= Solution
Use the usual <fractional Dirichlet domain scale>: $A=-\Delta_D$ on $H=L^2(\Omega)$, and $H_\rho=D(A^\rho)$. Its <Sobolev embedding theorem> gives
$$
H_\delta\hookrightarrow H^{2\delta}(\Omega)\hookrightarrow C^{0,\beta}(\overline\Omega),\qquad
0<\beta<\min(1,2\delta-n/2).
$$
Such a positive $\beta$ exists precisely because the assumed $\delta$ is greater than $n/4$. Consequently $v=u_t$ is continuous in time with values in this <Hölder space> on every compact positive-time interval. This embedding alone gives no two spatial derivatives for $v$; those must come from the equation.
First establish a bound that will also justify the nonlinear regularity steps. Put $F(s)=\lambda s-s^3$ and
$$
\boxed{K=\max\{|m|,|M|,\sqrt{\max(\lambda,0)}\}.}
$$
The constants $K$ and $-K$ enclose both the initial data and the zero <Dirichlet boundary data>. Moreover $F(K)\leq0$ and $F(-K)\geq0$, so they are a supersolution and a subsolution for the <parabolic comparison principle>. A direct energy proof avoids assuming pointwise continuity at time zero. Let $h=(u-K)_+$. This truncation has zero trace under the <Sobolev trace operator>; testing the evolution with $h$ and using <integration by parts> gives
$$
\frac12\frac d{dt}\|h\|_2^2=-\|\nabla h\|_2^2+\int_\Omega F(u)h\,dx.
$$
For $u>K$, the <mean value theorem> and $F'(s)=\lambda-3s^2\leq\lambda$ imply $F(u)\leq F(K)+\lambda(u-K)$. Therefore
$$
\frac12\frac d{dt}\|h\|_2^2\leq\lambda\|h\|_2^2.
$$
The <Gronwall inequality> gives $h=0$, since its initial <L2 norm> vanishes. For the initially rough datum, apply the inequality from $\varepsilon>0$ and let $\varepsilon\downarrow0$, using $u(\varepsilon)\to u_0$ in <L2 space>. The same argument for $-u$, which satisfies the same equation because $F$ is odd, proves
$$
\boxed{-K\leq u(x,t)\leq K\quad\text{throughout its existence interval}.}
$$
This is the <invariant interval for a cubic reaction-diffusion equation>; the energy calculation is valid by the <Sobolev chain rule> and approximation even before the stronger pointwise regularity is established.
Now fix $0<\tau<t_0$ and, if necessary, reduce the positive <Hölder exponent> $\beta$. Rearranging the evolution gives the elliptic identity
$$
\Delta u(t)=v(t)-F(u(t)),\qquad u(t)|_{\partial\Omega}=0.
$$
Its right side is bounded on $[\tau,t_0]\times\Omega$, because $v\in C([\tau,t_0],C^{0,\beta})$ and $|u|\leq K$. Boundary <elliptic regularity> first gives $u(t)\in W^{2,p}(\Omega)$ for every finite $p$, uniformly on this time interval. For $p>n$, the <Sobolev embedding theorem> makes $u(t)$ spatially continuously differentiable, with a positive <Hölder exponent>. Thus $F(u(t))$ lies in $C^{0,\beta}$ after reducing $\beta$ if necessary. The <global Schauder estimate> then gives $u(t)\in C^{2,\beta}(\overline\Omega)$.
These are continuous-in-time, not just separate-time, regularity statements. Since $v\in C(C^{0,\beta})$, the identity $u(t)-u(s)=\int_s^t v(r)\,dr$, initially an identity in <L2 space>, holds in $C^{0,\beta}$ once $u(s)$ is known to belong to that space. Hence $u$ is locally Lipschitz in time into $C^{0,\beta}$. Apply the <global Schauder estimate> to the difference of the two elliptic identities. Continuity of $v$ and of $F(u)$ into $C^{0,\beta}$ proves $u\in C([\tau,t_0],C^{2,\beta})$.
To gain the spatial regularity of the time derivative itself, take time <difference quotients> of the evolution. If $d_h(t)=[u(t+h)-u(t)]/h$, their equations are
$$
(d_h)_t=\Delta d_h+a_h(x,t)d_h,\qquad d_h|_{\partial\Omega}=0,
$$
where
$$
a_h=\lambda-[u(t+h)^2+u(t+h)u(t)+u(t)^2].
$$
The coefficients have uniformly bounded spatial <Hölder norms> and temporal <Hölder norms> on compact positive-time cylinders: this follows from the spatial regularity just proved and the local time Lipschitz bound in $C^{0,\beta}$. Also $d_h=\int_0^1v(t+sh)\,ds$ is uniformly bounded and tends to $v$ in $C^{0,\beta}$. Positive-time parabolic <Schauder estimates>, applied away from the lower time edge, give uniform $C^{1+\beta/2,2+\beta}$ bounds for $d_h$, with a slightly smaller exponent if needed. Passing to the limit proves that $v$ is continuous into $C^2(\overline\Omega)$ and satisfies
$$
v_t=\Delta v+(\lambda-3u^2)v,\qquad v|_{\partial\Omega}=0.
$$
The <Dirichlet boundary condition> for $v$ follows also by differentiating the identically zero boundary value in $C^0$. Since $\tau$ was arbitrary, this proves the requested, genuinely stronger regularity:
$$
\boxed{u\in C^1((0,t_0],C^2(\overline\Omega)).}
$$
This is <positive-time smoothing for a semilinear heat equation>; it is the extra parabolic step for $v$, not just the <Sobolev embedding theorem>, that gives the $C^1$ topology asserted here.
Finally, a uniform bound excludes a finite maximal existence time for this <semilinear partial differential equation>. Restart at any positive time $s$, using the <heat semigroup> on the <Banach space> $X=\{g\in C(\overline\Omega):g|_{\partial\Omega}=0\}$ with the supremum norm. Its <variation-of-constants formula> is
$$
u(s+t)=e^{-tA}u(s)+\int_0^te^{-(t-r)A}F(u(s+r))\,dr.
$$
The <heat semigroup> is a <contraction semigroup> on $X$. On the ball of radius $K+1$, $F$ is Lipschitz with constant at most $|\lambda|+3(K+1)^2$, and bounded by $|\lambda|(K+1)+(K+1)^3$. Choose a time length $\eta>0$ so that the integral is at most $1$ and its Lipschitz constant is less than $1$. The <Banach fixed-point theorem> then gives a local solution on $[s,s+\eta]$, with $\eta$ independent of $s$, because $\|u(s)\|_\infty\leq K$. If a finite maximal time existed, choosing $s$ less than $\eta$ before it would extend the same solution past it, a contradiction. Uniqueness identifies this continuation with the original solution, and positive-time smoothing restores the <classical solution> on each extended interval. \b[The solution is globally defined and remains between $-K$ and $K$.]
Back to article page
