = L > 0.
We have from the previous lemma
|uÝxÞ ? uÝyÞ| ² K |x ? y| 1/2
R3
for K = C||Au|| H
7
and WOLG we may suppose L = |uÝ0Þ|. Let R = ÝL/KÞ 2 and consider the open
ball, B R Ý0Þ Ð R 3 . For x 5 B R Ý0Þ
|uÝxÞ| > |uÝ0Þ| ? |uÝ0Þ ? uÝxÞ| ³ L ? K|x| 1/2 > L ? ÝK/LÞ = 0
Since u| @ = 0 this last estimate implies B R Ý0Þ Ð U and for x 5 B R Ý0Þ, |uÝxÞ| ³ L ? K|x| 1/2 .
Now the result follows from,
||u|| 2H ³ XB
R
|uÝxÞ| 2 dx ³ XB
Ý0Þ
2
R Ý0Þ
L ? K|x| 1/2 dx
1
³ 4^L 2 R 3 X Ý1 ? z 1/2 Þ 2 z 2 dz = CL 2 R 3 = CL 8 K ?6
0
i.e.,
L 4 ² CK 3 ||u|| H .n
Lemma 3 For 1 ³ J > 3/4, and -u, v 5 D A
1)
f : HJ í H
with
||fÝuÞ|| H ² C||A J u|| H ||A 1/2 u|| H
2) ||fÝuÞ ? fÝvÞ|| H ² CÝ||A J u|| H ||A 1/2 u ? A 1/2 v|| H + ||A 1/2 v|| H ||A J u ? A J v|| H Þ
Proof- Note that the embedding result asserts that for 1 ³ J > 3/4, H J is continuously
embedded in CÝŪÞ. This implies that there exists a constant C > 0, depending on U and J
such that for all u 5 D A , ||u|| K ² C ||A J u|| H . Then for u 5 D A , u 5 L K ÝUÞ
and /u//x i 5 L 2 ÝUÞ = H so fÝuÞ 5 H. Moreover
||fÝuÞ|| H ² ||u|| K ||4u|| H ² C ||A J u|| H ||4u|| H ² C||A J u|| H ||A 1/2 u|| H .
This proves 1). Now note that
||fÝuÞ ? fÝvÞ|| H ² ||u4u ? v4v|| H = ||u4Ýu ? vÞ ? Ýu ? vÞ4v|| H
² ||u|| K ||4Ýu ? vÞ|| H + ||u ? v|| K ||4v|| H
² C ||A J u|| H qA 1/2 u ? A 1/2 vq H + qA 1/2 vq H ||A J u ? A J v|| H .
This proves 2).n
Now we can show the results needed to establish existence for the solution of the IBVP.
Since D A = H 1 Ð H J Ð H 0 = H 0 ÝUÞ, 0 < J < 1, it follows that the mapping f can be
extended from H 1 to H J for 1 ³ J > 3/4. Moreover, H 3/4 Ð H 1/2 and
qA 1/2 u ? A 1/2 vq H ² qA 3/4 u ? A 3/4 vq H
It follows that f satisfies, for 1 ³ J > 3/4,
||fÝuÞ ? fÝvÞ|| H ² CÝ||A J u|| H + qA J vq H Þ ||A J u ? A J v|| H
8
i.e., f : H J í H is locally Lipschitz for 1 ³ J > 3/4. Then the IBVP has a unique mild
solution for every u 0 5 H J , 1 ³ J > 3/4. Since the semigroup S(t), generated by ?A is
analytic, this is also a strong solution.
9