Solving the heat equation with robin boundary conditions I have a coupled non-dimensional diffusion system in $v(z,\tau)$, formulated by the following equations
\begin{align}
\frac{\partial v}{\partial \tau} &= \Delta\frac{\partial^2 v}{\partial z^2},
%
\qquad &\text{for}\ z\in[0,1],\ \tau>0 \\ 
%%%
\frac{\partial v}{\partial z} &= Ev,
%
\qquad &\text{for}\ z=0,\ \tau>0,\\ 
%%%
\frac{\partial v}{\partial z} &= -D v,
%
\qquad &\text{for}\ z=1,\ \tau>0
\end{align}
where $\Delta,E,D>0$
We next proceed with separation of variables, let
\begin{align}
v = Z(z)T(\tau)
\end{align}
Substitution yields the following
\begin{align}
\frac{1}{\Delta}\frac{\dot{T}}{T} &= \frac{Z''}{Z} = -\lambda^2
\end{align}
Therefore we find
\begin{align}
T &\propto \exp{\left(-\Delta\lambda^2\tau\right)},\\
Z &= a \cos(\lambda z) + b\sin(\lambda z),\\
Z' &= \lambda \left( b\cos(\lambda z) -a \sin(\lambda z) \right)
\end{align}
WLOG we may set $a=1$, as we will later take a linear superposition of these solution functions. Therefore we have
\begin{align}
Z &= \cos(\lambda z) + b\sin(\lambda z),\\
Z' &= \lambda \left( b\cos(\lambda z) - \sin(\lambda z) \right)
\end{align}
Therefore via our boundary condition at $z=0$ we find
\begin{align}
\lambda b &= E
\quad\Rightarrow\quad
b = \frac{E}{\lambda}
\end{align}
and via our second
\begin{align}
\lambda \left( \frac{E}{\lambda}\cos(\lambda) - \sin(\lambda) \right) &= -D\left(\cos(\lambda) + \frac{E}{\lambda}\sin(\lambda)\right)\\
%%%
\Rightarrow\quad
E\lambda\cos(\lambda) - \lambda^2\sin(\lambda) &= -D\cos(\lambda) - ED\sin(\lambda)\\
%%%
\Rightarrow\quad
\left(E\lambda+D\right)\cos(\lambda)
&=
\left( \lambda^2- ED\right)\sin(\lambda)\\
%%%
\Rightarrow\quad
\tan(\lambda)
&=
\frac{E\lambda+D}{\lambda^2- ED}
\end{align}
This has countably infinite solutions $\lambda_i$ for $i\in\mathbb{N}$. Therefore we have the following solution for $v(z,\tau)$
\begin{align}
v(z,\tau) &=
\sum_{i=1}^\infty C_n
\left(
\cos(\lambda_i z)
+
\left(\frac{E}{\lambda_i}\right)\sin(\lambda_i z)
\right)
\exp{\left(-\Delta\lambda_i^2\tau\right)}
\end{align}
Therefore at $\tau=0$
\begin{align}
v(z,0) &=
\sum_{i=1}^\infty C_i
\left(
\cos(\lambda_i z)
+
\left(\frac{E}{\lambda_i}\right)\sin(\lambda_i z)
\right)
= 1
\end{align}
How can I find $C_i$?.
EDIT: If we define $Z_i(z)$ as follows
\begin{align}
Z_i(z) = 
\cos(\lambda_i z)
+
\left(\frac{E}{\lambda_i}\right)\sin(\lambda_i z)
\end{align}
then am I correct in thinking we use the following relation to find $C_i$?
\begin{align}
\int_0^1 Z_i(z)Z_j(z) \text{d}z = c_i\delta_{ij}
\end{align}
I'm not sure if this is the case, see this link. Does this mean my spatial basis is not orthogonal? 
 A: The equation in $Z$ is
$$
                       -Z'' = \lambda^2 Z, \\
                      Z'(0)-EZ(0)=0 \\
                      Z'(1)+DZ(1)=0.
$$
If $Z_1$ is a solution for $\lambda_1$ and $Z_2$ is a solution for $\lambda_2$, then
\begin{align}
     (\lambda_2^2-\lambda_1^2)\int_{0}^{1}Z_1(z)Z_2(z)dz & = \int_{0}^{1}Z_1Z_2''-Z_1''Z_2 dz \\
  & = \int_{0}^{1}\frac{d}{dz}(Z_1Z_2'-Z_1'Z_2)dz \\
  & = Z_1Z_2'-Z_1'Z_2|_{0}^{1} \\
  & = \left.\left|\begin{array}{cc}Z_1 & Z_2 \\ Z_1' & Z_2'\end{array}\right|\right|_{0}^{1} = 0.
\end{align}
The determinants at $0$ and at $1$ are individually $0$ because the matrices have non-trivial null spaces:
$$
      \left[\begin{array}{cc} Z_1(0) & Z_1'(0) \\ Z_2(0) & Z_2'(0)\end{array}\right]\left[\begin{array}{c}1 \\ -E\end{array}\right] = 0, \\
\left[\begin{array}{cc} Z_1(1) & Z_1'(1) \\ Z_2(1) & Z_2'(1)\end{array}\right]\left[\begin{array}{c}1 \\ \;\;D\end{array}\right] = 0.
$$
Therefore, if $\lambda_1\ne\lambda_2$,
$$
       \int_{0}^{1}Z_1(z)Z_2(z)dz = 0.
$$
