We could simply apply the chain rule, to avoid some confusions we let $ C(x,t) = C(x^* + vt,t^*) = C^*(x^*,t^*)$:
$$
\frac{\partial C}{\partial x} = \frac{\partial C^*}{\partial x^{\phantom{*}}}= \frac{\partial C^*}{\partial x^*} \frac{\partial x^*}{\partial x^{\phantom{*}}} + \frac{\partial C^*}{\partial t^*} \frac{\partial t^*}{\partial x^{\phantom{*}}} = \frac{\partial C}{\partial x^*}
$$
remember here in chain rule, the partial derivative is being taken wrt the first and second variable if not to confuse this wrt the total derivative, similary we could have $\displaystyle \frac{\partial^2 C}{\partial x^2} = \frac{\partial^2 C^*}{\partial {x^*}^2} $,
$$
\frac{\partial C}{\partial t} = \frac{\partial C^*}{\partial t} = \frac{\partial C^*}{\partial x^*} \frac{\partial x^*}{\partial t^{\phantom{*}}} + \frac{\partial C^*}{\partial t^*} \frac{\partial t^*}{\partial t^{\phantom{*}}} = -v\frac{\partial C^*}{\partial x^*} + \frac{\partial C^*}{\partial t^*}
$$
Plugging back to the original equation you will see the convection term is gone if we have done this velocity cone rescaling, you could think the original equation like a diffusion on a car with velocity $v$ measured by a standing person, after the change of variable it is just a pure diffusion measured on a car:
$$
\frac{\partial C^*}{\partial t^*} = D\frac{\partial^2 C^*}{\partial {x^*}^2}
$$
and the initial condition changes to $C^*(x^*,0) = C(x^*+vt^*,t^*)\Big\vert_{t^*=0}= f(x^*)$, the boundary condition remains the same.
Let's rewrite
$$ 2ku_t = u_{xx} - 2ku_x $$
where $k=50$. Separation of variables $u(x,t) = X(x)T(t)$ gives
$$ 2k\frac{T'}{T} = \frac{X''-2kX}{X} = -\lambda $$
The $X$ part has the characteristic polynomial
$$ r^2 - 2kr + \lambda = 0 $$
or $r = k \pm \sqrt{k^2-\lambda} $
Due to the periodicity, we need the roots to have an imaginary part, therefore $k^2-\lambda = -\mu^2$, or $\lambda = \mu^2+k^2$. This makes
$$ X(x) = e^{kx}\big[A\sin(\mu x) + B\cos(\mu x) \big] $$
The periodicity condition $X(x) = X(x+2\pi)$ requires $\mu = n$ where $n=1,2,3$.
The other boundary condition is $X'(0)=0$. Solving this gives $\mu A + kB = 0$, we can write
$$ X_n(x) = e^{kx}\big[k\sin(nx) - n\cos(nx)\big] $$
up to a multiplicative constant.
Solving for $T(t)$ and using superposition, we have the general solution
$$ u(x,t) = \sum_{n=1}^\infty c_n e^{-\frac{n^2+k^2}{2k}t}e^{kx}\big[k\sin(nx) - n\cos(nx)\big] $$
The initial condition is equivalent to solving for $c_n$ such that
$$ \sum_{n=1}^\infty c_n\big[k\sin(nx) - n\cos(nx)\big] = e^{-kx}\sin x $$
Using orthogonality, we have
$$ c_n = \frac{\int_0^{2\pi} e^{-kx}\sin x \big[k\sin(nx) - n\cos(nx)\big] dx}{\int_0^{2\pi} \big[k\sin(nx) - n\cos(nx)\big]^2 dx} $$
Best Answer
Consider the special case you mention, constant $a$ and $v$ and $f=1$.In view of the condition $w(0,t)=w(1,t)=0$ you can make an expansion in a Fourier series $$w(x,t)=∑_{n}d_{n}(t)exp[i2πnx]=∑_{n}d_{n}(t)exp[ik_{n}x]$$. Then you can obtain the values of the differential operators acting on the basis functions and the equations for the time-dependent Fourier coefficients.