use quarto, not Pluto to render pages

quarto/.gitignore

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+/.quarto/
+/_site/
+/_book/

quarto/ODEs/Project.toml

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+[deps]
+DiffEqBase = "2b5f629d-d688-5b77-993f-72d75c75574e"
+DifferentialEquations = "0c46a032-eb83-5123-abaf-570d42b7fbaa"
+ForwardDiff = "f6369f11-7733-5829-9624-2563aa707210"
+MonteCarloMeasurements = "0987c9cc-fe09-11e8-30f0-b96dd679fdca"
+NLsolve = "2774e3e8-f4cf-5e23-947b-6d7e65073b56"
+Plots = "91a5bcdd-55d7-5caf-9e0b-520d859cae80"
+QuadGK = "1fd47b50-473d-5c70-9696-f719f8f3bcdc"
+Roots = "f2b01f46-fcfa-551c-844a-d8ac1e96c665"
+SymPy = "24249f21-da20-56a4-8eb1-6a02cf4ae2e6"

quarto/ODEs/differential_equations.qmd

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+# The `DifferentialEquations` suite
+
+
+```{julia}
+#| echo: false
+
+import Logging
+Logging.disable_logging(Logging.Info) # or e.g. Logging.Info
+Logging.disable_logging(Logging.Warn)
+
+import SymPy
+function Base.show(io::IO, ::MIME"text/html", x::T) where {T <: SymPy.SymbolicObject}
+    println(io, "<span class=\"math-left-align\" style=\"padding-left: 4px; width:0; float:left;\"> ")
+    println(io, "\\[")
+    println(io, sympy.latex(x))
+    println(io, "\\]")
+    println(io, "</span>")
+end
+
+# hack to work around issue
+import Markdown
+import CalculusWithJulia
+function CalculusWithJulia.WeaveSupport.ImageFile(d::Symbol, f::AbstractString, caption; kwargs...)
+    nm = joinpath("..", string(d), f)
+    u = "![$caption]($nm)"
+    Markdown.parse(u)
+end
+
+nothing
+```
+
+This section uses these add-on packages:
+
+
+```{julia}
+using OrdinaryDiffEq
+using Plots
+using ModelingToolkit
+```
+
+```{julia}
+#| echo: false
+#| results: "hidden"
+using CalculusWithJulia.WeaveSupport
+
+const frontmatter = (
+        title = "The `DifferentialEquations` suite",
+        description = "Calculus with Julia: The `DifferentialEquations` suite",
+        tags = ["CalculusWithJulia", "odes", "the `differentialequations` suite"],
+);
+fig_size = (800, 600)
+nothing
+```
+
+---
+
+
+The [`DifferentialEquations`](https://github.com/SciML/DifferentialEquations.jl) suite of packages contains solvers for a wide range of various differential equations. This section just briefly touches touch on ordinary differential equations (ODEs), and so relies only on `OrdinaryDiffEq` part of the suite. For more detail on this type and many others covered by the suite of packages, there are many other resources, including the [documentation](https://diffeq.sciml.ai/stable/) and accompanying [tutorials](https://github.com/SciML/SciMLTutorials.jl).
+
+
+## SIR Model
+
+
+We follow along with an introduction to the SIR model for the spread of disease by [Smith and Moore](https://www.maa.org/press/periodicals/loci/joma/the-sir-model-for-spread-of-disease-introduction). This model received a workout due to the COVID-19 pandemic.
+
+
+The basic model breaks a population into three cohorts: The **susceptible** individuals, the **infected** individuals, and the **recovered** individuals. These add to the population size, $N$, which is fixed, but the cohort sizes vary in time. We name these cohort sizes $S(t)$, $I(t)$, and $R(t)$ and define $s(t)=S(t)/N$, $i(t) = I(t)/N$ and $r(t) = R(t)/N$ to be the respective proportions.
+
+
+The following *assumptions* are made about these cohorts by Smith and Moore:
+
+
+> No one is added to the susceptible group, since we are ignoring births and immigration. The only way an individual leaves the susceptible group is by becoming infected.
+
+
+
+This implies the rate of change in time of $S(t)$ depends on the current number of susceptibles, and the amount of interaction with the infected cohorts. The model *assumes* each infected person has $b$ contacts per day that are sufficient to spread the disease. Not all contacts will be with susceptible people, but if people are assumed to  mix within the cohorts, then there will be on average $b \cdot S(t)/N$  contacts with susceptible people per infected person. As each infected person is modeled identically, the time rate of change of $S(t)$ is:
+
+
+$$
+\frac{dS}{dt} = - b \cdot \frac{S(t)}{N} \cdot I(t) = -b \cdot s(t) \cdot I(t)
+$$
+
+It is negative, as no one is added, only taken off. After dividing by $N$, this can also be expressed as $s'(t) = -b s(t) i(t)$.
+
+
+> assume that a fixed fraction  $k$  of the infected group will recover during any given day.
+
+
+
+This means the change in time of the recovered depends on $k$ and the number infected, giving rise to the equation
+
+
+$$
+\frac{dR}{dt} = k \cdot I(t)
+$$
+
+which can also be expressed in proportions as $r'(t) = k  \cdot i(t)$.
+
+
+Finally,  from $S(t) + I(T) + R(t) = N$ we have $S'(T) + I'(t) + R'(t) = 0$ or $s'(t) + i'(t) + r'(t) = 0$.
+
+
+Combining, it is possible to express the rate of change of the infected population through:
+
+
+$$
+\frac{di}{dt} = b   \cdot s(t)   \cdot i(t) - k   \cdot i(t)
+$$
+
+The author's apply this model to  flu statistics from Hong Kong where:
+
+
+$$
+\begin{align*}
+S(0) &= 7,900,000\\
+I(0) &= 10\\
+R(0) &= 0\\
+\end{align*}
+$$
+
+In `Julia` we define these, `N` to model the total population, and `u0` to be the proportions.
+
+
+```{julia}
+S0, I0, R0 = 7_900_000, 10, 0
+N = S0 + I0 + R0
+u0 = [S0, I0, R0]/N       # initial proportions
+```
+
+An *estimated* set of values for $k$ and $b$ are $k=1/3$, coming from the average period of infectiousness being estimated at three days and $b=1/2$, which seems low in normal times, but not for an infected person who may be feeling quite ill and staying at home. (The model for COVID would certainly have a larger $b$ value).
+
+
+Okay, the mathematical modeling is done; now we try to solve for the unknown functions using `DifferentialEquations`.
+
+
+To warm up, if $b=0$ then $i'(t) = -k   \cdot i(t)$ describes the infected. (There is no circulation of people in this case.) The solution would be achieved through:
+
+
+```{julia}
+#| hold: true
+k = 1/3
+
+f(u,p,t) = -k * u  # solving  u′(t) = - k u(t)
+time_span = (0.0, 20.0)
+
+prob = ODEProblem(f, I0/N, time_span)
+sol = solve(prob, Tsit5(), reltol=1e-8, abstol=1e-8)
+
+plot(sol)
+```
+
+The `sol` object is a set of numbers with a convenient `plot` method. As may have been expected, this graph shows exponential decay.
+
+
+A few comments are in order. The problem we want to solve is
+
+
+$$
+\frac{di}{dt} = -k   \cdot i(t) = F(i(t), k, t)
+$$
+
+where $F$ depends on the current value ($i$),  a parameter ($k$), and the time ($t$). We did not utilize $p$ above for the parameter, as it was easy not to, but could have, and will in the following. The time variable $t$ does not appear by itself in our equation, so only `f(u, p, t) = -k * u` was used, `u` the generic name for a solution which in this case is $i$.
+
+
+The problem we set up needs an initial value (the $u0$) and a time span to solve over. Here we want time to model real time, so use floating point values.
+
+
+The plot shows steady decay, as there is no mixing of infected with others.
+
+
+Adding in the interaction requires a bit more work. We now have what is known as a *system* of equations:
+
+
+$$
+\begin{align*}
+\frac{ds}{dt} &= -b   \cdot s(t)  \cdot  i(t)\\
+\frac{di}{dt} &= b  \cdot  s(t)  \cdot  i(t) - k  \cdot  i(t)\\
+\frac{dr}{dt} &= k  \cdot  i(t)\\
+\end{align*}
+$$
+
+Systems of equations can be solved in a similar manner as a single ordinary differential equation, though adjustments are made to accommodate the multiple functions.
+
+
+We use a style that updates values in place, and note that `u` now holds $3$ different functions at once:
+
+
+```{julia}
+function sir!(du, u, p, t)
+    k, b = p
+    s, i, r = u[1], u[2], u[3]
+
+    ds = -b * s * i
+    di =  b * s * i - k * i
+    dr =              k * i
+
+    du[1], du[2], du[3] = ds, di, dr
+end
+```
+
+The notation `du` is suggestive of both the derivative and a small increment. The mathematical formulation follows the derivative, the numeric solution uses a time step and increments the solution over this time step. The `Tsit5()` solver, used here, adaptively chooses a time step, `dt`; were the `Euler` method used, this time step would need to be explicit.
+
+
+:::{.callout-note}
+## Mutation not re-binding
+The `sir!` function has the trailing `!` indicating – by convention – it *mutates* its first value, `du`. In this case, through an assignment, as in `du[1]=ds`. This could use some explanation. The *binding* `du` refers to the *container* holding the $3$ values, whereas `du[1]` refers to the first value in that container. So `du[1]=ds` changes the first value, but not the *binding* of `du` to the container. That is, `du` mutates. This would be quite different were the call `du = [ds,di,dr]` which would create a new *binding* to a new container and not mutate the values in the original container.
+
+:::
+
+With the update function defined, the problem is setup and a solution found with in the same manner:
+
+
+```{julia}
+p = (k=1/3, b=1/2)        # parameters
+time_span = (0.0, 150.0)  # time span to solve over, 5 months
+
+prob = ODEProblem(sir!, u0, time_span, p)
+sol = solve(prob, Tsit5())
+
+plot(sol)
+plot!(x -> 0.5, linewidth=2) # mark 50% line
+```
+
+The lower graph shows the number of infected at each day over the five-month period displayed. The peak is around 6-7% of the population at any one time. However, over time the recovered part of the population reaches over 50%, meaning more than half the population is modeled as getting sick.
+
+
+Now we change the parameter $b$ and observe the difference. We passed in a value `p` holding our two parameters, so we just need to change that and run the model again:
+
+
+```{julia}
+#| hold: true
+p = (k=1/2, b=2) # change b from 1/2 to 2 -- more daily contact
+prob = ODEProblem(sir!, u0, time_span, p)
+sol = solve(prob, Tsit5())
+
+plot(sol)
+```
+
+The graphs are somewhat similar, but the steady state is reached much more quickly and nearly everyone became infected.
+
+
+What about if $k$ were bigger?
+
+
+```{julia}
+#| hold: true
+p = (k=2/3, b=1/2)
+prob = ODEProblem(sir!, u0, time_span, p)
+sol = solve(prob, Tsit5())
+
+plot(sol)
+```
+
+The graphs show that under these conditions the infections never take off; we have $i' = (b\cdot s-k)i = k\cdot((b/k) s - 1) i$ which is always negative, since `(b/k)s < 1`, so infections will only decay.
+
+
+The solution object is indexed by time, then has the `s`, `i`, `r` estimates. We use this structure below to return the estimated proportion of recovered individuals at the end of the time span.
+
+
+```{julia}
+function recovered(k,b)
+    prob = ODEProblem(sir!, u0, time_span, (k,b));
+    sol = solve(prob, Tsit5());
+    s,i,r = last(sol)
+    r
+end
+```
+
+This function makes it easy to see the impact of changing the parameters. For example, fixing $k=1/3$ we have:
+
+
+```{julia}
+f(b) = recovered(1/3, b)
+plot(f, 0, 2)
+```
+
+This very clearly shows the sharp dependence on the value of $b$; below some level, the proportion of people who are ever infected (the recovered cohort) remains near $0$; above that level it can climb quickly towards $1$.
+
+
+The function `recovered` is of two variables returning a single value. In subsequent sections we will see a few $3$-dimensional plots that are common for such functions, here we skip ahead and show how to visualize multiple function plots at once using "`z`" values in a graph.
+
+
+```{julia}
+#| hold: true
+k, ks = 0.1, 0.2:0.1:0.9  # first `k` and then the rest
+bs = range(0, 2, length=100)
+zs = recovered.(k, bs)    # find values for fixed k, each of bs
+p = plot(bs, k*one.(bs), zs, legend=false)  # k*one.(ks) is [k,k,...,k]
+for k in ks
+    plot!(p, bs, k*one.(bs), recovered.(k, bs))
+end
+p
+```
+
+The 3-dimensional graph with `plotly` can have its viewing angle adjusted with the mouse. When looking down on the $x-y$ plane, which code `b` and `k`, we can see the rapid growth along a line related to $b/k$.
+
+
+Smith and Moore point out that  $k$  is roughly the reciprocal of the number of days an individual is sick enough to infect others. This can be estimated during a breakout. However, they go on to note that there is no direct way to observe  $b$,  but there is an indirect way.
+
+
+The ratio $c = b/k$ is the number of close contacts per day times the number of days infected which is the number of close contacts per infected individual.
+
+
+This can be estimated from the curves once steady state has been reached (at the end of the pandemic).
+
+
+$$
+\frac{di}{ds} = \frac{di/dt}{ds/dt} = \frac{b \cdot s(t)  \cdot i(t) - k  \cdot i(t)}{-b \cdot  s(t) \cdot  i(t)} = -1 + \frac{1}{c \cdot  s}
+$$
+
+This equation does not depend on $t$; $s$ is the dependent variable. It could be solved numerically, but in this case affords an algebraic solution: $i = -s + (1/c) \log(s) + q$, where $q$ is some constant. The quantity $q = i + s - (1/c) \log(s)$ does not depend on time, so is the same at time $t=0$ as it is as $t \rightarrow \infty$. At $t=0$ we have $s(0) \approx 1$ and $i(0) \approx 0$, whereas $t \rightarrow \infty$, $i(t) \rightarrow 0$ and $s(t)$ goes to the steady state value, which can be estimated. Solving with $t=0$, we see $q=0 + 1 - (1/c)\log(1) = 1$. In the limit them $1 = 0 + s_{\infty} - (1/c)\log(s_\infty)$ or $c = \log(s_\infty)/(1-s_\infty)$.
+
+
+## Trajectory with drag
+
+
+We now solve numerically the problem of a trajectory with a drag force from air resistance.
+
+
+The general model is:
+
+
+$$
+\begin{align*}
+x''(t) &= - W(t,x(t), x'(t), y(t), y'(t)) \cdot x'(t)\\
+y''(t) &= -g - W(t,x(t), x'(t), y(t), y'(t)) \cdot y'(t)\\
+\end{align*}
+$$
+
+with initial conditions: $x(0) = y(0) = 0$ and $x'(0) = v_0 \cos(\theta), y'(0) = v_0 \sin(\theta)$.
+
+
+This into an ODE by a standard trick. Here we define our function for updating a step. As can be seen the vector `u` contains both $\langle x,y \rangle$ and $\langle x',y' \rangle$
+
+
+```{julia}
+function xy!(du, u, p, t)
+	g, γ = p.g, p.k
+	x, y = u[1], u[2]
+	x′, y′ = u[3], u[4]  # unicode \prime[tab]
+
+    W = γ
+
+	du[1] = x′
+	du[2] = y′
+	du[3] =  0 - W * x′
+	du[4] = -g - W * y′
+end
+```
+
+This function $W$ is just a constant above, but can be easily modified as desired.
+
+
+:::{.callout-note}
+## A second-order ODE is a coupled first-order ODE
+The "standard" trick is to take a second order ODE like $u''(t)=u$ and turn this into two coupled ODEs by using a new name: $v=u'(t)$ and then $v'(t) = u(t)$. In this application, there are $4$ equations, as we have *both* $x''$ and $y''$ being so converted. The first and second components of $du$ are new variables, the third and fourth show the original equation.
+
+:::
+
+The initial conditions are specified through:
+
+
+```{julia}
+θ = pi/4
+v₀ = 200
+xy₀ = [0.0, 0.0]
+vxy₀ = v₀ * [cos(θ), sin(θ)]
+INITIAL = vcat(xy₀, vxy₀)
+```
+
+The time span can be computed using an *upper* bound of no drag, for which the classic physics formulas give (when $y_0=0$) $(0, 2v_{y0}/g)$
+
+
+```{julia}
+g = 9.8
+TSPAN = (0, 2*vxy₀[2] / g)
+```
+
+This allows us to define an `ODEProblem`:
+
+
+```{julia}
+trajectory_problem = ODEProblem(xy!, INITIAL, TSPAN)
+```
+
+When $\gamma = 0$ there should be no drag and we expect to see a parabola:
+
+
+```{julia}
+#| hold: true
+ps = (g=9.8, k=0)
+SOL = solve(trajectory_problem, Tsit5(); p = ps)
+
+plot(t -> SOL(t)[1], t -> SOL(t)[2], TSPAN...; legend=false)
+```
+
+The plot is a parametric plot of the $x$ and $y$ parts of the solution over the time span. We can see the expected parabolic shape.
+
+
+On a *windy* day, the value of $k$ would be positive. Repeating the above with $k=1/4$ gives:
+
+
+```{julia}
+#| hold: true
+ps = (g=9.8, k=1/4)
+SOL = solve(trajectory_problem, Tsit5(); p = ps)
+
+plot(t -> SOL(t)[1], t -> SOL(t)[2], TSPAN...; legend=false)
+```
+
+We see that the $y$ values have gone negative. The `DifferentialEquations` package can adjust for that with a *callback* which terminates the problem once $y$ has gone negative. This can be implemented as follows:
+
+
+```{julia}
+#| hold: true
+condition(u,t,integrator) = u[2] # called when `u[2]` is negative
+affect!(integrator) = terminate!(integrator) # stop the process
+cb = ContinuousCallback(condition, affect!)
+
+ps = (g=9.8, k = 1/4)
+SOL = solve(trajectory_problem, Tsit5(); p = ps, callback=cb)
+
+plot(t -> SOL(t)[1], t -> SOL(t)[2], TSPAN...; legend=false)
+```
+
+Finally, we note that the `ModelingToolkit` package provides symbolic-numeric computing. This allows the equations to be set up symbolically, as in `SymPy` before being passed off to `DifferentialEquations` to solve numerically. The above example with no wind resistance could be translated into the following:
+
+
+```{julia}
+#| hold: true
+@parameters t γ g
+@variables x(t) y(t)
+D = Differential(t)
+
+eqs = [D(D(x)) ~ -γ * D(x),
+       D(D(y)) ~ -g - γ * D(y)]
+
+@named sys = ODESystem(eqs)
+sys = ode_order_lowering(sys) # turn 2nd order into 1st
+
+u0 = [D(x) => vxy₀[1],
+      D(y) => vxy₀[2],
+      x => 0.0,
+      y => 0.0]
+
+p  = [γ => 0.0,
+      g => 9.8]
+
+prob = ODEProblem(sys, u0, TSPAN, p, jac=true)
+sol = solve(prob,Tsit5())
+
+plot(t -> sol(t)[3], t -> sol(t)[4], TSPAN..., legend=false)
+```
+
+The toolkit will automatically generate fast functions and can perform transformations (such as is done by `ode_order_lowering`) before passing along to the numeric solves.
+
+

quarto/ODEs/euler.qmd

@@ -0,0 +1,845 @@
+# Euler's method
+
+
+```{julia}
+#| echo: false
+
+import Logging
+Logging.disable_logging(Logging.Info) # or e.g. Logging.Info
+Logging.disable_logging(Logging.Warn)
+
+import SymPy
+function Base.show(io::IO, ::MIME"text/html", x::T) where {T <: SymPy.SymbolicObject}
+    println(io, "<span class=\"math-left-align\" style=\"padding-left: 4px; width:0; float:left;\"> ")
+    println(io, "\\[")
+    println(io, sympy.latex(x))
+    println(io, "\\]")
+    println(io, "</span>")
+end
+
+# hack to work around issue
+import Markdown
+import CalculusWithJulia
+function CalculusWithJulia.WeaveSupport.ImageFile(d::Symbol, f::AbstractString, caption; kwargs...)
+    nm = joinpath("..", string(d), f)
+    u = "![$caption]($nm)"
+    Markdown.parse(u)
+end
+
+nothing
+```
+
+This section uses these add-on packages:
+
+
+```{julia}
+using CalculusWithJulia
+using Plots
+using SymPy
+using Roots
+```
+
+```{julia}
+#| echo: false
+#| results: "hidden"
+using CalculusWithJulia.WeaveSupport
+
+
+const frontmatter = (
+        title = "Euler's method",
+        description = "Calculus with Julia: Euler's method",
+        tags = ["CalculusWithJulia", "odes", "euler's method"],
+);
+fig_size = (800, 600)
+nothing
+```
+
+---
+
+
+The following section takes up the task of numerically approximating solutions to differential equations. `Julia` has a huge set of state-of-the-art tools for this task starting with the [DifferentialEquations](https://github.com/SciML/DifferentialEquations.jl) package. We don't use that package in this section, focusing on  simpler methods and implementations for pedagogical purposes, but any further exploration should utilize the tools provided therein. A brief introduction to the package follows in an upcoming [section](./differential_equations.html).
+
+
+---
+
+
+Consider the differential equation:
+
+
+$$
+y'(x) = y(x) \cdot  x, \quad y(1)=1,
+$$
+
+which can be solved with `SymPy`:
+
+
+```{julia}
+@syms x, y, u()
+D = Differential(x)
+x0, y0 = 1, 1
+F(y,x) = y*x
+
+dsolve(D(u)(x) - F(u(x), x))
+```
+
+With the given initial condition, the solution becomes:
+
+
+```{julia}
+out = dsolve(D(u)(x) - F(u(x),x), u(x), ics=Dict(u(x0) => y0))
+```
+
+Plotting this solution over the slope field
+
+
+```{julia}
+p = plot(legend=false)
+vectorfieldplot!((x,y) -> [1, F(x,y)], xlims=(0, 2.5), ylims=(0, 10))
+plot!(rhs(out),  linewidth=5)
+```
+
+we see that the vectors that are drawn seem to be tangent to the graph of the solution. This is no coincidence, the tangent lines to integral curves are in the direction of the slope field.
+
+
+What if the graph of the solution were not there, could we use this fact to *approximately* reconstruct the solution?
+
+
+That is, if we stitched together pieces of the slope field, would we get a curve that was close to the actual answer?
+
+
+```{julia}
+#| hold: true
+#| echo: false
+#| cache: true
+## {{{euler_graph}}}
+function make_euler_graph(n)
+    x, y = symbols("x, y")
+    F(y,x) = y*x
+    x0, y0 = 1, 1
+
+    h = (2-1)/5
+    xs = zeros(n+1)
+    ys = zeros(n+1)
+    xs[1] = x0   # index is off by 1
+    ys[1] = y0
+    for i in 1:n
+        xs[i + 1] = xs[i] + h
+        ys[i + 1] = ys[i] + h * F(ys[i], xs[i])
+    end
+
+	p = plot(legend=false)
+    vectorfieldplot!((x,y) -> [1, F(y,x)], xlims=(1,2), ylims=(0,6))
+
+    ## Add Euler soln
+    plot!(p, xs, ys, linewidth=5)
+    scatter!(p, xs, ys)
+
+    ## add function
+    out = dsolve(u'(x) - F(u(x), x), u(x), ics=(u, x0, y0))
+    plot!(p, rhs(out), x0, xs[end], linewidth=5)
+
+    p
+end
+
+
+
+
+n = 5
+anim = @animate for i=1:n
+    make_euler_graph(i)
+end
+
+imgfile = tempname() * ".gif"
+gif(anim, imgfile, fps = 1)
+
+
+caption = """
+Illustration of a function stitching together slope field lines to
+approximate the answer to an initial-value problem. The other function drawn is the actual solution.
+"""
+
+ImageFile(imgfile, caption)
+```
+
+The illustration suggests the answer is yes, let's see. The solution is drawn over $x$ values $1$ to $2$. Let's try piecing together $5$ pieces between $1$ and $2$ and see what we have.
+
+
+The slope-field vectors are *scaled* versions of the vector `[1, F(y,x)]`. The `1` is the part in the direction of the $x$ axis, so here we would like that to be $0.2$ (which is $(2-1)/5$. So our vectors would be `0.2 * [1, F(y,x)]`. To allow for generality, we use `h` in place of the specific value $0.2$.
+
+
+Then our first pieces would be the line connecting $(x_0,y_0)$ to
+
+
+$$
+\langle x_0, y_0 \rangle + h \cdot \langle 1, F(y_0, x_0) \rangle.
+$$
+
+The above uses vector notation to add the piece scaled by $h$ to the starting point. Rather than continue with that notation, we will use subscripts. Let $x_1$, $y_1$ be the postion of the tip of the vector. Then we have:
+
+
+$$
+x_1 = x_0 + h, \quad y_1 = y_0 + h F(y_0, x_0).
+$$
+
+With this notation, it is easy to see what comes next:
+
+
+$$
+x_2 = x_1 + h, \quad y_2 = y_1 + h F(y_1, x_1).
+$$
+
+We just shifted the indices forward by $1$. But graphically what is this? It takes the tip of the first part of our "stitched" together solution, finds the slope filed there (`[1, F(y,x)]`) and then uses this direction to stitch together one more piece.
+
+
+Clearly, we can repeat. The $n$th piece will end at:
+
+
+$$
+x_{n+1} = x_n + h, \quad y_{n+1} = y_n + h F(y_n, x_n).
+$$
+
+For our example, we can do some numerics. We want $h=0.2$ and $5$ pieces, so values of $y$ at $x_0=1, x_1=1.2, x_2=1.4, x_3=1.6, x_4=1.8,$ and $x_5=2$.
+
+
+Below we do this in a loop. We have to be a bit careful, as in `Julia` the vector of zeros we create to store our answers begins indexing at $1$, and not $0$.
+
+
+```{julia}
+n=5
+h = (2-1)/n
+xs = zeros(n+1)
+ys = zeros(n+1)
+xs[1] = x0   # index is off by 1
+ys[1] = y0
+for i in 1:n
+  xs[i + 1] = xs[i] + h
+  ys[i + 1] = ys[i] + h * F(ys[i], xs[i])
+end
+```
+
+So how did we do? Let's look graphically:
+
+
+```{julia}
+plot(exp(-1/2)*exp(x^2/2), x0, 2)
+plot!(xs, ys)
+```
+
+Not bad. We wouldn't expect this to be exact - due to the concavity of the solution, each step is an underestimate. However, we see it is an okay approximation and would likely be better with a smaller $h$. A topic we pursue in just a bit.
+
+
+Rather than type in the above command each time, we wrap it all up in a function.  The inputs are $n$, $a=x_0$, $b=x_n$, $y_0$, and, most importantly, $F$. The output is massaged into a function through a call to `linterp`, rather than two vectors. The `linterp` function we define below just finds a function that linearly interpolates between the points and is `NaN` outside of the range of the $x$ values:
+
+
+```{julia}
+function linterp(xs, ys)
+    function(x)
+        ((x < xs[1]) || (x > xs[end])) && return NaN
+        for i in 1:(length(xs) - 1)
+            if xs[i] <= x < xs[i+1]
+                l = (x-xs[i]) / (xs[i+1] - xs[i])
+                return (1-l) * ys[i] + l * ys[i+1]
+            end
+        end
+        ys[end]
+    end
+end
+```
+
+With that, here is our function to find an approximate solution to $y'=F(y,x)$ with initial condition:
+
+
+```{julia}
+function euler(F, x0, xn, y0, n)
+  h = (xn - x0)/n
+  xs = zeros(n+1)
+  ys = zeros(n+1)
+  xs[1] = x0
+  ys[1] = y0
+  for i in 1:n
+    xs[i + 1] = xs[i] + h
+    ys[i + 1] = ys[i] + h * F(ys[i], xs[i])
+  end
+  linterp(xs, ys)
+end
+```
+
+With `euler`, it becomes easy to explore different values.
+
+
+For example, we thought the solution would look better with a smaller $h$ (or larger $n$). Instead of $n=5$, let's try $n=50$:
+
+
+```{julia}
+u₁₂ = euler(F, 1, 2, 1, 50)
+plot(exp(-1/2)*exp(x^2/2), x0, 2)
+plot!(u₁₂, x0, 2)
+```
+
+It is more work for the computer, but not for us, and clearly a much better approximation to the actual answer is found.
+
+
+## The Euler method
+
+
+```{julia}
+#| hold: true
+#| echo: false
+imgfile ="figures/euler.png"
+caption = """
+Figure from first publication of Euler's method. From [Gander and Wanner](http://www.unige.ch/~gander/Preprints/Ritz.pdf).
+"""
+
+ImageFile(:ODEs, imgfile, caption)
+```
+
+The name of our function reflects the [mathematician](https://en.wikipedia.org/wiki/Leonhard_Euler) associated with the iteration:
+
+
+$$
+x_{n+1} = x_n + h, \quad y_{n+1} = y_n + h \cdot F(y_n, x_n),
+$$
+
+to approximate a solution to the first-order, ordinary differential equation with initial values: $y'(x) = F(y,x)$.
+
+
+[The Euler method](https://en.wikipedia.org/wiki/Euler_method) uses linearization. Each "step" is just an approximation of the function value $y(x_{n+1})$ with the value from the tangent line tangent to the point $(x_n, y_n)$.
+
+
+Each step introduces an error. The error in one step is known as the *local truncation error* and can be shown to be about equal to $1/2 \cdot h^2 \cdot f''(x_{n})$ assuming $y$ has $3$ or more derivatives.
+
+
+The total error, or more commonly, *global truncation error*, is the error between the actual answer and the approximate answer at the end of the process. It reflects an accumulation of these local errors. This error is *bounded* by a constant times $h$. Since it gets smaller as $h$ gets smaller in direct proportion, the Euler method is called *first order*.
+
+
+Other, somewhat more complicated, methods have global truncation errors that involve higher powers of $h$ - that is for the same size $h$, the error is smaller. In analogy is the fact that Riemann sums have error that depends on $h$, whereas other methods of approximating the integral have smaller errors. For example, Simpson's rule had error related to $h^4$. So, the Euler method may not be employed if there is concern about total resources (time, computer, ...), it is important for theoretical purposes in a manner similar to the role of the Riemann integral.
+
+
+In the examples, we will see that for many problems the simple Euler method is satisfactory, but not always so. The task of numerically solving differential equations is not a one-size-fits-all one. In the following, a few different modifications are presented to the basic Euler method, but this just scratches the surface of the topic.
+
+
+#### Examples
+
+
+##### Example
+
+
+Consider the initial value problem $y'(x) = x + y(x)$ with initial condition $y(0)=1$. This problem can be solved exactly. Here we approximate over $[0,2]$ using Euler's method.
+
+
+```{julia}
+𝑭(y,x) = x + y
+𝒙0, 𝒙n, 𝒚0 = 0, 2, 1
+𝒇 = euler(𝑭, 𝒙0, 𝒙n, 𝒚0, 25)
+𝒇(𝒙n)
+```
+
+We graphically compare our approximate answer with the exact one:
+
+
+```{julia}
+plot(𝒇, 𝒙0, 𝒙n)
+𝒐ut = dsolve(D(u)(x) - 𝑭(u(x),x), u(x), ics = Dict(u(𝒙0) => 𝒚0))
+plot(rhs(𝒐ut), 𝒙0, 𝒙n)
+plot!(𝒇, 𝒙0, 𝒙n)
+```
+
+From the graph it appears our value for `f(xn)` will underestimate the actual value of the solution slightly.
+
+
+##### Example
+
+
+The equation $y'(x) = \sin(x \cdot y)$ is not separable, so need not have an easy solution. The default method will fail. Looking at the available methods with `sympy.classify_ode(𝐞qn, u(x))` shows a power series method which can return a power series *approximation* (a Taylor polynomial). Let's look at comparing an approximate answer given by the Euler method to that one returned by `SymPy`.
+
+
+First, the `SymPy` solution:
+
+
+```{julia}
+𝐅(y,x) = sin(x*y)
+𝐞qn = D(u)(x) - 𝐅(u(x), x)
+𝐨ut = dsolve(𝐞qn, hint="1st_power_series")
+```
+
+If we assume $y(0) = 1$, we can continue:
+
+
+```{julia}
+𝐨ut1 = dsolve(𝐞qn, u(x), ics=Dict(u(0) => 1), hint="1st_power_series")
+```
+
+The approximate value given by the Euler method is
+
+
+```{julia}
+𝐱0, 𝐱n, 𝐲0 = 0, 2, 1
+
+plot(legend=false)
+vectorfieldplot!((x,y) -> [1, 𝐅(y,x)], xlims=(𝐱0, 𝐱n), ylims=(0,5))
+plot!(rhs(𝐨ut1).removeO(),  linewidth=5)
+
+𝐮 = euler(𝐅, 𝐱0, 𝐱n, 𝐲0, 10)
+plot!(𝐮, linewidth=5)
+```
+
+We see that the answer found from using a polynomial series matches that of Euler's method for a bit, but as time evolves, the approximate solution given by Euler's method more closely tracks the slope field.
+
+
+##### Example
+
+
+The [Brachistochrone problem](http://www.unige.ch/~gander/Preprints/Ritz.pdf) was posed by Johann Bernoulli in 1696. It asked for the curve between two points for which an object will fall faster along that curve than any other. For an example, a bead sliding on a wire will take  a certain amount of time to get from point $A$ to point $B$, the time depending on the shape of the wire. Which shape will take the least amount  of time?
+
+
+```{julia}
+#| hold: true
+#| echo: false
+imgfile = "figures/bead-game.jpg"
+caption = """
+
+A child's bead game. What shape wire will produce the shortest time for a bed to slide from a top to the bottom?
+
+"""
+ImageFile(:ODEs, imgfile, caption)
+```
+
+Restrict our attention to the $x$-$y$ plane, and consider a path, between the point $(0,A)$ and $(B,0)$. Let $y(x)$ be the distance from $A$, so $y(0)=0$ and at the end $y$ will be $A$.
+
+
+[Galileo](http://www-history.mcs.st-and.ac.uk/HistTopics/Brachistochrone.html) knew the straight line was not the curve, but incorrectly thought the answer was a part of a circle.
+
+
+```{julia}
+#| hold: true
+#| echo: false
+imgfile = "figures/galileo.gif"
+caption = """
+As early as 1638, Galileo showed that an object falling along `AC` and then `CB` will fall faster than one traveling along `AB`, where `C` is on the arc of a circle.
+From the [History of Math Archive](http://www-history.mcs.st-and.ac.uk/HistTopics/Brachistochrone.html).
+"""
+ImageFile(:ODEs, imgfile, caption)
+```
+
+This simulation also suggests that a curved path is better than the shorter straight one:
+
+
+```{julia}
+#| hold: true
+#| echo: false
+#| cache: true
+##{{{brach_graph}}}
+
+function brach(f, x0, vx0, y0, vy0, dt, n)
+    m = 1
+    g = 9.8
+
+    axs = Float64[0]
+    ays = Float64[-g]
+    vxs = Float64[vx0]
+    vys = Float64[vy0]
+    xs = Float64[x0]
+    ys = Float64[y0]
+
+    for i in 1:n
+        x = xs[end]
+        vx = vxs[end]
+
+        ax = -f'(x) * (f''(x) * vx^2 + g) / (1 + f'(x)^2)
+        ay = f''(x) * vx^2 + f'(x) * ax
+
+        push!(axs, ax)
+        push!(ays, ay)
+
+        push!(vxs, vx + ax * dt)
+        push!(vys, vys[end] + ay * dt)
+        push!(xs, x       + vxs[end] * dt)# + (1/2) * ax * dt^2)
+        push!(ys, ys[end] + vys[end] * dt)# + (1/2) * ay * dt^2)
+    end
+
+    [xs ys vxs vys axs ays]
+
+end
+
+
+fs = [x -> 1 - x,
+      x -> (x-1)^2,
+      x -> 1 - sqrt(1 - (x-1)^2),
+      x ->  - (x-1)*(x+1),
+      x -> 3*(x-1)*(x-1/3)
+      ]
+
+
+MS = [brach(f, 1/100, 0, 1, 0, 1/100, 100) for f in fs]
+
+
+function make_brach_graph(n)
+
+    p = plot(xlim=(0,1), ylim=(-1/3, 1), legend=false)
+    for (i,f) in enumerate(fs)
+        plot!(f, 0, 1)
+        U = MS[i]
+        x = min(1.0, U[n,1])
+        scatter!(p, [x], [f(x)])
+    end
+    p
+
+end
+
+
+
+n = 4
+anim = @animate for i=[1,5,10,15,20,25,30,35,40,45,50,55,60]
+    make_brach_graph(i)
+end
+
+imgfile = tempname() * ".gif"
+gif(anim, imgfile, fps = 1)
+
+
+caption = """
+The race is on. An illustration of beads falling along a path, as can be seen, some paths are faster than others. The fastest path would follow a cycloid. See [Bensky and Moelter](https://pdfs.semanticscholar.org/66c1/4d8da6f2f5f2b93faf4deb77aafc7febb43a.pdf) for details on simulating a bead on a wire.
+"""
+
+ImageFile(imgfile, caption)
+```
+
+Now, the natural question is which path is best?  The solution can be [reduced](http://mathworld.wolfram.com/BrachistochroneProblem.html) to solving this equation for a positive $c$:
+
+
+$$
+1 + (y'(x))^2  = \frac{c}{y}, \quad c > 0.
+$$
+
+Reexpressing, this becomes:
+
+
+$$
+\frac{dy}{dx} = \sqrt{\frac{C-y}{y}}.
+$$
+
+This is a separable equation and can be solved, but even `SymPy` has trouble with this integral. However, the result has been known to be a piece of a cycloid since the insightful Jacob Bernoulli used an analogy from light bending to approach the problem. The answer is best described parametrically through:
+
+
+$$
+x(u) = c\cdot u - \frac{c}{2}\sin(2u), \quad y(u) = \frac{c}{2}( 1- \cos(2u)), \quad 0 \leq u \leq U.
+$$
+
+The values of $U$ and $c$ must satisfy $(x(U), y(U)) = (B, A)$.
+
+
+Rather than pursue this, we will solve it numerically for a fixed value of $C$ over a fixed interval to see the shape.
+
+
+The equation can be written in terms of $y'=F(y,x)$, where
+
+
+$$
+F(y,x) = \sqrt{\frac{c-y}{y}}.
+$$
+
+But as $y_0 = 0$, we immediately would have a problem with the first step, as there would be division by $0$.
+
+
+This says that for the optimal solution, the bead picks up speed by first sliding straight down before heading off towards $B$. That's great for the physics, but runs roughshod over our Euler method, as the first step has an infinity.
+
+
+For this, we can try the *backwards Euler* method which uses the slope at $(x_{n+1}, y_{n+1})$, rather than $(x_n, y_n)$. The update step becomes:
+
+
+$$
+y_{n+1} = y_n + h \cdot F(y_{n+1}, x_{n+1}).
+$$
+
+Seems innocuous, but the value we are trying to find, $y_{n+1}$, is now on both sides of the equation, so is only *implicitly* defined. In this code, we use the `find_zero` function from the `Roots` package. The caveat is, this function needs a good initial guess, and the one we use below need not be widely applicable.
+
+
+```{julia}
+function back_euler(F, x0, xn, y0, n)
+    h = (xn - x0)/n
+    xs = zeros(n+1)
+    ys = zeros(n+1)
+    xs[1] = x0
+    ys[1] = y0
+    for i in 1:n
+        xs[i + 1] = xs[i] + h
+        ## solve y[i+1] = y[i] + h * F(y[i+1], x[i+1])
+        ys[i + 1] = find_zero(y -> ys[i] + h * F(y, xs[i + 1]) - y, ys[i]+h)
+    end
+  linterp(xs, ys)
+end
+```
+
+We then have with $C=1$ over the interval $[0,1.2]$ the following:
+
+
+```{julia}
+𝐹(y, x; C=1) = sqrt(C/y - 1)
+𝑥0, 𝑥n, 𝑦0 = 0, 1.2, 0
+cyc = back_euler(𝐹, 𝑥0, 𝑥n, 𝑦0, 50)
+plot(x -> 1 - cyc(x), 𝑥0, 𝑥n)
+```
+
+Remember, $y$ is the displacement from the top, so it is non-negative. Above we flipped the graph to make it look more like expectation. In general, the trajectory may actually dip below the ending point and come back up. The above won't see this, for as written $dy/dx \geq 0$, which need not be the case, as the defining equation is in terms of $(dy/dx)^2$, so the derivative could have any sign.
+
+
+##### Example: stiff equations
+
+
+The Euler method is *convergent*, in that as $h$ goes to $0$, the approximate solution will converge to the actual answer. However, this does not say that for a fixed size $h$, the approximate value will be good. For example, consider the differential equation $y'(x) = -5y$. This has solution $y(x)=y_0 e^{-5x}$. However, if we try the Euler method to get an answer over $[0,2]$ with $h=0.5$ we don't see this:
+
+
+```{julia}
+ℱ(y,x) = -5y
+𝓍0, 𝓍n, 𝓎0 = 0, 2, 1
+𝓊 = euler(ℱ, 𝓍0, 𝓍n, 𝓎0, 4)     # n =4 => h = 2/4
+vectorfieldplot((x,y) -> [1, ℱ(y,x)], xlims=(0, 2), ylims=(-5, 5))
+plot!(x -> y0 * exp(-5x), 0, 2, linewidth=5)
+plot!(𝓊, 0, 2, linewidth=5)
+```
+
+What we see is that the value of $h$ is too big to capture the decay scale of the solution. A smaller $h$, can do much better:
+
+
+```{julia}
+𝓊₁ = euler(ℱ, 𝓍0, 𝓍n, 𝓎0, 50)    # n=50 => h = 2/50
+plot(x -> y0 * exp(-5x), 0, 2)
+plot!(𝓊₁, 0, 2)
+```
+
+This is an example of a [stiff equation](https://en.wikipedia.org/wiki/Stiff_equation). Such equations cause explicit methods like the Euler one problems, as small $h$s are needed to good results.
+
+
+The implicit, backward Euler method does not have this issue, as we can see here:
+
+
+```{julia}
+𝓊₂ = back_euler(ℱ, 𝓍0, 𝓍n, 𝓎0, 4)     # n =4 => h = 2/4
+vectorfieldplot((x,y) -> [1, ℱ(y,x)],  xlims=(0, 2), ylims=(-1, 1))
+plot!(x -> y0 * exp(-5x), 0, 2, linewidth=5)
+plot!(𝓊₂, 0, 2, linewidth=5)
+```
+
+##### Example: The pendulum
+
+
+The differential equation describing the simple pendulum is
+
+
+$$
+\theta''(t) = - \frac{g}{l}\sin(\theta(t)).
+$$
+
+The typical approach to solving for $\theta(t)$ is to use the small-angle approximation that $\sin(x) \approx x$, and then the differential equation simplifies to: $\theta''(t) = -g/l \cdot \theta(t)$, which is easily solved.
+
+
+Here we try to get an answer numerically. However, the problem, as stated, is not a first order equation due to the $\theta''(t)$ term. If we let $u(t) = \theta(t)$ and $v(t) = \theta'(t)$, then we get *two* coupled first order equations:
+
+
+$$
+v'(t) = -g/l \cdot \sin(u(t)), \quad u'(t) = v(t).
+$$
+
+We can try the Euler method here. A simple approach might be this iteration scheme:
+
+
+$$
+\begin{align*}
+x_{n+1} &= x_n + h,\\
+u_{n+1} &= u_n + h v_n,\\
+v_{n+1} &= v_n - h \cdot g/l \cdot \sin(u_n).
+\end{align*}
+$$
+
+Here we need *two* initial conditions: one for the initial value $u(t_0)$ and the initial value of $u'(t_0)$. We have seen if we start at an angle $a$ and release the bob from rest, so $u'(0)=0$ we get a sinusoidal answer to the linearized model. What happens here? We let $a=1$, $L=5$ and $g=9.8$:
+
+
+We write a function to solve this starting from $(x_0, y_0)$ and ending at $x_n$:
+
+
+```{julia}
+function euler2(x0, xn, y0, yp0, n; g=9.8, l = 5)
+  xs, us, vs = zeros(n+1), zeros(n+1), zeros(n+1)
+  xs[1], us[1], vs[1] = x0, y0, yp0
+  h = (xn - x0)/n
+  for i = 1:n
+    xs[i+1] = xs[i] + h
+	us[i+1] = us[i] + h * vs[i]
+	vs[i+1] = vs[i] + h * (-g / l) * sin(us[i])
+	end
+	linterp(xs, us)
+end
+```
+
+Let's take $a = \pi/4$ as the initial angle, then the approximate solution should be $\pi/4\cos(\sqrt{g/l}x)$ with period $T = 2\pi\sqrt{l/g}$. We try first to plot then over 4 periods:
+
+
+```{julia}
+𝗅, 𝗀 = 5, 9.8
+𝖳 = 2pi * sqrt(𝗅/𝗀)
+𝗑0, 𝗑n, 𝗒0, 𝗒p0 = 0, 4𝖳, pi/4, 0
+plot(euler2(𝗑0, 𝗑n, 𝗒0, 𝗒p0, 20), 0, 4𝖳)
+```
+
+Something looks terribly amiss. The issue is the step size, $h$, is too large to capture the oscillations. There are basically only $5$ steps to capture a full up and down motion. Instead, we try to get $20$ steps per period so $n$ must be not $20$, but $4 \cdot 20 \cdot T \approx 360$. To this graph, we add the approximate one:
+
+
+```{julia}
+plot(euler2(𝗑0, 𝗑n, 𝗒0, 𝗒p0, 360), 0, 4𝖳)
+plot!(x -> pi/4*cos(sqrt(𝗀/𝗅)*x), 0, 4𝖳)
+```
+
+Even now, we still see that something seems amiss, though the issue is not as dramatic as before. The oscillatory nature of the pendulum is seen, but in the Euler solution, the amplitude grows, which would necessarily mean energy is being put into the system.  A familiar instance of a pendulum would be a child on a swing. Without pumping the legs - putting energy in the system - the height of the swing's arc will not grow.  Though we now have oscillatory motion, this growth indicates the solution is still not quite right. The issue is likely due to each step mildly overcorrecting and resulting in an overall growth.  One of the questions pursues this a bit further.
+
+
+## Questions
+
+
+##### Question
+
+
+Use Euler's method with $n=5$ to approximate $u(1)$ where
+
+
+$$
+u'(x) = x - u(x), \quad u(0) = 1
+$$
+
+```{julia}
+#| hold: true
+#| echo: false
+F(y,x) = x - y
+x0, xn, y0 = 0, 1, 1
+val = euler(F, x0, xn, y0, 5)(1)
+numericq(val)
+```
+
+##### Question
+
+
+Consider the equation
+
+
+$$
+y' = x \cdot \sin(y), \quad y(0) = 1.
+$$
+
+Use Euler's method with $n=50$ to find the value of $y(5)$.
+
+
+```{julia}
+#| hold: true
+#| echo: false
+F(y, x) = x * sin(y)
+x0, xn, y0 = 0, 5, 1
+n = 50
+u = euler(F, x0, xn, y0, n)
+numericq(u(xn))
+```
+
+##### Question
+
+
+Consider the ordinary differential equation
+
+
+$$
+\frac{dy}{dx} = 1 - 2\frac{y}{x}, \quad y(1) = 0.
+$$
+
+Use Euler's method to solve for $y(2)$ when $n=50$.
+
+
+```{julia}
+#| hold: true
+#| echo: false
+F(y, x) = 1 - 2y/x
+x0, xn, y0 = 1, 2, 0
+n = 50
+u = euler(F, x0, xn, y0, n)
+numericq(u(xn))
+```
+
+##### Question
+
+
+Consider the ordinary differential equation
+
+
+$$
+\frac{dy}{dx} = \frac{y \cdot \log(y)}{x}, \quad y(2) = e.
+$$
+
+Use Euler's method to solve for $y(3)$ when $n=25$.
+
+
+```{julia}
+#| hold: true
+#| echo: false
+F(y, x) = y*log(y)/x
+x0, xn, y0 = 2, 3, exp(1)
+n = 25
+u = euler(F, x0, xn, y0, n)
+numericq(u(xn))
+```
+
+##### Question
+
+
+Consider the first-order non-linear ODE
+
+
+$$
+y' = y \cdot (1-2x), \quad y(0) = 1.
+$$
+
+Use Euler's method with $n=50$ to approximate the solution $y$ over $[0,2]$.
+
+
+What is the value at $x=1/2$?
+
+
+```{julia}
+#| hold: true
+#| echo: false
+F(y, x) = y * (1-2x)
+x0, xn, y0 = 0, 2, 1
+n = 50
+u = euler(F, x0, xn, y0, n)
+numericq(u(1/2))
+```
+
+What is the value at $x=3/2$?
+
+
+```{julia}
+#| hold: true
+#| echo: false
+F(y, x) = y * (1-2x)
+x0, xn, y0 = 0, 2, 1
+n = 50
+u = euler(F, x0, xn, y0, n)
+numericq(u(3/2))
+```
+
+##### Question: The pendulum revisited.
+
+
+The issue with the pendulum's solution growing in amplitude can be addressed using a modification to the Euler method attributed to [Cromer](http://astro.physics.ncsu.edu/urca/course_files/Lesson14/index.html). The fix is to replace the term `sin(us[i])` in the line `vs[i+1] = vs[i] + h * (-g / l) * sin(us[i])` of the `euler2` function with `sin(us[i+1])`, which uses the updated angular velocity in the $2$nd step in place of the value before the step.
+
+
+Modify the `euler2` function to implement the Euler-Cromer method. What do you see?
+
+
+```{julia}
+#| hold: true
+#| echo: false
+choices = [
+"The same as before - the amplitude grows",
+"The solution is identical to that of the approximation found by linearization of the sine term",
+"The solution has a constant amplitude, but its period is slightly *shorter* than that of the approximate solution found by linearization",
+"The solution has a constant amplitude, but its period is slightly *longer* than that of the approximate solution found by linearization"]
+answ = 4
+radioq(choices, answ, keep_order=true)
+```
+

quarto/ODEs/process.jl

@@ -0,0 +1,27 @@
+using WeavePynb
+using Mustache
+
+mmd(fname) = mmd_to_html(fname, BRAND_HREF="../toc.html", BRAND_NAME="Calculus with Julia")
+## uncomment to generate just .md files
+mmd(fname) = mmd_to_md(fname, BRAND_HREF="../toc.html", BRAND_NAME="Calculus with Julia")
+
+fnames = [
+          "odes",
+          "euler"
+          ]
+
+
+
+function process_file(nm, twice=false)
+    include("$nm.jl")
+    mmd_to_md("$nm.mmd")
+    markdownToHTML("$nm.md")
+    twice && markdownToHTML("$nm.md")
+end
+
+process_files(twice=false) = [process_file(nm, twice) for nm in fnames]
+
+"""
+## TODO ODEs
+
+"""

quarto/ODEs/solve.qmd

@@ -0,0 +1,295 @@
+# The problem-algorithm-solve interface
+
+
+```{julia}
+#| echo: false
+
+import Logging
+Logging.disable_logging(Logging.Info) # or e.g. Logging.Info
+Logging.disable_logging(Logging.Warn)
+
+import SymPy
+function Base.show(io::IO, ::MIME"text/html", x::T) where {T <: SymPy.SymbolicObject}
+    println(io, "<span class=\"math-left-align\" style=\"padding-left: 4px; width:0; float:left;\"> ")
+    println(io, "\\[")
+    println(io, sympy.latex(x))
+    println(io, "\\]")
+    println(io, "</span>")
+end
+
+# hack to work around issue
+import Markdown
+import CalculusWithJulia
+function CalculusWithJulia.WeaveSupport.ImageFile(d::Symbol, f::AbstractString, caption; kwargs...)
+    nm = joinpath("..", string(d), f)
+    u = "![$caption]($nm)"
+    Markdown.parse(u)
+end
+
+nothing
+```
+
+This section uses these add-on packages:
+
+
+```{julia}
+using Plots
+using MonteCarloMeasurements
+```
+
+```{julia}
+#| echo: false
+#| results: "hidden"
+using CalculusWithJulia.WeaveSupport
+
+const frontmatter = (
+        title = "The problem-algorithm-solve interface",
+        description = "Calculus with Julia: The problem-algorithm-solve interface",
+        tags = ["CalculusWithJulia", "odes", "the problem-algorithm-solve interface"],
+);
+fig_size = (800, 600)
+nothing
+```
+
+---
+
+
+The [DifferentialEquations.jl](https://github.com/SciML) package is an entry point to a suite of `Julia` packages for numerically solving differential equations in `Julia` and other languages. A common interface is implemented that flexibly adjusts to the many different problems and algorithms covered by this suite of packages. In this section, we review a very informative [post](https://discourse.julialang.org/t/function-depending-on-the-global-variable-inside-module/64322/10) by discourse user `@genkuroki` which very nicely demonstrates the usefulness of the problem-algorithm-solve approach used with `DifferentialEquations.jl`. We slightly modify the presentation below for our needs, but suggest a perusal of the original post.
+
+
+##### Example: FreeFall
+
+
+The motion of an object under a uniform gravitational field is of interest.
+
+
+The parameters that govern the equation of motions are the gravitational constant, `g`; the initial height, `y0`; and the initial velocity, `v0`. The time span for which a solution is sought is `tspan`.
+
+
+A problem consists of these parameters. Typical `Julia` usage would be to create a structure to hold the parameters, which may be done as follows:
+
+
+```{julia}
+struct Problem{G, Y0, V0, TS}
+    g::G
+    y0::Y0
+    v0::V0
+    tspan::TS
+end
+
+Problem(;g=9.80665, y0=0.0, v0=30.0, tspan=(0.0,8.0)) = Problem(g, y0, v0, tspan)
+```
+
+The above creates a type, `Problem`, *and* a default constructor with default values. (The original uses a more sophisticated setup that allows the two things above to be combined.)
+
+
+Just calling `Problem()` will create a problem suitable for the earth, passing different values for `g` would be possible for other planets.
+
+
+To solve differential equations there are many different possible algorithms. Here is the construction of two types to indicate two algorithms:
+
+
+```{julia}
+struct EulerMethod{T}
+  dt::T
+end
+EulerMethod(; dt=0.1) = EulerMethod(dt)
+
+struct ExactFormula{T}
+  dt::T
+end
+ExactFormula(; dt=0.1) = ExactFormula(dt)
+```
+
+The above just specifies a type for dispatch –- the directions indicating what code to use to solve the problem. As seen, each specifies a size for a time step with default of `0.1`.
+
+
+A type for solutions is useful for different `show` methods or other methods. One can be created through:
+
+
+```{julia}
+struct Solution{Y, V, T, P<:Problem, A}
+    y::Y
+    v::V
+    t::T
+    prob::P
+    alg::A
+end
+```
+
+The different algorithms then can be implemented as part of a generic `solve` function. Following the post we have:
+
+
+```{julia}
+solve(prob::Problem) = solve(prob, default_algorithm(prob))
+default_algorithm(prob::Problem) = EulerMethod()
+
+function solve(prob::Problem, alg::ExactFormula)
+    g, y0, v0, tspan = prob.g, prob.y0, prob.v0, prob.tspan
+    dt = alg.dt
+    t0, t1 = tspan
+    t = range(t0, t1 + dt/2; step = dt)
+
+    y(t) = y0 + v0*(t - t0) - g*(t - t0)^2/2
+    v(t) = v0 - g*(t - t0)
+
+    Solution(y.(t), v.(t), t, prob, alg)
+end
+
+function solve(prob::Problem, alg::EulerMethod)
+    g, y0, v0, tspan = prob.g, prob.y0, prob.v0, prob.tspan
+    dt = alg.dt
+    t0, t1 = tspan
+    t = range(t0, t1 + dt/2; step = dt)
+
+    n = length(t)
+    y = Vector{typeof(y0)}(undef, n)
+    v = Vector{typeof(v0)}(undef, n)
+    y[1] = y0
+    v[1] = v0
+
+    for i in 1:n-1
+        v[i+1] = v[i] - g*dt     # F*h step of Euler
+        y[i+1] = y[i] + v[i]*dt  # F*h step of Euler
+    end
+
+    Solution(y, v, t, prob, alg)
+end
+```
+
+The post has a more elegant means to unpack the parameters from the structures, but for each of the above, the parameters are unpacked, and then the corresponding algorithm employed. As of version `v1.7` of `Julia`, the syntax `(;g,y0,v0,tspan) = prob` could also be employed.
+
+
+The exact formulas, `y(t) = y0 + v0*(t - t0) - g*(t - t0)^2/2` and `v(t) = v0 - g*(t - t0)`, follow from well-known physics formulas. Each answer is wrapped in a `Solution` type so that the answers found can be easily extracted in a uniform manner.
+
+
+For example, plots of each can be obtained through:
+
+
+```{julia}
+earth = Problem()
+sol_euler = solve(earth)
+sol_exact = solve(earth, ExactFormula())
+
+plot(sol_euler.t, sol_euler.y;
+     label="Euler's method (dt = $(sol_euler.alg.dt))", ls=:auto)
+plot!(sol_exact.t, sol_exact.y; label="exact solution", ls=:auto)
+title!("On the Earth"; xlabel="t", legend=:bottomleft)
+```
+
+Following the post, since the time step `dt = 0.1` is not small enough, the error of the Euler method is rather large. Next we change the algorithm parameter, `dt`, to be smaller:
+
+
+```{julia}
+earth₂ = Problem()
+sol_euler₂ = solve(earth₂, EulerMethod(dt = 0.01))
+sol_exact₂ = solve(earth₂, ExactFormula())
+
+plot(sol_euler₂.t, sol_euler₂.y;
+     label="Euler's method (dt = $(sol_euler₂.alg.dt))", ls=:auto)
+plot!(sol_exact₂.t, sol_exact₂.y; label="exact solution", ls=:auto)
+title!("On the Earth"; xlabel="t", legend=:bottomleft)
+```
+
+It is worth noting that only the first line is modified, and only the method requires modification.
+
+
+Were the moon to be considered, the gravitational constant would need adjustment. This parameter is part of the problem, not the solution algorithm.
+
+
+Such adjustments are made by passing different values to the `Problem` constructor:
+
+
+```{julia}
+moon = Problem(g = 1.62, tspan = (0.0, 40.0))
+sol_eulerₘ = solve(moon)
+sol_exactₘ = solve(moon, ExactFormula(dt = sol_euler.alg.dt))
+
+plot(sol_eulerₘ.t, sol_eulerₘ.y;
+     label="Euler's method (dt = $(sol_eulerₘ.alg.dt))", ls=:auto)
+plot!(sol_exactₘ.t, sol_exactₘ.y; label="exact solution", ls=:auto)
+title!("On the Moon"; xlabel="t", legend=:bottomleft)
+```
+
+The code above also adjusts the time span in addition to the graviational constant. The algorithm for exact formula is set to use the `dt` value used in the `euler` formula, for easier comparison. Otherwise, outside of the labels, the patterns are the same. Only those things that need changing are changed, the rest comes from defaults.
+
+
+The above shows the benefits of using a common interface. Next, the post illustrates how *other* authors could extend this code, simply by adding a *new* `solve` method. For example,
+
+
+```{julia}
+struct Symplectic2ndOrder{T}
+    dt::T
+end
+Symplectic2ndOrder(;dt=0.1) = Symplectic2ndOrder(dt)
+
+function solve(prob::Problem, alg::Symplectic2ndOrder)
+    g, y0, v0, tspan = prob.g, prob.y0, prob.v0, prob.tspan
+    dt = alg.dt
+    t0, t1 = tspan
+    t = range(t0, t1 + dt/2; step = dt)
+
+    n = length(t)
+    y = Vector{typeof(y0)}(undef, n)
+    v = Vector{typeof(v0)}(undef, n)
+    y[1] = y0
+    v[1] = v0
+
+    for i in 1:n-1
+        ytmp = y[i] + v[i]*dt/2
+        v[i+1] = v[i] - g*dt
+        y[i+1] = ytmp + v[i+1]*dt/2
+    end
+
+    Solution(y, v, t, prob, alg)
+end
+```
+
+Had the two prior methods been in a package, the other user could still extend the interface, as above,  with just a slight standard modification.
+
+
+The same approach works for this new type:
+
+
+```{julia}
+
+earth₃ = Problem()
+sol_sympl₃ = solve(earth₃, Symplectic2ndOrder(dt = 2.0))
+sol_exact₃ = solve(earth₃, ExactFormula())
+
+plot(sol_sympl₃.t, sol_sympl₃.y; label="2nd order symplectic (dt = $(sol_sympl₃.alg.dt))", ls=:auto)
+plot!(sol_exact₃.t, sol_exact₃.y; label="exact solution", ls=:auto)
+title!("On the Earth"; xlabel="t", legend=:bottomleft)
+```
+
+Finally, the author of the post shows how the interface can compose with other packages in the `Julia` package ecosystem. This example uses the external package `MonteCarloMeasurements` which plots the behavior of the system for perturbations of the initial value:
+
+
+```{julia}
+
+earth₄ = Problem(y0 = 0.0 ± 0.0, v0 = 30.0 ± 1.0)
+sol_euler₄ = solve(earth₄)
+sol_sympl₄ = solve(earth₄, Symplectic2ndOrder(dt = 2.0))
+sol_exact₄ = solve(earth₄, ExactFormula())
+
+ylim = (-100, 60)
+P = plot(sol_euler₄.t, sol_euler₄.y;
+         label="Euler's method (dt = $(sol_euler₄.alg.dt))", ls=:auto)
+title!("On the Earth"; xlabel="t", legend=:bottomleft, ylim)
+
+Q = plot(sol_sympl₄.t, sol_sympl₄.y;
+         label="2nd order symplectic (dt = $(sol_sympl₄.alg.dt))", ls=:auto)
+title!("On the Earth"; xlabel="t", legend=:bottomleft, ylim)
+
+R = plot(sol_exact₄.t, sol_exact₄.y; label="exact solution", ls=:auto)
+title!("On the Earth"; xlabel="t", legend=:bottomleft, ylim)
+
+plot(P, Q, R; size=(720, 600))
+```
+
+The only change was in the problem, `Problem(y0 = 0.0 ± 0.0, v0 = 30.0 ± 1.0)`, where a different number type is used which accounts for uncertainty. The rest follows the same pattern.
+
+
+This example, shows the flexibility of the problem-algorithm-solver pattern while maintaining a consistent pattern for execution.
+
+

quarto/README.md

@@ -0,0 +1,11 @@
+# CalculusWithJulia via quarto
+
+To compile the pages through quarto
+
+
+* author in `.jmd` files (to be run through pluto)
+* run `julia make_qmd.jl` to convert to `.qmd` files
+* compile with `quarto`
+
+The files in subdirectories are generated, they should not be edited
+The files in this main directory are quarto specific.

quarto/_freeze/derivatives/curve_sketching/figure-html/cell-35-output-1.svg

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