2.1. $L(t)=L_0$

For a fixed domain length $L_0$, we have $\ddot {L}L^{3}=\gamma _0=0$ and the separable cases are those where

(2.16)\begin{equation} A(t)=\frac{\gamma_{1}}{2L_0^{3}} t^{2}+c t+d \qquad \textrm{for some } c, d. \end{equation}

The separable solutions then have the form

(2.17)\begin{align} u_n(\xi,t)& =\exp(\sigma_n t)g_n(\xi)\exp\left(f_0t-\frac{1}{4D}\left(\frac{\gamma_{1}^{2}}{3L_0^{6}} t^{3}+\frac{c\gamma_{1}}{L_0^{3}} t^{2}+c^{2} t\right)\right)\nonumber\\ & \quad\times \exp\left(-\frac{\xi}{2DL_0}\left(\frac{\gamma_{1}}{L_0^{2}} t+cL_0\right)\right). \end{align}

If $\gamma _{1}\neq 0$ then, as $t\rightarrow \infty$, $u(\xi,t)\rightarrow 0$ since the behaviour is dominated by the term

(2.18)\begin{equation} \exp\left(-\frac{\gamma_{1}^{2}}{12DL_0^{6}} t^{3}\right). \end{equation}

If $\gamma _{1}=0$ then, as $t\rightarrow \infty$, there is exponential growth or decay in the cases $f_0>\frac {D\pi ^{2}}{L_0^{2}}+\frac {c^{2}}{4D}$ or $f_0<\frac {D\pi ^{2}}{L_0^{2}}+\frac {c^{2}}{4D}$ respectively. Indeed, the long time behaviour of $u_n$ is governed by

(2.19)\begin{equation} \exp\left(\sigma_n t+f_0 t-\frac{1}{4D}c^{2} t \right) \end{equation}

where $\sigma _n =-\frac {Dn^{2}\pi ^{2}}{L_0^{2}}$.

2.2. $L(t)=L_0+\alpha t$ with $\alpha \neq 0$

When $L(t)=L_0+\alpha t$ then again $\ddot {L}L^{3}=\gamma _0=0$, but now the separable cases are those where

(2.20)\begin{equation} A(t)=\frac{\gamma_{1}}{2\alpha^{2}(L_0+\alpha t)}+c t+d \qquad \textrm{for some } c, d. \end{equation}

The separable solutions then have the form

(2.21)\begin{align} u_n(\xi, t)& =\exp\left(\frac{\sigma_n L_0 t}{L_0+\alpha t}\right)g_n(\xi)\left(\frac{L_0}{L_0+\alpha t}\right)^{1/2}\exp(f_0t) \nonumber\\ & \quad\times \exp\left(-\frac{1}{4D}\left(c^{2} t-\frac{c\gamma_{1} t}{\alpha L_0(L_0+\alpha t)}-\frac{\gamma_{1}^{2}}{12\alpha^{3}}\left(\frac{1}{(L_0+\alpha t)^{3}} -\frac{1}{L_0^{3}} \right)\right)\right)\nonumber\\ & \quad\times \exp\left(-\frac{\xi^{2}\alpha(L_0+\alpha t)}{4DL_0^{2}}-\frac{\xi c(L_0+\alpha t)}{2DL_0}+\frac{\xi\gamma_{1}}{4DL_0\alpha(L_0+\alpha t)}\right). \end{align}

If $\alpha >0$ then as $t\rightarrow \infty$, the behaviour is asymptotically governed by

(2.22)\begin{equation} \exp\left( f_0t-\frac{1}{4D}\left(c+\frac{\xi\alpha}{L_0}\right)^{2} t \right). \end{equation}

Recall that we defined

(2.23)\begin{equation} c_* = 2\sqrt{Df_0}. \end{equation}

Thus, if $-\alpha -c_* < c < c_*$, then there is a region of $\xi$ in which there is exponential growth: namely, where

(2.24)\begin{equation} \max\left(0,\frac{L_0}{\alpha}({-}c_*-c)\right)<\xi<\min\left(L_0, \frac{L_0}{\alpha}(c_*-c)\right). \end{equation}

Otherwise, $u(\xi,t)$ decays to zero everywhere in $(0,L_0)$.

If instead $\alpha <0$, then $L(t)\rightarrow 0$ as $t\rightarrow -L_0/\alpha$, and in this limit, $u(\xi,t)\rightarrow 0$. Indeed, if $\gamma _{1}\neq 0$, then the behaviour is governed by

(2.25)\begin{equation} \exp\left(\frac{\gamma_{1}^{2}}{48 D\alpha^{3}(L_0+\alpha t)^{3}}\right), \end{equation}

which decays exponentially since $\alpha <0$. If $\gamma _{1}=0$ then the governing term is

(2.26)\begin{equation} \exp\left(\frac{\sigma_n L_0 t}{L_0+\alpha t}\right) \end{equation}

where $\sigma _n =-\frac {Dn^{2}\pi ^{2}}{L_0^{2}}<0$ and so, again, $u(\xi,t)\rightarrow 0$.

2.3. $L(t)=\sqrt {L_0^{2}+2\rho t}$ with $\rho \neq 0$

If $L(t)^{2}=L_0^{2}+2\rho t$ then $\ddot {L}L^{3}=\gamma _{0}=-\rho ^{2}<0$ and the separable cases are those where

(2.27)\begin{equation} A(t)=\frac{-\gamma_{1}\sqrt{L_0^{2}+2\rho t}}{\rho^{2}}+c t+d \text{ for some } c, d. \end{equation}

The separable solutions then have the form

(2.28)\begin{align} u_n(\xi, t)& =\left(\frac{L_0^{2}+2\rho t}{L_0^{2}}\right)^{{\frac{\sigma_n L_0^{2}}{2\rho}-\frac{1}{4}-\frac{\gamma_{1}^{2}}{8\rho^{3} D}}}g_n(\xi)\exp\left(f_0 t-\frac{c^{2}}{4D} t\right)\nonumber\\ & \quad\times\exp\left(\frac{c\gamma_{1}}{2\rho^{2}D}\left(\sqrt{L_0^{2}+2\rho t}-L_0\right)-\frac{\xi^{2}\rho}{4DL_0^{2}}+\frac{\xi\gamma_{1}}{2DL_0\rho}-\frac{\xi c\sqrt{L_0^{2}+2\rho t}}{2DL_0}\right). \end{align}

Therefore, if $\rho >0$ then, as $t\rightarrow \infty$, there is exponential growth or decay in the cases $f_0>\frac {c^{2}}{4D}$ or $f_0<\frac {c^{2}}{4D}$ respectively. Indeed, the long-time behaviour in this case is governed by

(2.29)\begin{equation} \exp\left(f_0t-\frac{c^{2}}{4D}t\right). \end{equation}

If instead $\rho <0$ then $L(t)\rightarrow 0$ as $t\rightarrow -L_0^{2}/2\rho$, and in this limit $u(\xi,t)\rightarrow 0$. This follows because the behaviour of each $u_n$ is governed by

(2.30)\begin{equation} \left(\frac{L_0^{2}+2\rho t}{L_0^{2}}\right)^{{\frac{\sigma_n L_0^{2}}{2\rho}-\frac{1}{4}-\frac{\gamma_{1}^{2}}{8\rho^{3} D}}}. \end{equation}

This implies that $u(\xi,t)\rightarrow 0$, by using the bound in equation (2.12).

2.4. $L(t)=\sqrt {a t^{2} + 2b t + L_0^{2}}$ with $a\neq 0$ and $aL_0^{2}-b^{2} \neq 0$

If $L(t)^{2}=a t^{2} + 2b t + L_0^{2}$ then $\ddot {L}L^{3}=\gamma _{0}=aL_0^{2}-b^{2}$, and for $aL_0^{2}-b^{2} \neq 0$, the separable cases are those where

(2.31)\begin{equation} A(t)=\frac{-\gamma_{1}}{b^{2}-aL_0^{2}}\sqrt{a t^{2}+2b t + L_0^{2}} + ct +d \text{ for some } c, d. \end{equation}

The separable solutions have the form

(2.32)\begin{align} u_n(\xi, t)& =\Theta_n(t)g_n(\xi)\left(\frac{L_0^{2}}{a t^{2}+2b t+L_0^{2}}\right)^{1/4} \exp\left(f_0 t-\frac{1}{4D}\left(\frac{\gamma_{1}^{2}a}{(b^{2}-aL_0^{2})^{2}}+c^{2}\right) t\right)\nonumber\\ & \quad\times \exp\left(\frac{c\gamma_{1}}{2D(b^{2}-aL_0^{2})}\left(\sqrt{a t^{2}+2b t+L_0^{2}}-L_0\right)\right) \nonumber\\ & \quad\times \exp\left(-\frac{\xi^{2}(a t+b) }{4DL_0^{2}}+ \frac{\xi\gamma_{1}(a t+b)}{2DL_0(b^{2}-aL_0^{2})} -\frac{\xi c}{2DL_0}\sqrt{a t^{2}+2b t+L_0^{2}}\right), \end{align}

where if $\gamma _0=aL_0^{2}-b^{2} < 0$,

(2.33)\begin{equation} \Theta_n(t)=\left(\frac{\left(a t+b-\sqrt{b^{2}-aL_0^{2}}\right)\left(b+\sqrt{b^{2}-aL_0^{2}}\right)}{\left(b-\sqrt{b^{2}-aL_0^{2}}\right)\left(a t+b+\sqrt{b^{2}-aL_0^{2}}\right)}\right)^{{\frac{\sigma_n L_0^{2}}{2\sqrt{b^{2}-aL_0^{2}}} - \frac{\gamma_{1}^{2}}{8D(b^{2}-aL_0^{2})^{3/2}}}}, \end{equation}

and if $\gamma _0=aL_0^{2}-b^{2} >0$ then

(2.34)\begin{equation} \Theta_n(t)=e^{ \left(\frac{\sigma_n L_0^{2}}{\sqrt{aL_0^{2}-b^{2}}} +\frac{\gamma_{1}^{2}}{4D(aL_0^{2}-b^{2})^{3/2}} \right) \left(\arctan\left(\frac{a t+b}{\sqrt{aL_0^{2}-b^{2}}}\right)-\arctan\left(\frac{b}{\sqrt{aL_0^{2}-b^{2}}}\right)\right)}. \end{equation}

If $L(t)$ remains positive for all $t>0$, then $a>0$ and the behaviour as $t\rightarrow \infty$ is governed by

(2.35)\begin{equation} \exp\left(f_0t-\frac{1}{4D}\left(c-\frac{\gamma_{1}\sqrt{a}}{(b^{2}-aL_0^{2})}+\frac{\xi\sqrt{a}}{L_0}\right)^{2} t\right). \end{equation}

So, if $\displaystyle {-c_*+\frac {\gamma _{1}\sqrt {a}}{(b^{2}-aL_0^{2})}-\sqrt {a}< c< c_*+\frac {\gamma _{1}\sqrt {a}}{(b^{2}-aL_0^{2})}}$, then there is a region of $\xi$ in which there is exponential growth: namely, where

(2.36)\begin{align} \max\left(0,\frac{L_0}{\sqrt{a}}\left({-}c_*-c+\frac{\gamma_{1}\sqrt{a}}{(b^{2}-aL_0^{2})}\right)\right)<\xi<\min\left(L_0,\frac{L_0}{\sqrt{a}}\left(c_*-c+\frac{\gamma_{1}\sqrt{a}}{(b^{2}-aL_0^{2})}\right)\right). \end{align}

Otherwise, there is exponential decay everywhere.

If instead $L(t)\rightarrow 0$ in a finite time, then it must be that $aL_0^{2}-b^{2} < 0$ and that $L(t)\rightarrow 0$ as

(2.37)\begin{equation} t\rightarrow -\frac{1}{a}\sqrt{b^{2}-aL_0^{2}}-\frac{b}{a}. \end{equation}

In this limit, $u(\xi,t)\rightarrow 0$ since the behaviour is governed by

(2.38)\begin{equation} \left(\frac{b+\sqrt{b^{2}-aL_0^{2}}}{a t+b+\sqrt{b^{2}-aL_0^{2}}}\right)^{{\frac{\sigma_n L_0^{2}}{2\sqrt{b^{2}-aL_0^{2}}} - \frac{\gamma_{1}^{2}}{8D(b^{2}-aL_0^{2})^{3/2}}+\frac{1}{4}}}. \end{equation}

This implies that $u(\xi,t)\rightarrow 0$, by using the bound in equation (2.12) with $\rho ^{2}=b^{2}- aL_0^{2}=-\gamma _{0}$.

2.5. Observed properties of the solutions

These expressions are very instructive in understanding the effects of a time-dependent domain on the solution. From them, one can observe the ways in which the exact nature of the time-dependence influences the solution in both short and long time.

One may choose specific parameter values and, after solving the relevant Sturm–Liouville problem, plot the solutions. An illustrative example is given in Figure 1, with $D=L_0=1$, $f_0=10$, $L(t)=\sqrt {1+6t}$ and $A(t)\equiv 0$. Using the notation of Lemma 1, the principal eigenfunction is proportional to $e^{-\frac {1}{2}\eta ^{2}}H_3(\eta )$ where $\eta =\sqrt {\frac {3}{2}}\frac {\xi }{L_0}$ and $H_3$ is the third Hermite polynomial. We use a multiple of this for our initial condition: $v(\xi,0)=g_1(\xi )=e^{-\frac {3}{4}\xi ^{2}}(\xi -\xi ^{3})$. The principal eigenvalue is $\sigma _1=-\frac {21}{2}$. The solution is calculated using the exact expression in Equation (2.28) and plotted both on the fixed domain (as a function of the scaled variable $\xi$) and on the time-dependent domain (as a function of the original variable $x$). It is noted that, at small positive times, the solution decreases relative to its initial value, before the eventual exponential growth. This can be inferred from the formula in Equation (2.28) with the chosen parameter values. This is in contrast to the (more familiar) behaviour on a fixed domain, for which (in a separable solution) each of the eigenfunctions simply either grows or decays exponentially for $t>0$.

Figure 1. Exact solution on a time-dependent interval $0< x<\sqrt {1+6t}$, calculated using Equation (2.28). Left: solution $u(\xi,t)$ as a function of the scaled variable $\xi$. Right: solution $\psi (x,t)$ on the time-dependent domain.

Although the formulae in Equations (2.17), (2.21), (2.28), (2.32), (2.33), and (2.34) differ, we note some common behaviour of these exact solutions in the asymptotic large time (or finite time) limit.

Firstly, whenever the domain length tends to zero in a finite time, the solution also tends to zero uniformly in $\xi$ (see § 2.3 and § 2.4). In each case, this follows from an upper bound on the eigenvalue $\sigma _1$.

Note, also, that in the separable cases with $L(t)\rightarrow \infty$ as $t\rightarrow \infty$, the long-time behaviour does not depend on the eigenvalue $\sigma _1$. In these cases, $s(t)=o(t)$ and the term $\exp (\sigma _1 s(t))$ is not of leading order.

Next, note that the separable solutions share the property that there is exponential growth at any $\xi \in (0,L_0)$ such that

(2.39)\begin{equation} -c_*< \lim_{t\rightarrow\infty}\frac{x(\xi,t)}{t} < c_*, \end{equation}

whereas there is exponential decay if

(2.40)\begin{equation} \left\vert \lim_{t\rightarrow\infty} \frac{x(\xi,t)}{t} \right\vert > c_*. \end{equation}

(Here $x(\xi,t)=\frac {\xi }{L_0}L(t)+A(t)$ is the original variable.) This is, in some sense, similar to the behaviour of the solution, $\tilde {\psi }$ say, on the whole real line with initial conditions compactly supported in $[a,b]$:

(2.41)\begin{equation} \tilde{\psi}(x,t)= \frac{1}{\sqrt{4\pi Dt}}\int_a^{b} \tilde{\psi}(y,0) \exp\left(f_0t -\frac{(x-y)^{2}}{4Dt} \right)dy. \end{equation}

This spreads at the asymptotic speed $c_*$, in the sense that for $\vert c\vert < c_*$, $\tilde {\psi }(ct,t)\rightarrow \infty$ whereas $\sup _{\vert x\vert \geq c_*t}\tilde {\psi }(x,t)\rightarrow 0$ as $t\rightarrow \infty$. It is well known that $c_*$ is also the minimum wave speed for travelling wave solutions to the nonlinear FKPP equation and that it is the asymptotic spreading speed for solutions to the same equation on the real line with compactly supported initial conditions (see [Reference Aronson and Weinberger2, Reference Petrowsky, Kolmogorov and Piskunov21]). Much work has been done on determining the exact behaviour associated with this spreading and especially with respect to Bramson's logarithmic correction term (see [Reference Bramson11, Reference Bramson12, Reference Hamel, Nolen, Roquejoffre and Ryzhik19]). It is natural, therefore, to be interested in the exact behaviour of our solution $\psi (x,t)$ (in terms of the original variable $x$), at this critical interface between growth and decay. This is considered in § 3.

Finally, note that in § 2.1 and the particular sub-case where $\gamma _1=0$, the problem has become that of an interval of fixed length $L_0$ moving at a constant speed $c$. Our result is in agreement with [Reference Berestycki, Diekmann, Nagelkerke and Zegeling5, Reference Potapov and Lewis22] in deriving a critical domain length, which is defined by the equation $f_0=\frac {D\pi ^{2}}{L_0^{2}}+\frac {c^{2}}{4D}$, and represents a threshold between decay and growth.

2.6. Applications to more general $A(t)$, $L(t)$

The preceding results are relevant not only to those specific forms of $A(t)$, $L(t)$ which led to the exact solutions. The explicit expressions can also be used to deduce bounds on the solution for other, more general, forms of $A(t)$ and $L(t)$.

The parabolic comparison principle leads to the following result.

This, therefore, provides an explicit lower [or upper] bound for the solution $\psi$, whenever the domain contains [or is contained by] one of the separable cases.

A rather different extension of the method is to consider cases for which $\ddot {L}L^{3}$ and $\ddot {A}L^{3}$ are each bounded above or bounded below. In this case, we can bound the solution by expressions involving $A(t)$ and $L(t)$, together with the same Sturm–Liouville eigenfunctions and eigenvalues that occurred in the preceding sections.

Proposition 2 Suppose that

(2.42)\begin{equation} \ddot{L}(t)L(t)^{3}\leq \gamma_{0}^{+}\text{ and }\ddot{A}(t)L(t)^{3} \leq \gamma_{1}^{+} \end{equation}

for some constants $\gamma _{0}^{+},$ $\gamma _{1}^{+}$. Then, for any given initial conditions $u(\xi,0)$ in $L^{2}([0,L_0]),$ the solution $u(\xi,t)$ can be bounded above by a sum of the $u_n(\xi,t)$ in equation (2.3), where now the $g_n(\xi ),$ $\sigma _n$ satisfy the Sturm–Liouville problem in Equations (2.9), (2.10) with $\gamma _{0}=\gamma _{0}^{+}$ and $\gamma _{1}=\gamma _{1}^{+}$. If equation (2.42) holds with both inequalities reversed, then the solution $u(\xi,t)$ can instead be bounded below by a sum of the $u_n(\xi,t)$.

For the sake of completeness, we make the remark that, in the level of generality considered (i.e. $A(t)$ and $L(t)$ twice continuously differentiable), the domains — and consequently the solutions — will be hard to describe in any very general terms. Indeed, examples can be constructed with alternating growth and decay, such that the solution becomes both arbitrarily large and arbitrarily small over time. Such examples are not the focus of this paper.

2.7. Exact solutions on a ball in higher dimension

To conclude this section, we demonstrate that a similar process can lead to exact solutions to the problem on a ball in $\mathbb {R}^{n}$ with radius $R(t)$ and centre $\textbf {A}(t)$. Consider the problem

(2.43)\begin{align} \frac{\partial \psi}{\partial t} & = D \nabla^{2} \psi+ f_0\psi \qquad \textrm{in } \vert \textbf{x}-\textbf{A}(t)\vert < R(t) \end{align}

(2.44)\begin{align} \psi& =0 \text{ on } \vert \textbf{x}-\textbf{A}(t)\vert =R(t). \end{align}

A change of variables from $\textbf {x}$ to $\textbf {z}=\frac {(\textbf {x}-\textbf {A}(t))}{R(t)}R_0$, with $R_0=R(0)$, and from $\psi (\textbf {x},t)$ to

(2.45)\begin{align} w(\textbf{z}, t) = \psi(\textbf{x},t)\left(\frac{R(t)}{R_0}\right)^{\frac{n}{2}}\exp{\left({-}f_0 t +\int_0^{t} \frac{\vert\dot{\textbf{A}}(\zeta)\vert^{2} }{4D}d\zeta+\frac{ \dot{R}(t)R(t)}{4DR_0^{2}}\vert \textbf{z}\vert^{2}+\frac{R(t)}{2DR_0}\textbf{z} \mathbf{\cdot \dot{A}}(t)\right)} , \end{align}

followed by $s(t)= \int_0^{t} \frac {R_0^{2}}{R(\zeta )^{2}}d\zeta$, and $v(\textbf {z}, s)=w(\textbf {z}, t)$, leads to the equation

(2.46)\begin{align} \frac{\partial v}{\partial s} & = D \nabla^{2} v +\left(\frac{\vert \textbf{z}\vert ^{2} \ddot{R}(t(s))R(t(s))^{3}}{4DR_0^{4}} + \frac{(\textbf{z}\cdot\ddot{\textbf{A}}(t(s))) R(t(s))^{3}}{2DR_0^{3}} \right) v \text{ for } \vert \textbf{z} \vert < R_0 \end{align}

(2.47)\begin{align} v& =0 \text{ at } \vert \textbf{z} \vert =R_0. \end{align}

This is separable in $s$, $r=\vert \textbf {z} \vert$, and $\theta$ (the angular co-ordinates) if $\ddot {R}R^{3}=\gamma _{0}=$constant and $\ddot {\textbf {A}}R^{3}=\textbf {0}$. This corresponds to $R(t)^{2}=at^{2}+2bt+R_0^{2}$ for some constants $a$,$b$ and $\gamma _{0}=aR_0^{2}-b^{2}$; and $\textbf{A}(t)=\textbf{A}_\textbf{0}+\textbf{c}t$ for some constant vectors $\textbf{A}_\textbf{0}$ and $\textbf {c}$. The solutions can then be expressed in terms of a sum of the eigenfunctions $v_l(r,\theta )=H_l(\theta )X_l(r)$ of

(2.48)\begin{align} \sigma_l v_l(r,\theta) & = D \nabla^{2} v_l +\frac{r ^{2} \gamma_{0}}{4DR_0^{4}} v_l \text{ on } r < R_0 \end{align}

(2.49)\begin{align} v_l& =0 \text{ at } r =R_0 \end{align}

which satisfy the correct periodicity in $\theta$ and which are non-singular at the origin $r=0$. This leads to the following theorem.

The explicit expressions are similar to the one-dimensional case, but note the dependence on $n$ in the factor $\left (\frac {R(t)}{R_0}\right )^{\frac {n}{2}}$ in Equation (2.45) as well as, of course, the dependence on $n$ in the eigenfunctions and eigenvalues. In $n$ dimensions, we have the bound $\sigma _1 <-\frac {n\rho }{2R_0^{2}}$ on the principal eigenvalue when $\gamma _0=-\rho ^{2}<0$.

3.1. Behaviour near the critical speed

In this section, we take up the question mentioned in § 2.5, regarding the exact behaviour of our solution $\psi (x,t)$ at the critical interface between growth and decay. Recall that we defined

(3.1)\begin{equation} c_*=2\sqrt{Df_0}, \end{equation}

and that, in the separable solutions, there was an exponential growth of $u(\xi,t)$ at any $\xi \in (0,L_0)$ such that Equation (2.39) held, and exponential decay at any $\xi$ where Equation (2.40) held (i.e. $x(\xi,t)$ travelling slower or faster than $c_*$, respectively). It is natural to seek a more precise description of this changeover between regions of growth and decay. Given $A(t)$, $L(t)$, are we able to track the position $x(t)$ at which the solution $\psi (x,t)$ is equal to some a constant, $O(1)$, value? For which choices of $A(t)$, $L(t)$ will the solution be exactly of order $1$ near the boundary (neither growing to $\infty$ nor decaying to $0$)?

Initially, let us make use of an exact solution from § 2. Let $\hat {\psi }(x,t)$ be the solution on the interval

(3.2)\begin{equation} -c_*t - \frac{L_0}{2} < x <\frac{L_0}{2}+c_*t. \end{equation}

This is given by Equation (2.21) with $c=-c_*$, $\alpha =2c_*$, $\gamma _1=0$, $\sigma _n=-\frac {Dn^{2}\pi ^{2}}{L_0^{2}}$ and $g_n(\xi )=\sin \left (\frac {n\pi \xi }{L_0}\right )$. Without loss of generality (and for simplicity) take the initial conditions to be $\sin \left (\frac {\pi \xi }{L_0}\right )$. Then (recalling that $f_0=\frac {c_*^{2}}{4D}$) this exact solution is:

(3.3)\begin{align} \hat{u}(\xi, t)& =\exp\left(-\frac{D\pi^{2} t}{L_0( L_0+2c_* t) }\right)\sin\left(\frac{\pi\xi}{L_0}\right)\left(\frac{L_0}{L_0+2c_* t}\right)^{1/2}\nonumber\\ & \quad\times \exp\left(\frac{\xi c_*}{2DL_0}(L_0+2c_* t) \left(1-\frac{\xi}{L_0}\right)\right). \end{align}

Therefore at $x=-c_*t - L_0/2 +y$, we have

(3.4)\begin{align} \hat{\psi}({-}c_*t - L_0/2+y,t)& =\hat{u}\left(\frac{yL_0}{L_0+2c_*t},t\right)\nonumber\\ & =\exp\left(-\frac{D\pi^{2} t}{L_0( L_0+2c_* t) }\right) \sin\left(\frac{\pi y}{L_0+2c_*t} \right) \left(\frac{L_0}{L_0+2c_*t}\right)^{1/2}\nonumber\\ & \quad\times\exp\left(\frac{yc_*}{2D}\left(1-\frac{y}{L_0+2c_*t}\right)\right). \end{align}

If $y=O(1)$ then as $t\rightarrow \infty$,

(3.5)\begin{equation} \hat{\psi}({-}c_*t - L_0/2+y,t)=O\left( \frac{y}{t}\times \frac{1}{t^{1/2}}\times \exp\left(\frac{yc_*}{2D}\right)\right) = O(t^{{-}3/2})\rightarrow 0. \end{equation}

Observe that the choice $y(t)=\frac {3D}{c_*}\log (t+1)$ in equation (3.3) removes all the powers of $t$ and gives (as $t\rightarrow \infty$):

(3.6)\begin{align} \hat{\psi}\left({-}c_*t - \frac{L_0}{2}+\frac{3D}{c_*}\log(t+1),t\right)= O\left( \frac{\log t}{t}\times \frac{1}{t^{1/2}}\times (t+1)^{3/2} \right)=O( \log t) \rightarrow \infty. \end{align}

The form of this exact solution for $\hat {\psi }$ suggests that the critical choices of $A(t)$, $L(t)$ (where the solution near to the boundary remains exactly of order $1$) may occur when the endpoints move as $\pm c_* t$ plus a logarithmic term (plus smaller order corrections). Furthermore, the fact that the choice $y(t)=\frac {3D}{c_*}\log (t+1)$ removes all the powers of $t$ suggests the likely coefficient of such a term. The following section will give the precise statement of the behaviour on an interval which does include a logarithmic adjustment to the endpoints.

3.2. Precise behaviour in the critical case

From now on, we restrict attention to cases where $A(t)=-\frac {L(t)}{2}$. Our change of variables from Equation (2.5) becomes

(3.7)\begin{equation} w(\xi, t)=u(\xi, t)\left(\frac{L(t)}{L_0}\right)^{1/2}\exp{\left({-}f_0 t + \int_0^ t \frac{\dot{L}(\zeta)^{2} }{16D} d\zeta+\xi (\xi-L_0) \frac{\dot{L}(t)L(t)}{4DL_0^{2}}\right)}. \end{equation}

Let us give a precise definition of the behaviour we are interested in and the notation we shall use for it.

The following theorem is the main result which is proved in the remainder of this section. The proof relies on the construction of a supersolution and a subsolution, both having the specified behaviour.

Theorem 3 Let

(3.8)\begin{equation} A(t)=\frac{-L(t)}{2}={-}c_*t + \alpha\log(t+1) +\eta(t), \end{equation}

where $c_*=2\sqrt {Df_0},$ $\alpha >0,$ and

(3.9)\begin{equation} \eta(t)=O(1), \quad \dot{\eta}(t)=o(1/t), \ \ddot{\eta}(t)=o(1/t^{2}) \ \dddot{\eta}(t)=o(1/t^{3}) \text{ as } t\rightarrow\infty. \end{equation}

Then

(3.10)\begin{equation} \psi(A(t)+y,t)=\overline{\underline{O}}\left(y t^{-\frac{3}{2}+\frac{\alpha c_*}{2D}} \right)\text{ as }t\rightarrow\infty \text{, for }y=O(1). \end{equation}

This theorem gives asymptotic bounds on $\psi (x,t)$ for $x$ within $O(1)$ of the moving boundary. Hence, it also bounds the asymptotic behaviour of the gradient at the moving boundary itself:

(3.11)\begin{equation} \frac{\partial\psi}{\partial x}\left(\frac{-L(t)}{2},t\right)=\overline{\underline{O}}((t+1)^{-\frac{3}{2}+\frac{\alpha c_{*}}{2D}})\text{ as }t\rightarrow\infty. \end{equation}

In particular, at the critical value

(3.12)\begin{equation} \alpha=\alpha_{crit} = \frac{3D}{c_{*}}, \end{equation}

we have that $\psi \left (\frac {-L(t)}{2}+y,t\right )= \overline {\underline {O}}(y)$ as $t\rightarrow \infty$, and that the gradient at the boundary is bounded above and below independently of time: $\frac {\partial \psi }{\partial x}\left (\frac {-L(t)}{2},t\right )=\overline {\underline {O}}(1)$.

The derivation of the $\frac {3D}{c_{*}}\log (t+1)$ term in this context is completely different from the proofs in the other settings in which such a term arises. In this case (of a linear equation on a finite interval with moving boundaries), our derivation of the term is relatively straightforward, or accessible, being based solely on explicit super- and sub-solutions to a linear equation. Moreover, the bulk of our proof is in fact taken up in showing that the function $w(\xi,t)$ is exactly of order $\xi$ (or $y/t$). The other factor, $t^{-\frac {1}{2}+\frac {\alpha c_*}{2D}}$, in the critical behaviour comes straight from the change of variables. This observation and the exact expression used in the change of variables (Equation (3.7)) may therefore help to give insight into the source of the logarithmic term in other settings.

Recall that the function $w$ now satisfies

(3.13)\begin{align} & \frac{\partial w}{\partial t} = D \frac{L_0^{2} }{L(t)^{2}} \left( \frac{\partial^{2} w}{\partial \xi^{2}}+P(t) \frac{\xi}{L_0}\left( \frac{\xi}{L_0}-1\right)\frac{w}{L_0^{2}}\right) \text{ in } 0< \xi< L_0 \end{align}

(3.14)\begin{align} & w=0 \text{ at } \xi=0 \text{ and } \xi=L_0, \end{align}

where

(3.15)\begin{equation} P(t)= \frac{\ddot{L}(t)L(t)^{3}}{4D^{2}}. \end{equation}

The following two propositions give a supersolution and a subsolution for $w(\xi,t)$ under certain conditions on $P(t)$. It is worth noting that Proposition 3 and Proposition 4 apply in general whenever $w(\xi,t)$ satisfies Equations (3.13), (3.14) for any function $P(t)$ (satisfying the conditions of the proposition). They do not rely at all on the specific form of $P(t)$ that we are interested in here, given by Equation (3.15). In the case where $P(t)$ is given by Equation (3.15), the condition (3.21) in Proposition 4 becomes simply

(3.16)\begin{equation} \int_0^{\infty} \ddot{L}(\zeta)^{2/3} d\zeta < \infty . \end{equation}

Proposition 3 (Supersolution)

Let $w(\xi,t)\geq 0$ satisfy Equations (3.13), (3.14). If $P(t)\geq 0$ then (up to multiplication by a constant) $w(\xi,t)\leq \overline {w}(\xi,t)$ where

(3.17)\begin{equation} \overline{w}(\xi,t) =\sin \left(\frac{\pi\xi}{L_0}\right) \exp\left( -\int_0^{t} \frac{D\pi^{2}}{L(\zeta)^{2}}d\zeta \right). \end{equation}

Moreover, if

(3.18)\begin{equation} \int_0^{\infty} \frac{1}{L(\zeta)^{2}} d\zeta < \infty \end{equation}

then $w(\xi,t)=O(\xi )$ independently of time as $t\rightarrow \infty$, in the sense that given $B_1\in (0,L_0),$ there exists $\beta _1$ such that

(3.19)\begin{equation} w(\xi,t)\leq \beta_1 \xi \text{ as } t\rightarrow\infty, \text{ for all }0\leq\xi\leq B_1. \end{equation}

Proof. The function $\overline {w}(\xi,t)$ satisfies the boundary conditions and, since $P(t)\geq 0$, it satisfies the inequality

(3.20)\begin{align} \frac{\partial \overline{w}}{\partial t} & = D \frac{L_0^{2} }{L(t)^{2}}\frac{\partial^{2} \overline{w}}{\partial \xi^{2}} \nonumber\\ & \geq D \frac{L_0^{2} }{L(t)^{2}} \left( \frac{\partial^{2} \overline{w}}{\partial \xi^{2}}+P(t) \frac{\xi}{L_0}\left( \frac{\xi}{L_0}-1\right)\frac{\overline{w}}{L_0^{2}}\right) , \end{align}

and so it is a supersolution for $w(\xi,t)$. Hence, up to multiplication by a constant, $w(\xi,t)\leq \overline {w}(\xi,t)$. Moreover, if equation (3.18) holds, then $\overline {w}(\xi,t)=O(\xi )$ (independently of time as $t\rightarrow \infty$) and so equation (3.19) is proved.

Next, we construct a subsolution using the Airy function $\textrm{Ai}$ and its tangent at the position $\textrm{Ai} (0)$.

Proposition 4 (Subsolution)

Let $w(\xi,t)\geq 0$ satisfy Equations (3.13), (3.14). If $P(t)\rightarrow \infty$ as $t\rightarrow \infty$ and $\dot {P}(t)\geq 0,$ then (up to multiplication by a constant) $w(\xi,t)\geq \tilde {w}(\xi,t)= \underline {w}(\xi,t)a(t)$ where $\underline {w}(\xi,t)$ and $a(t)$ are given by equations (3.23) and (3.33). Moreover, if

(3.21)\begin{equation} \int_0^{\infty} \frac{P(\zeta)^{2/3}}{L(\zeta)^{2}} d\zeta < \infty \end{equation}

then for $P(t)^{1/3} \frac {\xi }{L_0}$ sufficiently small, $w(\xi,t)$ can be bounded below by a positive multiple of $\xi$ (independently of $t$) as $t\rightarrow \infty$. In other words, for $B_0>0$ small enough, there exists $\beta _0>0$ such that

(3.22)\begin{equation} \beta_0 \xi \leq w(\xi,t) \text{ as } t\rightarrow\infty,\quad \text{for all }0\leq\xi\leq B_0P(t)^{{-}1/3}L_0 . \end{equation}

Proof. Let $c_1$ be the largest real zero of the Airy function $\textrm{Ai}$. Note, for reference, the facts that $c_1<0$, $\textrm{Ai} '(c_1)>0$, $\textrm{Ai} (0)>0$, $\textrm{Ai} '(0)<0$, and $\textrm{Ai} ''(0)=0$. Define $\underline {w}(\xi,t)$ by:

(3.23)\begin{equation} \underline{w}(\xi,t)=\begin{cases}\displaystyle{\frac{1}{P(t)^{1/3}}}\textrm{Ai}\left( P(t)^{1/3} \frac{\xi}{L_0} +c_1\right) \\ \qquad\qquad \text{for }0\leq\xi\leq{-}c_1 P(t)^{{-}1/3}L_0 & \textrm{: Region I}\\ \displaystyle{\frac{1}{P(t)^{1/3}}}\left( \textrm{Ai}(0)+\textrm{Ai}'(0)\left( P(t)^{1/3} \frac{\xi}{L_0} +c_1\right)\right) \\\qquad\qquad\text{for } -c_1 P(t)^{{-}1/3}L_0 \leq \xi \leq {-}\left( \dfrac{\textrm{Ai}(0)}{\textrm{Ai}'(0)} +c_1 \right)P(t)^{{-}1/3}L_0 & \textrm{: Region II}\\ 0 \enspace\quad\qquad \text{for } -\left( \dfrac{\textrm{Ai}(0)}{\textrm{Ai}'(0)} +c_1 \!\right)\! P(t)^{{-}1/3}L_0\leq \xi \leq L_0 & \textrm{: Region III}.\\ \end{cases} \end{equation}

Note that $\underline {w}$ is continuous and non-negative on $[0,L_0]$ and satisfies the boundary conditions. Furthermore, both $\frac {\partial \underline {w}}{\partial \xi }$ and $\frac {\partial ^{2} \underline {w}}{\partial \xi ^{2}}$ are continuous across Regions I-II, including at the point where they meet, since at this point, the left and right limits both give $\frac {\partial \underline {w}}{\partial \xi }=\frac {\textrm{Ai} '(0)}{L_0}$ and $\frac {\partial ^{2} \underline {w}}{\partial \xi ^{2}}=0$. In each Region I and Region II, $\frac {\partial \underline {w}}{\partial t}$ satisfies

(3.24)\begin{equation} \frac{\partial \underline{w}}{\partial t} =\frac{\dot{P}(t)}{3P(t)}\left(-\underline{w} + \xi\frac{\partial \underline{w}}{\partial \xi}\right), \end{equation}

and so it follows from the continuity of each term that $\frac {\partial \underline {w}}{\partial t}$ is also continuous across Regions I–II.

In Region I:

(3.25)\begin{align} & \frac{\partial \underline{w}}{\partial t} - D \frac{L_0^{2} }{L(t)^{2}} \left( \frac{\partial^{2} \underline{w}}{\partial \xi^{2}}+P(t) \frac{\xi}{L_0}\left( \frac{\xi}{L_0}-1\right)\frac{\underline{w}}{L_0^{2}}\right) \nonumber\\ & \quad={-}\frac{\dot{P}(t)}{3P(t)}\underline{w} + \frac{\dot{P}(t)}{3P(t)} \frac{\xi}{L_0} \textrm{Ai}'\left( P(t)^{1/3} \frac{\xi}{L_0} +c_1 \right) \nonumber\\ & \qquad- \frac{DP(t)^{1/3}}{L(t)^{2}}\textrm{Ai}''\left( P(t)^{1/3} \frac{\xi}{L_0} +c_1 \right) - \frac{DP(t)}{L(t)^{2}}\frac{\xi^{2}}{L_0^{2}}\underline{w}+\frac{DP(t)}{L(t)^{2}}\frac{\xi}{L_0}\underline{w} \end{align}

(3.26)\begin{align} & = \frac{\dot{P}(t)}{3P(t)}\left(-\underline{w} + \xi\frac{\partial \underline{w}}{\partial \xi}\right)- \frac{D}{L(t)^{2}}P(t)^{2/3}\left( P(t)^{1/3} \frac{\xi}{L_0} +c_1 \right)\underline{w} \nonumber\\ & \qquad- \frac{DP(t)}{L(t)^{2}}\frac{\xi^{2}}{L_0^{2}}\underline{w}+\frac{DP(t)}{L(t)^{2}}\frac{\xi}{L_0}\underline{w} \end{align}

(3.27)\begin{align} & = \frac{\dot{P}(t)}{3P(t)}\left(-\underline{w} + \xi\frac{\partial \underline{w}}{\partial \xi}\right) -\frac{DP(t)^{2/3}}{L(t)^{2}}c_1\underline{w} - \frac{DP(t)}{L(t)^{2}}\frac{\xi^{2}}{L_0^{2}}\underline{w} . \end{align}

Note that $\frac {\partial ^{2} \underline {w}}{\partial \xi ^{2}}\leq 0$ in Region I, since $\textrm{Ai} ''(x)=x\textrm{Ai} (x)\leq 0$ on $[c_1,0]$. Therefore, using $\underline {w}(0,t)=0$, it holds that

(3.28)\begin{equation} \xi\frac{\partial \underline{w}}{\partial \xi}(\xi,t) \leq \underline{w}(\xi,t) \text{ in Region I.} \end{equation}

Thus, Equation (3.27) together with the assumption that $P(t)\geq 0$ and $\dot {P}(t)\geq 0$ implies that, in Region I,

(3.29)\begin{equation} \frac{\partial \underline{w}}{\partial t} - D \frac{L_0^{2} }{L(t)^{2}} \left( \frac{\partial^{2} \underline{w}}{\partial \xi^{2}}+P(t) \frac{\xi}{L_0}\left( \frac{\xi}{L_0}-1\right)\frac{\underline{w}}{L_0^{2}}\right) \leq{-}c_1\frac{DP(t)^{2/3}}{L(t)^{2}}\underline{w}. \end{equation}

In Region II, since $P(t)\geq 0$, $\dot {P}(t)\geq 0$, and $\textrm{Ai} '(0)<0$,

(3.30)\begin{align} & \frac{\partial \underline{w}}{\partial t} - D \frac{L_0^{2} }{L(t)^{2}} \left( \frac{\partial^{2} \underline{w}}{\partial \xi^{2}}+P(t) \frac{\xi}{L_0}\left( \frac{\xi}{L_0}-1\right)\frac{\underline{w}}{L_0^{2}}\right)\nonumber\\ & \quad={-}\frac{\dot{P}(t)}{3 P(t)}\underline{w}+ \frac{\dot{P}(t)}{3 P(t)}\frac{\xi}{L_0}\textrm{Ai}'(0)- \frac{DP(t)}{L(t)^{2}}\frac{\xi^{2}}{L_0^{2}}\underline{w} + \frac{DP(t)}{L(t)^{2}}\frac{\xi}{L_0}\underline{w} \end{align}

(3.31)\begin{align} & \quad\leq \frac{DP(t)}{L(t)^{2}}\frac{\xi}{L_0}\underline{w} \end{align}

(3.32)\begin{align} & \quad\leq \left( -\frac{\textrm{Ai}(0)}{\textrm{Ai}'(0)} -c_1 \right) \frac{DP(t)^{2/3}}{L(t)^{2}} \underline{w}. \end{align}

This leads us to define $\tilde {w}(\xi,t)= \underline {w}(\xi,t)a(t)$ where

(3.33)\begin{equation} a(t)=\exp {\left( \left( \frac{\textrm{Ai}(0)}{\textrm{Ai}'(0)} +c_1 \right)\int_0^{t} \frac{DP(\zeta)^{2/3}}{L(\zeta)^{2}} d\zeta \right)}. \end{equation}

Then in Regions I–II, the function $\tilde {w}(\xi,t)$ is $C^{2}$ in $\xi$, $C^{1}$ in $t$ and it satisfies

(3.34)\begin{equation} \frac{\partial \tilde{w}}{\partial t} - D \frac{L_0^{2} }{L(t)^{2}} \left( \frac{\partial^{2} \tilde{w}}{\partial \xi^{2}}+P(t) \frac{\xi}{L_0}\left( \frac{\xi}{L_0}-1\right)\frac{\tilde{w}}{L_0^{2}}\right) \leq 0, \end{equation}

so it is a classical subsolution for $0\leq \xi \leq -\left ( \frac {\textrm{Ai} (0)}{\textrm{Ai} '(0)} +c_1 \right ) P(t)^{-1/3}L_0$ (Regions I-II).

It is clear that $\tilde {w}\equiv 0$ is also a classical subsolution in Region III. At the point where Region II and Region III meet, $\tilde {w}$ is continuous, it is a classical subsolution on either side, and $\frac {\partial \tilde {w}}{\partial \xi }$ has a jump discontinuity from a negative value on the left (Region II) to zero on the right (Region III). It follows that $\tilde {w}(\xi,t)$ is a weak subsolution to the parabolic problem on $(0,L_0)$. Therefore, up to multiplication by some constant,

(3.35)\begin{equation} \tilde{w}(\xi,t)\leq w(\xi,t). \end{equation}

If Equation (3.21) holds, then $a(t)$ converges to a strictly positive value as $t\rightarrow \infty$. Then, since

(3.36)\begin{equation} \underline{w}(\xi,t) \sim \frac{\textrm{Ai}'(c_1)}{L_0}\xi \text{ as }P(t)^{1/3} \frac{\xi}{L_0}\rightarrow 0, \end{equation}

it follows that for $P(t)^{1/3} \frac {\xi }{L_0}$ sufficiently small, $\tilde {w}(\xi,t)$ can be bounded above and below by positive multiples of $\xi$ (independently of time as $t\rightarrow \infty$). In particular, for $B_0>0$ small enough, there exists $\beta _0>0$ such that

(3.37)\begin{equation} \beta_0 \xi \leq \tilde{w}(\xi,t) \textrm{ as } t\rightarrow\infty,\quad\text{ for all }0\leq\xi\leq B_0P(t)^{{-}1/3}L_0 . \end{equation}

Equation (3.22) follows and the proposition is proved.

Next, we use the super- and sub-solutions for $w(\xi,t)$ to prove Theorem 3:

Proof of Theorem 3. Recall from Equation (3.8) that

(3.38)\begin{equation} L(t)=2(c_*t - \alpha\log(t+1) -\eta(t)). \end{equation}

Thus, as $t\rightarrow \infty$ the function $\displaystyle {P(t)=\frac {\ddot {L}(t)L(t)^{3}}{4D^{2}}}$ obeys

(3.39)\begin{equation} P(t)\sim \frac{4 \alpha c_{*}^{3} }{D^{2}}t \rightarrow\infty \text{ and } \dot{P}(t) \sim \frac{4\alpha c_{*}^{3} }{D^{2}} >0. \end{equation}

Moreover, since $L(t)\sim 2c_* t$ and $\ddot {L}(t)\sim 2\alpha t^{-2}$ as $t\rightarrow \infty$, it also holds that

(3.40)\begin{equation} \int_0^{\infty} \frac{1}{L(\zeta)^{2}} d\zeta < \infty \text{ and } \int_0^{\infty} \ddot{L}(\zeta)^{2/3} d\zeta < \infty. \end{equation}

So, both Proposition 3 and 4 apply to this case, giving that (for some positive constants) $C_1\underline {w}(\xi,t)a(t)\leq w(\xi,t)\leq C_2\overline {w}(\xi,t)$, and that for $B_0>0$ small enough, there exist $0<\beta _0\leq \beta _1$ such that

(3.41)\begin{equation} \beta_0 \xi \leq \tilde{w}(\xi,t)\leq \beta_1\xi \text{ as } t\rightarrow\infty, \text{ for all }0\leq\xi\leq B_0P(t)^{{-}1/3}L_0 . \end{equation}

Hence, we have shown that $w(\xi,t)$ is exactly of order $\xi$:

(3.42)\begin{equation} w(\xi,t)=\overline{\underline{O}}(\xi)\qquad\textrm{ as }\xi=O(P(t)^{{-}1/3} )\rightarrow 0. \end{equation}

In terms of the original function $\psi (x,t)$, recall that

(3.43)\begin{equation} \psi(x,t)=u(\xi, t)=w(\xi, t)\left(\frac{L_0}{L(t)}\right)^{1/2}\exp{\left(f_0 t - \int_0^ t \frac{\dot{L}(\zeta)^{2} }{16D} d\zeta - \xi (\xi-L_0) \frac{\dot{L}(t)L(t)}{4DL_0^{2}}\right)} \end{equation}

and note that with $L(t)$ given by Equation (3.38),

(3.44)\begin{align} f_0 t - \int_0^ t \frac{\dot{L}(\zeta)^{2} }{16D} d\zeta & = \frac{c_{*}^{2}}{4D}t - \int_0^ t \left( \frac{c_{*}^{2}}{4D} -\frac{\alpha c_{*}}{2D(\zeta+1)} +O\left(\frac{1}{(\zeta+1)^{2}}\right) + O(\dot{\eta}(\zeta)) \right) d\zeta \end{align}

(3.45)\begin{align} & =\frac{\alpha c_{*}}{2D}\log(t+1) + O(1) \qquad \textrm{as } t\rightarrow\infty. \end{align}

Consider $x=\frac {-L(t)}{2}+y$ with $y=O(1)$. Then $\xi =\frac {yL_0}{L(t)}=O\left (\frac {1}{t+1}\right )$ is certainly $O(P(t)^{-1/3} )$ as $t\rightarrow \infty$, and so

(3.46)\begin{align} \psi\left(\frac{-L(t)}{2}+y,t\right) & =w\left(\frac{y}{L(t)}, t\right)\left(\frac{L_0}{L(t)}\right)^{1/2}\nonumber\\ & \quad\times\exp{\left(f_0 t - \int_0^ t \frac{\dot{L}(\zeta)^{2} }{16D} d\zeta - \frac{y}{L(t)}\left(\frac{y}{L(t)}-1\right) \frac{\dot{L}(t)L(t)}{4D}\right)} \nonumber\\ & =\overline{\underline{O}}\left(\frac{y}{t+1}\right) \times \frac{1}{(t+1)^{1/2}} \times \exp{\left(\frac{\alpha c_{*}}{2D}\log(t+1) - \frac{y^{2}}{4D(t+1)} +\frac{y c_{*}}{2D}\right)} \nonumber\\ & = \overline{\underline{O}}\left( y (t+1)^{-\frac{3}{2}+\frac{\alpha c_{*}}{2D}} \exp{\left(\frac{y c_{*}}{2D} \right)}\right), \end{align}

which concludes the proof of Theorem 3.

3.3. Critical case in higher dimensions

To conclude this section, we note that a similar analysis can also be applied to a ball in $\mathbb {R}^{n}$.

Theorem 4 Let $\psi$ satisfy

(3.47)\begin{align} & \frac{\partial \psi}{\partial t} = D \nabla^{2} \psi+ f_0\psi \text{ in } \{ \vert \textbf{x}\vert < R(t)\} \subset \mathbb{R}^{n} \end{align}

(3.48)\begin{align} & \psi=0 \text{ on } \vert \textbf{x}\vert= R(t), \end{align}

where

(3.49)\begin{equation} R(t)=c_{*}t-\alpha \log(t+1)-\eta(t) \end{equation}

with $\alpha >0,$ and $\eta$ satisfying Equation (3.9), and where $n\leq 3$. Then

(3.50)\begin{equation} \psi(\textbf{x}, t) = \overline{\underline{O}} (y (t+1)^{{-}1-\frac{n}{2}+\frac{\alpha c_{*}}{2D}}) \text{ as }t\rightarrow\infty \text{, for }y = R(t)-\vert \textbf{x} \vert=O(1). \end{equation}

Remark 2 Hence, the ‘critical value’ of $\alpha$, for which the solution behaves exactly as order $y$, is now

(3.51)\begin{equation} \alpha_{crit}=\frac{(2+n)D}{c_{*}}. \end{equation}

As in the one-dimensional case, this appears to match the coefficient of the logarithmic correction term in the nonlinear FKPP problem on $\mathbb {R}^{n}$, with compactly supported initial conditions (see [Reference Gärtner18, Reference Roquejoffre, Rossi and Roussier-Michon24]).

Proof. The change of variables $\textbf {z}=\frac {\textbf {x}}{R(t)}R_0$ and

(3.52)\begin{equation} W(\textbf{z}, t) = \psi(\textbf{x},t)\left(\frac{R(t)}{R_0}\right)^{\frac{n}{2}}\exp{\left({-}f_0 t +\int_0^{t} \frac{\vert \dot{R}(\zeta)\vert^{2} }{4D}d\zeta+\frac{ \dot{R}(t)R(t)}{4DR_0^{2}}(r^{2}-R_0^{2})\right)} \end{equation}

leads to the equation

(3.53)\begin{align} & \frac{\partial W}{\partial t} = D \frac{R_0^{2} }{R(t)^{2}} \left( \nabla^{2} W + Q(t)\left(\frac{r^{2}}{R_0^{2}}-1\right)\frac{W}{R_0^{2}} \right) \text{ on } r < R_0 \end{align}

(3.54)\begin{align} & W=0 \text{ at } r=R_0, \end{align}

where $r=\vert \textbf {z} \vert$ and

(3.55)\begin{equation} Q(t)=\frac{\ddot{R}(t)R(t)^{3}}{4D^{2}}, \end{equation}

which satisfies $Q(t)>0$, $Q(t)\rightarrow \infty$, $\dot {Q}\geq 0$ as $t\rightarrow \infty$.

Let $h_0(r)$ be the radially symmetric principal eigenfunction of

(3.56)\begin{align} \lambda h & ={-}\nabla^{2} h \text{ for } r < 1 \end{align}

(3.57)\begin{align} h & =0 \text{ at } r=1 \end{align}

in the $n$-dimensional ball, with eigenvalue $\lambda _0$. Then the function

(3.58)\begin{equation} \overline{W}(r,t)=h_0\left(\frac{r}{R_0}\right)\exp{\left(\int_0^{t} -\frac{D\lambda_0}{R(\zeta)^{2}}d\zeta \right)} \end{equation}

is a supersolution for $W$. Thus, up to multiplication by a constant, $W \leq \overline {W}$.

Next consider the function

(3.59)\begin{equation} w_1(r,t)=\tilde{w}(R_0-r,t), \end{equation}

where $\tilde {w}(\xi,t)= \underline {w}(\xi,t)a(t)$ is given in equations (3.23) and (3.33) with $L(t)=2R(t)$, $L_0=2R_0$, $\xi =R_0-r$, and $P(t)=\frac {\ddot {L}(t)L(t)^{3}}{4D^{2}}$. Note that

(3.60)\begin{align} \left(\frac{r^{2}}{R_0^{2}}-1\right)\frac{Q(t)}{R_0^{2}} & = \left(\frac{\xi-R_0}{R_0}+1\right)\left(\frac{\xi-R_0}{R_0}-1\right)\frac{1}{4R_0^{2}D^{2}}\ddot{R}(t)R(t)^{3} \nonumber\\ & =\frac{2\xi}{L_0}\left(\frac{2\xi}{L_0}-2\right) \frac{1}{ L_0^{2}D^{2}}\frac{\ddot{L}(t)L(t)^{3}}{16} \nonumber\\ & = \frac{\xi}{L_0}\left(\frac{\xi}{L_0}-1\right) \frac{1}{L_0^{2}}\frac{\ddot{L}(t)L(t)^{3}}{4D^{2}} \nonumber\\ & = \frac{\xi}{L_0}\left(\frac{\xi}{L_0}-1\right) \frac{P(t)}{L_0^{2}}. \end{align}

Therefore (for $t$ large enough), this function $w_1(r,t)$ satisfies

(3.61)\begin{equation} \frac{\partial w_1}{\partial t} \leq D \frac{R_0^{2} }{R(t)^{2}} \left( \frac{\partial^{2} w_1}{\partial r^{2}} + Q(t) \left(\frac{r^{2}}{R_0^{2}}-1\right)\frac{w_1}{R_0^{2}}\right) \text{ on } r < R_0. \end{equation}

Next, using the form of the Laplacian in $n$ dimensions, we claim that the function

(3.62)\begin{equation} \hat{w}(r,t)=\frac{w_1(r,t)}{r^{\frac{n-1}{2}} } \end{equation}

is a subsolution in the $n$-dimensional case when $n\leq 3$. Certainly the boundary condition (at $r=R_0$) and the non-singular condition (at $r=0$) will be satisfied, since $w_1(R_0,t)=0$, and $w_1(r,t)=0$ on some neighbourhood $[0,r_0)$ of $r=0$. Moreover,

(3.63)\begin{align} \frac{\partial \hat{w}}{\partial t} & = \frac{1}{r^{\frac{n-1}{2}} }\frac{\partial w_1}{\partial t} \leq \frac{1}{r^{\frac{n-1}{2}} }D \frac{R_0^{2} }{R(t)^{2}}\left( \frac{\partial^{2} w_1}{\partial r^{2}} + Q(t) \left(\frac{r^{2}}{R_0^{2}}-1\right)\frac{w_1}{R_0^{2}}\right) \end{align}

(3.64)\begin{align} & = D \frac{R_0^{2} }{R(t)^{2}}\left( \frac{1}{r^{\frac{n-1}{2}} }\frac{\partial^{2} }{\partial r^{2}} \left(r^{\frac{n-1}{2}}\hat{w} \right) +Q(t)\left(\frac{r^{2}}{R_0^{2}}-1\right)\frac{\hat{w}}{R_0^{2}} \right) \end{align}

(3.65)\begin{align} & = D \frac{R_0^{2} }{R(t)^{2}} \left( \nabla^{2} \hat{w} + \left(\frac{n-1}{2}\right)\left(\frac{n-3}{2}\right)\frac{\hat{w}}{r^{2}} +Q(t)\left(\frac{r^{2}}{R_0^{2}}-1\right)\frac{\hat{w}}{R_0^{2}}\right) \end{align}

(3.66)\begin{align} & \leq D \frac{R_0^{2} }{R(t)^{2}} \left( \nabla^{2} \hat{w} + Q(t)\left(\frac{r^{2}}{R_0^{2}}-1\right)\frac{\hat{w}}{R_0^{2}}\right), \end{align}

where the equality follows from the form of the Laplacian in $n$ dimensions and the final inequality holds for $n=1$,$2$,$3$. Thus, $\hat {w}$ is a subsolution and, up to multiplication by a constant, we obtain $\hat {w}\leq W$.

We are interested in $y = R(t)-\vert \textbf {x} \vert =O(1)$. This corresponds to $\vert \textbf {z} \vert = r=R_0-\frac {yR_0}{R(t)}$ with $y=O(1)$, for which the above supersolution and subsolution ($C_1\hat {w}\leq W \leq C_2\overline {W}$) provide the bounds $w(\textbf {z}, t)=\overline {\underline {O}}(R_0-\vert \textbf {z} \vert ) = \overline {\underline {O}}\left (\frac {yR_0}{R(t)}\right )$, independently of $t$. Therefore, the same calculations as in the one-dimensional case give:

(3.67)\begin{align} \psi(\textbf{x} ,t)& =w(\textbf{z}, t)\left(\frac{R_0}{R(t)}\right)^{n/2}\exp{\left(f_0 t -\int_0^{t} \frac{\vert \dot{R}(\zeta)\vert^{2} }{4D}d\zeta-\frac{ \dot{R}(t)R(t)}{4DR_0^{2}}(r^{2}-R_0^{2})\right)} \nonumber\\ & =\overline{\underline{O}} \left(\frac{y}{t+1}\right) \times \frac{1}{(t+1)^{n/2}} \times \exp{\left(\frac{\alpha c_{*}}{2D}\log(t+1) - \frac{y^{2}}{4D(t+1)} +\frac{y c_{*}}{2D} \right)} \nonumber\\ & = \overline{\underline{O}}\left( y (t+1)^{{-}1-\frac{n}{2}+\frac{\alpha c_{*}}{2D}} \exp{\left(\frac{y c_{*}}{2D} \right)} \right). \end{align}

Remark 3 The proof of Theorem 4 shows that, in any dimension $n$,

(3.68)\begin{equation} \psi(\textbf{x}, t) = O (y (t+1)^{{-}1-\frac{n}{2}+\frac{\alpha c_{*}}{2D}}) \qquad \textrm{as }t\rightarrow\infty \text{, for }y = R(t)-\vert \textbf{x} \vert=O(1). \end{equation}

This follows from the supersolution. However, the subsolution $\hat {w}$ used in the proof only satisfies the required inequality when $n\leq 3$. One may conjecture that the full result of Theorem 4 actually applies in all dimensions $n$.