Construction of
solutions
$f’(x)=af(\lambda x),$ $\lambda>1$.
東大数理
米田剛(Tsuyoshi
Yoneda)
Graduate School of Mathematical
Sciences
The University
of
Tokyo
1. MAIN
RESULTWe denote
by
$C^{\infty}(\mathbb{R})$the
set of all infinitely differentiable functions
on
$\mathbb{R}$.
Let
$C_{0}^{\infty}(\mathbb{R})=\{f\in C^{\infty}(\mathbb{R})$
:
$\frac{d^{p}f}{dx^{p}}(x)arrow 0$as
$|x|arrow\infty,$ $P=0,1,2,$ $\cdots\}$,
$C_{comp}^{\infty}(\mathbb{R})=$
{
$f\in C^{\infty}(\mathbb{R})$:
$f$has
a
compact
support}.
We denote by
$x_{[r,\ell)}$the
characteristic
function of
the
interval
$[r, s$).Let
$X$be the set of
all step
functions
(1.1)
$p= \sum_{j=1}^{m}c_{j}\chi_{[r_{j-1},r_{j})}$,
such that
$m=1,2,$$\cdots$and
$c_{j}\in \mathbb{R}$ $(j=1,2, \cdots m)$
,
$c_{1}=1$,
$r_{0}=0$,
$0<r_{1}<r_{2}<\cdots<r_{m}<\infty$,
$\sum_{j=1}^{m}c_{j}(r_{j}-r_{j-1})=1$
.
We note that
$\int_{\mathbb{R}}p(x)dx=1$for
$p\in X$.
For
$\lambda>1$and
$p\in X$,
we
define
an
operator
$T=T_{\lambda,p}$:
$L^{1}(\mathbb{R})arrow$$L^{1}(\mathbb{R})$
(by
[17]
(p.
149, Remark
2)
and
[2],
we
can
say
that
$T_{\lambda,p}$:
$B_{1,\infty}^{8}(\mathbb{R})arrow B_{1,\infty}^{\epsilon+1}(\mathbb{R})$
for
$s\in \mathbb{R}$,
since
$\lambda p(\lambda\cdot)\in B_{1,\infty}^{1}(\mathbb{R}))$as
follows:
(1.2)
$Tf(x)=T_{\lambda,p}f(x)=\lambda(p*f)(\lambda x)$.
We
note that
$Tf\in L^{1}(\mathbb{R})\cap C(\mathbb{R})$, since
$f\in L^{1}(\mathbb{R})$and
$p\in L_{\infty mp}^{\infty}(\mathbb{R})$.
If
$f\in B_{1,\infty}^{s}(\mathbb{R})$we
have
$\hat{f}(1+|\cdot|^{2})^{s/2}\in L^{\infty}(\mathbb{R})$(see
[15]).
Therefore
Theorem 1.1.
For
$\lambda>1$and
$p\in X$,let
$T$be
the
operator
defined
by
(1.2).
Then there exists
a
function
$u=u_{\lambda,p}\in C_{comp}^{\infty}(\mathbb{R})$such that,
for
all
$f\in L^{1}(\mathbb{R})$(or
for
all
$f \in\bigcup_{s\in \mathbb{R}}B_{1,\infty}^{s}(\mathbb{R})_{f}$which
function
set
is
strt
ctly
bigger
than
$L^{1}(\mathbb{R}))$,
$\lim_{\ellarrow\infty}T^{\ell}f(x)=c_{f}u(x)$
uniformly with
resPect
to
$x\in \mathbb{R}$,
where
$c_{f}=\hat{f}(0)$.
Mooeover,
(1)
$Tu=u$;
(2)
$\int_{R}u(x)dx=1$;
(3)
$\frac{d}{dx}pu(0)=0k$for
all
$k=0,1,2,$$\cdots$;
(4)
if
$p\geq 0_{f}$then
$u\geq 0$;and,
(5)
if
supp
$p\subset[0, r]$, then supp
$u\subset[0, r/(\lambda-1)]$.
In the
above
statement supp
$f$denotes
the support
of
$f$.
Remark 1.1. If
$p_{1}\neq p_{2}$,
then
$u_{\lambda,p_{1}}\neq u_{\lambda,p_{2}}$.
2.
APPLICATIONTo construct solutions for
the
$foUowing$equation,
(2.1)
$\{\begin{array}{ll}f’(x)=\lambda^{2}f(\lambda x), x\in \mathbb{R},f(0)=0, \end{array}$we
use
$u=u_{\lambda,p}$in
Theorem
1.1 with
$\lambda>1$and
$p= \sum_{j=1}^{m}c_{j}\chi[rr$ ) $\in$$X$
, and
we
define
a
sequence
of functions
$\{v_{\ell}\}$inductively
as
follows:
$\{\begin{array}{ll}v_{0}(x)=u(x)+\sum_{j=1}^{m}\overline{c}_{j}u(x-r_{j}), v_{\ell+1}(x)=v_{\ell}(x)+\sum_{j=1}^{m}\tilde{c}_{j}v_{\ell}(x-\lambda^{l+1}r_{j}), \ell=0,1,2, \cdots,\end{array}$
where
$\tilde{c}_{j}=-(c_{j}-c_{j+1}),$ $j=1,2,$ $\cdots m-1,\tilde{c}_{m}=-c_{m}$.
Then
$v_{\ell}\in C_{\infty mp}^{\infty}(\mathbb{R}),$ $\ell=0,1,2,$ $\cdots$,
and
$vp(x)=v_{\ell+1}(x)$
,
$x\in(-\infty, \lambda^{\ell+1}r_{1}$],
$\ell=0,1,2,$ $\cdots$.
Therefore
we
get
a
function
$f=f_{\lambda p}\in C^{\infty}(\mathbb{R})$such
that
$f(x)=v_{\ell}(x)$
,
$x\in(-\infty, \lambda^{\ell+1}r_{1}$],
$\ell=0,1,2,$ $\cdots$.
Theorem 2.1.
Let
$\lambda>1$.
For every
$p\in X$, the
function
$f=f_{\lambda,p}$given
by the
above
method
is
a
solution
for
(2.1).
Remark
2.1.
By the
property
(5) in Theorem 1.1 and the
definition
of
$v_{\ell}$
,
supp
$u\subset[0, r_{m}/(\lambda-1)]$, $supp(v_{0}-u)\subset[r_{1}, r_{m}\lambda/(\lambda-1)]$,
supp
$v_{\ell}\subset[0, r_{m}\lambda^{\ell+1}/(\lambda-1)]$,
$supp(v_{\ell+1}-v_{\ell})\subset[r_{1}\lambda^{l+1},r_{m}\lambda^{\ell+2}/(\lambda-1)]$
.
Therefore,
if
$r_{m}/(\lambda-1)<r_{1}$,
then
$f(x)=0$
on
$[r_{m}\lambda^{\ell}/(\lambda-1), r_{1}\lambda^{\ell}]$,
$P=0,1,2,$ $\ldots$
, and
$f$is
bounded
(See
the
graph (S12)). Let
$n_{4}$
$p_{i}= \sum_{j=1}c_{i,j}\chi_{[r:_{l-1r_{i.j})}}.,\in X$
,
$i=1,2$.
If
$p_{1}\neq p_{2}$and
there
exists
$R>0$such
taht
$r_{i,m_{1}}/(\lambda-1)<R<r_{i,1}$,
$i=1,2$
,
then
$f_{\lambda,p_{1}}\neq f_{\lambda,p_{2}}$by
Remark 1.1, while
$\int_{0}^{R}f_{\lambda,ps}(x)dx=$$\int_{0}^{R}u_{\lambda,p:}(x)dx=1,$ $i=1,2$
.
Remark 2.2. In the definition
(1.1),
let
$m=\infty$with
the
condition
$\sum_{j=1}^{\infty}|c_{j+1}-c_{j}|<\infty$
.
Then the derivative of
$p$in
the
sense
of
dis-tribution
is
a
finite Radon
measure
and
the
Fourier
transform of
the
derivative is
an
almost
periodic
function. With
some
conditions
we
can
ako
construct solutions for
(2.1)
from such functions. For the Fourier
preimage
of the
space
of
all
finite
Radon measures, see, for
example,
$[6, 7]$.
In the rest of this
section,
we
give graphs
of
solutions for
$f’(x)=$$4f(2x),$ $f’(x)=(3/2)^{2}f(3x/2)$
and
$f’(x)=9f(3x)$
with
$f(O)=0$.
Let
$p$be
as
in
Table
1.
Then, calculating
$T_{\lambda,p}^{\ell}\chi_{[0_{y}1)}$numerically
by
computer,
we
get
the graphs of
solutions
$(S1)-(S7)$in Figure 1,
$(S8)-$(Sll)
in Figure
2 and
$(S12)-(S13)$in Figure
3.
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(S1)
(S2)
(S3)
(S4)
(S5)
FIGURE 1.
Solutions
of
$f’(x)=4f(2x)$
.
TABLE 1.
$\lambda,$ $p$and the
solution
$f_{p,\lambda}$.
FIGURE 2.
Solutions
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$f’(x)=(3/2)^{2}f(3x/2)$.
(S13)
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GRADUATESCHOOL OFMATHEMATICALSCIENCES, THEUNIVERSITYOFTOKYO,
3-8-1 KOMABA, MEGURO-KU TOKYO 153-8914, JAPAN E-mail address: yonedadms.u-tokyo.ac.jp