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.
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- One or more equations did not get rendered due to their size.
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Sequential continuity on U means sequential continuity at every point of
U.
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- sequentialContinuous U F = ∀ (p : ↑U), sequentialContinuousAt U F p
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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].
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.
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.
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.
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.
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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.