Euler Equations — Question 6

Question 6 – Solution

Strategy. Integrate the weighted derivative first, then use regularity at zero to remove the logarithmic homogeneous mode.

Step 1: Integrate the Euler operator. Dividing by $x$ gives $(xy')'=x/(1+x^2)$. Hence $$xy'=\tfrac 12\ln(1+x^2)+C,\qquad y=\frac 12\int_1^x\frac{\ln(1+s^2)}s\,ds+C\ln x+D.$$ These formulas display both constants and solve the equation for every $x>0$.

Step 2: Apply the singular-endpoint conditions. Since $\ln(1+s^2)/s\sim s$ at zero, its integral there converges. Boundedness therefore forces $C=0$; the value $y(0)=0$ fixes the remaining constant. The unique selected solution is $$\boxed{y(x)=\frac 12\int_0^x\frac{\ln(1+s^2)}s\,ds.}$$ Any two bounded solutions differ by a constant, and their common value at zero makes that constant zero.

Step 3: Expand the integral with justification. The divided logarithmic series $\sum_{n\ge 1}(-1)^{n-1}s^{2n-1}/n$ is uniformly convergent on $|s|\le\rho<1$, by comparison with $\sum_{n\ge 1}\rho^{2n-1}/n$. Termwise integration therefore yields, for $|x|<1$, $$\boxed{y(x)=\frac 14\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^{2n}}{n^2} =\frac{x^2}{4}-\frac{x^4}{16}+\frac{x^6}{36}-\cdots.}$$ The ratio of successive nonzero terms tends to $x^2$, so the radius is $1$. At $x=\pm 1$ the series converges absolutely by comparison with $\sum 1/n^2$. The series and the even integral extension agree there by continuity: the series is uniformly convergent on $[-1,1]$ by the same comparison.

Step 4: Recover derivatives and a certificate. The local power series supplies a smooth even extension with $y'(0)=0$ and $y''(0)=1/2$. With the forcing defined continuously as zero at $0$, the original equation holds there as well. On $[0,1]$, the alternating term magnitudes decrease, so for $P_4=x^2/4-x^4/16$, $$\boxed{0\le y(x)-P_4(x)\le\frac{x^6}{36}.}$$ The bound includes the endpoint $x=1$; it is a certificate, not a claim that the two-term approximation is highly accurate there.