一阶线性常微分方程解非唯一性的实例咨询

Awesome question—this is a super common point of confusion when contrasting linear and nonlinear ODEs, so let's break it down step by step.

First off, here's the key takeaway: For the first-order linear homogeneous ODE ( \frac{d}{dt}y(t) = f(t)y(t) ) with initial condition ( y(t_0)=p_0 ), there are no cases where the solution is non-unique. That's exactly why you haven't been able to find an example—they don't exist!

Why is this different from nonlinear ODEs? #

You probably know the classic nonlinear counterexample: ( \frac{dy}{dt} = 3y^{2/3} ) with ( y(0)=0 ) has two distinct solutions: ( y(t)=0 ) everywhere, and ( y(t)=t^3 ). But linear ODEs have a rigid structure that eliminates this ambiguity.

Let's look at how we construct solutions for your linear equation. Even if ( f(t) ) isn't continuous, as long as it's Riemann integrable over an interval containing ( t_0 ), we can use the integrating factor method to write the unique solution:
[ y(t) = p_0 \exp\left( \int_{t_0}^t f(s) ds \right) ]

This solution is unique because:

• The integral of a Riemann integrable function is uniquely defined, even if the function has discontinuities.
• The exponential function is one-to-one (injective), so there's no way to get two different outputs from the same initial condition and integral.

What if ( f(t) ) isn't integrable? #

If ( f(t) ) is so pathological that it can't be integrated (like a highly discontinuous function that fails Riemann integrability), then the equation might not have a well-defined solution at all. But this is a problem of existence, not uniqueness—if a solution does exist, it's guaranteed to be the only one.

To recap: #

• For your linear ODE, no non-unique solution examples exist. The linear structure ensures that as long as a solution exists (i.e., ( f(t) ) is integrable), the initial condition pins down exactly one solution.
• The continuity of ( f(t) ) is a sufficient condition for existence and uniqueness, but not a necessary one. Even discontinuous but integrable ( f(t) ) still give unique solutions.

内容的提问来源于stack exchange，提问作者Viktor Jeppesen