From Wikipedia:

Explicit methods calculate the state of a system at a later time from the state of the system at the current time, while implicit methods find a solution by solving an equation involving both the current state of the system and the later one.

Then why the Crank-Nicolson is considered implicit in time, if a later time is calculated by solving a system of linear equations that depend only on the previous time?


2 Answers 2


A simplification - the Crank-Nicolson method uses the average of the forward and backward Euler methods.

The backward Euler method is implicit, so Crank-Nicolson, having this as one of its components, is also implicit.

More accurately, this method is implicit because $u^{n+1}_i$ depends on $F^{n+1}_i$, not just $F^{n}_i$

This is means that the state at time $t=n+1$ is defined in terms of itself, and not just the previous time step.

Or as stated in another answer, if the $F^{n+1}_i$ term is non-zero, the method is implicit.

  • $\begingroup$ But $F_i^{n+1}$ is spatial dependence and not temporal, no? $\endgroup$
    – Sparkler
    Commented Oct 9, 2015 at 6:22
  • $\begingroup$ $F^{n+1}_{i}$ depends on $u^{n+1}_{i}$ meaning it depends on the next state of the system and as a result it depends on space and time. This is not surprising given that there are both temporal $n+1$ and spatial $i$ indicies $\endgroup$
    – James
    Commented Oct 9, 2015 at 6:31
  • $\begingroup$ No, $n+1$ is the temporal dependence. You can see this on wikipedia, $u=(i∆x, n∆t)$ $\endgroup$ Commented Oct 9, 2015 at 6:39

The Crank-Nicolson method is:

$\frac{u^{n+1}_{i}-u^{n}_{i}}{dt} = \frac{1}{2}(F^{n+1}_{i}+F^{n}_{i})$

This method calculates the next state of the system, i.e. $u^{n+1}_{i}$, by solving an equation involving the previous states and the next state. In the case of the heat equation for example we would get a linear system and if we are using finite elements this system would look like:

$M(\frac{u^{n+1}-u^{n}}{dt}) = -\frac{1}{2}K(u^{n+1}+u^{n})$


$(M+\frac{1}{2}dtK)u^{n+1} = (M-\frac{1}{2}dtK)u^{n}$

where M is the mass matrix and K is the stiffness matrix. Because the heat equation is linear we could separate the $u^{n+1}_{i}$'s from the $u^{n}_{i}$'s however the cost of this separation is that we must solve a linear system. In the case of a nonlinear $F^{n+1}_{i}$ and $F^{n}_{i}$ this separation would not be possible and instead we would have to use something like Newtons method to iterate our way to a solution.

More generally, if we have a timestepping scheme of the form:

$\frac{u^{n+1}_{i}-u^{n}_{i}}{dt} = \alpha_{-1}F^{n+1}_{i}+\alpha_{0}F^{n}_{i}+\alpha_{1}F^{n-1}_{i}+\alpha_{2}F^{n-2}_{i}+...$

it is implicit if $\alpha_{-1}$ is non-zero.


If you are more familiar with finite differences rather than finite elements then the Crank-Nicolson solution to the heat equation would look like:

$\frac{u^{n+1}-u^{n}}{dt} = \frac{1}{2}A(u^{n+1}+u^{n})$


$(I-\frac{1}{2}dtA)u^{n+1} = (I+\frac{1}{2}dtA)u^{n}$

where $I$ is the identity matrix and $A$ is the finite difference tridiagonal discretization matrix.


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