Documentation

Books.IntroductiontoRealAnalysisVolumeI_JiriLebl_2025.Chapters.Chap06.section03

def sequentialContinuousAt (U : Set ( × )) (F : U) (p : U) :

Definition 6.3.1: For a set U ⊆ ℝ², a function F : U → ℝ is continuous at (x, y) ∈ U if for every sequence (xₙ, yₙ) of points of U with xₙ → x and yₙ → y, we have F (xₙ, yₙ) → F (x, y). It is continuous if it is continuous at every point of U.

Equations
  • One or more equations did not get rendered due to their size.
Instances For
    def sequentialContinuous (U : Set ( × )) (F : U) :

    Sequential continuity on U means sequential continuity at every point of U.

    Equations
    Instances For
      theorem tendsto_subtype_coord_iff {U : Set ( × )} {p : U} {s : U} :
      Filter.Tendsto s Filter.atTop (nhds p) Filter.Tendsto (fun (n : ) => (↑(s n)).1) Filter.atTop (nhds (↑p).1) Filter.Tendsto (fun (n : ) => (↑(s n)).2) Filter.atTop (nhds (↑p).2)
      theorem sequentialContinuousAt_iff {U : Set ( × )} {F : U} {p : U} :

      On ℝ × ℝ, sequential continuity at a point agrees with the usual notion of continuity at that point.

      Sequential continuity on U is equivalent to the usual notion of continuity.

      theorem picard_existence_uniqueness {a b c d x₀ : } (hx₀ : x₀ Set.Ioo a b) {y₀ : } (hy₀ : y₀ Set.Ioo c d) (F : ) (hcont : ContinuousOn (fun (p : × ) => F p.1 p.2) (Set.Icc a b ×ˢ Set.Icc c d)) (hLip : ∃ (L : ), 0 L xSet.Icc a b, ySet.Icc c d, zSet.Icc c d, |F x y - F x z| L * |y - z|) :
      ∃ (h : ), 0 < h Set.Icc (x₀ - h) (x₀ + h) Set.Icc a b ∃ (f : ), ContinuousOn f (Set.Icc (x₀ - h) (x₀ + h)) (∀ xSet.Icc (x₀ - h) (x₀ + h), f x Set.Icc c d) (∀ xSet.Ioo (x₀ - h) (x₀ + h), HasDerivAt f (F x (f x)) x) f x₀ = y₀ ∀ (g : ), ContinuousOn g (Set.Icc (x₀ - h) (x₀ + h))(∀ xSet.Icc (x₀ - h) (x₀ + h), g x Set.Icc c d)(∀ xSet.Ioo (x₀ - h) (x₀ + h), HasDerivAt g (F x (g x)) x)g x₀ = y₀Set.EqOn g f (Set.Icc (x₀ - h) (x₀ + h))

      Theorem 6.3.2 (Picard's theorem on existence and uniqueness). Let I = [a,b] and J = [c,d] be closed bounded intervals with interiors (a,b) and (c,d), and take (x₀, y₀) ∈ (a,b) × (c,d). If F : ℝ → ℝ → ℝ is continuous on I × J and Lipschitz in the second variable with constant L ≥ 0, meaning |F x y - F x z| ≤ L * |y - z| for all x ∈ I and y z ∈ J, then there is h > 0 with [x₀ - h, x₀ + h] ⊆ I and a unique differentiable function f : ℝ → ℝ with values in J satisfying f x₀ = y₀ and f' x = F x (f x) on [x₀ - h, x₀ + h].

      theorem picard_exponential_example :
      ∃ (h : ), 0 < h h < 1 / 2 ∃ (f : ), (∀ xSet.Icc (-h) h, HasDerivAt f (f x) x) f 0 = 1 ∀ (g : ), (∀ xSet.Icc (-h) h, HasDerivAt g (g x) x) g 0 = 1Set.EqOn g f (Set.Icc (-h) h)

      Example 6.3.3: Applying Picard's theorem to the initial value problem f' = f with f 0 = 1 (that is, F x y = y) produces some h > 0 with h < 1/2 and a function f : ℝ → ℝ defined on [-h, h] such that f' x = f x on that interval and f 0 = 1, and any other solution with the same initial value agrees with f on [-h, h]. The function extends globally as exp, so the exponential is the unique global solution with that initial value.

      theorem exp_unique_solution :
      ∃! f : , (∀ (x : ), HasDerivAt f (f x) x) f 0 = 1

      The exponential is the unique globally defined solution of the ODE f' = f satisfying f 0 = 1.

      theorem one_sub_ne_zero_of_lt {x : } (hx : x < 1) :
      1 - x 0
      theorem hasDerivAt_inv_one_sub {x : } (hx : x < 1) :
      HasDerivAt (fun (x : ) => (1 - x)⁻¹) ((1 - x) ^ 2)⁻¹ x
      theorem quadratic_ode_example :
      ∃ (f : ), (∀ x < 1, HasDerivAt f (f x ^ 2) x) f 0 = 1 x < 1, f x = (1 - x)⁻¹

      Example 6.3.4: For the ODE f' = f ^ 2 with initial condition f 0 = 1, the solution is f x = (1 - x)⁻¹, defined on the interval (-∞, 1). The nonlinearity y ↦ y^2 is not globally Lipschitz, so the guaranteed existence interval from Picard's theorem shrinks as the initial value approaches x = 1.

      theorem sqrt_abs_ode_nonunique :
      ∃ (f : ) (g : ), f g f 0 = 0 g 0 = 0 (∀ (x : ), HasDerivAt f (2 * |f x|) x) ∀ (x : ), HasDerivAt g (2 * |g x|) x

      Example 6.3.5: For the initial value problem f' x = 2 * √|f x| with f 0 = 0, the right-hand side F x y = 2 * √|y| is continuous but not Lipschitz in y, so uniqueness can fail. The piecewise function f x = x^2 for x ≥ 0 and f x = -x^2 for x < 0 is a solution, while the zero function is another, so a solution exists but is not unique.

      noncomputable def dirichletPhi :

      Example 6.3.6: Let φ x = 0 when x is rational and φ x = 1 otherwise. The ODE y' = φ has no solution for any initial condition, since the right-hand side is discontinuous and derivatives have the Darboux property, so no differentiable function can have φ as its derivative.

      Equations
      Instances For
        theorem dirichletPhi_value_rational {x : } (hx : x Set.range fun (r : ) => r) :
        theorem dirichletPhi_value_irrational {x : } (hx : xSet.range fun (r : ) => r) :
        theorem no_solution_dirichlet_ode :
        ¬∃ (f : ), Differentiable f ∀ (x : ), HasDerivAt f (dirichletPhi x) x

        No differentiable function can solve the discontinuous ODE y' = φ where φ is the Dirichlet function. In particular, there is no solution satisfying any prescribed initial value.