Question: 2. Assume $p$ and $q$ are continuous functions on some open interval $I$ and $y_1$ and $y_2$ are two solutions on $I$ of
y'' + p(x)y' + q(x)y = 0.
Abel’s Theorem then implies that either $W([y_1, y_2]) = 0$ for all $x \in I$ or $W([y_1, y_2]) \neq 0$ for all $x \in I$ (where $W([y_1, y_2])$ is the Wronskian).
Use this to show the following:
(a) (5 points) If $y_1$ and $y_2$ have a maximum at the same point $x = x_0 \in I$, then they are linearly dependent.
(b) (5 points) If $y_1$ has a maximal value of $0$, obtained at some point $x_0 \in I$, then $y_1$ and $y_2$ are linearly dependent.
1. **Problem statement:** Given the second order linear homogeneous differential equation $$y'' + p(x)y' + q(x)y = 0$$ with continuous functions $p$ and $q$ on an open interval $I$, and two solutions $y_1$ and $y_2$ on $I$, Abel's theorem states that the Wronskian $W([y_1,y_2])(x)$ is either identically zero or never zero on $I$.
We need to show:
(a) If $y_1$ and $y_2$ have a maximum at the same point $x_0 \in I$, then $y_1$ and $y_2$ are linearly dependent.
(b) If $y_1$ has a maximal value of $0$ at some $x_0 \in I$, then $y_1$ and $y_2$ are linearly dependent.
2. **Recall the Wronskian definition and Abel's theorem:**
The Wronskian of $y_1$ and $y_2$ is defined as
$$W(x) = \begin{vmatrix} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \end{vmatrix} = y_1(x)y_2'(x) - y_2(x)y_1'(x).$$
Abel's theorem states that
$$W(x) = W(x_0) \exp\left(-\int_{x_0}^x p(t) dt\right),$$
so $W(x)$ is either identically zero or never zero on $I$.
3. **Part (a): If $y_1$ and $y_2$ have a maximum at the same point $x_0$**
- At a maximum point $x_0$, the derivative of each function is zero:
$$y_1'(x_0) = 0, \quad y_2'(x_0) = 0.$$
- Substitute into the Wronskian:
$$W(x_0) = y_1(x_0) y_2'(x_0) - y_2(x_0) y_1'(x_0) = y_1(x_0) \cdot 0 - y_2(x_0) \cdot 0 = 0.$$
- Since $W(x_0) = 0$, by Abel's theorem, $W(x) = 0$ for all $x \in I$.
- A zero Wronskian everywhere implies $y_1$ and $y_2$ are linearly dependent.
4. **Part (b): If $y_1$ has a maximal value of 0 at $x_0$**
- Since $y_1$ has a maximum at $x_0$, we have
$$y_1'(x_0) = 0,$$
and the maximal value is zero, so
$$y_1(x_0) = 0.$$
- Evaluate the Wronskian at $x_0$:
$$W(x_0) = y_1(x_0) y_2'(x_0) - y_2(x_0) y_1'(x_0) = 0 \cdot y_2'(x_0) - y_2(x_0) \cdot 0 = 0.$$
- Again, by Abel's theorem, $W(x) = 0$ for all $x \in I$.
- Hence, $y_1$ and $y_2$ are linearly dependent.
**Final conclusion:** In both cases, the Wronskian is zero everywhere, so the solutions $y_1$ and $y_2$ are linearly dependent.