Explicit and implicit methods

In applied mathematics, explicit and implicit methods are approaches for mathematical simulation ofphysical 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 after one interval of 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 system state and the future one. To put it in symbols, if ${\displaystyle Y(t)}$ is the current system state and ${\displaystyle Y(t+\Delta t)}$ is the state at the next instance in time (${\displaystyle \Delta t}$ is a small time step), then, for an explicit method

${\displaystyle Y(t+\Delta t)=F(Y(t))\,}$

while for an implicit method one solves an equation

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

to find ${\displaystyle Y(t+\Delta t).}$

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 which arise in real life are stiff problems, for which the use of an explicit method requires impracticably small time steps ${\displaystyle \Delta t}$ if the error in the result is not to tend to infinity (see numerical stability). For a given error on the result, 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. The choice of method depends upon the problem to be solved.

Illustration using the forward and backward Euler methods

Consider the ordinary differential equation

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

with the initial condition ${\displaystyle y(0)=1.}$  Consider a grid ${\displaystyle t_{k}=ka/n}$  for 0≤k≤n, that is, the time step is ${\displaystyle \Delta t=a/n,}$  and denote ${\displaystyle y_{k}=y(t_{k})}$  for each ${\displaystyle k}$ .

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

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

yields

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

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

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

one finds the equation

${\displaystyle y_{k+1}+\Delta ty_{k+1}^{2}=y_{k}}$ 

for ${\displaystyle y_{k+1}}$ . 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 ${\displaystyle y}$  at the next time step is given by

${\displaystyle y_{k+1}={\frac {-1+{\sqrt {1+4y_{k}}}}{4}}\quad \quad (4)}$ 

(compare this with formula (3)).

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.