Let \(f,g:\mathbb R\to\mathbb R\) be smooth, with \[ f(0)=0, \qquad f'(0)\ne0. \] Consider \[ f(x)=t\,g(x). \]
- Show that for \(|t|<\delta\), for some \(\delta>0\), there is a unique continuous solution \(x(t)\) with \(x(0)=0\).
- Find the first-order Taylor expansion of \(x(t)\) at \(t=0\).
Solution
Set \[ H(x,t)\coloneqq f(x)-t g(x). \]
1. There are \(r,\delta>0\) such that, for every \(\abs{t}<\delta\), theequation \(H(x,t)=0\) has exactly one solution in \((-r,r)\).
Because \(f'(0)\neq0\), choose \(r>0\) so small that \(f'\) has a constant nonzero sign on \([-r,r]\). In particular, \[ f(-r)f(r)<0. \] Since \[ \frac{\partial H}{\partial x}(x,t) = f'(x)-t g'(x), \] continuity on the compact interval \([-r,r]\) allows \(\delta>0\) to be chosen so small that \[ \frac{\partial H}{\partial x}(x,t) \] has the same nonzero sign as \(f'(x)\) whenever \(\abs{x}\leq r\) and \(\abs{t}<\delta\). Shrinking \(\delta\) once more, the two numbers \[ H(-r,t) \qquad\text{and}\qquad H(r,t) \] retain the opposite signs of \(f(-r)\) and \(f(r)\).
For each \(\abs{t}<\delta\), the intermediate value theorem therefore gives a root of the function \(x\mapsto H(x,t)\) in \((-r,r)\), while the nonvanishing derivative with respect to \(x\) makes this function strictly monotone there. The root is consequently unique.
2. These roots define the unique continuous solution \(x(t)\) with\(x(0)=0\) on \(\abs{t}<\delta\), and in fact \(x\) is smooth.
At every root found in step 1, \[ \frac{\partial H}{\partial x}\neq0. \] The implicit function theorem therefore makes the unique root a smooth function of \(t\) locally. The local functions agree on overlaps by uniqueness in step 1, so they form a smooth function \[ x:(-\delta,\delta)\longrightarrow(-r,r) \] satisfying \[ f(x(t))=t g(x(t)). \] At \(t=0\), the unique root in \((-r,r)\) is \(x=0\), so \(x(0)=0\).
If \(y\) is any continuous solution on \((-\delta,\delta)\) with \(y(0)=0\), then \(y\) cannot leave \((-r,r)\): otherwise continuity would give some \(t_0\) with \(y(t_0)=r\) or \(y(t_0)=-r\), contradicting the fact from step 1 that neither boundary point is a root for \(\abs{t}<\delta\). Thus \(y(t)\in(-r,r)\) for every \(t\), and uniqueness in step 1 gives \(y=x\).
3. The first-order Taylor expansion is\[ \boxed{ x(t)=\frac{g(0)}{f'(0)}t+O(t^2) } \qquad (t\to0). \]
Differentiate \[ f(x(t))=t g(x(t)). \] The chain and product rules give \[ f'(x(t))x'(t) = g(x(t))+t g'(x(t))x'(t). \] Evaluating at \(t=0\) and using \(x(0)=0\) yields \[ f'(0)x'(0)=g(0), \] so \[ x'(0)=\frac{g(0)}{f'(0)}. \] Since \(x\) is smooth, Taylor’s theorem gives \[ x(t) = x(0)+x'(0)t+O(t^2) = \frac{g(0)}{f'(0)}t+O(t^2). \]
4. Q.E.D.
Step 2 proves Part (1), and step 3 gives the expansion required in Part (2).