Hello Class,

Here is some more specifics on how to handle the boundary conditions for
the Runge-Kutta part of the PPDE computer problem:

*General Dirichlet Boundary Conditions:

u(x_0,t)=B_0(t) and u(x_m,t)=B_m(t) specified ("_" means subscript).

*Runge-Kutta formula for the new time t_{j+1}, given U_{ij} at prior time t_j:

U_{i,j+1} = U_{ij} + k*(RK1+2 (RK2+RK3)+RK4)_{ij}/6, for i=1 to m-1 
interior points only ("*" means multiplication), where

RK1_{ij} = F(x_i,t_j,RU1_{ij},DRU1_{ij},DDRU1_{ij}), for i=1 to m-1;
RU1_{0j}=B_{0j}=B_0(t_j);
RU1_{ij}=U_{ij} if i=1 to m-1;
RU1_{mj}=B_{mj}=B_m(t_j);
CFD: DRU2_{i,j}=(RU1_{i+1,j}-RU1_{i-1,j})/(2*h), for i=1 to m-1 interior only;
CFD: DDRU2_{i,j}=(RU1_{i+1,j}-2*RU1_{ij}+RU1_{i-1,j})/h^2, for i=1 to m-1;
    ("^" means super-script)

RK2_{ij} = F(x_i,t_j+0.5,RU2_{ij},DRU2_{ij},DDRU2_{ij}), for i=1 to m-1;
RU2_{0j}=B_{0j}=B_0(t_j);
RU2_{ij}=U_{ij}+(k/2)*RK1_{ij}, if i=1 to m-1;
RU2_{mj}=B_{mj}=B_m(t_j);
CFD: DRU2_{i,j}=(RU2_{i+1,j}-RU2_{i-1,j})/(2*h), for i=1 to m-1;
CFD: DDRU2_{i,j}=(RU2_{i+1,j}-2*RU2_{ij}+RU1_{i-1,j})/h^2, for i=1 to m-1;
    
RK3_{ij} = F(x_i,t_j+0.5,RU3_{ij},DRU3_{ij},DDRU3_{ij}), for i=1 to m-1;
RU3_{0j}=B_{0j}=B_0(t_j);
RU3_{ij}=U_{ij}+(k/2)*RK2_{ij}, if i=1 to m-1;
RU3_{mj}=B_{mj}=B_m(t_j);
CFD: DRU3_{i,j}=(RU3_{i+1,j}-RU3_{i-1,j})/(2*h), for i=1 to m-1;
CFD: DDRU3_{i,j}=(RU3_{i+1,j}-2*RU3_{ij}+RU1_{i-1,j})/h^2, for i=1 to m-1;

RK4_{ij} = F(x_i,t_j+1,RU4_{ij},DRU4_{ij},DDRU4_{ij}), for i=1 to m-1;
RU4_{0j}=B_{0j}=B_0(t_j);
RU4_{ij}=U_{ij}+k*RK3_{ij}, if i=1 to m-1;
RU3_{mj}=B_{mj}=B_m(t_j);
CFD: DRU4_{i,j}=(RU4_{i+1,j}-RU4_{i-1,j})/(2*h), for i=1 to m-1;
CFD: DDRU4_{i,j}=(RU4_{i+1,j}-2*RU4_{ij}+RU1_{i-1,j})/h^2, for i=1 to m-1;

Note that the only forms that change are those ofr RK[l]_{ij} and RU[l]_{ij}
for "[l]" = 1 to 4.

SR/FBH

PS:  Ignore anything in the original computer problem that mentions
     derivative boundary conditions