Solving heat equation using Bender-Schmidt method. I was trying to solve this heat equation using Bender-Schmidt method $$u^{j+1} = A u^{j}+rB$$
The heat equation is $$u_{xx} = 32u_{t},0<x<1,u(x,0)=0,u(0,t)=0,u(1,t)=10+t$$taking $$h = 0.25,k=0.025$$ for $$0 \leq t \leq 1$$ Where $h$ is the step length for $x$ axis and $k$ is the step size in time direction.
I see that Bender Schmidt method doesnot take into account the boundary conditions?
So number of points in $x$ axis is $N = \frac{b-a}{h} = 4$
so I got $ A =            
              \begin{bmatrix} 
                         0.975 & 0.0125 &     0 \\
                         0.0125 & 0.975 & 0.0125\\
                         0     &  0.0125 & 0.975\\ 
              \end{bmatrix}     
$
and matrix $B = [0,0,0]'$
and $r = \frac{kc^2}{h^2} = 0.0125 < 0.5$ so the solution system is stable.
but since $u^{(0)} = [0,0,0]'$ , implying $u^{(1)},u^{(2)},...$ are all zero vectors,where am i doing wrong?
I am not using the boundary conditions anywhere?why is that so?
EDIT:-
Basically this is the matrix representation of the explicit Forward finite difference method.Like $u_{j}^{n+1}  = ru_{j-1}^{n}+(1-2r)u_{j}^{n}+ru_{j+1}^{n}$ where $j =1,2,...,(N-1)$.
$ A =            
              \begin{bmatrix} 
                         (1-2r) & r &     0 \\
                         r & (1-2r)& r\\
                         0     &  r & (1-2r)\\ 
              \end{bmatrix}     
$
and matrix $B$ is  $[f(x_{0}),0,f(x_{N})]'$
 A: The heat equation $u_t = \alpha u_{xx}$ where $\alpha = 1/32$ m²/s is considered, with zero initial conditions in the interior of the domain, and mixed boundary conditions (homogeneous on one side, and time-dependent non-homogeneous on the other). The domain $(x,t)\in [0,1]^2$ is discretized, such that $x_j = jh$ and $t_n = nk$, and $u_j^n \simeq u(x_j, t_n)$. An explicit forward finite-difference method writes
$$
u_j^{n+1} = (1-2r)\, u_j^n + r\, (u_{j-1}^n + u_{j+1}^n) \, ,
$$
where $r = \alpha k / h^2 \leq 1/2$. The initial conditions in the interior of the domain are accounted for by setting $u_1^0 = \dots = u_{1/h - 1}^0 = 0$.
The homogeneous boundary condition at $x=0$ is accounted for by setting $u_0^{n+1} = u_0^{n} = 0$ for all $n$, whereas the non-homogeneous boundary at $x=1$ is accounted for by setting $u_{1/h}^{0} = 10$ and
$u_{1/h}^{n+1} = u_{1/h}^{n} + k.$
In matrix form, the method writes
$$
{\bf u}^{n+1} = {\bf A}\, {\bf u}^{n} + {\bf b} \, , \tag{*}
$$
where
$$
{\bf A} = \left(\begin{array}{ccccc}
1 & & & & \\
r & 1-2r & r & \\
 & \ddots & \ddots & \ddots & \\
 & & r & 1-2r & r \\
 & & & & 1
\end{array}\right) ,\qquad
{\bf b} = \left(\begin{array}{c}
0 \\
0 \\
\vdots \\
0 \\
k
\end{array}\right) ,
$$
and ${\bf u}^n = (u_0^n,\dots ,u_{1/h}^n)^\top$. The first and last equations can be removed since $u_0^n = 0$ and $u_{1/h}^n = 10 + nk$ are known. Thus, the system $({^*})$ rewrites with respect to the unknowns ${\bf u}^n = (u_1^n,\dots ,u_{1/h -1}^n)^\top$, and one has
$$
{\bf A} = \left(\begin{array}{cccc}
 1-2r & r & \\
r & \ddots & \ddots & \\
& \ddots & \ddots & r \\
& & r & 1-2r \\
\end{array}\right) ,\qquad
{\bf b} = r\left(\begin{array}{c}
0 \\
\vdots \\
0 \\
10 + nk
\end{array}\right) .
$$

Usually, the terminology Bender-Schmidt method denotes the case $r=1/2$.
