Systems of Differential Equations — Question 4

Question 4 – Solution

Strategy. Retain the state coordinate that division by $t$ conceals at the singular point.

Step 1: Eliminate on the regular subintervals. Since $x''=tw$, for $t\ne 0$ we have $w=x''/t$. Differentiating gives $$\boxed{t x'''-x''=0.}$$ The leading coefficient vanishes at zero. This singularity came from the chosen scalar coordinate; the original system matrix is continuous everywhere.

Step 2: Solve before dividing. Integrating successively gives $$\boxed{w=c,\qquad z=b+ct^2/2,\qquad x=a+bt+ct^3/6.}$$ Thus $(x,x',x'')(0)=(a,b,0)$, while $x'''(0)=c$. The missing state coordinate is encoded in the third derivative, not in the usual second derivative.

Step 3: Check global equivalence. On either side of zero, $(x''/t)'=0$, so $x=a_\pm+b_\pm t+c_\pm t^3/6$. Continuity of $x,x'$ matches $a_\pm,b_\pm$, and continuity of $x'''$ matches $c_\pm$. All $C^3$ solutions therefore have the single global polynomial form above. Reconstruct $z=x'$ and $w=x''/t$ off zero, setting $w(0)=x'''(0)$. These are smooth and satisfy all three system equations. At zero, the scalar equation itself requires $x''(0)=0$.

Step 4: Explain the apparent nonuniqueness. The family $x=ct^3/6$ has $(x,x',x'')(0)=(0,0,0)$ for every $c$. It corresponds to distinct system data $(0,0,c)$, so it does not contradict uniqueness for the regular system. The plotted members use $c=-6,0,6$.