Solution

The substitution separates the uniform equilibrium from the decaying perturbation. Direct differentiation gives

u_{t} = e^{- βt} (v_{t} - β v), u_{xx} = e^{- βt} v_{xx}, β (u - u_{0}) = β e^{- βt} v .

v_{t} = v_{xx}

. Its data become

v (0, t) = e^{βt} (u_{1} - u_{0}), v (π, t) = e^{βt} (u_{2} - u_{0}), v (x, 0) = f (x) - u_{0} .

v

are homogeneous. Separation of variables gives

X^{''} + n^{2} X = 0

and

T^{'} = - n^{2} T

, selecting the normal modes

sin (n x) e^{- n^{2} t}

for

n \geq 1

. Thus

v (x, t) = \sum_{n = 1}^{\infty} b_{n} sin (n x) e^{- n^{2} t}, b_{n} = \frac{2}{π} \int_{0}^{π} [x (π - x) - u_{0}] sin (n x) d x .

\int_{0}^{π} x (π - x) sin (n x) d x = 2 [1 - (- 1)^{n}] / n^{3}

\int_{0}^{π} sin (n x) d x = [1 - (- 1)^{n}] / n

v (x, t) = \frac{4}{π} n \geq 1 n odd \sum (\frac{2}{n ^{3}} - \frac{u _{0}}{n}) e^{- n^{2} t} sin (n x) .

For

u_{0} \neq = 2

, its first coefficient is nonzero. Multiplying by

e^{- βt}

and retaining the slowest-decaying mode gives

u (x, t) = u_{0} - \frac{4}{π} (u_{0} - 2) e^{- (β + 1) t} sin x + O (e^{- (β + 9) t}) (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

β + 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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