Euler Equations — Question 3

Question 3 – Solution

Strategy. Use $t=\ln x$ to reveal a harmonic oscillator and then translate its time scale back to $x$.

Step 1: Solve the transformed oscillator. The transformed equation is $Y''+4Y=0$, so $y=A\cos(2\ln x)+B\sin(2\ln x)$. At $x=1$, $A=0$ and $2B=y'(1)=2$, hence $$\boxed{y(x)=\sin(2\ln x),\qquad x>0.}$$ The characteristic exponents are $m=\pm 2i$.

Step 2: Locate the geometric sequence of zeros. Zeros occur when $2\ln x=k\pi$: $$\boxed{x_k=e^{k\pi/2},\quad k\in\mathbb Z,\qquad x_{k+1}/x_k=e^{\pi/2}.}$$ They accumulate at $0$ as $k\to-\infty$ and grow without bound as $k\to\infty$. There is no smallest positive zero.

Step 3: Separate boundedness from a limit. The IVP solution has $|y|\le 1$, but sequences with $\ln x=\pi/4-k\pi$ and $\ln x=3\pi/4-k\pi$ tend to $0$ while giving values $1$ and $-1$. Thus no limit at zero exists. Any nonzero real solution is a nonzero-amplitude sinusoid in $\ln x$ and likewise attains two distinct limiting values along sequences approaching zero. Only the zero solution extends continuously there.

Step 4: Read the scaled phase curve. For the IVP, $z=\cos(2\ln x)$, so $\boxed{y^2+z^2=1}$. In logarithmic time, $\dot y=2z$, $\dot z=-2y$; at $(y,z)=(0,1)$ the motion is to the right. The circle is traversed clockwise as $x$ increases, because $\ln x$ increases with $x$. The horizontal coordinate here is $y$, not $x$; equal scales preserve the unit circle.