Euler Equations — Question 4

Question 4 – Solution

Strategy. Work with positive $-x$, retaining the sign in the chain rule.

Step 1: Choose a real branch. For $x=-e^t$, $dx/dt=x$, so $Y'=xy'$ and $Y''=xy'+x^2y''$ still hold. The equation becomes $Y''-Y/4=0$. A real pair is $\boxed{(-x)^{1/2},\ (-x)^{-1/2}}$ for $x<0$. The expression $x^{1/2}$ has no real value on this interval; powers of $|x|=-x$ avoid that problem.

Step 2: Translate the slope correctly. At $t=\ln 4$, $Y=3$ and $Y'=xy'=(-4)(-1/8)=1/2$. Writing $y=A\sqrt{-x}+B/\sqrt{-x}$ yields $2A+B/2=3$ and $A-B/4=1/2$, hence $A=1$, $B=2$. Thus $$\boxed{y=\sqrt{-x}+\frac 2{\sqrt{-x}},\qquad y'=-\frac 1{2\sqrt{-x}}+\frac 1{(-x)^{3/2}}.}$$ At $-4$ these give $3$ and $-1/8$, as required.

Step 3: State the maximal interval. The solution is smooth throughout $(-\infty,0)$, and its divergence at $0$ prevents extension through that point. This is the maximal real interval containing $-4$. As $x\to-\infty$, $y\sim\sqrt{-x}\to\infty$; as $x\to 0^-$, $y\sim 2/\sqrt{-x}\to\infty$.

Step 4: Locate the minimum and orient time. Put $r=-x>0$. The derivative is $(2-r)/(2r^{3/2})$, so it is negative for $x<-2$ and positive for $-2<x<0$. Therefore $$\boxed{x_*=-2,\qquad y(x_*)=2\sqrt 2}$$ is the unique global minimum. Since $dt/dx=1/x<0$, increasing $x$ corresponds to decreasing $t$ on this half-line. This reverses the relation between the two horizontal directions, not the derivative formulas.