Euler Equations — Question 7

Question 7 – Solution

Strategy. In logarithmic coordinates, the two endpoint conditions reduce to elementary trigonometric or hyperbolic equations.

Step 1: Exclude nonpositive parameters. Set $Y(t)=y(e^t)$. Then $Y''+\lambda Y=0$, with $Y(0)=Y(L)=0$. If $\lambda=0$, $Y=A+Bt$ and both constants must vanish. If $\lambda=-\mu^2<0$, $Y=A\cosh(\mu t)+B\sinh(\mu t)$. The first condition gives $A=0$; since $\sinh(\mu L)>0$, the second gives $B=0$.

Step 2: Find all nonzero families. For $\lambda=\omega^2>0$, the first condition leaves $Y=B\sin(\omega t)$. A nonzero $B$ requires $\sin(\omega L)=0$, so $$\boxed{\lambda_n=(n\pi/L)^2,\qquad y=C\sin\!\left(\frac{n\pi\ln x}{L}\right),\quad n=1,2,\ldots.}$$ There are no other real parameter values admitting nonzero solutions.

Step 3: Normalize and locate the zeros. At $x=1$, $y'(1)=Y'(0)=C n\pi/L$. Thus $$\boxed{y_n(x)=\frac{L}{n\pi}\sin\!\left(\frac{n\pi\ln x}{L}\right).}$$ Its zeros on the specified interval are $x_k=e^{kL/n}$ for $k=0,\ldots,n$. Exactly $n-1$ lie strictly inside; they are equally spaced in $\ln x$, not in $x$.

Step 4: Distinguish endpoint and initial data. The original endpoint conditions do not specify the initial slope. At an admissible parameter, every amplitude satisfies both endpoints, including zero; hence those data are not unique. Once $y'(1)=1$ is imposed, the usual IVP uniqueness applies. For $L=\pi$, the first three normalized profiles are $Y_n(t)=\sin(nt)/n$, $n=1,2,3$, with $0,1,2$ interior zeros, respectively. The figure uses logarithmic horizontal coordinates.