Ode check thread safety

russell_ode/README.md

@@ -16,6 +16,7 @@ _This crate is part of [Russell - Rust Scientific Library](https://github.com/cp
     * [Robertson's equation](#robertson)
     * [Van der Pol's equation](#van-der-pol)
     * [One-transistor amplifier](#amplifier1t)
+    * [PDE: discrete Laplacian (2D)](#laplacian)
 
 ## <a name="introduction"></a> Introduction
 
@@ -748,3 +749,121 @@ Total time                       = 97.951021ms
 The results are plotted below:
 
 ![One-transistor Amplifier - Radau5](data/figures/amplifier1t_radau5.svg)
+
+### <a name="laplacian"></a> PDE: discrete Laplacian operator in 2D
+
+For convenience (e.g., in benchmarks), `russell_ode` implements a discrete Laplacian operator (2D) based on the Finite Differences Method.
+
+This operator can be used to solve simple partial differential equation (PDE) problems.
+
+#### Laplace equation
+
+Approximate (with the Finite Differences Method, FDM) the solution of
+
+```text
+∂²ϕ     ∂²ϕ
+———  +  ——— = 0
+∂x²     ∂y²
+```
+
+on a (1.0 × 1.0) rectangle with the following essential (Dirichlet) boundary conditions:
+
+```text
+left:    ϕ(0.0, y) = 50.0
+right:   ϕ(1.0, y) =  0.0
+bottom:  ϕ(x, 0.0) =  0.0
+top:     ϕ(x, 1.0) = 50.0
+```
+
+See the code [pde_laplace_equation.rs](https://github.com/cpmech/russell/tree/main/russell_ode/examples/pde_laplace_equation.rs)
+
+The results are illustrated below:
+
+![Laplace equation](data/figures/pde_laplace_equation.svg)
+
+#### Poisson equation 1
+
+Approximate (with the Finite Differences Method, FDM) the solution of
+
+```text
+∂²ϕ   ∂²ϕ
+——— + ——— = 2 x (y - 1) (y - 2 x + x y + 2) exp(x - y)
+∂x²   ∂y²
+```
+
+on a (1.0 × 1.0) square with the homogeneous boundary conditions.
+
+The analytical solution is:
+
+```text
+ϕ(x, y) = x y (x - 1) (y - 1) exp(x - y)
+```
+
+See the code [test_pde_poisson_1.rs](https://github.com/cpmech/russell/tree/main/russell_ode/tests/test_pde_poisson_1.rs)
+
+The results are illustrated below:
+
+![Poisson equation 1](data/figures/test_pde_poisson_1.svg)
+
+#### Poisson equation 2
+
+Approximate (with the Finite Differences Method, FDM) the solution of
+
+```text
+∂²ϕ   ∂²ϕ
+——— + ——— = - π² y sin(π x)
+∂x²   ∂y²
+```
+
+on a (1.0 × 1.0) square with the following essential boundary conditions:
+
+```text
+left:    ϕ(0.0, y) = 0.0
+right:   ϕ(1.0, y) = 0.0
+bottom:  ϕ(x, 0.0) = 0.0
+top:     ϕ(x, 1.0) = sin(π x)
+```
+
+The analytical solution is:
+
+```text
+ϕ(x, y) = y sin(π x)
+```
+
+Reference: Olver PJ (2020) - page 210 - Introduction to Partial Differential Equations, Springer
+
+See the code [test_pde_poisson_2.rs](https://github.com/cpmech/russell/tree/main/russell_ode/tests/test_pde_poisson_2.rs)
+
+The results are illustrated below:
+
+![Poisson equation 2](data/figures/test_pde_poisson_2.svg)
+
+#### Poisson equation 3
+
+Approximate (with the Finite Differences Method, FDM) the solution of
+
+```text
+∂²ϕ     ∂²ϕ
+———  +  ——— =  source(x, y)
+∂x²     ∂y²
+```
+
+on a (1.0 × 1.0) square with homogeneous essential boundary conditions
+
+The source term is given by (for a manufactured solution):
+
+```text
+source(x, y) = 14y³ - (16 - 12x) y² - (-42x² + 54x - 2) y + 4x³ - 16x² + 12x
+```
+
+The analytical solution is:
+
+```text
+ϕ(x, y) = x (1 - x) y (1 - y) (1 + 2x + 7y)
+```
+
+See the code [test_pde_poisson_3.rs](https://github.com/cpmech/russell/tree/main/russell_ode/tests/test_pde_poisson_3.rs)
+
+The results are illustrated below:
+
+![Poisson equation 3](data/figures/test_pde_poisson_3.svg)