Explicit and implicit methods

<h2 id="computation">Computation</h2>
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 practice are <a href="/facts/Stiff_equation/wQls6Mo1">stiff</a>, for which the use of an explicit method requires impractically small time steps 
 
 
 
 Δ
 t
 
 
 {\displaystyle \Delta t}
 
 to keep the error in the result bounded (see <a href="/facts/Numerical_stability/jRDdo8zV">numerical stability</a>). 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.
Since the implicit method cannot be carried out for each kind of differential operator, it is sometimes advisable to make use of the so called operator splitting method, which means that the differential operator is rewritten as the sum of two complementary operators

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

while one is treated explicitly and the other implicitly.
For usual applications the implicit term is chosen to be linear while the explicit term can be nonlinear. This combination of the former method is called Implicit-Explicit Method (short IMEX,<a class="footnote-ref" id="fnref:1" href="#fn:1">1</a><a class="footnote-ref" id="fnref:2" href="#fn:2">2</a><a class="footnote-ref" id="fnref:3" href="#fn:3">3</a>).

<h2 id="illustration-using-the-forward-and-backward-euler-methods">Illustration using the forward and backward Euler methods</h2>
Consider the <a href="/facts/Ordinary_differential_equation/ygKGQ8kD">ordinary differential equation</a>

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

with the initial condition 
 
 
 
 y
 (
 0
 )
 =
 1.
 
 
 {\displaystyle y(0)=1.}
 
 Consider a grid 
 
 
 
 
 t
 
 k
 
 
 =
 a
 
 
 k
 n
 
 
 
 
 {\displaystyle t_{k}=a{\frac {k}{n}}}
 
 for 0 ≤ k ≤ n, that is, the time step is 
 
 
 
 Δ
 t
 =
 a
 
 /
 
 n
 ,
 
 
 {\displaystyle \Delta t=a/n,}
 
 and denote 
 
 
 
 
 y
 
 k
 
 
 =
 y
 (
 
 t
 
 k
 
 
 )
 
 
 {\displaystyle y_{k}=y(t_{k})}
 
 for each 
 
 
 
 k
 
 
 {\displaystyle k}
 
. <a href="/facts/Discretization/JvjSzwyk">Discretize</a> this equation using the simplest explicit and implicit methods, which are the forward Euler and backward Euler methods (see <a href="/facts/Numerical_ordinary_differential_equations/rbrYBco3">numerical ordinary differential equations</a>) and compare the obtained schemes.

Forward Euler method

The forward <a href="/facts/Euler_method/7yF5B4z9">Euler method</a>

(
            
              
                
                  d
                  y
                
                
                  d
                  t
                
              
            
            )
          
          
            k
          
        
        ≈
        
          
            
              
                y
                
                  k
                  +
                  1
                
              
              −
              
                y
                
                  k
                
              
            
            
              Δ
              t
            
          
        
        =
        −
        
          y
          
            k
          
          
            2
          
        
      
    
    {\displaystyle \left({\frac {dy}{dt}}\right)_{k}\approx {\frac {y_{k+1}-y_{k}}{\Delta t}}=-y_{k}^{2}}

yields

y
          
            k
            +
            1
          
        
        =
        
          y
          
            k
          
        
        −
        Δ
        t
        
          y
          
            k
          
          
            2
          
        
        
        
        
        (
        3
        )
        
      
    
    {\displaystyle y_{k+1}=y_{k}-\Delta ty_{k}^{2}\quad \quad \quad (3)\,}

for each 
 
 
 
 k
 =
 0
 ,
 1
 ,
 …
 ,
 n
 .
 
 
 {\displaystyle k=0,1,\dots ,n.}
 
 This is an explicit formula for 
 
 
 
 
 y
 
 k
 +
 1
 
 
 
 
 {\displaystyle y_{k+1}}
 
.

Backward Euler method

With the <a href="/facts/Backward_Euler_method/lccBRRYY">backward Euler method</a>

y
                
                  k
                  +
                  1
                
              
              −
              
                y
                
                  k
                
              
            
            
              Δ
              t
            
          
        
        =
        −
        
          y
          
            k
            +
            1
          
          
            2
          
        
      
    
    {\displaystyle {\frac {y_{k+1}-y_{k}}{\Delta t}}=-y_{k+1}^{2}}

one finds the implicit equation

y
          
            k
            +
            1
          
        
        +
        Δ
        t
        
          y
          
            k
            +
            1
          
          
            2
          
        
        =
        
          y
          
            k
          
        
      
    
    {\displaystyle y_{k+1}+\Delta ty_{k+1}^{2}=y_{k}}

for 
 
 
 
 
 y
 
 k
 +
 1
 
 
 
 
 {\displaystyle y_{k+1}}
 
 (compare this with formula (3) where 
 
 
 
 
 y
 
 k
 +
 1
 
 
 
 
 {\displaystyle y_{k+1}}
 
 was given explicitly rather than as an unknown in an equation).
This is a <a href="/facts/Quadratic_equation/9IMnkJAT">quadratic equation</a>, having one negative and one positive <a href="/facts/Root_of_a_function/mlK3dhoT">root</a>. The positive root is picked because in the original equation the initial condition is positive, and then 
 
 
 
 y
 
 
 {\displaystyle y}
 
 at the next time step is given by

y
          
            k
            +
            1
          
        
        =
        
          
            
              −
              1
              +
              
                
                  1
                  +
                  4
                  Δ
                  t
                  
                    y
                    
                      k
                    
                  
                
              
            
            
              2
              Δ
              t
            
          
        
        .
        
        
        (
        4
        )
      
    
    {\displaystyle y_{k+1}={\frac {-1+{\sqrt {1+4\Delta ty_{k}}}}{2\Delta t}}.\quad \quad (4)}

In the vast majority of cases, the equation to be solved when using an implicit scheme is much more complicated than a quadratic equation, and no analytical solution exists. Then one uses <a href="/facts/Root-finding_algorithm/IxTDkhBq">root-finding algorithms</a>, such as <a href="/facts/Newton%27s_method/VlRhI2FL">Newton's method</a>, to find the numerical solution.

Crank-Nicolson method

With the <a href="/facts/Crank-Nicolson_method/6RnaoWIP">Crank-Nicolson method</a>

y
                
                  k
                  +
                  1
                
              
              −
              
                y
                
                  k
                
              
            
            
              Δ
              t
            
          
        
        =
        −
        
          
            1
            2
          
        
        
          y
          
            k
            +
            1
          
          
            2
          
        
        −
        
          
            1
            2
          
        
        
          y
          
            k
          
          
            2
          
        
      
    
    {\displaystyle {\frac {y_{k+1}-y_{k}}{\Delta t}}=-{\frac {1}{2}}y_{k+1}^{2}-{\frac {1}{2}}y_{k}^{2}}

one finds the implicit equation

y
          
            k
            +
            1
          
        
        +
        
          
            1
            2
          
        
        
          Δ
          t
        
        
          y
          
            k
            +
            1
          
          
            2
          
        
        =
        
          y
          
            k
          
        
        −
        
          
            1
            2
          
        
        Δ
        t
        
          y
          
            k
          
          
            2
          
        
      
    
    {\displaystyle y_{k+1}+{\frac {1}{2}}{\Delta t}y_{k+1}^{2}=y_{k}-{\frac {1}{2}}\Delta ty_{k}^{2}}

for 
 
 
 
 
 y
 
 k
 +
 1
 
 
 
 
 {\displaystyle y_{k+1}}
 
 (compare this with formula (3) where 
 
 
 
 
 y
 
 k
 +
 1
 
 
 
 
 {\displaystyle y_{k+1}}
 
 was given explicitly rather than as an unknown in an equation). This can be numerically solved using <a href="/facts/Root-finding_algorithm/IxTDkhBq">root-finding algorithms</a>, such as <a href="/facts/Newton%27s_method/VlRhI2FL">Newton's method</a>, to obtain 
 
 
 
 
 y
 
 k
 +
 1
 
 
 
 
 {\displaystyle y_{k+1}}
 
.
Crank-Nicolson can be viewed as a form of more general IMEX (Implicit-Explicit) schemes.

Forward-Backward Euler method

In order to apply the IMEX-scheme, consider a slightly different differential equation:

d
              y
            
            
              d
              t
            
          
        
        =
        y
        −
        
          y
          
            2
          
        
        ,
         
        t
        ∈
        [
        0
        ,
        a
        ]
        
        
        (
        5
        )
      
    
    {\displaystyle {\frac {dy}{dt}}=y-y^{2},\ t\in [0,a]\quad \quad (5)}

It follows that

(
            
              
                
                  d
                  y
                
                
                  d
                  t
                
              
            
            )
          
          
            k
          
        
        ≈
        
          y
          
            k
            +
            1
          
        
        −
        
          y
          
            k
          
          
            2
          
        
        ,
         
        t
        ∈
        [
        0
        ,
        a
        ]
      
    
    {\displaystyle \left({\frac {dy}{dt}}\right)_{k}\approx y_{k+1}-y_{k}^{2},\ t\in [0,a]}

and therefore

y
          
            k
            +
            1
          
        
        =
        
          
            
              
                y
                
                  k
                
              
              (
              1
              −
              
                y
                
                  k
                
              
              Δ
              t
              )
            
            
              1
              −
              Δ
              t
            
          
        
        
        
        (
        6
        )
      
    
    {\displaystyle y_{k+1}={\frac {y_{k}(1-y_{k}\Delta t)}{1-\Delta t}}\quad \quad (6)}

for each 
 
 
 
 k
 =
 0
 ,
 1
 ,
 …
 ,
 n
 .
 
 
 {\displaystyle k=0,1,\dots ,n.}

<h2 id="see-also">See also</h2>
<ul><li><a href="/facts/Courant%E2%80%93Friedrichs%E2%80%93Lewy_condition/f4r00Wca">Courant–Friedrichs–Lewy condition</a></li>
<li><a href="/facts/SIMPLE_algorithm/Wo36xVHU">SIMPLE algorithm</a>, a semi-implicit method for pressure-linked equations</li></ul>
<h2 id="sources">Sources</h2>

<h2 id="references">References</h2>

<ol>
<li id="fn:1">U.M. Ascher, S.J. Ruuth, R.J. Spiteri: Implicit-Explicit Runge-Kutta Methods for Time-Dependent Partial Differential Equations, Appl Numer Math, vol. 25(2-3), 1997 <a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.48.1525&rep=rep1&type=pdf" target="_blank">http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.48.1525&rep=rep1&type=pdf</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></li>
<li id="fn:2">L.Pareschi, G.Russo: Implicit-Explicit Runge-Kutta schemes for stiff systems of differential equations, Recent Trends in Numerical Analysis, Vol. 3, 269-289, 2000 <a href="https://www.researchgate.net/profile/Lorenzo_Pareschi/publication/230865813_Implicit-Explicit_Runge-Kutta_schemes_for_stiff_systems_of_differential_equations/links/0046352a03ba3ee92a000000.pdf" target="_blank">https://www.researchgate.net/profile/Lorenzo_Pareschi/publication/230865813_Implicit-Explicit_Runge-Kutta_schemes_for_stiff_systems_of_differential_equations/links/0046352a03ba3ee92a000000.pdf</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></li>
<li id="fn:3">Sebastiano Boscarino, Lorenzo Pareschi, and Giovanni Russo: Implicit-Explicit Methods for Evolutionary Partial Differential Equations, SIAM, ISBN 978-1-61197-819-3 (2024). <a href="#fnref:3" class="footnote-back-ref">↩</a></li>
</ol>

Explicit and implicit methods open-in-new

Explicit and implicit methods