Basic Concepts for nth Order Linear Equations — Question 5

Question 5 – Solution

Strategy. Compare independence of arbitrary smooth functions with independence inside one regular homogeneous solution space.

Step 1: Prove independence by separated values. The supports are $[-5/2,-3/2]$, $[-1/2,1/2]$ and $[3/2,5/2]$, respectively. They are disjoint, and each function has value $e^{-1}$ at its center $-2,0,2$. If $c_1f_1+c_2f_2+c_3f_3=0$ identically, evaluation at those three centers gives $c_1=c_2=c_3=0$. The functions are therefore linearly independent.

Step 2: Compute the determinant structurally. Outside a function’s support, it and its derivatives vanish; at the support boundary, the supplied smoothness facts give the same conclusion. At every $x$, at most one column of $$\begin{pmatrix}f_1&f_2&f_3\\f_1'&f_2'&f_3'\\f_1''&f_2''&f_3''\end{pmatrix}$$ can be nonzero. Thus $\boxed{W(f_1,f_2,f_3)(x)=0\text{ for all }x}$.

Step 3: Test the common-equation hypothesis. Suppose all three solved the stated regular homogeneous equation. The function $f_1$ has $f_1(0)=f_1'(0)=f_1''(0)=0$. The zero solution has the same data, so uniqueness on $\mathbb R$ would force $f_1\equiv 0$. But $f_1(-2)=e^{-1}>0$, a contradiction. Such continuous coefficients cannot exist.

Step 4: State the theorem with its hypotheses. For three solutions of the same normalized third-order homogeneous linear equation with continuous coefficients on a connected interval, a zero Wronskian at one point gives a nonzero coefficient combination with zero initial derivative vector. Uniqueness then makes that combination identically zero, proving dependence. Our functions are arbitrary smooth functions, not solutions of one such equation. Their identically zero Wronskian alone therefore does not imply dependence.