Euler Equations — Question 8

Question 8 – Solution

Strategy. Normalize a symmetric sum and a divided difference before allowing the characteristic roots to merge.

Step 1: Identify the degenerating pair. The power equation is $(m-a)^2-\varepsilon^2=0$, giving $m=a\pm\varepsilon$. Both $x^{a+\varepsilon}$ and $x^{a-\varepsilon}$ tend to $x^a$ as $\varepsilon\to 0$. Their separate limits are the same function, so they cannot remain a fundamental pair.

Step 2: Normalize before taking the limit. For $\varepsilon>0$, set $t=\ln x$ and define $$\boxed{u_\varepsilon=x^a\cosh(\varepsilon t),\qquad v_\varepsilon=x^a\frac{\sinh(\varepsilon t)}{\varepsilon}.}$$ These are the half-sum and the difference divided by $2\varepsilon$ of the two powers. At $x=1$ they have the four required value/slope data by direct differentiation.

Step 3: Check the determinant exactly. Differentiation and $\cosh^2s-\sinh^2s=1$ give $$\boxed{W(u_\varepsilon,v_\varepsilon)=x^{2a-1}>0.}$$ At $\varepsilon=0$, define $u_0=x^a$ and $v_0=x^a\ln x$. They solve the repeated-root equation, have the same normalized data, and their Wronskian is again $x^{2a-1}$. Independence holds for every parameter value.

Step 4: Justify the limit uniformly. On $[r,R]$, let $T=\max(|\ln r|,|\ln R|)$ and $M=\max_{[r,R]}x^a$. For $0<\varepsilon\le 1$, Taylor’s theorem for $\cosh$ and $\sinh$ gives $$|u_\varepsilon-u_0|\le\frac{M\varepsilon^2T^2\cosh T}{2},\qquad |v_\varepsilon-v_0|\le\frac{M\varepsilon^2T^3\cosh T}{6}.$$ Both bounds tend to zero uniformly. The divided difference tends to $\partial_m x^m|_{m=a}=x^a\ln x$, recovering the independent logarithmic solution that the two unnormalized limits lose.