Okay, not giving up hope due to encouragement from @uvs I decided to snoop around the idea more. It turns out that in order to solve two inhomogeneous Neumann BCs, one must use Green's functions. Using this library from University of Nebraska, Lincoln, I was able to work out the solution for this post. I've had to change the problem to be
$$\dfrac{1}{\alpha}{\partial{\rm G}\left(x,t\right) \over \partial t}
=
{\partial^{2}{\rm G}\left(x,t\right) \over \partial x^{2}}$$
$$\partial_{x^*} G(0,t) = 0$$
$$\partial_{x^*} G(1,t) = 0$$
with the initial condition
$$T(x,0)=T_0(x)$$
and the forcing function
$$F(x') =p\space \delta(x')\space H(t') $$
where p is the power inserted upon the tissue ( which is equal to $q\space A/k$), $\delta$ is the dirac delta function, and H is the heaviside distribution function, and A is the x-sectional area of the material. This corresponds to problem type X23 in the GF library, and is solvable by the convolution
$$ T(x,t) = \int_0^t{\int_0^L{G(x,t|x',t')F(x',t') dx'}dt'} $$
which provides me an answer too ugly to be represented here, but I've solved this in matlab and arrived at some good results: here are some examples of the medium with
${
h = 900;
k = 2.5;
rho = 998;
L = 0.0024;
Cp = 860;
alpha = k/(rho*Cp);
Tb = 36;
Tinf = 20;
T0 = Tb;
tf = 10;
D = 0.01;
Area = .25*pi*(D*D);
q = 5;
q = q/Area;
Bi = h*L/k;
absorptionCoeff = 0.5$
When solving the homogeneous problem on a finite domain with time dependent boundary conditions you want to get the equilibrium temperature. So if we have
$$\begin{cases} u_{t} = k u_{xx} & 0 \leq x \leq L , t>0 \\ u(x,0) = f(x) , & 0 \leq x \leq L \\ u(0,t) = A(t) , u(L,t) = B(t) & t > 0 \end{cases}$$
In this case we need a candidate solution for the equilibrium temperature so we create a function called the reference temperature, $r(x,t)$ and we want the function to satisfy.
$$ r(0,t) = A(t) \\ r(L,t) = B(t) $$
we can create one like so
$$ r(x,t) = A(t) + \frac{x}{L}\big[B(t) - A(t) \big]$$
Now, instead we have
$$ v(x,t) = u(x,t) - r(x,t) $$
and this solution $v(x,t)$ will solve the homogeneous problem.
Ok, now that it has been set up we'll let
$$ r(x,t) = A(t) + \frac{x}{L}\big[B(t) - A(t) \big]$$
and now we replace $A(t) = \frac{1}{t+a^{-1}} + a$ and $B(t) = \frac{-1}{t+a^{-1}} + a $
$$ r(x,t) = \frac{1}{t+a^{-1}} + a + \frac{x}{L} \big [\frac{-2}{t+a^{-1}} \big]$$
Now, instead we have this homogeneous problem you should be able to solve
$$\begin{cases} v_{t} = k v_{xx} & 0 \leq x \leq L , t>0 \\ v(x,0) = f(x) - r(x) , & 0 \leq x \leq L \\ v(0,t) = 0 , v(L,t) = 0 & t > 0 \end{cases}$$
The solution to this is
$$ v(x,t) = \sum_{i=1}^{\infty} a_{n} \sin(\frac{n \pi x}{L}) e^{-k t(\frac{n \pi}{L})^{2} }$$
you solve for the coefficients $a_{n}$ and we get
$$ a_{n} = \frac{2}{L} \int_{0}^{L} \big[ f(x) - r(x) \big] \sin(\frac{n \pi x}{L}) dx $$
where $f(x)$ is your function
$$ f(x) = \begin{cases} u_{1} & x < 1 \\ 0 & x \geq 1 \end{cases}$$
which looks like the complement of a Heaviside function multiplied by the constant $u_1$. Try that and see what happens.
This about the reference temperature function $r(x,t)$. We need a function such that it obeys the boundary conditions.
If $u(x,t) = v(x,t) + r(x,t)$, we have already said that $v(x,t)$ is the homogeneous solution. So $u(0,t) = v(0,t) + r(0,t)$ and $u(L,t) = v(L,t) + r(L,t)$. But $v(0,t) = 0$ and $v(L,t) = 0$.
$ x = 0$ we get
$$ r(0,t) = A(t) + \frac{0}{L}[B(t) - A(t)] = A(t)$$
and at $x=L$ we have
$$ r(L,t) = A(t) + \frac{L}{L}[B(t) -A(t)] = B(t) + A(t) - A(t) = B(t) $$
which is what we wanted.
Best Answer
This is false. The cosine function has zeros at $\pi/2,\pi/2+\pi,\pi/2+2\pi,\cdots$, i.e. at $\frac{(2n+1)\pi}{2}$.