Merge fixing the 27pt and 9pt stencil solver example documentation

Fix some issues with documentation in the 27-pt and 7-pt stencil examples.

Related PR #306

examples/nine-pt-stencil-solver/doc/intro.dox

@@ -4,36 +4,36 @@
 This example solves a 2D Poisson equation:
 
 $
-    \Omega = (0,1)^2
-    \Omega_b = [0,1]^2   (with boundary)
-    \partial\Omega = \Omega_b \backslash \Omega
-    u : \Omega_b -> R
-    u'' = f in \Omega
-    u = u_D on \partial\Omega
+    \Omega = (0,1)^2 \\
+    \Omega_b = [0,1]^2 \text{  (with boundary)} \\
+    \partial\Omega = \Omega_b \backslash \Omega \\
+    u : \Omega_b -> R \\
+    u'' = f \in \Omega \\
+    u = u_D \in \partial\Omega \\
 $
 
 using a finite difference method on an equidistant grid with `K` discretization
 points (`K` can be controlled with a command line parameter). The discretization
 may be done by any order Taylor polynomial.
-For an equidistant grid with K "inner" discretization points (x1,y1), ...,
-(xk,y1),(x1,y2), ..., (xk,yk,z1) step size h = 1 / (K + 1) and a stencil \in
-\R^{3 x 3}, the formula produces a system of linear equations
+For an equidistant grid with K "inner" discretization points $ (x1,y1), ...,
+(xk,y1),(x1,y2), ..., (xk,yk,z1) $ step size $h = 1 / (K + 1)$ and a stencil $\in
+\R^{3 x 3}$, the formula produces a system of linear equations
 
-\sum_{a,b=-1}^1 stencil(a,b) * u_{(i+a,j+b} = -f_k h^2,  on any inner node with
+$\sum_{a,b=-1}^1 stencil(a,b) * u_{(i+a,j+b} = -f_k h^2 $,  on any inner node with
 a neighborhood of inner nodes
 
 On any node, where neighbor is on the border, the neighbor is replaced with a
-'-stencil(a,b) * u_{i+a,j+b}' and added to the right hand side vector. For
+$-stencil(a,b) * u_{i+a,j+b}$ and added to the right hand side vector. For
 example a node with a neighborhood of only edge nodes may look like this
 
-\sum_{a,b=-1}^(1,0) stencil(a,b) * u_{(i+a,j+b} = -f_k h^2 - \sum_{a=-1}^1
-stencil(a,1) * * u_{(i+a,j+1}
+$\sum_{a,b=-1}^(1,0) stencil(a,b) * u_{(i+a,j+b} = -f_k h^2 - \sum_{a=-1}^1
+stencil(a,1) * u_{(i+a,j+1}$
 
 which is then solved using Ginkgo's implementation of the CG method
 preconditioned with block-Jacobi. It is also possible to specify on which
 executor Ginkgo will solve the system via the command line.
-The function `f` is set to `f(x,y) = 6x + 6y` (making the solution `u(x,y) = x^3
-+ y^3`), but that can be changed in the `main` function. Also the stencil values
+The function `f` is set to $f(x,y) = 6x + 6y$ (making the solution $u(x,y) = x^3
++ y^3$), but that can be changed in the `main` function. Also the stencil values
 for the core, the faces, the edge and the corners can be changed when passing
 additional parameters.
 

examples/nine-pt-stencil-solver/nine-pt-stencil-solver.cpp

@@ -55,7 +55,7 @@ On any node, where neighbor is on the border, the neighbor is replaced with a
 example a node with a neighborhood of only edge nodes may look like this
 
 \sum_{a,b=-1}^(1,0) stencil(a,b) * u_{(i+a,j+b} = -f_k h^2 - \sum_{a=-1}^1
-stencil(a,1) * * u_{(i+a,j+1}
+stencil(a,1) * u_{(i+a,j+1}
 
 which is then solved using Ginkgo's implementation of the CG method
 preconditioned with block-Jacobi. It is also possible to specify on which

examples/twentyseven-pt-stencil-solver/doc/intro.dox

@@ -3,37 +3,37 @@
 This example solves a 3D Poisson equation:
 
 $
-    \Omega = (0,1)^3
-    \Omega_b = [0,1]^3 (with boundary)
-    \partial\Omega = \Omega_b \backslash \Omega
-    u : \Omega \rightarrow R\\
-    u'' = f in \Omega\\
-    u = u_D on \partial\Omega
+    \Omega = (0,1)^3 \\
+    \Omega_b = [0,1]^3 \text{ (with boundary)} \\
+    \partial\Omega = \Omega_b \backslash \Omega \\
+    u : \Omega \rightarrow R \\
+    u'' = f \in \Omega \\
+    u = u_D \in \partial\Omega \\
 $
 
 using a finite difference method on an equidistant grid with `K` discretization
 points (`K` can be controlled with a command line parameter). The discretization
 may be done by any order Taylor polynomial.
-For an equidistant grid with K "inner" discretization points (x1,y1,z1), ...,
-(xk,y1,z1),(x1,y2,z1), ..., (xk,yk,z1), (x1,y1,z2), ..., (xk,yk,zk), step size h
-= 1 / (K + 1) and a stencil \in \R^{3 x 3 x 3}, the formula produces a system of
+For an equidistant grid with K "inner" discretization points $(x1,y1,z1), ...,
+(xk,y1,z1),(x1,y2,z1), ..., (xk,yk,z1), (x1,y1,z2), ..., (xk,yk,zk)$, step size $h
+= 1 / (K + 1)$ and a stencil $\in \R^{3 x 3 x 3}$, the formula produces a system of
 linear equations
 
-\sum_{a,b,c=-1}^1 stencil(a,b,c) * u_{(i+a,j+b,k+c} = -f_k h^2,  on any inner
+$\sum_{a,b,c=-1}^1 stencil(a,b,c) * u_{(i+a,j+b,k+c} = -f_k h^2$,  on any inner
 node with a neighborhood of inner nodes
 
 On any node, where neighbor is on the border, the neighbor is replaced with a
-'-stencil(a,b,c) * u_{i+a,j+b,k+c}' and added to the right hand side vector.
+$-stencil(a,b,c) * u_{i+a,j+b,k+c}$ and added to the right hand side vector.
 For example a node with a neighborhood of only face nodes may look like this
 
-\sum_{a,b,c=-1}^(1,1,0) stencil(a,b,c) * u_{(i+a,j+b,k+c} = -f_k h^2 -
-\sum_{a,b=-1}^(1,1) stencil(a,b,1) * * u_{(i+a,j+b,k+1}
+$\sum_{a,b,c=-1}^(1,1,0) stencil(a,b,c) * u_{(i+a,j+b,k+c} = -f_k h^2 -
+\sum_{a,b=-1}^(1,1) stencil(a,b,1) * u_{(i+a,j+b,k+1}$
 
 which is then solved using Ginkgo's implementation of the CG method
 preconditioned with block-Jacobi. It is also possible to specify on which
 executor Ginkgo will solve the system via the command line.
-The function `f` is set to `f(x,y,z) = 6x + 6y + 6z` (making the solution
-`u(x,y,z) = x^3 + y^3 + z^3`), but that can be changed in the `main` function.
+The function `f` is set to $f(x,y,z) = 6x + 6y + 6z$ (making the solution
+$u(x,y,z) = x^3 + y^3 + z^3$), but that can be changed in the `main` function.
 Also the stencil values for the core, the faces, the edge and the corners can be
 changed when passing additional parameters.
 

examples/twentyseven-pt-stencil-solver/twentyseven-pt-stencil-solver.cpp

@@ -56,7 +56,7 @@ On any node, where neighbor is on the border, the neighbor is replaced with a
 For example a node with a neighborhood of only face nodes may look like this
 
 \sum_{a,b,c=-1}^(1,1,0) stencil(a,b,c) * u_{(i+a,j+b,k+c} = -f_k h^2 -
-\sum_{a,b=-1}^(1,1) stencil(a,b,1) * * u_{(i+a,j+b,k+1}
+\sum_{a,b=-1}^(1,1) stencil(a,b,1) * u_{(i+a,j+b,k+1}
 
 which is then solved using Ginkgo's implementation of the CG method
 preconditioned with block-Jacobi. It is also possible to specify on which