Examine a dynamic 2D heat equation $\dot{u} = \Delta u$ with zero boundary temperature. A standard finite difference approach is used on a rectangle using a $n\times n$ grid. For the resulting linear systems a banded Cholesky solver is used. Compare the three methods explicit, implicit and Crank-Nicolson for the time stepping.
- Explicit method
$$ \begin{gathered} \frac{u_{i+1,j} - u_{i,j}}{\Delta t} = \kappa \frac{u_{i,j-1} - 2u_{i,j} + u_{i,j+1}}{(\Delta x)^2} \\ u_{i+1,j} = u_{i,j} + \frac{\kappa \Delta t}{(\Delta x)^2}(u_{i,j-1} - 2u_{i,j} + u_{i,j+1})\\ \vec{u}_{i+1} = \vec{u}_i - \kappa \Delta t \textbf{A}_n \cdot \vec{u}_{i}\\ \end{gathered} $$
- Implicit method
$$ \begin{gathered} \frac{u_{i+1,j} - u_{i,j}}{\Delta t} = \kappa \frac{u_{i+1,j-1} - 2u_{i+1,j} + u_{i+1,j+1}}{(\Delta x)^2} \\ \vec{u}_{i+1} = \vec{u}_i - \kappa \Delta t \textbf{A}_n \cdot \vec{u}_{i+1}\\ u_i = (\mathbb{I}_n + \kappa \Delta t \textbf{A}_n)^{-i}\cdot \vec{u}_0 \end{gathered} $$
- Crank-Nicolson
$$ \begin{gathered} \frac{u_{i+1, j}-u_{i,j}}{\kappa \Delta t} = \frac{u_{i,j-1} - 2u_{i,j} + u_{i,j + 1}}{2(\Delta x)^2} + \frac{u_{i+1,j-1} - 2u_{i+1,j} + u_{i+1,j + 1}}{2(\Delta x)^2}\\ \vec{u}_{i+1} - \vec{u}_i = -\frac{\kappa \Delta t}{2}(\textbf{A}_n \cdot \vec{u}_{i+1} + \textbf{A}_n \cdot \vec{u}_i) \end{gathered} $$
Here is the question:
Does the implicit method require the least memory?
Here is my proposition for an answer:
I would say that the statement if FALSE
The implicit method requires more computational effort to give an answer, because the matrix $\textbf{A}_n$ needs to be inverted. However, I would say that this is not the reason why the statement is false: for the implicit method there is $\textbf{no extra/less storage needed}$ (compared to the explicit method for example) because there is no extra data generated from the computation (for example no other matrix generated). Is the answer correct as well as the reasoning behind it?
Following the comments of @zimbra314, I posted the question in Computational science beta
It depends on how do you apply inversion. If you use a direct method such as LU factorization, the inverted matrix will have more non-zero entries due to fill-in. For $n\times n$ 2D grid, $A$ matrix will have $\mathcal{O}(n^2)$ entries while factored matrix will have $\mathcal{O}(n^{2} \log n)$ entries; so it would require more memory than explicit method.