Basic Concepts for nth Order Linear Equations — Question 6

Question 6 – Solution

Strategy. Test the endpoint functionals on the cubic solution space instead of relying on their number.

Step 1: Form the coefficient system. Write $y=a+bx+cx^2+dx^3$. The conditions become $$\begin{pmatrix}1&0&0&0\\1&1&1&1\\0&2&2&3\\0&0&0&6\end{pmatrix} \begin{pmatrix}a\\b\\c\\d\end{pmatrix} =\begin{pmatrix}0\\0\\0\\\eta\end{pmatrix}.$$ The fourth row is six times the third minus twelve times the second plus twelve times the first. The first three rows are independent, so the rank is $\boxed{3}$.

Step 2: Classify the parameter completely. The first three equations imply $a=0$, $b+c+d=0$ and $2b+2c+3d=0$; subtracting twice the middle equation gives $d=0$. Thus the last equation can hold only if $\eta=0$. For that value, $c=-b$ and $$\boxed{y=Kx(1-x),\quad K\in\mathbb R.}$$ There are infinitely many solutions if $\eta=0$ and none if $\eta\ne 0$. No value of $\eta$ gives a unique solution under the original conditions.

Step 3: Replace the redundant information. The condition $y'(0)=1$ gives $b=1$. The first three conditions already imply $a=d=0$ and $c=-b$, so $$\boxed{y=x(1-x)}$$ is the unique solution after replacement. The new condition fixes the one remaining free parameter.

Step 4: Distinguish endpoint data from a full initial state. The IVP theorem concerns $y,y',y'',y'''$ specified at one ordinary point. These endpoint conditions mix values at two points and can be dependent or inconsistent on the solution space. In the compatible family, changing $K$ changes height and curvature while preserving zero endpoint values, opposite endpoint slopes, and equal endpoint curvatures. The figure shows three such members; their different initial derivative vectors explain why IVP uniqueness is not violated.