Construction of solutions $f'(x) = Title1(Banach spaces, function spaces, applications) Author(s) Yoneda, Tsuyoshi Citation 数理解析研究所講究録 (2007), 1570: 1-7 Issue Date 2007-10 URL http://hdlhandlenet/2433/81278 Right Type Departmental Bulletin Paper Textversion publisher Kyoto University
1570 2007 1-7 1 Construction of solutions $f (x)=af(\lambda x),$ $\lambda>1$ (Tsuyoshi Yoneda) Graduate School of Mathematical Sciences The University of Tokyo 1 MAIN RESULT on $\mathbb{r}$ $C^{\infty}(\mathbb{R})$ We denote by Let the set of all infinitely differentiable functions $C_{0}^{\infty}(\mathbb{R})=\{f\in C^{\infty}(\mathbb{R})$ : $\frac{d^{p}f}{dx^{p}}(x)arrow 0$ as $C_{comp}^{\infty}(\mathbb{R})=$ { $f\in C^{\infty}(\mathbb{R})$ : $f$ has a compact support} $ x arrow\infty,$ $P=0,1,2,$ $\cdots\}$, We denote by $x_{[r,\ell)}$ the characteristic function of the interval $[r, s$ ) be the set of all step functions Let $X$ (11) $p= \sum_{j=1}^{m}c_{j}\chi_{[r_{j-1},r_{j})}$, 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 $T_{\lambda,p}$ [17] (p 149, Remark 2) and [2], we can say that :, since $s\in \mathbb{r}$ $\lambda p(\lambda\cdot)\in B_{1,\infty}^{1}(\mathbb{R}))$ as follows: $B_{1,\infty}^{8}(\mathbb{R})arrow B_{1,\infty}^{\epsilon+1}(\mathbb{R})$ for (12) $Tf(x)=T_{\lambda,p}f(x)=\lambda(p*f)(\lambda x)$ We note that, since and $Tf\in L^{1}(\mathbb{R})\cap C(\mathbb{R})$ $f\in L^{1}(\mathbb{R})$ $p\in If we have (see [15]) Therefore $f\in B_{1,\infty}^{s}(\mathbb{R})$ $\hat{f}(1+ \cdot ^{2})^{s/2}\in L^{\infty}(\mathbb{R})$ we have for $k>-s-1$ if $T^{k}f\in L^{2}(\mathbb{R})\cap C(\mathbb{R})$ $f\in B_{1,\infty}^{s}(\mathbb{R})$ L_{\infty mp}^{\infty}(\mathbb{r})$
2 Theorem 11 For $\lambda>1$ and $p\in X$, let be the operator defined $T$ C_{comp}^{\infty}(\mathbb{R})$ such that, by (12) Then there exists a $u=u_{\lambda,p}\in function for all (or for all $f\in L^{1}(\mathbb{R})$ $f strtctly bigger than $L^{1}(\mathbb{R}))$, Mooeover, \in\bigcup_{s\in \mathbb{r}}b_{1,\infty}^{s}(\mathbb{r})_{f}$ which function set is $\lim_{\ellarrow\infty}t^{\ell}f(x)=c_{f}u(x)$ uniformly with respect to $x\in \mathbb{r}$, where $c_{f}=\hat{f}(0)$ (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 11 If $p_{1}\neq p_{2}$, then $u_{\lambda,p_{1}}\neq u_{\lambda,p_{2}}$ 2 APPLICATION To construct solutions for the $fouowing$ equation, (21) $\{\begin{array}{ll}f (x)=\lambda^{2}f(\lambda x), x\in \mathbb{r},f(0)=0, \end{array}$ we use $u=u_{\lambda,p}$ in Theorem 11 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 $\tilde{c}_{j}=-(c_{j}-c_{j+1}),$ $j=1,2,$ $\cdots m-1,\tilde{c}_{m}=-c_{m}$ C_{\infty mp}^{\infty}(\mathbb{r}),$ $\ell=0,1,2,$, and $\cdots$ Then $v_{\ell}\in $vp(x)=v_{\ell+1}(x)$, $x\in(-\infty, \lambda^{\ell+1}r_{1}$], $\ell=0,1,2,$ $\cdots$ Therefore we get a function $f=f_{\lambda p}\in C^{\infty}(\mathbb{R})$ such that $x\in(-\infty, $f(x)=v_{\ell}(x)$, \lambda^{\ell+1}r_{1}$], $\ell=0,1,2,$ $\cdots$ Since $f(x)=u(x)$ near $x=0,$ for all $k=0,1,2,$ $\frac{d}{dx}\tau kf(o)=0$ $\cdots$
3 Theorem 21 Let $\lambda>1$ For every $p\in X$, the function $f=f_{\lambda,p}$ given by the above method is a solution for (21) Remark 21 By the property (5) in Theorem 11 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 $v_{\ell}\subset[0, r_{m}\lambda^{\ell+1}/(\lambda-1)]$, $supp(v_{\ell+1}-v_{\ell})\subset[r_{1}\lambda^{l+1},r_{m}\lambda^{\ell+2}/(\lambda-1)]$ $[r_{m}\lambda^{\ell}/(\lambda-1), Therefore, if $r_{m}/(\lambda-1)<r_{1}$, then $f(x)=0$ on $P=0,1,2,$ $\ldots$, and is bounded (See the graph (S12)) Let $f$ r_{1}\lambda^{\ell}]$, $n_{4}$ $p_{i}= \sum_{j=1}c_{i,j}\chi_{[r:_{l-1r_{ij})}},\in X$, $i=1,2$ If and there exists $R>0$ such taht $p_{1}\neq p_{2}$ $r_{i,m_{1}}/(\lambda-1)<r<r_{i,1}$, $f_{\lambda,p_{1}}\neq f_{\lambda,p_{2}}$ $i=1,2$, then by Remark 11, while $\int_{0}^{r}f_{\lambda,ps}(x)dx=$ $\int_{0}^{r}u_{\lambda,p:}(x)dx=1,$ $i=1,2$ Remark 22 In the definition (11), let $m=\infty$ with the condition $\sum_{j=1}^{\infty} c_{j+1}-c_{j} <\infty$ Then the derivative of in the sense of distribution is a finite Radon measure and the Fourier transform of the $p$ derivative is an almost periodic function With some conditions we can ako construct solutions for (21) 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 be as in Table 1 Then, calculating $p$ $T_{\lambda,p}^{\ell}\chi_{[0_{y}1)}$ 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 REFERENCES [1] A Augustynowicz, H Leszczy\ oki and W Walter, On some nonlinear $ordina\eta$ differential equations Utth advanced arguments, Nonlinear Anal, 53 (2003), 495-505 [2] VI Burenkov, On estimates of Fourier transforrns and convolutions in Nikol skij-besov spaces, Proc Steklov Inst 3(1990), 35-44
4 (S1) (S2) (S3) (S4) (S5) FIGURE 1 Solutions of $f (x)=4f(2x)$
5 $\lambda,$ TABLE 1 $p$ $f_{p,\lambda}$ and the solution FIGURE 2 Solutions of $f (x)=(3/2)^{2}f(3x/2)$ (S13) FIGURE 3 Solutions of $f (x)=9f(3x)$
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