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We solve a PDE of $U(t, x, y)$ using finite difference method with following relation: $$\left(1 - \dfrac{k}{2}\Delta_x^2\right)\left(1 - \dfrac{k}{2}\Delta_y^2\right)U^{n+1}_{ij} = \left(1 + \dfrac{k}{2}\Delta_x^2\right)\left(1 + \dfrac{k}{2}\Delta_y^2\right)U^{n}_{ij}$$

but the book says the computational form as follow to be easy to solve \begin{eqnarray} &&\left(1 - \dfrac{k}{2}\Delta_x^2\right)U^{*}_{ij} = \left(1 + \dfrac{k}{2}\Delta_x^2\right)\left(1 + \dfrac{k}{2}\Delta_y^2\right)U^{n}_{ij}\\ &&\left(1 - \dfrac{k}{2}\Delta_y^2\right)U^{n+1}_{ij} = U^{*}_{ij} \end{eqnarray}

Here $U^n_{ij}$ is that difference $t$ as $n$ with length $k,$ $x,y$ with $i,j$ with length $h.$ And $$\Delta_x^2 U_{ij} = h^{-2}(U_{i+1,j} - 2U_{i,j} + U_{i-1,j})$$ $$\Delta_y^2 U_{ij} = h^{-2}(U_{i,j+1} - 2U_{i,j} + U_{i,j-1})$$

My question is why do we need the computational form instead of the first version i.e it is easy in where?

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    As far as I can tell, then computational form is the same as your discretisation, except for the fact that the computational form calculates an intermediate step before updating the solution at $n+1$. – 2017-01-22
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    @ Mattos. does that mean we only want to solve equation system AX = c instead of ABX = c? – 2017-01-22

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