# Unique Solution to Reaction-Diffusion Equation

I can neither confirm nor deny that only one solution (namely $u\equiv 0$) satisfies the following boundary value problem in $\Omega\times [0,T]\subset\mathbb{R}^{n+1}$.

\begin{cases} \displaystyle \frac{\partial u}{\partial t}-\Delta u = -u^3, &\text{in }\Omega \\ u(x,0)=0, &x\in\Omega \\ u(x,t)=0, &x\in\partial\Omega, t>0. \end{cases}

Any suggestions? My thoughts: there is a change of variables that changes the equation to one where the maximum principle holds; suppose $u$ and $v$ are solutions and consider some variation of their difference (e.g., $u-v$, $(u-v)^2$, $e^{\frac{1}{4}(u-v)^4}$, etc.); consider some energy functional. I can't get any of the "standard" techniques to work.

A natural energy functional is $$E(u) = \int_\Omega |u|^2 \, .$$ Then $E(u(\cdot,0)) = 0$ and $$\frac{d}{dt} E(u(\cdot,t)) = - 2 \int_\Omega (|\nabla u|^2 + u^4) \le 0$$ by integration by parts. Therefore $E(u(\cdot,t)) = 0$ for all $t$ and $u = 0$ is the only solution.

Just to be on the safe side, let me affirm at the outset that I assume we are dealing with classical solutions, i.e., sufficiently differentiable functions $u$ satisfying (1) below.

From the given equation

$u_t - \nabla^2u = -u^3, \tag{1}$

we obtain, upon multiplication by $u$,

$uu_t - u\nabla^2 u = -u^4; \tag{2}$

we have

$uu_t = \dfrac{1}{2}(u^2)_t; \tag{3}$

we also have

$\nabla \cdot (u \nabla u) = \langle \nabla u, \nabla u \rangle + u \nabla^2 u, \tag{4}$

from which

$u\nabla^2 u = \nabla \cdot (u\nabla u) - \Vert \nabla u \Vert^2, \tag{5}$

since

$\langle \nabla u, \nabla u \rangle = \Vert \nabla u \Vert^2. \tag{6}$

Inserting (3) and (5) into (2) yields

$\dfrac{1}{2}(u^2)_t - \nabla \cdot (u \nabla u) + \Vert \nabla u \Vert^2 = -u^4; \tag{7}$

for fixed $s \in [0, T]$, we integrate (7) over $\Omega$ and find

$\dfrac{1}{2}\int_\Omega (u^2(s))_t dV - \int_\Omega \nabla \cdot \ (u(s) \nabla u(s)) dV + \int_\Omega \Vert \nabla u(s) \Vert^2 dV = -\int_\Omega (u(s))^4 dV; \tag{8}$

also,

$\int_\Omega \nabla \cdot (u(s) \nabla u(s)) dV = \int_{\partial \Omega} u(s) \nabla u(s) \cdot dS = 0, \tag{9}$

by the divergence theorem and the boundary conditions. Note that in writing (8), (9), and following, I have adopted the convention that spatial variables are suppressed; thus $u(s)$ is shorthand for $u(\vec x, s)$ with $\vec x \in \Omega$ and $s \in [0, T]$; this notation is especially useful in cases such as the present in which the spatial variables are "integrated out".

We further have

$\int_\Omega (u^2(s))_t dV = \dfrac{d}{dt}\int_\Omega u^2(s) dV, \tag{10}$

and performing a few little algebraic maneuvers we find that (8) becomes

$\dfrac{1}{2} \dfrac{d}{dt}\int_\Omega u^2(s) dV = -\int_\Omega u^4(s) dV - \int_\Omega \Vert \nabla u(s) \Vert^2 dV. \tag{11}$

Integrating (11) 'twixt $t_1, t_2 \in [0, T]$, we obtain

$\dfrac{1}{2} (\int_\Omega u^2(t_2) dV - \int_\Omega u^2(t_1) dV) = \dfrac{1}{2} \int_{t_1}^{t_2} (\dfrac{d}{dt}\int_\Omega u^2 (s) dV) ds$ $= -\int_{t_1}^{t_2} (\int_\Omega (u^4(s) + \Vert \nabla u(s) \Vert^2 dV)ds. \tag{12}$

We use formula (12) to show that $u = 0$ is the unique solution to (1) satisfying the stated boundary conditions; choising $t_1 = 0$, we find

$\dfrac{1}{2}\int_\Omega u^2(t_2) dV = -\int_0^{t_2} (\int_\Omega (u^4(s) + \Vert \nabla u(s) \Vert^2)dV) ds. \tag{13}$

If now $u(p, t') \ne 0$ for some $(p, t') \in \Omega \times [0, T]$, then since $u$ is continuous, we must have $u \ne 0$ in some neighborhood $W \subset \Omega \times [0, T]$ of $(p, t')$.  Thus $u^2, u^4 > 0$ on $V$, while $\Vert \nabla u \Vert^2 \ge 0$ on $V$.  It follows that both

$\int_\Omega u^2(t') dV > 0 \tag{14}$

and

$\int_\Omega (u^4(t') + \Vert \nabla u(t') \Vert^2) dV > 0. \tag{15}$

Taken in concert, (13)-(15) form a rather unharmonious trio, bring as they are mutually contradictory.  Thus we must have $u(p, t') = 0$ for $(p, t') \in \Omega \times [0, T]$; the $0$ solution of (1) is indeed unique under the hypothesized boundary conditions.

Note Added Thursday 11 June 2015 3:34 PM PST: I wanted to leave a few remarks concerning possible generalizations of this result. First of all, for $0 < \alpha \in \Bbb R$ and sufficiently differentiable bounded $\beta(x) > 0$, $x \in \Omega$, the equation

$u_t - \alpha \nabla^2 u = -\beta(x) u^3 \tag{16}$

has precisely one solution $u \equiv 0$ satisfying the given boundary conditions; indeed, it is easy to see that such $u$ satisfy the analogs of (11) and (12):

$\dfrac{1}{2} \dfrac{d}{dt}\int_\Omega u^2(s) dV = -\int_\Omega \beta u^4(s) dV - \alpha\int_\Omega \Vert \nabla u(s) \Vert^2 dV, \tag{17}$

$\dfrac{1}{2} (\int_\Omega u^2(t_2) dV - \int_\Omega u^2(t_1) dV) = \dfrac{1}{2} \int_{t_1}^{t_2} (\dfrac{d}{dt}\int_\Omega u^2 (s) dV) ds$ $= -\int_{t_1}^{t_2} (\int_\Omega (\beta u^4(s) + \alpha \Vert \nabla u(s) \Vert^2 dV)ds; \tag{18}$

the step-by-step derivations of (17), (18) from (16) are simple extensions of those taking us from (1) to (11)-(12), and our choice of positivity for $\alpha$ and $\beta(x)$ force $u \equiv 0$ just as above. So the result is true for much more general coefficients than as originally stated. Second, the equation

$u_t - \alpha \nabla^2 u = -\beta(x) u^p, \tag{19}$

where $p$ is a positive odd integer leads to the same conclusion, and for essentially the same reasons; in this case (17) becomes

$\dfrac{1}{2} \dfrac{d}{dt}\int_\Omega u^2(s) dV = -\int_\Omega \beta u^{p + 1}(s) dV - \alpha\int_\Omega \Vert \nabla u(s) \Vert^2 dV, \tag{20}$

and since $p + 1$ is even, we have

$\int_\Omega \beta u^{p + 1}(s) dV > 0 \tag{21}$

unless $u = 0$; so the same result binds. This in contrast to the case of $p \ge 0$ even, whence $p + 1$ is odd and we cannot conclude that (21) holds, invalidating the presen line of argument.

It would be engaging to explore the cases $p < 0$ an integer, or even $p$ non-integer, but I think I've gone far enough for the present. End of Note.