Euler Equations — Question 2

Question 2 – Solution

Strategy. A repeated exponential in logarithmic time produces a logarithmic factor in the original variable.

Step 1: Construct a genuine fundamental pair. With $t=\ln x$, the equation becomes $Y''-4Y'+4Y=0$, whose characteristic polynomial is $(m-2)^2$. Thus $$y=x^2(A+B\ln x).$$ For $u=x^2$, $v=x^2\ln x$, the Wronskian is $W=x^2(2x\ln x+x)-2x(x^2\ln x)=x^3\ne 0$ on $x>0$.

Step 2: Use both initial values. At $1$, $y(1)=A$ and $y'(1)=2A+B$. Hence $A=B=1$, giving $$\boxed{y=x^2(1+\ln x).}$$ Because $x^2>0$, the only positive zero is $\boxed{x=e^{-1}}$.

Step 3: Test successive derivatives at zero. For this solution, $$y'=x(3+2\ln x),\qquad y''=5+2\ln x.$$ As $x\to 0^+$, $y\to 0$ and $y'\to 0$, but $y''\to-\infty$. Defining $y(0)=0$ gives a right-hand $C^1$ extension with derivative $0$: indeed $y(x)/x=x(1+\ln x)\to 0$. It can be joined to zero on the negative side as a $C^1$ function, but no $C^2$ extension is possible.

Step 4: Classify classical extensions. In general, $y''=2A+B(2\ln x+3)$, so a finite second-derivative limit requires $B=0$. On $x<0$ the general solution is $x^2(C+D\ln|x|)$. A $C^2$ join similarly forces $D=0$, and matching second derivatives forces $C=A$. Thus precisely $\boxed{y=Ax^2}$ extend through $0$ as $C^2$ solutions of the original equation. Each satisfies $y(0)=y'(0)=0$, so these data do not give uniqueness at the singular point.