 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 $\boxed{\alpha =-1/2,\beta =1/2}$. The drift term cancels the $-\beta \xi F^{\prime }$ contribution in $u_t$, leavingMultiply by $F^{\prime }$ and integrate. The limiting equilibrium has $F^{\prime }=0$, fixing the constant and giving $(F^{\prime }{)}^2=(1-F^2{)}^2/4$. The increasing branch from zero to one obeys $F^{\prime }=(1-F^2)/2$. ThereforeThe 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 $u_{xx}\le 0$. The equation would then give $0=u_{xx}-u^3/2$, so $u\le 0$. Hence 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 becomeswith $v(0,\tau )=0$, $v(\infty ,\tau )=1$ and prescribed $v(\xi ,0)=f(\xi )$. For two solutions let $w=v_1-v_2$. ThenThe bracket is at most $1/2$ since $a^2+ab+b^2\ge 0$. Thus $z=e^{-\tau /2}w$ solves $z_{\tau }=z_{\xi \xi }-Qz$ with $Q=(v_1^2+v_1{v}_2+v_2^2)/2\ge 0$, 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 becauseFinite-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.