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?
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 $\lambda_1\ne\lambda_2$, $$ \int_{0}^{1}Z_1(z)Z_2(z)dz = 0. $$