Euler Equations — Question 9

Question 9 – Solution

Strategy. Track both the power envelope and the nonvanishing oscillation amplitude after every differentiation.

Step 1: Differentiate the complex powers. The exponents are $\alpha\pm i\beta$, so the real solutions are $$y=x^\alpha\big(A\cos(\beta\ln x)+B\sin(\beta\ln x)\big).$$ The $j$th derivative of $x^{\alpha+i\beta}$ is $\prod_{\ell=0}^{j-1}(\alpha-\ell+i\beta)x^{\alpha-j+i\beta}$. Since $\beta>0$, none of these factors vanishes. A nonzero real solution therefore retains a nonzero sinusoidal amplitude at every derivative order.

Step 2: Establish the exact regularity threshold. If $\alpha>k$, all derivatives through order $k$ tend to zero. Their extensions are consistent derivatives: for $j<k$, $y^{(j)}(x)/x=O(x^{\alpha-j-1})\to 0$. If $\alpha=k$, the $k$th derivative has constant nonzero amplitude and no limit. If $\alpha<k$, that derivative has an unbounded envelope and attains unbounded values along a sequence approaching zero. Hence $$\boxed{\text{A nonzero solution extends as }C^k\text{ to }0\iff\alpha>k.}$$ This includes $k=0$; bounded oscillation at equality is not continuity.

Step 3: Inspect the concrete example. For $\alpha=5/2$, $\beta=1$, $$\boxed{y''=x^{1/2}\left(\frac{11}{4}\sin(\ln x)+4\cos(\ln x)\right).}$$ The solution extends as $C^0,C^1,C^2$, with all three center values zero, but not as $C^3$ or any higher integer order. The plot shows $y''(e^t)$ and its decaying envelope as $t\to-\infty$.

Step 4: Verify the joined classical solution. The equation is $x^2y''-4xy'+(29/4)y=0$. On $x>0$ the formula solves it; on $x<0$ the zero function does. The zero limits through the second derivative make the join $C^2$, and at $0$ every term in the original equation is zero. Both this nonzero joined solution and the identically zero solution have $y(0)=y'(0)=0$. Classical IVP uniqueness fails at this singular point.