Update tutorials: MAM continuation improvements, other tutorial adjustments

Author: ocots

docs/src/assets/Manifest.toml

@@ -1251,9 +1251,9 @@ version = "2.13.0"
 
 [[deps.LinearSolve]]
 deps = ["ArrayInterface", "ConcreteStructs", "DocStringExtensions", "EnumX", "GPUArraysCore", "InteractiveUtils", "Krylov", "Libdl", "LinearAlgebra", "MKL_jll", "Markdown", "OpenBLAS_jll", "PrecompileTools", "Preferences", "RecursiveArrayTools", "Reexport", "SciMLBase", "SciMLLogging", "SciMLOperators", "Setfield", "StaticArraysCore"]
-git-tree-sha1 = "06aa84654f44350d5cf3c678702ca48795c12e54"
+git-tree-sha1 = "42b5cb44317e89ef75dd841c9c8eba9045bf9ff0"
 uuid = "7ed4a6bd-45f5-4d41-b270-4a48e9bafcae"
-version = "3.74.0"
+version = "3.75.0"
 
     [deps.LinearSolve.extensions]
     LinearSolveAMDGPUExt = "AMDGPU"

docs/src/tutorial-discretisation.md

@@ -78,7 +78,7 @@ sol = solve(ocp; scheme=:midpoint, display=false)
 plot(sol; size=(800, 800))
 ```
 
-Let us now compare different discretization schemes to evaluate their accuracy and performance. See the [Strategy options](@id manual-solve-strategy-options) for information on default values for parameters like `tol`.
+Let us now compare different discretization schemes to evaluate their accuracy and performance. See the [Strategy options](@extref OptimalControl manual-solve-strategy-options) for information on default values for parameters like `tol`.
 
 ```@example main-disc
 # Solve the problem with different discretization methods

docs/src/tutorial-lqr.md

@@ -1,9 +1,5 @@
 # A simple Linear–quadratic regulator example
 
-```@meta
-Draft = false
-```
-
 ## Problem statement
 
 We consider the following Linear Quadratic Regulator (LQR) problem, which consists in minimizing

docs/src/tutorial-mam.md

@@ -145,8 +145,9 @@ function continuation_mam(Ts; init_guess=init)
     iterations_list = Int[]
     current_sol = init_guess
     
+    opts = (display=false, grid_size=1000, tol=1e-8, print_level=0)
     for T in Ts
-        current_sol = solve(ocp(T); display=false, init=current_sol, grid_size=1000, tol=1e-8)
+        current_sol = solve(ocp(T); init=current_sol, opts...)
         push!(objectives, objective(current_sol))
         push!(iterations_list, iterations(current_sol))
     end
@@ -155,7 +156,7 @@ function continuation_mam(Ts; init_guess=init)
 end
 
 # Perform continuation
-Ts = range(1, 100, 100)
+Ts = range(50, 100, 101)
 objectives, iters, final_sol = continuation_mam(Ts)
 
 # Display results
@@ -186,14 +187,6 @@ vline!(plt1, [T_min], label="Minimum at T=$(round(T_min, digits=1))", z_order=:b
 plot(plt1, size=(800,400))
 ```
 
-We now focus on the region around the minimum for a clearer view:
-
-```@example main-mam
-plt2 = scatter(Ts[40:100], log10.(objectives[40:100]), xlabel="Time", label="Objective (log10)")
-vline!(plt2, [T_min], label="Minimum", z_order=:back)
-plot(plt2, size=(800,400))
-```
-
 !!! note "Interpretation"
 
     The optimal transition time $T^*$ balances two competing effects: shorter times require larger deviations from the deterministic flow (higher action), while longer times allow the system to follow the flow more closely. The minimum represents the most probable transition time in the small noise limit.