## Mathematics Publications

Mathematics

Article

#### Publication Version

Published Version

1983

#### Journal or Book Title

SIAM Journal on Mathematical Analysis

14

6

1139

1153

10.1137/0514088

#### Abstract

In this paper we examine the initial-boundary value problems $(\alpha ):u_t = u_{xx}$, $0 < x < L,t > 0$, $u(0,t) = u(x,0) = 0$, $u_x (L,t) = \phi (u(L,t))$ and $(\beta ):u_{tt} = u_{xx}$, $0 < x < L,t > 0$, $u(0,t) = u(x,0) = u_t (x,0)$, $u_x (L,t) = \phi (u(L,t))$ where $\phi ( - \infty ,1) \to (0,\infty )$ is continuously differentiable, monotone increasing and $\lim _{u \to 1} - \phi (u) = + \infty$. For problem $(\alpha )$ we show that there is a positive number $L_0$ such that if $L \leq L_0$, $u(x,t) \leq 1 - \delta$ for some $\delta > 0$ for all $t > 0$, while if $L > L_0 ,u(L,t)$ reaches one in finite time while $u_t (L,t)$ becomes unbounded in that time. For problem $(\beta )$ it is shown that if L is sufficiently small, then $u(L,t) \leq 1 - \delta$ for all $t > 0$ while if L is sufficiently large and $\int_0^1 \phi (\eta )d\eta < \infty ,u(L,t)$ reaches one in finite time whereas if $\int_0^1 \phi (\eta )d\eta = \infty ,u(L,t)$ reaches one in finite or infinite time.

In either of the last two situations $u_t (L,t)$ becomes unbounded if the time interval is finite. If u reaches one in infinite time, then $\int_0^1 {u_x^2 } (x,t)dx$ and $u(x,t)$ are unbounded on the half line and half strip respectively.

This is an article from SIAM Journal on Mathematical Analysis 14 (1983): doi:10.1137/0514088. Posted with permission.

Society for Industrial and Applied Mathematics

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