fix github rendering problem for periodic bc readme

Author: tianjuxue

applications/periodic_bc/README.md

@@ -26,29 +26,29 @@ filters such as `Warp By Scalar`.
 
 The example solves
 
-$$
+```math
 -\nabla\cdot(\theta\nabla u)=f
 \quad\mathrm{in}\quad \Omega=(0,1)^2,
-$$
+```
 
 with homogeneous Dirichlet boundary conditions on the bottom and top edges,
 
-$$
+```math
 u(x,0)=u(x,1)=0,
-$$
+```
 
 and periodicity between the left and right edges,
 
-$$
+```math
 u(0,y)=u(1,y).
-$$
+```
 
 The source term is
 
-$$
+```math
 f(x,y)=x\sin(5\pi y)
 +\exp\left(-\frac{(x-0.5)^2+(y-0.5)^2}{0.02}\right).
-$$
+```
 
 The material parameter `theta` is treated as the differentiable scalar
 parameter in the AD check.
@@ -59,10 +59,10 @@ Let `u_full` be the full vector of nodal degrees of freedom and `u_red` be the
 reduced vector after eliminating duplicate periodic degrees of freedom. The
 periodic constraint is represented as
 
-$$
+```math
 \boldsymbol{u}_{\mathrm{full}}
 =\boldsymbol{P}\boldsymbol{u}_{\mathrm{red}}.
-$$
+```
 
 Here `P_mat` has shape `(N, M)`, where `N` is the number of full DOFs before
 applying periodic constraints and `M` is the number of independent DOFs after
@@ -75,55 +75,55 @@ or adjoint loads back to the independent reduced DOFs.
 For this example, every right-boundary node is matched to the corresponding
 left-boundary node shifted by
 
-$$
+```math
 (x,y)\mapsto(x+1,y).
-$$
+```
 
 `periodic_boundary_conditions` constructs `P_mat` by assigning the slave
 right-boundary DOF to the same reduced column as its master left-boundary DOF.
 The nonlinear residual and tangent are then assembled in reduced coordinates:
 
-$$
+```math
 \boldsymbol{R}_{\mathrm{red}}
 =\boldsymbol{P}^{T}\boldsymbol{R}_{\mathrm{full}},
 \qquad
 \boldsymbol{K}_{\mathrm{red}}
 =\boldsymbol{P}^{T}\boldsymbol{K}_{\mathrm{full}}\boldsymbol{P}.
-$$
+```
 
 Dirichlet rows are eliminated before applying the periodic projection.
 
 ## Toy problem
 
 Consider a tiny non-FEM system with four full unknowns,
 
-$$
+```math
 \boldsymbol{u}_{\mathrm{full}}
 =
 \begin{bmatrix}
 u_0 & u_1 & u_2 & u_3
 \end{bmatrix}^{T},
-$$
+```
 
 and impose one periodic constraint:
 
-$$
+```math
 u_3=u_0.
-$$
+```
 
 Only three independent unknowns remain:
 
-$$
+```math
 \boldsymbol{u}_{\mathrm{red}}
 =
 \begin{bmatrix}
 a & b & c
 \end{bmatrix}^{T}.
-$$
+```
 
 The expansion matrix is
 
-$$
+```math
 \boldsymbol{P}
 =
 \begin{bmatrix}
@@ -140,14 +140,14 @@ $$
 \begin{bmatrix}
 a\\ b\\ c\\ a
 \end{bmatrix}.
-$$
+```
 
 Thus `N = 4`, `M = 3`, and the slave DOF `u_3` reuses the same reduced column as
 the master DOF `u_0`.
 
 Suppose the full residual is
 
-$$
+```math
 \boldsymbol{R}_{\mathrm{full}}
 (\boldsymbol{u},\theta)
 =
@@ -168,11 +168,11 @@ $$
 \begin{bmatrix}
 1\\0\\0\\1
 \end{bmatrix}.
-$$
+```
 
 The reduced residual is
 
-$$
+```math
 \boldsymbol{R}_{\mathrm{red}}
 =
 \boldsymbol{P}^{T}
@@ -181,11 +181,11 @@ $$
 =
 \boldsymbol{K}_{\mathrm{red}}\boldsymbol{u}_{\mathrm{red}}
 -\theta\boldsymbol{f}_{\mathrm{red}},
-$$
+```
 
 where
 
-$$
+```math
 \boldsymbol{K}_{\mathrm{red}}
 =
 \boldsymbol{P}^{T}\boldsymbol{K}_{\mathrm{full}}\boldsymbol{P}
@@ -203,40 +203,40 @@ $$
 \begin{bmatrix}
 2\\0\\0
 \end{bmatrix}.
-$$
+```
 
 The first reduced residual component contains the sum of the master and slave
 full residual entries. For example, if
 
-$$
+```math
 \boldsymbol{R}_{\mathrm{full}}
 =
 \begin{bmatrix}
 R_0\\R_1\\R_2\\R_3
 \end{bmatrix},
-$$
+```
 
 then
 
-$$
+```math
 \boldsymbol{P}^{T}\boldsymbol{R}_{\mathrm{full}}
 =
 \begin{bmatrix}
 R_0+R_3\\R_1\\R_2
 \end{bmatrix}.
-$$
+```
 
 At `theta = 1`, solving
 
-$$
+```math
 \boldsymbol{K}_{\mathrm{red}}\boldsymbol{u}_{\mathrm{red}}
 =
 \boldsymbol{f}_{\mathrm{red}}
-$$
+```
 
 gives
 
-$$
+```math
 \boldsymbol{u}_{\mathrm{red}}
 =
 \begin{bmatrix}
@@ -250,39 +250,39 @@ $$
 \begin{bmatrix}
 1\\1\\1\\1
 \end{bmatrix}.
-$$
+```
 
 ## Sensitivity in reduced coordinates
 
 Let $\hat{\boldsymbol{u}}\in\mathbb{R}^{M}$ denote the independent reduced DOFs
 and $\boldsymbol{u}\in\mathbb{R}^{N}$ denote the full DOFs. The forward solve
 can be viewed as
 
-$$
+```math
 \boldsymbol{u}
 =
 \boldsymbol{P}\hat{\boldsymbol{u}},
 \qquad
 \boldsymbol{P}\in\mathbb{R}^{N\times M},
 \qquad
 N\ge M.
-$$
+```
 
 After applying periodic constraints, the full residual equation
 $\boldsymbol{R}(\boldsymbol{u},\theta)=\boldsymbol{0}$ becomes the reduced
 equation
 
-$$
+```math
 \boldsymbol{P}^{T}
 \boldsymbol{R}
 \left(\boldsymbol{P}\hat{\boldsymbol{u}},\theta\right)
 =
 \boldsymbol{0}.
-$$
+```
 
 Differentiating with respect to $\theta$ gives
 
-$$
+```math
 \boldsymbol{P}^{T}
 \frac{\partial \boldsymbol{R}}{\partial \boldsymbol{u}}
 \boldsymbol{P}
@@ -292,11 +292,11 @@ $$
 \frac{\partial \boldsymbol{R}}{\partial \theta}
 =
 \boldsymbol{0}.
-$$
+```
 
 Therefore
 
-$$
+```math
 \frac{d\hat{\boldsymbol{u}}}{d\theta}
 =
 -
@@ -307,12 +307,12 @@ $$
 \right)^{-1}
 \boldsymbol{P}^{T}
 \frac{\partial \boldsymbol{R}}{\partial \theta}.
-$$
+```
 
 For an objective $J(\boldsymbol{u},\theta)$, write gradients of $J$ as row
 vectors. The reduced and full gradients are related by
 
-$$
+```math
 \left(
 \frac{\partial J}{\partial \hat{\boldsymbol{u}}}
 \right)^{T}
@@ -321,12 +321,12 @@ $$
 \left(
 \frac{\partial J}{\partial \boldsymbol{u}}
 \right)^{T}.
-$$
+```
 
 The adjoint variable $\boldsymbol{\lambda}\in\mathbb{R}^{M}$ is defined in the
 reduced space:
 
-$$
+```math
 \left(
 \boldsymbol{P}^{T}
 \frac{\partial \boldsymbol{R}}{\partial \boldsymbol{u}}
@@ -337,22 +337,22 @@ $$
 \left(
 \frac{\partial J}{\partial \hat{\boldsymbol{u}}}
 \right)^{T}.
-$$
+```
 
 Start from the chain rule in reduced coordinates:
 
-$$
+```math
 \frac{dJ}{d\theta}
 =
 \frac{\partial J}{\partial \theta}
 +
 \frac{\partial J}{\partial \hat{\boldsymbol{u}}}
 \frac{d\hat{\boldsymbol{u}}}{d\theta}.
-$$
+```
 
 Using the adjoint definition,
 
-$$
+```math
 \frac{\partial J}{\partial \hat{\boldsymbol{u}}}
 =
 \boldsymbol{\lambda}^{T}
@@ -361,11 +361,11 @@ $$
 \frac{\partial \boldsymbol{R}}{\partial \boldsymbol{u}}
 \boldsymbol{P}
 \right),
-$$
+```
 
 so
 
-$$
+```math
 \frac{dJ}{d\theta}
 =
 \frac{\partial J}{\partial \theta}
@@ -377,11 +377,11 @@ $$
 \boldsymbol{P}
 \right)
 \frac{d\hat{\boldsymbol{u}}}{d\theta}.
-$$
+```
 
 The differentiated reduced residual gives
 
-$$
+```math
 \left(
 \boldsymbol{P}^{T}
 \frac{\partial \boldsymbol{R}}{\partial \boldsymbol{u}}
@@ -392,19 +392,19 @@ $$
 -
 \boldsymbol{P}^{T}
 \frac{\partial \boldsymbol{R}}{\partial \theta}.
-$$
+```
 
 Substituting this into the chain rule gives
 
-$$
+```math
 \frac{dJ}{d\theta}
 =
 \frac{\partial J}{\partial \theta}
 -
 \boldsymbol{\lambda}^{T}
 \boldsymbol{P}^{T}
 \frac{\partial \boldsymbol{R}}{\partial \theta}.
-$$
+```
 
 The important point is that both the tangent solve and the adjoint solve live
 in the reduced space. The full residual and full tangent are evaluated at
@@ -423,9 +423,9 @@ The script generates a Gmsh mesh, solves the forward problem, writes the VTU
 solution, and prints a finite-difference versus automatic-differentiation check
 for
 
-$$
+```math
 J(\theta)=\sum_i u_i(\theta)^2.
-$$
+```
 
 For derivative verification, prefer a direct linear solve such as
 `spsolve_solver`, or PETSc with `ksp_type="preonly"` and `pc_type="lu"`.