How to get a numerical and explicit answer? I have $iA_t=A_{xx}+A_{yy}+f(x,y,t)A$
$A(x,y,t)=0$ on the border of the unit square $(x,y)\in((0,1)(0,1))$
$A(x,y,0)=u(x,y)=$ ,for example, $=\sin(pix)\sin(piy)$
For numerical answer I have:
$i\frac{A^{n+0.5}_{k,l}-A^{n}_{k,l}}{\tau}=\sigma\frac{A^{n+0.5}_{k+1,l}-2A^{n+0.5}_{k,l}+A^{n+0.5}_{k-1,l}}{h^2}+(1-\sigma)\frac{A^{n}_{k+1,l}-2A^{n}_{k,l}+A^{n}_{k-1,l}}{h^2}+\frac{1}{2}f^{n}_{k,l}A^{n}_{k,l}$
$i\frac{A^{n+1}_{k,l}-A^{n+0.5}_{k,l}}{\tau}=\frac{A^{n+1}_{k,l+1}-2A^{n+1}_{k,l}+A^{n+1}_{k,l-1}}{h^2}+(1-\sigma)\frac{A^{n+0.5}_{k,l+1}-2A^{n+0.5}_{k,l}+A^{n+0.5}_{k,l-1}}{h^2}+\frac{1}{2}f^{n}_{k,l}A^{n}_{k,l}$
And therefore:
$\frac{1-\sigma}{h^2}A^{n}_{k+1,l}+(\frac{i}{\tau}-\frac{2(1-\sigma)}{h^2}+f^{n}_{k,l})A^{n}_{k,l}+\frac{1-\sigma}{h^2}A^{n}_{k-1,l}=\\ =-\frac{\sigma}{h^2}A^{n+0.5}_{k+1,l}+(\frac{i}{\tau}+\frac{2\sigma}{h^2})A^{n+0.5}_{k,l}-\frac{\sigma}{h^2}A^{n+0.5}_{k-1,l}$
$\frac{1-\sigma}{h^2}A^{n+0.5}_{k,l+1}+(\frac{i}{\tau}-\frac{2(1-\sigma)}{h^2})A^{n+0.5}_{k,l}+\frac{1-\sigma}{h^2}A^{n+0.5}_{k,l-1}=\\ =-\frac{\sigma}{h^2}A^{n+1}_{k,l+1}+(\frac{i}{\tau}+\frac{2\sigma}{h^2})A^{n+1}_{k,l}-\frac{\sigma}{h^2}A^{n+1}_{k,l-1}$
Without $f(x,y,t)A$ I get matrixes:
$M=\begin{pmatrix} \frac{i}{\tau}-\frac{2(1-\sigma)}{h^2} & \frac{1-\sigma}{h^2} & 0 & 0 & \ldots\\ \frac{1-\sigma}{h^2}& \frac{i}{\tau}-\frac{2(1-\sigma)}{h^2} & \frac{1-\sigma}{h^2} & 0 & \ldots\\ 0 & \frac{1-\sigma}{h^2}& \frac{i}{\tau}-\frac{2(1-\sigma)}{h^2} &\frac{1-\sigma}{h^2} & \ldots\\ 0 & 0 & \frac{1-\sigma}{h^2}& \frac{i}{\tau}-\frac{2(1-\sigma)}{h^2} & \ddots\\ \vdots & \vdots & \vdots & \ddots & \ddots \end{pmatrix}$
$N=\begin{pmatrix} \frac{i}{\tau}+\frac{2\sigma}{h^2} & -\frac{\sigma}{h^2} & 0 & 0 & \ldots\\ -\frac{\sigma}{h^2}& \frac{i}{\tau}+\frac{2\sigma}{h^2} & -\frac{\sigma}{h^2} & 0 & \ldots\\ 0 & -\frac{\sigma}{h^2}& \frac{i}{\tau}+\frac{2\sigma}{h^2} &-\frac{\sigma}{h^2}& \ldots\\ 0 & 0 &-\frac{\sigma}{h^2}& \frac{i}{\tau}+\frac{2\sigma}{h^2} & \ddots\\ \vdots & \vdots & \vdots & \ddots & \ddots \end{pmatrix}$
And I should solve two matrix equation for every $t_i$:
$A^{n+0.5} = N^{-1}MA^{n}$
$A^{n+1} = N^{-1}MA^{n+0.5}$
QUESTIONS
- How to get numerical answer with $f(x,y,t)\neq 0$? There is a suspicion that at each iteration it will be necessary to change the matrix M
- What is explicit answer?