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Finite time blow up solutions for heat equations with Neumann boundary conditions on R+4\mathbb{R}_{+}^{4}

Published 25 Nov 2025 in math.AP | (2511.20451v1)

Abstract: We consider the nonlinear heat equations with Neumann boundary conditions {ut=Δuamp;in R<em>+<sup>4</sup>×(0,T), dudx</em>4(x~,0,t) =u<sup>2(x~,</sup>0,t)amp;in R<sup>3</sup>×(0,T). \begin{cases} u_{t}=Δu &amp; \text{in}\ \mathbb{R}<em>{+}<sup>{4}</sup> \times(0, T) ,\ -\frac{d u}{d x</em>{4}}(\tilde{x}, 0, t) \ =u<sup>2(\tilde{x},</sup> 0, t)&amp; \text{in}\ \mathbb{R}<sup>{3}</sup> \times(0, T). \end{cases} We establish the existence of a finite-time blow-up solution. Specifically, for any sufficiently small $T&gt;0$ and any kk distinct points q1,,qkR<sup>3q_{1},\dots,q_{k}\in \mathbb{R}<sup>{3}, there exists an initial datum u0u_{0} such that the corresponding solution u(x,t)u(x,t) blows up exactly at q1,,qkq_{1},\dots,q_{k} as tTt\nearrow T. Furthermore, when tTt\nearrow T, the solution admits the asymptotic profile u(x,t)=j=1<sup>kUμ<em>j(t),ξ</em>j(t)(x)+Z0<sup>(x)+o(1)</sup></sup>as tT,u(x,t)=\sum_{j=1}<sup>{k}U_{μ<em>{j}(t),ξ</em>{j}(t)}(x)+Z_0<sup>*(x)+o(1)\quad</sup></sup> \text{as}~ t\nearrow T, where Uμ<em>j(t),ξ</em>j(t)(x):=μ<em>j<sup>1(t)</sup>U(xξ</em>j(t)μ<em>j(t)), xR</em>+<sup>4,U_{μ<em>{j}(t),ξ</em>{j}(t)}(x):=μ<em>{j}<sup>{-1}(t)</sup> U\left(\frac{x-ξ</em>{j}(t)}{μ<em>{j}(t)}\right),~ x\in \mathbb{R}</em>{+}<sup>{4}, and Z0<sup></sup>C0<sup>(R+<sup>4)Z_{0}<sup>{*}\in</sup> C_{0}<sup>{\infty}(\mathbb{R}_{+}<sup>{4}) satisfying $$Z_{0}<sup>{*}(q_{j},0)&lt;0\quad</sup> \text{for all}\ j=1,\dots,k.$$ Here, U(y)U(y) denotes the harmonic extension to R<em>+<sup>4\mathbb{R}<em>{+}<sup>{4} of the positive radially symmetric solution U~\widetilde{U} to the fractional Yamabe problem (Δ)<sup>12</sup>U~=U~<sup>2(-Δ)<sup>{\frac{1}{2}}</sup> \widetilde{U} = \widetilde{U}<sup>{2} in R<sup>3\mathbb{R}<sup>{3}. For some constants $β</em>{j}&gt;0$, the scaling parameters μj(t)μ{j}(t) and the translation parameters ξ<em>j(t)ξ<em>{j}(t) satisfy μ</em>j(t)=β<em>jlog2T(Tt)log(Tt)<sup>2(1</sup>+o(1))0, ξ</em>j(t)(qj,0)as tT.μ</em>{j}(t)=β<em>{j}\frac{|\log 2T|(T-t)}{|\log(T-t)|<sup>{2}}(1</sup> + o(1)) \to 0,~ξ</em>{j}(t)\to (q_{j},0)\quad \text{as} ~t\nearrow T.

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