In mathematics, the fictitious domain method is a method to find the solution of a partial differential equations on a complicated domain
, by substituting a given problem
posed on a domain
, with a new problem posed on a simple domain
containing
.
Assume in some area
we want to find solution
of the equation:

with boundary conditions:

The basic idea of fictitious domains method is to substitute a given problem
posed on a domain
, with a new problem posed on a simple shaped domain
containing
(
). For example, we can choose n-dimensional parallelotope as
.
Problem in the extended domain
for the new solution
:


It is necessary to pose the problem in the extended area so that the following condition is fulfilled:
![{\displaystyle u_{\epsilon }(x){\xrightarrow[{\epsilon \rightarrow 0}]{}}u(x),x\in D}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b631489e1686019104e60a79c137d8cb9fed2666)
Simple example, 1-dimensional problem
[edit]


Prolongation by leading coefficients
[edit]
solution of problem:

Discontinuous coefficient
and right part of equation previous equation we obtain from expressions:


Boundary conditions:

Connection conditions in the point
:
![{\displaystyle [u_{\epsilon }]=0,\ \left[k^{\epsilon }(x){\frac {du_{\epsilon }}{dx}}\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b5d0fff3e1d5a74c1314c40cbf5d89a80eb9c82)
where
means:
![{\displaystyle [p(x)]=p(x+0)-p(x-0)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/308458c63068288908dc190ddd0475940fcd2b31)
Equation (1) has analytical solution therefore we can easily obtain error:

Prolongation by lower-order coefficients
[edit]
solution of problem:

Where
we take the same as in (3), and expression for

Boundary conditions for equation (4) same as for (2).
Connection conditions in the point
:
![{\displaystyle [u_{\epsilon }(0)]=0,\ \left[{\frac {du_{\epsilon }}{dx}}\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cdf31b2d4d05429863841e47193564a63b8e4a9f)
Error:

- P.N. Vabishchevich, The Method of Fictitious Domains in Problems of Mathematical Physics, Izdatelstvo Moskovskogo Universiteta, Moskva, 1991.
- Smagulov S. Fictitious Domain Method for Navier–Stokes equation, Preprint CC SA USSR, 68, 1979.
- Bugrov A.N., Smagulov S. Fictitious Domain Method for Navier–Stokes equation, Mathematical model of fluid flow, Novosibirsk, 1978, p. 79–90