More on the Wronskian — Question 10

Question 10 – Solution

Strategy. Differentiate the moving coefficients too; pointwise invertibility is not a constant linear transformation of solution functions.

Step 1: Include the extra derivative terms. In the derivatives of $f,g$, the terms involving $u',v'$ contribute $W[u,v]$, as in a constant rotation. The remaining terms are $\theta'g$ in $f'$ and $-\theta'f$ in $g'$. Hence $$\boxed{W[f,g]=W[u,v]-\theta'(f^2+g^2) =W[u,v]-\theta'(u^2+v^2).}$$ The additional term is exactly what the constant-coefficient formula omits.

Step 2: Observe a complete collapse. For $u=\cos t$, $v=\sin t$, the addition formulas give $f=\cos(t-\theta)$, $g=\sin(t-\theta)$. At $\theta=t$, $f=1$ and $g=0$, so $W=0$. They are dependent, and $f''+f=1\ne 0$. Pointwise invertibility of the moving coefficient matrix does not preserve the solution space.

Step 3: Test a nonzero determinant. At $\theta=t/2$, the functions are $\cos(t/2),\sin(t/2)$ and $W=1/2$. Each has $y''+y=(3/4)y$, so neither solves the original equation identically. They instead form a fundamental set for $y''+y/4=0$.

Step 4: Classify all valid rotations. Put $\varphi=t-\theta$. The residuals for $f=\cos\varphi$, $g=\sin\varphi$ are $$(1-\varphi'^2)\cos\varphi-\varphi''\sin\varphi,\qquad (1-\varphi'^2)\sin\varphi+\varphi''\cos\varphi.$$ Both vanish exactly when $\varphi'^2=1$ and $\varphi''=0$, since the displayed rotation is invertible. On the connected real line this gives $\varphi=\pm t+C$, hence $$\boxed{\theta(t)=C\quad\text{or}\quad\theta(t)=2t+C,\qquad C\in\mathbb R.}$$ In the second case the final functions are still constant combinations of $\cos t,\sin t$, despite their time-dependent defining coefficients.