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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.

The transformation $\tau =\log t$, $\xi =x/\sqrt{t}$, $v=\sqrt{t}u$ gives $v_{\tau }=v_{\xi \xi }+(v-v^3)/2$. Thus the stationary half-line profile $v=\tanh (\xi /2)$ yields $u=t^{-1/2}\tanh [x/(2\sqrt{t})]$, with zero value at the origin and far-field value $t^{-1/2}$. For two bounded classical solutions, their difference has reaction coefficient $[1-(v_1^2+v_1{v}_2+v_2^2)]/2\le 1/2$. Multiplying it by $e^{-\tau /2}$ therefore produces a linear parabolic equation with nonnegative absorption. A positive growing quadratic spatial barrier and the maximum principle give uniqueness on the unbounded half-line. This proof identifies the finite-time boundedness or growth hypothesis needed at spatial infinity rather than assuming a supremum is automatically attained there.

A boundary condition at spatial infinity specifies limiting behaviour as the spatial coordinate leaves every bounded set, such as ${\lim }_{x\to +\infty }u(x,t)=g(t)$ on a half-line. A pointwise limit for each time is different from uniform control over a time interval. Maximum-principle and uniqueness proofs on unbounded domains must use a suitable uniform boundedness, growth restriction or barrier hypothesis to justify passage from finite domains to infinity.