B5 - implicit methods

linear ODEs

considering the previous example:

\frac{d y}{d t} = - γ y \to \frac{y^{j + 1} - y^{j}}{Δ t} = - γ y^{j}

writing the next time step on the RHS:

\begin{matrix} \frac{d y}{d t} = - γ y \to \frac{y^{j + 1} - y^{j}}{Δ t} = - γ y^{j + 1} \\ y^{j + 1} = (1 + γ Δ t)^{- 1} y^{j} \end{matrix}

$(1 + γ Δ t)^{- 1} \approx 1 - γ Δ t + \dots$ for $Δ t ≪ 1$
as $γ Δ t > 0 :$ $(1 + γ Δ t)^{- 1} < 1$ , ie. always stable

non linear ODEs

\frac{d y}{d t} = f (y) \to \frac{y^{j + 1} - y^{j}}{Δ t} = f (y^{j + 1}) ⟹ y^{j + 1} = y^{j} + f (y^{j + 1}) Δ t

as there is no explicit formula for the value, $y^{j + 1}$ , it needs to be linearised
taylor expanding $f$ around $y^{j} :$

f (y^{j + 1}) = f (y^{j}) + f^{'} (y^{j}) (y^{j + 1} - y^{j}) + O [(Δ y)^{2}]

y^{j + 1} = y^{j} + [f (y^{j}) + f^{'} (y^{j}) (y^{j + 1} - y^{j})] Δ t = y^{j} + [1 - Δ t f^{'} (y^{j})]^{- 1} f (y^{j}) Δ t

for a damped ODE: $f^{'} < 0$ , and the method is always stable
this is a generic method of solving nonlinear ODEs

PDEs

\frac{\partial y}{\partial t} = κ \frac{\partial^{2} y}{\partial x^{2}}

trying to express RHS as an averaged value of the current and the next time steps:

\frac{y_{k}^{j + 1} - y_{k}^{j}}{Δ t} = \frac{1}{2} κ (\frac{\partial^{2} y}{\partial x^{2}} |_{x_{k}, t_{j + 1}} + \frac{\partial^{2} y}{\partial x^{2}} |_{x_{k}, t_{j}})

substituting finite difference approximations of the first and second derivatives:

- α y_{k + 1}^{j + 1} + (1 + 2 α) y_{k}^{j + 1} - α y_{k - 1}^{j + 1} = y_{k}^{j} + α (y_{k + 1}^{j} - 2 y_{k}^{j} + y_{k - 1}^{j})

where, $α = κ Δ t / 2 (Δ x)^{2}$

$k$ couples with $k - 1$ and $k + 1$