Proving Unique Solution of $y''+e^{-x}f(y)=0$ with $y(0)=y'(0)=0$

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Discussion Overview

The discussion revolves around proving the uniqueness of the solution to the differential equation \( y'' + e^{-x}f(y) = 0 \) with initial conditions \( y(0) = y'(0) = 0 \). The context includes exploring the implications of the function \( f \) being continuous and satisfying \( tf(t) \geq 0 \) for all \( t \).

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • Some participants note that since \( tf(t) \geq 0 \) for all \( t \), it implies \( f(t) \geq 0 \) for \( t \geq 0 \) and \( f(t) \leq 0 \) for \( t \leq 0 \).
  • It is suggested that for \( y \geq 0 \), \( f(y) \geq 0 \) leads to \( y'' \leq 0 \), while for \( y \leq 0 \), \( f(y) \leq 0 \) results in \( y'' \geq 0 \).
  • One participant proposes assuming \( y'' > 0 \) for \( y < 0 \) and \( y'' < 0 \) for \( y > 0 \) to work towards a contradiction.
  • Another participant suggests that if \( y \) is not zero in some interval, then it must be either positive or negative in that interval, and recommends looking into the Strong Maximum Principle for further insights.
  • There is a question about whether the problem is homework-related, with participants confirming that it is indeed from a past course.

Areas of Agreement / Disagreement

Participants generally agree on the nature of the function \( f \) and its implications for the differential equation, but there are differing approaches and no consensus on the proof of uniqueness or the next steps to take.

Contextual Notes

Participants express uncertainty about the implications of their assumptions and the specific steps needed to prove the uniqueness of the solution. The discussion reflects a range of interpretations of the conditions imposed by the function \( f \) and the behavior of the solution \( y \).

Malmstrom
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Let [tex]f \in \mathcal{C}(\mathbb{R})[/tex] be a continuous function such that [tex]tf(t) \geq 0[/tex] [tex]\forall t[/tex]. I must prove that
[tex]y''+e^{-x}f(y)=0[/tex]
[tex]y(0)=y'(0)=0[/tex]

has [tex]y \equiv 0[/tex] as unique solution. No idea whatsoever up to this moment, so... thanks in adv.
 
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I haven't worked this through, but here are some thoughts.

Since tf(t) >= 0 for all t, it must be true that f(t) >= 0 for t >= 0, and f(t) <= 0 for t <= 0. For this reason, and assuming that y is some twice differentiable function of x, g(x), f(y) >= 0 for y >= 0 and f(y) <=0 for y <= 0.

Looking at the differential equation, e-x > 0 for all x. For any y >= 0, f(y) >= 0, hence y'' must be <= 0 (because y'' plus a positive number has to equal zero).

In a similar vein, for any y <= 0, f(y) <= 0, so y'' must be >= 0.

Where I would go next is to assume that y'' > 0 for y (= g(x)) < 0, and y'' < 0 for y > 0, and work toward a contradiction.
 
Malmstrom said:
Let [tex]f \in \mathcal{C}(\mathbb{R})[/tex] be a continuous function such that [tex]tf(t) \geq 0[/tex] [tex]\forall t[/tex]. I must prove that
[tex]y''+e^{-x}f(y)=0[/tex]
[tex]y(0)=y'(0)=0[/tex]

has [tex]y \equiv 0[/tex] as unique solution. No idea whatsoever up to this moment, so... thanks in adv.
Is this homework?
 
Redbelly98 said:
Is this homework?

This is homework of a past course I did not attend, so I don't *have to* do this stuff. I'm doing it 'cause I'm attending some different stuff about ODEs and lack some of the prerequisites. Anyway I have absolutely no problem in posting any future question in the homework section if you tell me to.
 
Hi, yes please use the homework section in the future. Our "homework" rules apply to independent study problems from textbooks as well.

Oh, and ... welcome to Physics Forums! :smile:
 
Mark44 said:
I haven't worked this through, but here are some thoughts.

Since tf(t) >= 0 for all t, it must be true that f(t) >= 0 for t >= 0, and f(t) <= 0 for t <= 0. For this reason, and assuming that y is some twice differentiable function of x, g(x), f(y) >= 0 for y >= 0 and f(y) <=0 for y <= 0.

Looking at the differential equation, e-x > 0 for all x. For any y >= 0, f(y) >= 0, hence y'' must be <= 0 (because y'' plus a positive number has to equal zero).

In a similar vein, for any y <= 0, f(y) <= 0, so y'' must be >= 0.

Where I would go next is to assume that y'' > 0 for y (= g(x)) < 0, and y'' < 0 for y > 0, and work toward a contradiction.

Hi Mark. You were very helpful indeed but I can't figure out what makes the whole thing go wrong and forces y to be constant.
 
Perhaps you might want to try the following. Suppose y is not zero all around. Then there must exist an interval (x1,x2) where either y > 0 or y < 0.

Without loss of generality, assume y < 0.

Look up the Strong Maximum Principle and that will help you out.
 

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