Basic Concepts for nth Order Linear Equations — Question 7

Question 7 – Solution

Strategy. Solve separately where the leading coefficient is nonzero, then impose the actual regularity required at the join.

Step 1: Solve on both regular components. For $x\ne 0$, $xz'-2z=0$ gives $z=Cx^2$. Two integrations, with the constant absorbed into the quartic coefficient, yield $$y=\begin{cases}A_+x^4+B_+x+C_+,&x>0,\\ A_-x^4+B_-x+C_-,&x<0. \end{cases}$$ The constants on the two components are initially independent.

Step 2: Impose a classical join. Continuity of $y$ and $y'$ forces $C_+=C_-=C$ and $B_+=B_-=B$. The second and third derivatives are $12A_\pm x^2$ and $24A_\pm x$, both tending to zero, so $C^3$ regularity imposes no relation between $A_+$ and $A_-$. Every global solution is therefore $$\boxed{y(x)=C+Bx+\begin{cases}A_+x^4,&x\ge 0,\\A_-x^4,&x<0,\end{cases}}$$ with four independent constants. At zero the equation reduces to $-2y''(0)=0$, which these functions satisfy. This is a four-dimensional solution space.

Step 3: Classify the initial data. Every solution has $y(0)=C$, $y'(0)=B$ and $y''(0)=0$. Thus $\boxed{\gamma=0}$ is necessary and sufficient. When it holds, $C=\alpha$, $B=\beta$ and both $A_+,A_-$ remain arbitrary. When $\gamma\ne 0$, no classical solution exists.

Step 4: Check the extra derivative. The one-sided fourth derivatives are $24A_+$ and $24A_-$. A $C^4$ join requires and is guaranteed by $A_+=A_-=A$, giving the polynomial $C+Bx+Ax^4$. One free parameter $A$ still survives for the compatible initial data, so uniqueness is not restored. The leading coefficient vanishes at $0$, and normalization introduces $-2/x$. The hypotheses guaranteeing dimension three and arbitrary unique initial data hold only on the separate half-lines.