证明sin(x²)与cos(x²)无法同为二阶线性齐次常微分方程的解

Let's tackle this proof head-on with contradiction, using a tool that's perfect for this scenario: Abel's Identity. You already have the Wronskian calculated, so we just need to connect that to the properties of linear ODEs.

First, let's lay out our starting assumption for the contradiction:

• Suppose there exist continuous functions $p(x)$ and $q(x)$ defined on all $\mathbb{R}$ such that both $y_1(x) = \sin(x^2)$ and $y_2(x) = \cos(x^2)$ satisfy the second-order linear homogeneous ODE:
  $$L[y] = \frac{d^2 y}{dx^2} + p(x)\frac{dy}{dx} + q(x)y = 0$$
• We already know their Wronskian is $W(x) = -2x$; for $x \neq 0$, this is non-zero, so the two functions are linearly independent (which aligns with ODE solution set properties).

Step 1: Use Abel's Identity (No Differentiability of $p(x)$ Needed!) #

A key property of second-order linear homogeneous ODEs is Abel's Identity, which applies as long as $p(x)$ is continuous (exactly what we're assuming). For any two solutions $y_1, y_2$, their Wronskian satisfies:
$$W'(x) = -p(x)W(x)$$
This is the linchpin here—we don't need to know anything about the derivative of $p(x)$, just that it's continuous.

Step 2: Compute Derivatives and Solve for $p(x)$ #

First, calculate the derivative of our Wronskian:
$$W(x) = -2x \implies W'(x) = -2$$
Plug $W(x)$ and $W'(x)$ into Abel's Identity:
$$-2 = -p(x) \cdot (-2x)$$
Simplify this to solve for $p(x)$:
$$-2 = 2x p(x) \implies p(x) = -\frac{1}{x}$$

Step 3: Contradiction is Clear #

The problem states $p(x)$ must be continuous on all $\mathbb{R}$, but $p(x) = -\frac{1}{x}$ has a discontinuity at $x=0$—this directly violates our initial assumption.

Why This Avoids the Uniqueness Theorem #

We didn't need to invoke the uniqueness theorem at all here. Abel's Identity only requires continuity of $p(x)$ (given) and that $y_1, y_2$ are solutions (our starting assumption). The contradiction arises purely from forcing $p(x)$ to have a discontinuity, which breaks the problem's requirements.

内容的提问来源于stack exchange，提问作者jackson5