Add columns math vs code

.gitignore

@@ -11,6 +11,7 @@
 *.jld2
 *.json
 !.zenodo.json
+!.markdownlint.json
 
 # System-specific files and directories generated by the BinaryProvider and BinDeps packages
 # They contain absolute paths specific to the host computer, and so should not be committed

.markdownlint.json

@@ -0,0 +1,4 @@
+{
+    "MD046": false,
+    "MD013": false
+}

docs/make.jl

@@ -129,8 +129,6 @@ Draft = false
 =#
 makedocs(;
     draft=false,
-    #draft=true,
-    #warnonly=[:cross_references, :autodocs_block],
     sitename="OptimalControl.jl",
     format=Documenter.HTML(;
         repolink="https://" * repo_url,
@@ -141,6 +139,7 @@ makedocs(;
         assets=[
             asset("https://control-toolbox.org/assets/css/documentation.css"),
             asset("https://control-toolbox.org/assets/js/documentation.js"),
+            "assets/custom.css",
         ],
     ),
     pages=[

docs/src/assets/Manifest.toml

@@ -1677,7 +1677,7 @@ version = "0.5.6+0"
 deps = ["ADNLPModels", "CTBase", "CTDirect", "CTFlows", "CTModels", "CTParser", "CommonSolve", "DocStringExtensions", "ExaModels"]
 path = "/Users/ocots/Research/logiciels/dev/control-toolbox/OptimalControl"
 uuid = "5f98b655-cc9a-415a-b60e-744165666948"
-version = "1.1.2"
+version = "1.1.3"
 
 [[deps.Opus_jll]]
 deps = ["Artifacts", "JLLWrappers", "Libdl"]

docs/src/assets/custom.css

@@ -0,0 +1,36 @@
+/* Responsive layout for two-column content */
+.responsive-columns-left-priority {
+    display: flex;
+    gap: 1rem;
+    align-items: flex-start;
+    margin-bottom: 1em;
+}
+
+.responsive-columns-left-priority > div {
+    flex: 1;
+    min-width: 0;
+    transition: opacity 0.5s ease-in-out, flex 0.5s ease-in-out, max-width 0.5s ease-in-out, max-height 0.5s ease-in-out;
+}
+
+.responsive-columns-left-priority > div:last-child {
+    opacity: 1;
+    max-width: 100%;
+    max-height: none;
+}
+
+/* Media query for screens smaller than 700px */
+@media (max-width: 700px) {
+    .responsive-columns-left-priority > div:last-child {
+        opacity: 0;
+        max-width: 0;
+        max-height: 0;
+        flex: 0 0 0;
+        overflow: hidden;
+        margin: 0;
+        padding: 0;
+    }
+
+    .responsive-columns-left-priority > div:first-child {
+        flex: 1 1 100%;
+    }
+}

docs/src/example-double-integrator-energy.md

@@ -13,18 +13,15 @@ We assume that the mass is constant and equal to one, and that there is no frict
     \dot x_1(t) = x_2(t), \quad \dot x_2(t) = u(t),\quad u(t) \in \R,
 ```
 
-which is simply the [double integrator](https://en.wikipedia.org/w/index.php?title=Double_integrator&oldid=1071399674) system.
-Let us consider a transfer starting at time $t_0 = 0$ and ending at time $t_f = 1$, for which we want to minimise the transfer energy
+which is simply the [double integrator](https://en.wikipedia.org/w/index.php?title=Double_integrator&oldid=1071399674) system. Let us consider a transfer starting at time $t_0 = 0$ and ending at time $t_f = 1$, for which we want to minimise the transfer energy
 
 ```math
     \frac{1}{2}\int_{0}^{1} u^2(t) \, \mathrm{d}t
 ```
 
 starting from $x(0) = (-1, 0)$ and aiming to reach the target $x(1) = (0, 0)$.
 
-First, we need to import the [OptimalControl.jl](https://control-toolbox.org/OptimalControl.jl) package to define the 
-optimal control problem, [NLPModelsIpopt.jl](https://jso.dev/NLPModelsIpopt.jl) to solve it, 
-and [Plots.jl](https://docs.juliaplots.org) to visualise the solution.
+First, we need to import the [OptimalControl.jl](https://control-toolbox.org/OptimalControl.jl) package to define the optimal control problem, [NLPModelsIpopt.jl](https://jso.dev/NLPModelsIpopt.jl) to solve it, and [Plots.jl](https://docs.juliaplots.org) to visualise the solution.
 
 ```@example main
 using OptimalControl
@@ -36,6 +33,11 @@ using Plots
 
 Let us define the problem with the [`@def`](@ref) macro:
 
+```@raw html
+<div class="responsive-columns-left-priority">
+<div>
+```
+
 ```@example main
 t0 = 0
 tf = 1
@@ -47,12 +49,35 @@ ocp = @def begin
     u ∈ R, control
     x(t0) == x0
     x(tf) == xf
-    ẋ(t) == [x₂(t), u(t)]
+    ẋ(t) == [x₂(t), u(t)]
     0.5∫( u(t)^2 ) → min
 end
 nothing # hide
 ```
 
+```@raw html
+</div>
+<div>
+```
+
+### Mathematical formulation
+
+```math
+    \begin{aligned}
+        & \text{Minimise} && \frac{1}{2}\int_0^1 u^2(t) \,\mathrm{d}t \\
+        & \text{subject to} \\
+        & && \dot{x}_1(t) = x_2(t), \\[0.5em]
+        & && \dot{x}_2(t) = u(t), \\[1.0em]
+        & && x(0) = (-1,0), \\[0.5em] 
+        & && x(1) = (0,0).
+    \end{aligned}
+```
+
+```@raw html
+</div>
+</div>
+```
+
 !!! note "Nota bene"
 
     For a comprehensive introduction to the syntax used above to define the optimal control problem, see [this abstract syntax tutorial](@ref manual-abstract-syntax). In particular, non-Unicode alternatives are available for derivatives, integrals, *etc.*
@@ -142,12 +167,12 @@ plot(sol)
 
 ## State constraint
 
-### Direct method
+### Direct method: constrained case
 
 We add the path constraint
 
 ```math
-x_2(t) \le 1.2.
+    x_2(t) \le 1.2.
 ```
 
 Let us model, solve and plot the optimal control problem with this constraint.
@@ -175,7 +200,7 @@ sol = solve(ocp)
 plt = plot(sol; label="Direct", size=(800, 600))
 ```
 
-### Indirect method
+### Indirect method: constrained case
 
 The pseudo-Hamiltonian is (considering the normal case):
 
@@ -186,13 +211,13 @@ H(x, p, u, \mu) = p_1 x_2 + p_2 u - \frac{u^2}{2} + \mu\, c(x),
 with $c(x) = x_2 - a$. Along a boundary arc we have $c(x(t)) = 0$. Differentiating, we obtain:
 
 ```math
-	\frac{\mathrm{d}}{\mathrm{d}t}c(x(t)) = \dot{x}_2(t) = u(t) = 0.
+    \frac{\mathrm{d}}{\mathrm{d}t}c(x(t)) = \dot{x}_2(t) = u(t) = 0.
 ```
 
-The zero control is maximising; hence, $p_2(t) = 0$ along the boundary arc. 
+The zero control is maximising; hence, $p_2(t) = 0$ along the boundary arc.
 
 ```math
-	\dot{p}_2(t) = -p_1(t) - \mu(t) \quad \Rightarrow \mu(t) = -p_1(t).
+    \dot{p}_2(t) = -p_1(t) - \mu(t) \quad \Rightarrow \mu(t) = -p_1(t).
 ```
 
 Since the adjoint vector is continuous at the entry time $t_1$ and the exit time $t_2$, we have four unknowns: the initial costate $p_0 \in \mathbb{R}^2$ and the times $t_1$ and $t_2$. We need four equations: the target condition provides two, reaching the constraint at time $t_1$ gives $c(x(t_1)) = 0$, and finally $p_2(t_1) = 0$.
@@ -253,4 +278,4 @@ flow_sol = φ((t0, tf), x0, p0; saveat=range(t0, tf, 100))
 
 # plot the solution on the previous plot
 plot!(plt, flow_sol; label="Indirect", color=2, linestyle=:dash)
-```
+```

docs/src/example-double-integrator-time.md

@@ -32,6 +32,60 @@ using Plots
 
 Let us define the problem:
 
+```@raw html
+<div class="responsive-columns-left-priority">
+<div>
+```
+
+```@example main
+ocp = @def begin
+
+    tf ∈ R,          variable
+    t ∈ [0, tf],     time
+    x = (q, v) ∈ R², state
+    u ∈ R,           control
+
+    -1 ≤ u(t) ≤ 1
+
+    q(0)  == -1
+    v(0)  == 0
+    q(tf) == 0
+    v(tf) == 0
+
+    ẋ(t) == [v(t), u(t)]
+
+    tf → min
+
+end
+nothing # hide
+```
+
+```@raw html
+</div>
+<div>
+```
+
+### Mathematical formulation
+
+```math
+    \begin{aligned}
+        & \text{Minimise} && t_f \\[0.5em]
+        & \text{subject to} \\[0.5em]
+        & && \dot q(t) = v(t), \\
+        & && \dot v(t) = u(t), \\[0.5em]
+        & && -1 \le u(t) \le 1, \\[0.5em]
+        & && q(0) = -1, \\[0.5em]
+        & && v(0) = 0, \\[0.5em]
+        & && q(t_f) = 0, \\[0.5em]
+        & && v(t_f) = 0.
+    \end{aligned}
+```
+
+```@raw html
+  </div>
+</div>
+```
+
 ```@example main
 ocp = @def begin