Boundary Value Problems — Question 4

Question 4 – Solution

Strategy. Inspect the denominator in the endpoint equation and distinguish a chosen parameter path from a unique limiting response.

Step 1: Solve and classify the singular case. The left boundary gives $y=x^2+cx$. The mixed condition becomes $(1-h)c=b-2+h$. For $h\ne 1$, $$\boxed{y=x^2+\left(\frac{b-1}{1-h}-1\right)x.}$$ Its second derivative is two, its left value is zero, and substitution verifies the mixed condition. At $h=1$, the condition reduces to $b=1$ independently of $c$: infinitely many solutions if $b=1$, and none otherwise.

Step 2: Quantify sensitivity. At fixed $h\ne 1$, the change in the response is $$\boxed{\Delta y(x)=\frac{\delta x}{1-h},\qquad \max_{[0,1]}|\Delta y|=\frac{|\delta|}{|1-h|}.}$$ Thus arbitrarily small boundary errors can be strongly amplified near $h=1$.

Step 3: Examine one compatible path. If $b=1$ and $h\ne 1$, then $c=-1$, so every solution on this path is $y=x^2-x$. Its limit is one member of the infinite family at $(h,b)=(1,1)$. The path selects this member; the limiting boundary conditions alone do not.

Step 4: Compare different approaches. For the proposed sequences, $(b_n-1)/(1-h_n)=k$, so $$\boxed{y_n=x^2+(k-1)x\quad\text{for every }n.}$$ All parameter pairs approach $(1,1)$, but their solution limits depend on $k$. They range over the entire limiting family. Consequently no choice of a single solution at $(1,1)$ can make the solution map continuous there along all paths.