3. Does an Etch-A-Sketch exist? Etch-A-Sketch is (allegedly) a drawing mechanism where a curve is drawn by toying with two knobs that continuously vary curvature. So, in order for an Etch-A-Sketch to exist, the graph of any function y = y(x) must be drawable by such means. Is this true? Is a curve uniquely determined by how it curves?

The curvature of a curve y = y(x) at (x, y(x)) is given by
κ(x) = y′′(x) .

(1 + (y′(x))^2)^3/2 The goal in this problem to try to recover y(x) if we are given κ(x).

(a) Let Y = y′. Under what conditions (on κ(x) or (x0, y1) if any) can we guarantee that there exists a unique

solution to
passing through the point (x0, y1)?

Y′ = κ(x)·(1+Y2)^3/2

 

 

(b) Find an explicit solution to the ODE
subject to the initial condition Y (x0) = 0. Your answer will be in terms of the function

Y′ = κ(x)·(1+Y2)3/2 ∫x

K(x) = Integral(κ(t) dt)

 

 

(c) Find a function y(x) satisfying y′(0) = 0 and y(0) = 1 whose curvature is κ(x) = cos(x) on [0, π/2).

 

 

 

June 14, 2016 MAT224: Differential Equations
Existence and Uniqueness of Solutions
Consider the Initial Value Problem:
′ y (x) = f (x, y) y(x0 ) = y0 . It would be good to know if a solution to this IVP exists
before going to look for it. So we devote some time now
to that issue. The main tool that gives us sufficient conditions under which we can be guaranteed a (unique) solution is Picard’s Theorem:
Theorem 1 (Picard). Let f (x, y) and ∂f /∂y be continuous functions of x and y on a closed rectangle R with sides
parallel to the x and y axes. If (x0 , y0 ) is on the interior of
R, then there exists h > 0 with the property that the initial
value problem
y ′ (x) = f (x, y) y(x0 ) = y0 has a unique solution on the interval |x − x0 | ≤ h.
Sketch of Proof. First, we convert to an integral equation:
∫ x
y(x) = y0 +
f (t, y) dt. Solution
Note that f (x, y) = y · |y|
where. Also note that 2y
∂f
= 0 ∂y −2y is continuous everyy>0
y=0
y < 0. So ∂f /∂y is continuous everywhere as well. So for
every point (x0 , y0 ), the IVP y ′ = y · |y|, y(x0 ) =
y0 has a unique solution on some interval containing x0 . HW: Due Wednesday, 15-Jun
1. For what points (x0 , y0 ) does the ODE
y2
1 + x2
have a solution passing through (x0 , y0 )?
y′ = x0 We wish to find a continuous solution y(x) to the above
integral equation. Let R = [a, b] × [c, d], and equip
C[a, b], the Vector Space of continuous functions on
[a, b], with the norm
∥g∥ = max |g(x)|.
x∈[a,b] Consider the operator T : C[a, b] → C[a, b] defined by
∫ x
T (g) = y0 +
f (t, g) dt.
x0 The main idea is that the hypotheses on the continuity of f
and its partial derivative force the operator T to be a contraction as viewed on [x0 − h, x0 + h] for h sufficiently 2. For what points (x0 , y0 ) does the ODE
small:
y − x2
.
y′ = 2
y −x
This means that there exhave a solution passing through (x0 , y0 )?
ists a constant K < 1 such
T (D)
that ∥T (g1 ) − T (g2 )∥ ≤
K∥g1 − g2 ∥:
D
A contracting map must have a unique fixed point. In
this case, this means there exists a function y(x) such that
T (y) = y. This is the solution to our differential equation.

Example: For what points (x0 , y0 ) does Picard’s Theorem
guarantee that the initial value problem
y ′ = y · |y|, y(x0 ) = y0 , has a unique solution on some interval |x − x0 | ≤ h? 2 3. Does an Etch-A-Sketch exist? Etch-A-Sketch is (allegedly) a drawing mechanism where a curve is drawn by toying
with two knobs that continuously vary curvature. So, in order for an Etch-A-Sketch to exist, the graph of any
function y = y(x) must be drawable by such means. Is this true? Is a curve uniquely determined by how it curves?
The curvature of a curve y = y(x) at (x, y(x)) is given by
κ(x) = y ′′ (x)
.
(1 + (y ′ (x))2 )3/2 The goal in this problem to try to recover y(x) if we are given κ(x).
(a) Let Y = y ′ . Under what conditions (on κ(x) or (x0 , y1 ) if any) can we guarantee that there exists a unique
solution to
Y ′ = κ(x) · (1 + Y 2 )3/2
passing through the point (x0 , y1 )? (b) Find an explicit solution to the ODE
Y ′ = κ(x) · (1 + Y 2 )3/2
subject to the initial condition Y (x0 ) = 0. Your answer will be in terms of the function
∫ x
κ(t) dt.
K(x) =
x0 (c) Find a function y(x) satisfying y ′ (0) = 0 and y(0) = 1 whose curvature is κ(x) = cos(x) on [0, π/2).

Attachments:

• 1492938401-3 Existence_Uniqueness.pdf