Definitions — Question 5

PDF ↗

Question 5

Consider the initial value problem y′=2y,y(0)=0,y'=2\sqrt{y},\qquad y(0)=0, where the square root is the nonnegative real root. For each real parameter a≥0a\ge 0, define on ℝ\mathbb R ya(x)={0,x≤a,(x−a)2,x>a.y_a(x)=\begin{cases} 0,&x\le a,\\ (x-a)^2,&x>a. \end{cases} Tasks

  1. Prove that yay_a is continuously differentiable at its joining point x=ax=a.

  2. Verify the differential equation everywhere, including at x=ax=a, and verify the initial value.

  3. Show that different parameters give distinct solutions of the same IVP. Identify an additional solution that never leaves 00.

  4. Sketch y0y_0, y1y_1, and y2y_2 in your solution and explain what this example shows about uniqueness. Does it mean that every first-order IVP is nonunique?

Original worksheet page 1: question and worked solution for 1-1-005
Show solutionHide solution

Question 5 – Solution

Strategy. Check the joining point directly; solving on the two sides alone is insufficient. Then compare solutions with the same initial value.

Step 1: Smooth joining. Both pieces tend to 00 as x→ax\to a. At aa, the difference quotient is 00 for h<0h<0 and h2/h=hh^2/h=h for h>0h>0. Hence ya′(a)=0y_a'(a)=0 and ya′(x)={0,x≤a,2(x−a),x>a.y_a'(x)=\begin{cases}0,&x\le a,\\2(x-a),&x>a.\end{cases} This derivative is continuous at aa, so yay_a is C1C^1 on ℝ\mathbb R.

Step 2: Equation and initial value. For x≤ax\le a, both ya′y_a' and 2ya2\sqrt{y_a} are 00. For x>ax>a, x−a>0x-a>0, so 2(x−a)2=2|x−a|=2(x−a)=ya′.2\sqrt{(x-a)^2}=2|x-a|=2(x-a)=y_a'. Since a≥0a\ge 0, ya(0)=0y_a(0)=0. Thus

See the diagram in the original worksheet below.

Step 3: Distinctness. If 0≤a<b0\le a<b, choose a<x<ba<x<b. Then ya(x)>0y_a(x)>0 but yb(x)=0y_b(x)=0. Also y≡0y\equiv 0 is a solution distinct from every finite-parameter yay_a.

Step 4: Meaning. This IVP has infinitely many solutions: the initial value does not determine when a solution leaves 00. It disproves automatic uniqueness for all first-order IVPs; it does not assert nonuniqueness for every IVP. No general uniqueness theorem is needed for this counterexample.

Original worksheet page 2: question and worked solution for 1-1-005

Original worksheet layout. Use Enlarge or open the PDF for a closer view.