Let \(p,q,r\) be continuous real-valued functions on \(\mathbb R\) with \(p>0\). Prove that \[ p(t)x''(t)+q(t)x'(t)+r(t)x(t)=0 \] has exactly the same solutions as an equation of the form \[ (a(t)x'(t))'+b(t)x(t)=0, \] where \(a\) is continuously differentiable and \(b\) is continuous.
Solution
1. The function\[ c(t)\coloneqq\frac{q(t)}{p(t)} \]is continuous on \(\RR\).
Both \(p\) and \(q\) are continuous, and the hypothesis \[ p(t)>0 \] for every \(t\) implies that the denominator never vanishes.
2. Define\[ a(t) \coloneqq \exp\left( \int_0^t c(s)\,ds \right). \]Then \(a\) is continuously differentiable, strictly positive, and satisfies\[ a'(t)=a(t)\frac{q(t)}{p(t)}. \]
By step 1, the fundamental theorem of calculus gives \[ \frac{d}{dt} \int_0^t c(s)\,ds = c(t). \] The chain rule therefore yields \[ a'(t) = a(t)c(t) = a(t)\frac{q(t)}{p(t)}. \] Continuity of \(c\) also gives \(a\in C^1(\RR)\), and an exponential is strictly positive.
3. Define\[ b(t)\coloneqq a(t)\frac{r(t)}{p(t)}. \]Then \(b\) is continuous.
The functions \(a\) and \(r\) are continuous, while \(p\) is continuous and nowhere zero. Hence their displayed product and quotient is continuous.
4. For every twice differentiable function \(x\),\[ (a(t)x'(t))'+b(t)x(t) = \frac{a(t)}{p(t)} \left( p(t)x''(t)+q(t)x'(t)+r(t)x(t) \right). \]
By the product rule and step 2, \[ \begin{aligned} (ax')'+bx &= ax''+a'x'+bx\\ &= ax'' + a\frac{q}{p}x' + a\frac{r}{p}x\\ &= \frac ap \left( px''+qx'+rx \right). \end{aligned} \]
5. The two differential equations have exactly the same solutions.
By step 2 and the hypothesis \(p>0\), \[ \frac{a(t)}{p(t)}>0 \] for every \(t\). Thus the right-hand side in step 4 vanishes at every \(t\) if and only if \[ p(t)x''(t)+q(t)x'(t)+r(t)x(t)=0 \] at every \(t\). Therefore \[ \boxed{ (a(t)x'(t))'+b(t)x(t)=0 } \] has exactly the same solution set as the original equation.
6. Q.E.D.
Steps 2 and 3 give the required regularity of \(a\) and \(b\), and step 5 proves equivalence of the equations.