= Solution
Solve the bank equation first:
$$
x(T)=W-\int_0^T e^{r(T-t)}u(t)dt,\qquad
W=e^{rT}x(0)+\frac A r(e^{rT}-1)>0.
$$
For $\mu>0$ the objective separates into pointwise strictly concave functions of positive $u$. Their derivative is $e^{-\beta t}/u-\mu e^{-\beta T}e^{r(T-t)}$. Setting it to zero gives the unique optimum
$$
\boxed{u^*(t;\mu)=\mu^{-1}e^{(\beta-r)(T-t)}}.
$$
Let $B=\int_0^T e^{\beta s}ds$, equal to $(e^{\beta T}-1)/\beta$ when $\beta\ne0$ and $T$ otherwise. Then $x^*(T;\mu)=W-B/\mu$ is strictly increasing from negative infinity to $W$ as $\mu$ increases from zero to infinity. Consequently
$$
\boxed{\mu^*=B/W}
$$
is the unique positive multiplier with zero terminal balance.
For the constrained problem with no terminal reward, $x(T)\ge0$ is exactly the weighted spending budget $\int e^{r(T-t)}u(t)dt\le W$. The candidate $u^*(\cdot;\mu^*)$ spends that budget exactly. Concavity gives $\log u-\log u^*\le(u-u^*)/u^*$. Since $e^{-\beta t}/u^*=\mu^*e^{-\beta T}e^{r(T-t)}$, integrating this inequality makes the <utility function> difference nonpositive for every feasible control. This proves the requested optimality of the <logarithmic spending with a terminal multiplier> solution directly, including uniqueness up to changes on measure-zero sets.
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