= Solution
Set $\xi=x/t^\beta$ and $u=t^\alpha F(\xi)$. Balancing $u_t$, <diffusion> and the cubic gives $\alpha-1=\alpha-2\beta=3\alpha$, hence $\boxed{\alpha=-1/2,\quad\beta=1/2}$. The drift term cancels the $-\beta\xi F'$ contribution in $u_t$, leaving
$$
F''+\tfrac12F-\tfrac12F^3=0,\qquad F(0)=0,\quad F(+\infty)=1.
$$
Multiply by $F'$ and integrate. The limiting <equilibrium> has $F'=0$, fixing the constant and giving $(F')^2=(1-F^2)^2/4$. The increasing branch from zero to one obeys $F'=(1-F^2)/2$. Therefore
$$
\boxed{F(\xi)=\tanh(\xi/2),\qquad
u(x,t)=t^{-1/2}\tanh\!\left(\frac{x}{2\sqrt t}\right).}
$$
The <first integral> also shows why another turning branch cannot reach the required limit from zero.
At a space-time interior maximum, $u_t=u_x=0$ and $u_{xx}\leq0$. The equation would then give $0=u_{xx}-u^3/2$, so $u\leq0$. Hence \b[a positive interior maximum is impossible]. In particular, a <global maximum> cannot be attained at an interior point, since the far-field boundary has positive values. If “maximum” instead means every <local maximum> of an arbitrary signed solution, the calculation only excludes positive local maxima; nonnegativity must be added for the broader usual interpretation. Positive boundary values alone do not impose nonnegativity on arbitrary <initial data>.
For uniqueness in the standard bounded <classical solution> class on finite time strips, set $\tau=\log t$, $\xi=x/\sqrt t$, and $v(\xi,\tau)=\sqrt t\,u(x,t)$. The equation becomes
$$
v_\tau=v_{\xi\xi}+\tfrac12v-\tfrac12v^3,
$$
with $v(0,\tau)=0$, $v(\infty,\tau)=1$ and prescribed $v(\xi,0)=f(\xi)$. For two solutions let $w=v_1-v_2$. Then
$$
w_\tau=w_{\xi\xi}+\left[\tfrac12-\tfrac12(v_1^2+v_1v_2+v_2^2)\right]w.
$$
The bracket is at most $1/2$ since $a^2+ab+b^2\geq0$. Thus $z=e^{-\tau/2}w$ solves $z_\tau=z_{\xi\xi}-Qz$ with $Q=(v_1^2+v_1v_2+v_2^2)/2\geq0$, zero <initial data> and zero boundary difference.
For completeness on the unbounded interval, compare $\pm z$ with $\eta e^{3\tau}(1+\xi^2)$ on a large rectangle. This is a strict positive <supersolution> because
$$
(\partial_\tau-\partial_{\xi\xi}+Q)e^{3\tau}(1+\xi^2)
=e^{3\tau}[3(1+\xi^2)-2+Q(1+\xi^2)]>0.
$$
Finite-strip boundedness lets its right boundary dominate $z$ when the rectangle is sufficiently wide. The <weak parabolic maximum principle> then bounds $|z|$ by this barrier. Send the width to infinity and then $\eta$ to zero. Hence $z=w=0$, giving $\boxed{\text{at most one bounded classical solution for }t>1}$. The boundedness or a suitable uniform far-field/growth condition is the usual hypothesis that makes this unbounded-domain argument valid; pointwise limits at <spatial infinity> should not silently replace it. This is <similarity reduction for a cubic absorption equation with dilating drift>.
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