= Solution
The substitution separates the uniform equilibrium from the decaying perturbation. Direct <differentiation> gives
$$
u_t=e^{-\beta t}(v_t-\beta v),\qquad u_{xx}=e^{-\beta t}v_{xx},\qquad\beta(u-u_0)=\beta e^{-\beta t}v.
$$
Consequently the <reaction-diffusion equation> reduces to the <heat equation> $v_t=v_{xx}$. Its data become
$$
\boxed{v(0,t)=e^{\beta t}(u_1-u_0),\quad v(\pi,t)=e^{\beta t}(u_2-u_0),\quad v(x,0)=f(x)-u_0.}
$$
When the three constants coincide, the <Dirichlet boundary conditions> for $v$ are homogeneous. <Separation of variables> gives $X''+n^2X=0$ and $T'=-n^2T$, selecting the <normal modes> $\sin(nx)e^{-n^2t}$ for $n\geq1$. Thus
$$
v(x,t)=\sum_{n=1}^\infty b_n\sin(nx)e^{-n^2t},\qquad b_n=\frac2\pi\int_0^\pi[x(\pi-x)-u_0]\sin(nx)\,dx.
$$
Two <integrations by parts> give $\int_0^\pi x(\pi-x)\sin(nx)dx=2[1-(-1)^n]/n^3$, while the constant term uses $\int_0^\pi\sin(nx)dx=[1-(-1)^n]/n$. Therefore the <Fourier sine series> is
$$
\boxed{v(x,t)=\frac4\pi\sum_{\substack{n\geq1\\n\text{ odd}}}\left(\frac2{n^3}-\frac{u_0}n\right)e^{-n^2t}\sin(nx).}
$$
For $u_0\ne2$, its first coefficient is nonzero. Multiplying by $e^{-\beta t}$ and retaining the slowest-decaying mode gives
$$
\boxed{u(x,t)=u_0-\frac4\pi(u_0-2)e^{-(\beta+1)t}\sin x+O(e^{-(\beta+9)t})\quad(t\to\infty).}
$$
The remainder estimate holds uniformly in $x$ for sufficiently large positive time. If $u_0=2$, the fundamental vanishes and the next possible mode has decay rate $\beta+9$, explaining the exclusion.
There is an initial endpoint compatibility issue unless $u_0=0$: the given polynomial profile vanishes at the endpoints, whereas the boundary values equal $u_0$. The series realizes the initial data in <L2 space> and pointwise in the open interval, imposing the prescribed boundary values for $t>0$. For nonzero $u_0$ it cannot also be continuous at both initial endpoint corners.
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