# Constant solution

1. May 4, 2010

### 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.

2. May 4, 2010

### Staff: Mentor

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.

3. May 4, 2010

### Redbelly98

Staff Emeritus
Is this homework?

4. May 5, 2010

### Malmstrom

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.

5. May 5, 2010

### Redbelly98

Staff Emeritus
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!

6. May 5, 2010

### Malmstrom

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.

7. May 5, 2010

### l'Hôpital

Perhaps you might wanna 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.