updates

Author: sveinlin

doc/.src/book/.dict4spell.txt

@@ -47,6 +47,7 @@ Yapi
 discretization
 eco
 fdm
+hplgit
 io
 refactor
 reide

doc/.src/chapters/advec/advec.do.txt

@@ -117,7 +117,7 @@ centered differences in space:
 [D_t^+ u + vD_{2x} u = 0]^n_i
 \end{equation}
 !et
-Written out, we see that this expression reads
+Written out, we see that this expression implies that
 
 !bt
 \[ u^{n+1} = u^n - \half C (u^n_{i+1}-u_{i-1}^n),\]
@@ -265,7 +265,7 @@ which results in the updating formula
 !et
 A special scheme is needed to compute $u^1$, but we leave that problem for
 now. Anyway, this special scheme can be found in 
-"`advec1D.py`": "${src_advec}/advec/advec1D.py".
+"`advec1D.py`": "$https://github.com/hplgit/fdm-book/blob/master/src/advec/advec1D.py".
 
 === Implementation ===
 
@@ -424,12 +424,12 @@ error.
 
 FIGURE: [fig-advec/gaussian_UP_C08, width=800 frac=1] Advection of a Gaussian function with a forward in time, upwind in space scheme and $C=0.8$, $\Delta t = 0.01$ (left) and $\Delta t=0.001$ (right). label{advec:1D:UP:fig1:C08}
 
-Advection of the Gaussian function with a with a forward in time, upwind in space scheme, using $C=0.8$ and $\Delta t = 0.01$ can be seen in a "movie file":"${docraw}/mov-advec/gaussian/UP/C08_dt001/movie.ogg". Alternatively, with $\Delta t = 0.005$, we get this "movie file":
+Advection of the Gaussian function with a forward in time, upwind in space scheme, using $C=0.8$ and $\Delta t = 0.01$ can be seen in a "movie file":"${docraw}/mov-advec/gaussian/UP/C08_dt001/movie.ogg". Alternatively, with $\Delta t = 0.005$, we get this "movie file":
 "${docraw}/mov-advec/gaussian/UP/C08_dt0005/movie.ogg".
 
 FIGURE: [fig-advec/cosinehat_UP_08, width=800 frac=1] Advection of half a cosine function with a forward in time, upwind in space scheme and $C=0.8$, $\Delta t = 0.001$ (left) and $\Delta t=0.01$ (right). label{advec:1D:UP:fig2:C08}
 
-Advection of the cosine hat function with a with a forward in time, upwind in space scheme, using $C=0.8$ and $\Delta t = 0.01$ can be seen in a "movie file":"${docraw}/mov-advec/cosinehat/UP/C08_dt01.ogg". Alternatively, with $\Delta t = 0.001$, we get this "movie file":
+Advection of the cosine hat function with a forward in time, upwind in space scheme, using $C=0.8$ and $\Delta t = 0.01$ can be seen in a "movie file":"${docraw}/mov-advec/cosinehat/UP/C08_dt01.ogg". Alternatively, with $\Delta t = 0.001$, we get this "movie file":
 "${docraw}/mov-advec/cosinehat/UP/C08_dt001.ogg".
 
 The amplification factor can be computed using the
@@ -447,7 +447,7 @@ For $C<1$ there is, unfortunately,
 non-physical damping of discrete Fourier components, giving rise to reduced
 amplitude of $u^n_i$ as in Figures ref{advec:1D:UP:fig1:C08}
 and ref{advec:1D:UP:fig2:C08}. The damping seen
-in this figure is quite severe. Stability requires $C\leq 1$.
+in these figures is quite severe. Stability requires $C\leq 1$.
 
 !bnotice Interpretation of upwind difference as artificial diffusion
 One can interpret the upwind difference as extra, artificial diffusion
@@ -724,7 +724,7 @@ def run(scheme='UP', case='gaussian', C=1, dt=0.01):
     print 'Integral of u:', integral.max(), integral.min()
 !ec
 The complete code is found in the file
-"`advec1D.py`": "${src_advec}/advec/advec1D.py".
+"`advec1D.py`": "$https://github.com/hplgit/fdm-book/blob/master/src/advec/advec1D.py".
 
 ===== A Crank-Nicolson discretization in time and centered differences in space =====
 label{advec:1D:CN}
@@ -882,13 +882,8 @@ This means that $|A|=1$ and also that we have an exact solution if $C=1$!
 ===== Analysis of dispersion relations =====
 label{advec:1D:disprel}
 
-idx{`dispersion_analysis.py`}
-
 We have developed expressions for $A(C,p)$ in the exact solution
-$u_q^n=A^ne^{ikq\Delta x}$ of the discrete equations. These
-expressions are valuable for investigating the quality of the numerical
-solutions, see the file
-"`dispersion_analysis.py`": "${src_advec}/advec/dispersion_analysis.py".
+$u_q^n=A^ne^{ikq\Delta x}$ of the discrete equations. 
 Note that the Fourier component that solves the original
 PDE problem has no damping and moves with constant velocity $v$. There
 are two basic errors in the numerical Fourier component: there may be

doc/.src/chapters/diffu/diffu_analysis.do.txt

@@ -659,8 +659,8 @@ which has roots
 !bt
 \[ A = -2F\sin^2 p \pm \sqrt{4F^2\sin^4 p + 1}\tp\]
 !et
-Both roots have $|A|>1$ so the always amplitude grows, which is not in
-accordance with physics of the problem.
+Both roots have $|A|>1$ so the amplitude always grows, which is not in
+accordance with the physics of the problem.
 However, for a PDE with a first-order derivative in space, instead of
 a second-order one, the Leapfrog scheme performs very well.
 % if BOOK == 'book':

doc/.src/chapters/diffu/diffu_fd1.do.txt

@@ -1009,7 +1009,7 @@ and collecting all unknown terms on the left-hand side, we get
 \begin{align}
 u^{n+1}_i - \half F(u^{n+1}_{i-1} - 2u^{n+1}_i + u^{n+1}_{i+1})
 &= u^{n}_i + \half F(u^{n}_{i-1} - 2u^{n}_i + u^{n}_{i+1})\nonumber\\
-&\qquad \half f_i^{n+1} + \half f_i^n\tp
+&\qquad + \half f_i^{n+1} + \half f_i^n\tp
 \end{align}
 !et
 

doc/.src/chapters/diffu/diffu_fd2.do.txt

@@ -344,7 +344,7 @@ coordinates, which with axi-symmetry takes the form
 
 Let us assume that $u$ does not change along the tube axis so it
 suffices to compute variations in a cross section. Then $\partial u/\partial
-z = 0$ and the we have a 1D diffusion equation in the radial coordinate
+z = 0$ and we have a 1D diffusion equation in the radial coordinate
 $r$ and time $t$. In particular, we shall address the initial-boundary
 value problem
 
@@ -509,7 +509,7 @@ The associated discrete form is then
 
 !bt
 \begin{equation}
-[D_t u = \half (\gamma+1)\dfc([D_rD_r \overline{u}^t + \overline{f}^t]^n_i,
+[D_t u = \half (\gamma+1)\dfc D_rD_r \overline{u}^t + \overline{f}^t]^{n+\frac{1}{2}}_i,
 \end{equation}
 !et
 for a Crank-Nicolson scheme.

doc/.src/chapters/diffu/diffu_rw.do.txt

@@ -124,7 +124,7 @@ $\bar x_{2,k}$, and so on. The empirical estimate of $\E{\bar X_k}$ is the
 average,
 
 !bt
-\[ \E{\bar X_k} \approx = \frac{1}{W}\sum_{j=0}^{W-1} \bar x_{j,k},\]
+\[ \E{\bar X_k} \approx \frac{1}{W}\sum_{j=0}^{W-1} \bar x_{j,k},\]
 !et
 while an empirical estimate of $\Var{\bar X_k}$ is
 
@@ -255,7 +255,7 @@ idx{verification}
 
 When we have a scalar and a vectorized code, it is always a good idea to
 develop a unit test for checking that they produce the same result.
-A problem in the present context is that the two versions apply to different
+A problem in the present context is that the two versions apply two different
 random number generators. For a test to be meaningful, we need to fix
 the seed and use the same generator. This means that the scalar version
 must either use `np.random` or have this as an option. An option

doc/.src/chapters/nonlin/nonlin_ode.do.txt

@@ -230,7 +230,7 @@ while $u^{(1)}$ is the value of the unknown at the previous time level
 (in general, $u^{(\ell)}$ is the value of the unknown $\ell$ levels
 back in time). The notation will be frequently used in later
 sections. What is meant by $u$ should be evident from the context: $u$
-may be 1) the exact solution of the ODE/PDE problem,
+may either be 1) the exact solution of the ODE/PDE problem,
 2) the numerical approximation to the exact solution, or 3) the unknown
 solution at a certain time level.
 
@@ -291,7 +291,7 @@ $u\rightarrow u^{(1)}$, which is the expected result.
 For those who are not well experienced with approximating mathematical
 formulas by series expansion, an alternative method of investigation
 is simply to compute the limits of the two roots as $\Delta t\rightarrow 0$
-and see if a limit unreasonable:
+and see if a limit appears unreasonable:
 
 !bc pyshell
 >>> print r1.limit(dt, 0)
@@ -748,8 +748,8 @@ the Picard iterations, that is visibly much larger than the
 time discretization error due to a large $\Delta t$. This is illustrated
 by comparing the upper two plots in
 Figure ref{nonlin:timediscrete:logistic:impl:fig:u}. The one to
-the right has a stricter tolerance $\epsilon = 10^{-3}$, which leads
-to all the curves corresponding to Picard and Newton iteration to be
+the right has a stricter tolerance $\epsilon = 10^{-3}$, which causes
+all the curves corresponding to Picard and Newton iteration to be
 on top of each other (and no changes can be visually observed by
 reducing $\epsilon_r$ further). The reason why Newton's method does
 much better than Picard iteration in the upper left plot is that

doc/.src/chapters/nonlin/nonlin_pde1D.do.txt

@@ -639,7 +639,7 @@ With Picard iteration we get
 
 !bt
 \begin{align*}
-\frac{1}{2\Delta x^2}(& -(\dfc(u^-_{-1}) + 2\dfc(u^-_{0}
+\frac{1}{2\Delta x^2}(& -(\dfc(u^-_{-1}) + 2\dfc(u^-_{0})
 + \dfc(u^-_{1}))u_1\, +\\
 &(\dfc(u^-_{-1}) + 2\dfc(u^-_{0}) + \dfc(u^-_{1}))u_0
  + au_0\\
@@ -672,7 +672,7 @@ occurrences of $u_2$ by $D$:
 !bt
 \begin{align*}
 \frac{1}{2\Delta x^2}(&-(\dfc(u^-_{0}) + \dfc(u^-_{1}))u_{0}\, +\\
-& (\dfc(u^-_{0}) + 2\dfc(u^-_{1}) + \dfc(D))u_1 + au_1\\
+& (\dfc(u^-_{0}) + 2\dfc(u^-_{1}) + \dfc(D)))u_1 + au_1\\
 &=f(u^-_1) + \frac{1}{2\Delta x^2}(\dfc(u^-_{1}) + \dfc(D))D\tp
 \end{align*}
 !et

doc/.src/chapters/softeng2/softeng2.do.txt

@@ -788,7 +788,7 @@ for n in It[1:-1]:                  # time loop
 === Efficiency ===
 
 For a mesh consisting of $120\times 120$ cells, the scalar Python code
-require 1370 CPU time units, the vectorized version requires 5.5,
+requires 1370 CPU time units, the vectorized version requires 5.5,
 while the Cython version requires only 1! For a smaller mesh with
 $60\times 60$ cells Cython is about 1000 times faster than the scalar
 Python code, and the vectorized version is about 6 times slower than

doc/.src/chapters/trunc/trunc.do.txt

@@ -1367,7 +1367,7 @@ a slightly smaller $\omega$ since the numerics will make it larger:
 # Ref to exercise
 
 !bt
-\[ [ u'' + (\omega(1 - \frac{1}{24}\omega^2\Delta t^2))^2 u = 0\tp\]
+\[ [ u'' + (\omega(1 - \frac{1}{24}\omega^2\Delta t^2))^2 u ]^n = 0\tp\]
 !et
 Expanding the squared term and omitting the higher-order term $\Delta t^4$
 gives exactly the ODE (ref{trunc:vib:undamped:corr:ode}). Experiments
@@ -2042,7 +2042,7 @@ and (ref{trunc:table:avg:arith:eq})-(ref{trunc:table:avg:arith}) we have
 &= \lambda(x_{i+\half})\uexd{x}(x_{i+\half},t_n)
 + \lambda(x_{i+\half})
 \frac{1}{24}\uexd{xxx}(x_{i+\half},t_n)\Delta x^2 + \\
-&\quad \uexd{x}(x_{i+\half})
+&\quad \uexd{x}(x_{i+\half},t_n)
 \frac{1}{8}\lambda''(x_{i+\half})\Delta x^2
 +\Oof{\Delta x^4}\\
 &= [\lambda \uexd{x}]^n_{i+\half} + G^n_{i+\half}\Delta x^2
@@ -2053,9 +2053,9 @@ where we have introduced the short form
 
 !bt
 \[ G^n_{i+\half} =
-(\frac{1}{24}\uexd{xxx}(x_{i+\half},t_n)\lambda((x_{i+\half})
+\frac{1}{24}\uexd{xxx}(x_{i+\half},t_n)\lambda(x_{i+\half})
 + \uexd{x}(x_{i+\half},t_n)
-\frac{1}{8}\lambda''(x_{i+\half}))\Delta x^2\tp\]
+\frac{1}{8}\lambda''(x_{i+\half})\tp\]
 !et
 Similarly, we find that
 
@@ -2276,7 +2276,7 @@ the $u_{xx}$ term:
 
 !bt
 \[ [D_t u]^{n+\half}_i = \dfc\half([D_xD_x u]^n_i +
-[D_xD_x u]^{n+1}_i + f^{n+\half}_i\tp\]
+[D_xD_x u]^{n+1}_i) + f^{n+\half}_i\tp\]
 !et
 The equation for the truncation error is