Dtjytyk

I have a coupled non-dimensional diffusion system in $v(z,tau)$, formulated by the following equations

begin{align}
frac{partial v}{partial tau} &= Deltafrac{partial^2 v}{partial z^2},
%
qquad &text{for} zin[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(-Deltalambda^2tauright)},\
Z &= a cos(lambda z) + bsin(lambda z),\
Z' &= lambda left( bcos(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) + bsin(lambda z),\
Z' &= lambda left( bcos(lambda z) - sin(lambda z) right)
end{align}

Therefore via our boundary condition at $z=0$ we find
begin{align}
lambda b &= E
quadRightarrowquad
b = frac{E}{lambda}
end{align}

and via our second
begin{align}
lambda left( frac{E}{lambda}cos(lambda) - sin(lambda) right) &= -Dleft(cos(lambda) + frac{E}{lambda}sin(lambda)right)\
%%%
Rightarrowquad
Elambdacos(lambda) - lambda^2sin(lambda) &= -Dcos(lambda) - EDsin(lambda)\
%%%
Rightarrowquad
left(Elambda+Dright)cos(lambda)
&=
left( lambda^2- EDright)sin(lambda)\
%%%
Rightarrowquad
tan(lambda)
&=
frac{Elambda+D}{lambda^2- ED}
end{align}

This has countably infinite solutions $lambda_i$ for $iinmathbb{N}$. 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(-Deltalambda_i^2tauright)}
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_idelta_{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:
$$
left[begin{array}{cc} Z_1(0) & Z_1'(0) \ Z_2(0) & Z_2'(0)end{array}right]left[begin{array}{c}1 \ -Eend{array}right] = 0, \
left[begin{array}{cc} Z_1(1) & Z_1'(1) \ Z_2(1) & Z_2'(1)end{array}right]left[begin{array}{c}1 \ ;;Dend{array}right] = 0.
$$
Therefore, if $lambda_1nelambda_2$,
$$
int_{0}^{1}Z_1(z)Z_2(z)dz = 0.
$$