Boundary Value Problems — Question 7

Question 7 – Solution

Strategy. Solve on the open interval first, then apply each proposed endpoint condition to the possible singular term.

Step 1: Find the family on the actual domain. Integration gives $x^2y'=C$ and $y=A-C/x$. Incorporating $y(1)=1$, write $$\boxed{y=1-k+k/x,\qquad k\in\mathbb R.}$$ For $x>0$ the normalized equation is $y''+(2/x)y'=0$. Its coefficient $2/x$ is singular at zero; the original leading coefficient also vanishes there.

Step 2: Test boundedness and energy. Boundedness at zero forces $k=0$, giving $y\equiv 1$. Since $y'=-k/x^2$, $$\int_0^1x^2(y')^2dx=\lim_{t\downarrow 0}k^2\left(\frac 1t-1\right).$$ This is infinite for $k\ne 0$ and zero for $k=0$. Thus boundedness and finite weighted energy select the same single solution in this family.

Step 3: Prescribe a weighted endpoint value. For the displayed family, $$\lim_{x\downarrow 0}xy(x)=k,\qquad \boxed{\lim_{x\downarrow 0}x^2y'(x)=-k.}$$ Therefore each real prescribed weighted limit selects exactly one solution, including unbounded solutions when $k\ne 0$. A weighted boundary condition is not the same as an ordinary finite value $y(0)$.

Step 4: Test the proposed continuous boundary data. A continuous extension is bounded, so it must be $y\equiv 1$ and must have $y(0)=1$. Consequently no continuous solution satisfies $y(0)=0,y(1)=1$. One cannot assign an unrelated value at zero and retain continuity. Regular endpoint theory requires hypotheses on the normalized coefficients that fail here; the explicit family and its limits decide this problem instead.