## Solvability of a parabolic boundary value problem in the presence of power singularities in the right sides.(Russian)Zbl 0657.35067

Let G be a smooth bounded domain in $$R^ n$$, $$\Omega =\bar G\times [0,T)$$, $$b>0$$ an integer, $$S=\partial G\times [0,T)$$, $${\mathcal H}^ s(\Omega)={\mathcal H}^ s_ x{}_ t^{s/2b}(\Omega)$$ and $${\mathcal H}^ s(S)={\mathcal H}_ x^ s{}_ t^{s/2b}(S)$$ (anisotropic Sobolev- Slobodetskii’s spaces), P a subset of $$Z\times Z$$, $$Z_ a=\{z\in Z:$$ $$z>0$$, $$z/a=$$ (mod 1)$$\}$$. For the definition of spaces $$\tilde H^ s(G)$$ and $$\tilde {\mathcal H}_{(\alpha,\tau,P)}(G)$$ $$(s\not\in Z_ 1\cup Z_{2b})$$ see the present author [Math. USSR, Sb. 56, 447-471 (1987); translation from Mat. Sb., Nov. Ser. 128(176), No.4, 451-473 (1985; Zbl 0609.35045)]. Let $${\mathcal L}(x,t,D,D_ t)$$ be a parabolic operator in $${\bar \Omega}$$ of order $$2m=2br$$, $$b_ j(x',t,D,D_ t)$$ $$(j=1,...,m)$$ operators of weighted order $$m_ j<2m$$ connected with L on S by the Lopatinskii’s (complementarity) condition. Let us consider the following problem: Find a solution u in $$\tilde {\mathcal H}^ s_{(r,2m,P)}(\Omega)$$ to the boundary value problem ${\mathcal L}(x,t,D,D_ t)u(x,t)=f(x,t)\in {\mathcal H}^{s-2m}(\Omega),$
${\mathcal B}_ j(x',t,D,D_ t)u(x,t)_{| S}=\phi_ j(x',t)\in {\mathcal H}^{s-m_ j-}(S)\quad (j=1,...,m),$
$D_ t^{\lambda -1}u(x,t)_{| t=0}=\psi_{\lambda}(x)\in \tilde H_{(P_{\lambda})}^{s-2b\lambda +b}(G)\quad (\lambda =1,...,r),$ where $$P_{\lambda}=\{k: (k,\lambda)\in P\}.$$
If $$f(x,t)=h(x,t)$$ $$(| y'|^{2b}+| t-t_ 0|)^{-\mu}$$, where $$h\in L_ 2(\Omega)$$, 2b$$\mu\geq q/2+b$$, $$F_{\mu}(x,t)=reg f(x,t)$$ is a regularization of f(x,t) with $$s=2b\mu -q/2-b+\epsilon$$ $$(\epsilon >0)$$, $$\phi_ j\in {\mathcal H}^{2m-2b\mu -m_ j- +\epsilon}(S),$$ $$\psi_{\lambda}(x)\in \tilde H^{2m-2b\lambda -2b\mu +b+\epsilon}(G)$$ and the compatibility condition is satisfied for $$t=0$$, then the problem ${\mathcal L}u=F_{\mu}(x,t),\quad {\mathcal B}_ ju_{| S}=\phi_ j(x',t)\quad (j=1,...,m),\quad D_ t^{\lambda - 1}u(x,0)=\psi_{\lambda}(x)\quad (\lambda =1,...,r)$ is solvable and the solution lies in $$\tilde {\mathcal H}^{2m-2b\mu - \epsilon}_{(r,2m,P)}(\Omega)$$.
Reviewer: J.Danesova

### MSC:

 35K35 Initial-boundary value problems for higher-order parabolic equations 35R05 PDEs with low regular coefficients and/or low regular data 46E35 Sobolev spaces and other spaces of “smooth” functions, embedding theorems, trace theorems

Zbl 0609.35045
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