Fundamental Sets of Solutions — Question 6

Question 6 – Solution

Strategy. Each connected component has its own integration constants because zero is excluded and no matching rule is imposed.

Step 1: Verify and test independence. Both functions are constant on each component, so both have zero second derivative on $D$. If $Au+Bv=0$ on $D$, the positive side gives $A+B=0$ and the negative side gives $A-B=0$. Thus $A=B=0$.

Step 2: Test spanning. Every combination $Au+Bv$ is constant on each side, so it cannot represent the solution $y=t$. The pair is independent on $D$ but does not span its solution space. On either individual component, $v=u$ or $v=-u$, so the restricted pair is dependent.

Step 3: Exhibit all four freedoms. Integrating separately gives $$\boxed{y(t)=\begin{cases}A_-+B_-t,&t<0,\\A_++B_+t,&t>0.\end{cases}}$$ Let $\chi_+$ equal $1$ for $t>0$ and $0$ for $t<0$, and let $\chi_-=1-\chi_+$. An independent spanning set is $$\{\chi_-,\ t\chi_-,\ \chi_+,\ t\chi_+\}.$$ Restriction to either side proves independence by the independence of $1,t$. Each of the four displayed constants can be chosen freely, so the space has dimension four and no set with fewer than four functions spans it. All four functions are smooth on $D$.

Step 4: Impose data on one component. On the positive side, $B_+=1$ and $A_+=-1$. Thus $y=t-1$ for $t>0$, while $A_-,B_-$ remain arbitrary for $t<0$. Initial-value uniqueness controls the connected interval containing the initial point; it supplies no information across the excluded point.