Limit of the solution of a differential equation

[Ressurection]

Homework Statement

Given the differential equation : $\frac{dy}{dt}$ + a(t)*y = f(t)
in which a and f are continuous functions in ℝ and verify:
a(t) > c > 0 $\forall$t , lim f(t) = 0

Show that any solution of the differential equation satisfies:
lim y(t) = 0

Homework Equations

The Attempt at a Solution

My first thought was to apply the limit to the equation right away, so that would give me

lim (y') + lim(a) * lim(y) = 0

Now this means both the limits of y and y' must exist, and since a(t) > c > 0, either they are both 0 or one is positive and the other negative.

When lim y is a finite constant, then lim y' = 0, which only allows for the case of both limits being 0. (Side note: does lim y being finite imply that lim y' = 0?)

When limit y is +∞, this would require limit y' = -∞. Is this case possible?
My intuition tells me it isn't, but I'm not absolutely sure..

Also, if lim (a) = ∞, then I don't see any restriction that imposes lim y = 0, which makes me believe there may be a more simpler way to solve this problem.

Note: All limits refer to t → ∞.

 

[Robert1986]

I think the intuition goes something like this:

$$\frac{dy}{dx} = -a(t)y(t) + f(t)$$

OK, so $a(t)>0$ all the time, and $\lim f(t) = 0$. So, if $y(t)<0$, then the derivative is positive (let's assume $f$ is zero for the intuition, but you're going to have to mention the fact that its limit is zero but it is not zero) and if $y(t)>0$ then the derivative is less than zero. So, basically, if $y$ is positive, it is going down, and if $y$ is negative, it is going up. Now, that's just the intuition, so you will have to clean it up to make it rigorous (depending on how rigorous you need to make it.) But, do you understand the general idea?

 

[SammyS]

Homework Statement

Given the differential equation : $\frac{dy}{dt}$ + a(t)*y = f(t)
in which a and f are continuous functions in ℝ and verify:
a(t) > c > 0 $\forall$t , lim f(t) = 0

Show that any solution of the differential equation satisfies:
lim y(t) = 0

Homework Equations

The Attempt at a Solution

My first thought was to apply the limit to the equation right away, so that would give me

lim (y') + lim(a) * lim(y) = 0

Now this means both the limits of y and y' must exist, and since a(t) > c > 0, either they are both 0 or one is positive and the other negative.

When lim y is a finite constant, then lim y' = 0, which only allows for the case of both limits being 0. (Side note: does lim y being finite imply that lim y' = 0?)

When limit y is +∞, this would require limit y' = -∞. Is this case possible?
My intuition tells me it isn't, but I'm not absolutely sure..

Also, if lim (a) = ∞, then I don't see any restriction that imposes lim y = 0, which makes me believe there may be a more simpler way to solve this problem.

Note: All limits refer to t → ∞.

Hello Ressurection. Welcome to PF !

Consider the function $\displaystyle g(t)=\frac{\sin(t^2)}{t}$.

$\displaystyle \lim_{t\to\infty}\,g(t)=0\ .$

$\displaystyle g'(t)$ oscillates between -2 & 2 with increasing frequency as t → ∞.

 

[Ressurection]

I think the intuition goes something like this:

$$\frac{dy}{dx} = -a(t)y(t) + f(t)$$

OK, so $a(t)>0$ all the time, and $\lim f(t) = 0$. So, if $y(t)<0$, then the derivative is positive (let's assume $f$ is zero for the intuition, but you're going to have to mention the fact that its limit is zero but it is not zero) and if $y(t)>0$ then the derivative is less than zero. So, basically, if $y$ is positive, it is going down, and if $y$ is negative, it is going up. Now, that's just the intuition, so you will have to clean it up to make it rigorous (depending on how rigorous you need to make it.) But, do you understand the general idea?

I think I do, because when taking the limit, the only 2 possibilities that satisfy that relation are:
- the limits do not exist (which would be the case of a sine or cosine)
- the limit is 0, which is the point that prevents the function from oscillating

Hello Ressurection. Welcome tom PF !

Consider the function $\displaystyle g(t)=\frac{\sin(t^2)}{t}$.

$\displaystyle \lim_{t\to\infty}\,g(t)=0\ .$

$\displaystyle g'(t)$ oscillates between -2 & 2 with increasing frequency as t → ∞.

That's a nice example, but taking the limit of the differential equation proves that the 2 limits must exist doesn't it? In that case, the limit of g' does not exist.


Sorry for the mess that my first post may be, I have zero experience regarding TeX

 

[klondike]

Try integrating factor.

 

[SammyS]

...

That's a nice example, but taking the limit of the differential equation proves that the 2 limits must exist doesn't it? In that case, the limit of g' does not exist.

I believe that you are correct in that if the two limits exist and y(t) → 0, then also y'(t) → 0 .

Sorry for the mess that my first post may be, I have zero experience regarding TeX

You'll get the hang of using TeX.

What you wrote was pretty clear.