Basic Concepts for nth Order Linear Equations — Question 9

Question 9 – Solution

Strategy. Construct actual delayed solutions; continuity of a nonlinear right-hand side alone is not a uniqueness proof.

Step 1: Check linearity, not just the leading coefficient. The equation is third order but nonlinear because $\sqrt{|y|}$ is not a linear function of the unknown. A leading coefficient of $1$ and continuity do not make the linear existence-and-uniqueness theorem applicable.

Step 2: Balance exponents and amplitudes. For $s=x-\tau>0$, the proposed branch has $y'''=Cp(p-1)(p-2)s^{p-3}$ and $\sqrt{|y|}=\sqrt C\,s^{p/2}$. Since $p>3$ and $C>0$, equality requires $p-3=p/2$, giving $p=6$. Then $120C=\sqrt C$, so $\boxed{C=1/14400}$.

Step 3: Verify the join and the initial state. Define $$\boxed{y_\tau(x)=\begin{cases}0,&x\le\tau,\\(x-\tau)^6/14400,&x>\tau.\end{cases}}$$ Derivatives through order five approach zero from the right and equal zero on the left; difference quotients at the join give the same derivatives. Thus $y_\tau\in C^5$, and the equation holds at the join too, since $y'''=\sqrt{|y|}=0$ there. The sixth derivative jumps from $0$ to $6!/14400=1/20$, so it is not $C^6$ at $\tau$. Because $0\le\tau$, all three initial data vanish, including when $\tau=0$.

Step 4: Identify the precise contrast. Distinct delays give distinct solutions with identical initial data; the identically zero function is another solution. This proves nonuniqueness constructively. For the linear equation $y'''-y=0$, continuous normalized coefficients and the same zero initial vector force the unique zero solution. Here $\sqrt h/h=1/\sqrt h\to\infty$, so the nonlinear right-hand side is not locally Lipschitz in $y$ at zero. Nonlinearity alone does not always cause nonuniqueness; the displayed family establishes the failure in this specific case.