Euler Equations — Question 10

Question 10 – Solution

Strategy. Remove the repeated exponent, integrate in logarithmic time, and match the finite-forcing transition.

Step 1: Derive the integral response. With $D=d/dt$, the transformed equation is $(D-1)^2Y=g(e^t)$, so $Z''=e^{-t}g(e^t)$. The data at $t=0$ give $Z(0)=Z'(0)=0$. Thus $$Z(t)=\int_0^t(t-s)e^{-s}g(e^s)\,ds,\qquad \boxed{y(x)=x\int_1^x\ln(x/r)\frac{g(r)}{r^2}\,dr.}$$ The second formula follows from $r=e^s$, including $ds=dr/r$.

Step 2: Evaluate the finite forcing interval. Here $Z''=1$ for $0<t<1$ and $Z''=0$ for $t>1$. Integrating with the zero initial data gives $Z=t^2/2$ up to $1$, then $Z=t-1/2$. Therefore $$\boxed{y(x)=\begin{cases} \tfrac 12x(\ln x)^2,&1\le x\le e,\\ x(\ln x-\tfrac 12),&x\ge e. \end{cases}}$$ Both formulas have the same value at $e$, and the first satisfies both initial conditions at $1$.

Step 3: Check the join and the derivative jump. For $Y=e^tZ$, $y'=Z+Z'$ and $y''=(Z'+Z'')/x$. At $t=1$, both sides have $Z=1/2$ and $Z'=1$, so $$y(e)=e/2,\qquad y'(e)=3/2,\qquad y''(e^-)=2/e,\quad y''(e^+)=1/e.$$ The finite jump in forcing produces a jump $-1/e$ in $y''$, while $y,y'$ remain continuous. The unequal one-sided limits rule out $C^2$ regularity, irrespective of any value assigned to $g(e)$.

Step 4: Interpret the surviving homogeneous motion. After $e$, the solution is the nonzero homogeneous combination $x\ln x-x/2$. Thus $\boxed{y(x)/(x\ln x)\to 1}$ as $x\to\infty$. The completed forcing interval leaves nonzero value and slope at the switch. Those become initial data for the homogeneous equation; zero subsequent forcing does not reset them to zero.