Final Thoughts: Question 1-3.6 & Solution

Question 6

Compare an equation with its squared version: $\text{(A)}\quad y'=-\sqrt{1-y^2},\qquad |y|\le 1,$ $\text{(B)}\quad (y')^2=1-y^2.$ Both are assigned $y(0)=0$ . Consider $u(x)=\sin x$ and $v(x)=-\sin x$ .

Tasks

Verify that both candidates satisfy (B) and the initial value. Which satisfies (A) on $(-\pi/2,\pi/2)$ ?
State the sign restriction lost by squaring. Give a condition that, together with (B), recovers (A) for real differentiable functions.
Explain why $v=-\sin x$ does not satisfy (A) on all of $\mathbb R$ .
Extend the accepted middle branch to a $C^1$ solution of (A) on all of $\mathbb R$ by adjoining constant pieces outside $[-\pi/2,\pi/2]$ . Verify the joins and sketch the extension in your solution.

Original worksheet page 1: question and worked solution for 1-3-006

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Question 6 – Solution

Strategy. Squaring forgets a sign. Restore that sign when selecting a branch, and check differentiability when extending it.

Step 1: Compare the candidates. Both satisfy $(y')^2=\cos^2x=1-y^2$ and have value zero at $0$ . On $(-\pi/2,\pi/2)$ , $\cos x>0$ , so the right-hand side of (A) is $-|\cos x|=-\cos x$ . Hence $\boxed{v=-\sin x\text{ satisfies (A), but }u=\sin x\text{ does not.}}$

Step 2: Recover the lost information. Equation (A) forces $y'\le 0$ . Conversely, (B) and $y'\le 0$ imply $|y|\le 1$ and $y'=-\sqrt{1-y^2}$ . Outside the middle interval, $v'=-\cos x$ is positive wherever $\cos x<0$ , so $v$ fails (A) there.

Step 3: A global extension. Define $\boxed{Y(x)=\begin{cases} 1,&x\le-\pi/2,\\ -\sin x,&-\pi/2<x<\pi/2,\\ -1,&x\ge\pi/2. \end{cases}}$

See the diagram in the original worksheet below.

At either join, the middle value matches the adjoining constant and its derivative $-\cos x$ tends to zero, the derivative of the constant piece. Thus $Y$ is $C^1$ . On the constant pieces and at the joins, both sides of (A) are zero. On the middle piece, the verification above applies. Therefore $Y$ solves the original IVP on $\mathbb R$ ; extension is possible here because both endpoint values and derivatives remain finite.

Original worksheet page 2: question and worked solution for 1-3-006

Final Thoughts — Question 6

Question 6

Question 6 – Solution