Explicit and implicit methods

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In applied mathematics, explicit and implicit methods are approaches for mathematical simulation of physical processes, or in other words, they are numerical methods for solving time-variable ordinary and partial differential equations.

Explicit methods calculate the state of a system at a later time from the state of the system at the current time, while an implicit method finds it by solving an equation involving both the current state of the system and the later one. To put it in symbols, if $Y (t)$ is the current system state and $Y (t + Δ t)$ is the state at the later time ( $Δ t$ is a small time step), then, for an explicit method

$Y(t+\Delta t) = F(Y(t))\,$

while for an implicit method one solves an equation

$G(Y(t), Y(t+\Delta t))=0 \quad\quad (1)\,$

to find $Y (t + Δ t).$

It is clear that implicit methods require an extra computation (solving the above equation), and they can be much harder to implement. Implicit methods are used because many problems arising in real life are stiff, for which the use of an explicit method requires impractically small time steps $Δ t$ to keep the error in the result bounded (see numerical stability). For such problems, to achieve given accuracy, it takes much less computational time to use an implicit method with larger time steps, even taking into account that one needs to solve an equation of the form (1) at each time step. That said, whether one should use an explicit or implicit method depends upon the problem to be solved.

[edit] Illustration using the forward and backward Euler methods

Consider the ordinary differential equation

$\frac{dy}{dt} = -y^2, \ y\in [0, a]\quad \quad (2)$

with the initial condition $y (0) = 1.$ Consider a grid $t k = k a / n$ for 0≤k≤n, that is, the time step is $Δ t = a / n,$ and denote $y k = y (t k)$ for each $k$ .

Discretize this equation using the simplest explicit and implicit methods, which are the forward Euler and backward Euler methods (see numerical ordinary differential equations). The forward Euler method

$\frac{y_{k+1}-y_k}{\Delta t} = - y_k^2$

yields

$y_{k+1}=y_k-\Delta t y_k^2 \quad \quad \quad(3)\,$

for each $k=0, 1, \dots, n,$ while with the backward Euler method

$\frac{y_{k+1}-y_k}{\Delta t} = - y_{k+1}^2$

one finds the equation

$y_{k+1}+\Delta t y_{k+1}^2=y_k$

(compare this with formula (3))

for $y k + 1$ . This is a quadratic equation, having one negative and one positive root. The positive root is picked because in the original equation the initial condition is positive, and then $y$ at the next time step is given by

$y_{k+1}=\frac{-1+\sqrt{1+4\Delta t y_k}}{2 \Delta t} \quad \quad (4)$ .

In the vast majority of cases the equation to be solved for is much more complicated than a quadratic equation, and no exact solution exists. Then one uses root-finding algorithms, such as Newton's method.

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Category: Numerical differential equations

Explicit and implicit methods

From Wikipedia, the free encyclopedia

[edit] Illustration using the forward and backward Euler methods

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