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

[Malmstrom]

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

has $$y \equiv 0$$ as unique solution. No idea whatsoever up to this moment, so... thanks in adv.

 

[Mark44]

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.

 

[Redbelly98]

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

has $$y \equiv 0$$ as unique solution. No idea whatsoever up to this moment, so... thanks in adv.

Is this homework?

 

[Malmstrom]

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.

 

[Redbelly98]

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!

 

[Malmstrom]

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.

 

[l'Hôpital]

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.