I am dealing with the following set of equations:
$\alpha w(i) ~= ~a w(i-1)+bw(i+1),~i=1,2,\ldots,B-1$ and $w(0) = 1,~w(1)= \frac{\alpha-a}{b}w(0),~\alpha w(B)=w(B-1)\left(\frac{a}{1-b}\right)$,
which needs to be solved for $B+2$ unknowns $\alpha,\{w(i)\}_{i=0}^{B}$.
I am confused on how to proceed. I know that solving the characteristic equation
$b\lambda^2 + a\lambda -\alpha = 0$ will yield me solutions of the form
$w(i) = K_1 \lambda^i+K_2\lambda^i$. But I am confused about how to find $K_1$ and $K_2$ and the $\alpha$.
Because the recurrence holds from $i=1$, where $w(i-1) = w(0)$ to $i = B - 1$, where $w(i+1) = w(B)$, the formula $w(i) =K_1 \lambda_1^i+K_2\lambda_2^i$ holds for $0 \le i \le B$.
Let $\beta = \frac{\alpha}{2b}, \gamma = \sqrt{\beta^2 - \frac ab}, r = \frac ab$, then $$w(1) = \frac {\alpha - a}b = 2\beta-r\\\lambda_1 = \beta + \gamma\\\lambda_2 = \beta - \gamma$$ This at least will make the equations look a little nicer. Now, $$1 = w(0) = K_1(\beta + \gamma)^0 + K_2(\beta - \gamma)^0 = K_1 + K_2\\2\beta - r = w(1) = K_1(\beta + \gamma) + K_2(\beta - \gamma) = \beta + (K_1 - K_2)\gamma$$ So $$K_1 - K_2 = \frac {\beta - r}\gamma$$ and $$K_1 = \frac{\gamma + \beta - r}{2\gamma}\\K_2 = \frac{\gamma - \beta + r}{2\gamma}\\w(i) = \frac{(\beta + \gamma - r)(\beta + \gamma)^i - (\beta - \gamma - r)(\beta - \gamma)^i}{2\gamma}$$
Note that if we define $$v(i) = \frac{(\beta + \gamma)^i - (\beta - \gamma)^i}{2\gamma}$$ then $$w(i) = v(i+1) -rv(i)$$
Now, the final equation that must be satisfied is $$\alpha w(B) = \frac{a}{1-b}w(B-1)\\\alpha(v(B+1) - rv(B))= \frac{a}{1-b}(v(B) -rv(B-1))\\ \alpha v(B+1) -\frac {\alpha a}b v(B) - \frac{a}{1-b} v(B) + \frac{a^2}{b(1-b)}v(B-1) = 0\\ \alpha b(1-b)\left((\beta + \gamma)^{B+1} - (\beta - \gamma)^{B+1}\right) + (\alpha a - (\alpha + 1)ab)\left((\beta + \gamma)^{B} - (\beta - \gamma)^{B}\right) + a^2\left((\beta + \gamma)^{B-1} - (\beta - \gamma)^{B-1}\right) = 0$$
Since $\beta, \gamma$ are functions of $\alpha$, this is a very ugly equation in the single unknown $\alpha$. It may be possible to reduce it, but likely this would require numerical methods to find, if given actual values.