Update TeX formatting

Fix up whitespace and delimiters in both source and render

Author: abhro

lib/BVProblemLibrary/src/linear.jl

@@ -36,7 +36,7 @@ Linear boundary value problem with analytical solution, given by
 with boundary condition
 
 ```math
-y_1(0)=1, y_1(1)=0
+y_1(0)=1, \;\; y_1(1)=0
 ```
 
 # Solution
@@ -94,7 +94,7 @@ Linear boundary value problem with analytical solution, given by
 with boundary condition
 
 ```math
-y_1(0)=1, y_1(1)=0
+y_1(0)=1, \;\; y_1(1)=0
 ```
 
 # Solution
@@ -155,13 +155,13 @@ Linear boundary value problem with analytical solution, given by
 where
 
 ```math
-f(t, y_1, y_2) = -(2+\cos(\pi t))y_2 + y_1 -(1+\lambda \pi^2)\cos(\pi t) - (2+\cos(\pi t))\pi\sin(\pi t)
+f(t, y_1, y_2) = -[2+\cos(\pi t)]y_2 + y_1 -(1+\lambda \pi^2)\cos(\pi t) - [2+\cos(\pi t)]\pi\sin(\pi t)
 ```
 
 with boundary condition
 
 ```math
-y_1(-1)=-1, y_1(1)=-1
+y_1(-1)=-1,\;\; y_1(1)=-1
 ```
 
 # Solution
@@ -219,7 +219,7 @@ Linear boundary value problem with analytical solution, given by
 where
 
 ```math
-f(y_1, y_2)=-y2+(1+\lambda)y1
+f(y_1, y_2) = -y_2 + (1+\lambda) y_1
 ```
 
 with boundary condition
@@ -288,13 +288,13 @@ Linear boundary value problem with analytical solution, given by
 where
 
 ```math
-f(t, y_1, y_2)=ty_2+y_1-(1+\lambda\pi^2)\cos(\pi t)+\pi t\sin(\pi t)
+f(t, y_1, y_2) = ty_2 + y_1 - (1+\lambda\pi^2)\cos(\pi t) + \pi t\sin(\pi t)
 ```
 
 with boundary condition
 
 ```math
-y_1(-1)=-1, y_1(1)=-1
+y_1(-1)=-1, \;\; y_1(1)=-1
 ```
 
 # Solution
@@ -353,13 +353,13 @@ Linear boundary value problem with analytical solution, given by
 where
 
 ```math
-f(t, y_2)=ty_2 - \lambda\pi^2\cos(\pi t)-\pi t\sin(\pi t)
+f(t, y_2) = ty_2 - \lambda\pi^2\cos(\pi t)-\pi t\sin(\pi t)
 ```
 
 with boundary condition
 
 ```math
-y_1(-1)=-2, y_1(1)=0
+y_1(-1)=-2, \;\; y_1(1)=0
 ```
 
 # Solution
@@ -425,13 +425,13 @@ Linear boundary value problem with analytical solution, given by
 where
 
 ```math
-f(t, y_1, y_2)=ty_2+y_1-(1+\lambda\pi^2)\cos(\pi t)+\pi t\sin(\pi t)
+f(t, y_1, y_2) = t y_2 + y_1 - (1+\lambda\pi^2)\cos(\pi t) + \pi t\sin(\pi t)
 ```
 
 with boundary condition
 
 ```math
-y_1(-1)=-1, y_1(1)=1
+y_1(-1)=-1, \;\; y_1(1)=1
 ```
 
 # Solution
@@ -489,7 +489,7 @@ Linear boundary value problem with analytical solution, given by
 with boundary condition
 
 ```math
-y_1(0)=1, y_1(1)=2
+y_1(0)=1, \;\; y_1(1)=2
 ```
 
 # Solution
@@ -498,7 +498,7 @@ The analytical solution for ``t \in [0, 1]`` is
 
 ```math
 \begin{align*}
-y_1(t) &= (2 - \exp(-1/\lambda) - \exp(-t/\lambda))/(1 - \exp(-1/\lambda)) \\
+y_1(t) &= \frac{2 - \exp(-1/\lambda) - \exp(-t/\lambda)}{1 - \exp(-1/\lambda)} \\
 y_2(t) &= y_1'(t)
 \end{align*}
 ```
@@ -612,7 +612,7 @@ Linear boundary value problem with analytical solution, given by
 with boundary condition
 
 ```math
-y_1(-1)=0, y_1(1)=2
+y_1(-1)=0, \;\; y_1(1)=2
 ```
 
 # Solution
@@ -670,13 +670,13 @@ Linear boundary value problem with analytical solution, given by
 where
 
 ```math
-f(t, y_1)=y_1-(1+\lambda\pi^2)\cos(\pi t)
+f(t, y_1) = y_1 - (1+\lambda\pi^2)\cos(\pi t)
 ```
 
 with boundary condition
 
 ```math
-y_1(-1)=0, y_1(1)=2
+y_1(-1)=0, \;\; y_1(1)=2
 ```
 
 # Solution
@@ -736,7 +736,7 @@ Linear boundary value problem with analytical solution, given by
 where
 
 ```math
-f(t, y_1)=y_1-(1+\lambda\pi^2)\cos(\pi t)
+f(t, y_1) = y_1 - (1+\lambda\pi^2)\cos(\pi t)
 ```
 
 with boundary condition
@@ -801,13 +801,13 @@ Linear boundary value problem with analytical solution, given by
 where
 
 ```math
-f(t, y_1)=y_1-(1+\lambda\pi^2)\cos(\pi t)
+f(t, y_1) = y_1 - (1+\lambda\pi^2)\cos(\pi t)
 ```
 
 with boundary condition
 
 ```math
-y_1(-1)=0, y_1(1)=-1
+y_1(-1)=0, \;\; y_1(1)=-1
 ```
 
 # Solution
@@ -872,7 +872,8 @@ f(t, y_1)=y_1-(1+\lambda\pi^2)\cos(\pi t)
 with boundary condition
 
 ```math
-y_1(-1)=\exp(-2/\sqrt{\lambda}, y_1(1)=\exp(-2/\sqrt{\lambda})
+y_1(-1) = \exp(-2/\sqrt{\lambda}), \;\;
+y_1(1)  = \exp(-2/\sqrt{\lambda})
 ```
 
 # Solution
@@ -927,7 +928,7 @@ Linear boundary value problem with analytical solution, given by
 with boundary condition
 
 ```math
-y_1(-1)=1, y_1(1)=1
+y_1(-1)=1, \;\; y_1(1)=1
 ```
 
 # Solution
@@ -978,13 +979,13 @@ Linear boundary value problem with analytical solution, given by
 where
 
 ```math
-f(t, y_1)=-π^2y_1/4
+f(t, y_1) = -π^2 y_1/4
 ```
 
 with boundary condition
 
 ```math
-y_1(0)=0, y_1(1)=\sin(\pi/(2*\lambda))
+y_1(0)=0, \;\; y_1(1)=\sin(\pi/(2\lambda))
 ```
 
 # Solution
@@ -1041,13 +1042,14 @@ Linear boundary value problem with analytical solution, given by
 where
 
 ```math
-f(t, y_1)=-3\lambda y_1/(\lambda+t^2)^2
+f(t, y_1) = -3\lambda y_1 / (\lambda+t^2)^2
 ```
 
 with boundary condition
 
 ```math
-y_1(-0.1)=-0.1/\sqrt{\lambda+0.01}, y_1(0.1)=0.1/\sqrt{\lambda+0.01}
+y_1(-0.1) = \frac{-0.1}{\sqrt{\lambda+0.01}},
+y_1(0.1)  = \frac{ 0.1}{\sqrt{\lambda+0.01}}
 ```
 
 # Solution
@@ -1105,7 +1107,7 @@ Linear boundary value problem with analytical solution, given by
 with boundary condition
 
 ```math
-y_1(0)=1, y_1(1)=0.1/\sqrt{\lambda+0.01}
+y_1(0)=1, \;\; y_1(1)=\frac{0.1}{\sqrt{\lambda+0.01}}
 ```
 
 # Solution

lib/BVProblemLibrary/src/nonlinear.jl

@@ -164,7 +164,7 @@ Nonlinear boundary value problem with analytical solution, given by
 where
 
 ```math
-f(y_1)=y_1+y_1^2-\exp(-2t/\sqrt{\lambda})
+f(y_1) = y_1 + y_1^2 - \exp(-2t/\sqrt{\lambda})
 ```
 
 with boundary condition
@@ -279,7 +279,7 @@ Nonlinear boundary value problem with no analytical solution, given by
 where
 
 ```math
-f(y_1)=\lambda\sinh(\lambda z)
+f(y_1) = \lambda\sinh(\lambda z)
 ```
 
 with boundary condition
@@ -339,13 +339,13 @@ The steady state Navier-Stokes equations generate a second order differential eq
 where
 
 ```math
-f(t, y_1, y_2)=(\frac{1+\gamma}{2}-\lambda A'(t))y_1y_2-\frac{y_2}{y_1}-\frac{A'(t)}{A(t)}(1-(\frac{\gamma-1}{2})y_1^2)
+f(t, y_1, y_2) = \left(\frac{1+\gamma}{2}-\lambda A'(t)\right) y_1y_2 - \frac{y_2}{y_1} - \frac{A'(t)}{A(t)} \left(1 - \frac{\gamma-1}{2} y_1^2\right)
 ```
 
 with boundary condition
 
 ```math
-y_1(0)=0.9129, y_1(1)=0.375
+y_1(0)=0.9129, \;\; y_1(1)=0.375
 ```
 
 # Solution
@@ -393,13 +393,13 @@ Nonlinear boundary value problem with no analytical solution, given by
 where
 
 ```math
-f(y_1, y_2)=-y_1y_2+y_1
+f(y_1, y_2) = -y_1 y_2 + y_1
 ```
 
 with boundary condition
 
 ```math
-y_1(0)=-1/3, y_1(1)=1/3
+y_1(0)=-1/3, \;\; y_1(1)=1/3
 ```
 
 # Solution
@@ -453,7 +453,7 @@ f(y_1, y_2)=-y_1y_2+y_1
 with boundary condition
 
 ```math
-y_1(0)=1, y_1(1)=-1/3
+y_1(0)=1, \;\; y_1(1)=-1/3
 ```
 
 # Solution
@@ -561,7 +561,7 @@ f(y_1, y_2)=-y_1y_2+y_1
 with boundary condition
 
 ```math
-y_1(0)=1, y_1(1)=3/2
+y_1(0)=1, \;\; y_1(1)=3/2
 ```
 
 # Solution
@@ -615,7 +615,7 @@ f(y_1, y_2)=-y_1y_2+y_1
 with boundary condition
 
 ```math
-y_1(0)=0, y_1(1)=3/2
+y_1(0)=0, \;\; y_1(1)=3/2
 ```
 
 # Solution
@@ -663,13 +663,13 @@ Nonlinear boundary value problem with no analytical solution, given by
 where
 
 ```math
-f(y_1, y_2)=-y_1y_2+y_1
+f(y_1, y_2) = -y_1y_2+y_1
 ```
 
 with boundary condition
 
 ```math
-y_1(0)=-7/6, y_1(1)=3/2
+y_1(0)=-7/6, \;\; y_1(1)=3/2
 ```
 
 # Solution
@@ -725,13 +725,14 @@ Nonlinear boundary value problem with no analytical solution, given by
 where
 
 ```math
-f(z, \theta, M, Q)=\frac{1}{\lambda}((z-1)\cos\theta-M\sec\theta)+\lambda Q\tan\theta
+f(z, \theta, M, Q) = \frac{1}{\lambda} [(z-1)\cos\theta-M\sec\theta] + \lambda Q\tan\theta
 ```
 
 with boundary condition
 
 ```math
-y_1(0)=0, y_3(0)=0, y_1(1)=0, y_3(1)=0
+y_1(0)=0, \;\; y_3(0)=0, \;\;
+y_1(1)=0, \;\; y_3(1)=0
 ```
 
 # Solution
@@ -787,7 +788,8 @@ Nonlinear boundary value problem with no analytical solution, given by
 with boundary condition
 
 ```math
-y_1(0)=0, y_2(0)=0, y_1(1)=1, y_2(1)=0
+y_1(0)=0, \;\; y_2(0)=0, \;\;
+y_1(1)=1, \;\; y_2(1)=0
 ```
 
 # Solution

lib/DAEProblemLibrary/src/DAEProblemLibrary.jl

@@ -69,32 +69,34 @@ du0 = [
 @doc doc"""
 The Transistor Amplifier model
 
-M\frac{dy}{dt}=f(t,y),\quad y(0)=y_0,\quad y'(0)=y_0'
+```math
+M\frac{dy}{dt} = f(t,y), \quad y(0)=y_0,\quad y'(0)=y_0'
+```
 
 ```math
-M=\left(\begin{array}{cccccccc}
--C_{1} & C_{1} & 0 & 0 & 0 & 0 & 0 & 0 \\
-C_{1} & -C_{1} & 0 & 0 & 0 & 0 & 0 & 0 \\
-0 & 0 & -C_{2} & 0 & 0 & 0 & 0 & 0 \\
-0 & 0 & 0 & -C_{3} & C_{3} & 0 & 0 & 0 \\
-0 & 0 & 0 & C_{3} & -C_{3} & 0 & 0 & 0 \\
-0 & 0 & 0 & 0 & 0 & -C_{4} & 0 & 0 \\
-0 & 0 & 0 & 0 & 0 & 0 & -C_{5} & C_{5} \\
-0 & 0 & 0 & 0 & 0 & 0 & C_{5} & -C_{5}
-\end{array}\right)
+M = \begin{pmatrix}
+-C_1 &  C_1 &    0 &    0 &    0 &    0 &    0 &    0 \\
+ C_1 & -C_1 &    0 &    0 &    0 &    0 &    0 &    0 \\
+   0 &    0 & -C_2 &    0 &    0 &    0 &    0 &    0 \\
+   0 &    0 &    0 & -C_3 &  C_3 &    0 &    0 &    0 \\
+   0 &    0 &    0 &  C_3 & -C_3 &    0 &    0 &    0 \\
+   0 &    0 &    0 &    0 &    0 & -C_4 &    0 &    0 \\
+   0 &    0 &    0 &    0 &    0 &    0 & -C_5 &  C_5 \\
+   0 &    0 &    0 &    0 &    0 &    0 &  C_5 & -C_5
+\end{pmatrix}
 ```
 
 ```math
-f(t, y)=\left(\begin{array}{c}
--\frac{U_{e}(t)}{R_{0}}+\frac{y_{1}}{R_{0}} \\
--\frac{U_{b}}{R_{2}}+y_{2}\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)-(\alpha-1) g\left(y_{2}-y_{3}\right) \\
--g\left(y_{2}-y_{3}\right)+\frac{y_{3}}{R_{3}} \\
--\frac{U_{b}}{R_{4}}+\frac{y_{4}}{R_{4}}+\alpha g\left(y_{2}-y_{3}\right) \\
--\frac{U_{b}}{R_{6}}+y_{5}\left(\frac{1}{R_{5}}+\frac{1}{R_{6}}\right)-(\alpha-1) g\left(y_{5}-y_{6}\right) \\
--g\left(y_{5}-y_{6}\right)+\frac{y_{6}}{R_{7}} \\
--\frac{U_{b}}{R_{8}}+\frac{y_{7}}{R_{8}}+\alpha g\left(y_{5}-y_{6}\right) \\
-\frac{y_{8}}{R_{9}}
-\end{array}\right)
+f(t, y)=\begin{pmatrix}
+-\frac{U_e(t)}{R_0} + \frac{y_1}{R_0} \\
+-\frac{U_b}{R_2} + y_2\left(\frac{1}{R_1}+\frac{1}{R_2}\right) - (\alpha-1) g\left(y_2-y_3\right) \\
+-g\left(y_2-y_3\right) + \frac{y_3}{R_3} \\
+-\frac{U_b}{R_4} + \frac{y_4}{R_4} + \alpha g\left(y_2-y_3\right) \\
+-\frac{U_b}{R_6} + y_5\left(\frac{1}{R_5}+\frac{1}{R_6}\right) - (\alpha-1) g\left(y_5-y_6\right) \\
+-g\left(y_5-y_6\right) + \frac{y_6}{R_7} \\
+-\frac{U_b}{R_8} + \frac{y_7}{R_8} + \alpha g\left(y_5-y_6\right) \\
+\frac{y_8}{R_9}
+\end{pmatrix}
 ```
 
 ## Reference