How could we calculate the limit?

[evinda]

Hello! (Wave)Let the (linear) differential equation $y'+ay=b(x)$ where $a>0, b$ continuous on $[0,+\infty)$ and $\lim_{x \to +\infty} b(x)=l \in \mathbb{R}$.

Show that each solution of the differential equation goes to $\frac{l}{a}$ while $x \to +\infty$,

i.e. if $\phi$ is any solution of the differential equation, show that $\lim_{x \to +\infty} \phi(x)=\frac{l}{a}$.

That's what I have tried:

The solution of the differential equation will be of the form:

$\phi(x)=ce^{-ax}+e^{-ax} \int_0^x e^{at}b(t) dt$

$\lim_{x \to +\infty} c e^{-ax}=0$

So, $\lim_{x \to +\infty} \phi(x)=\lim_{x \to +\infty} e^{-ax} \int_0^x e^{at}b(t)dt$How can we calculate the limit $\lim_{x \to +\infty} e^{-ax} \int_0^x e^{at}b(t)dt$ ? (Thinking)

 

[chisigma]

Hello! (Wave)Let the (linear) differential equation $y'+ay=b(x)$ where $a>0, b$ continuous on $[0,+\infty)$ and $\lim_{x \to +\infty} b(x)=l \in \mathbb{R}$.

Show that each solution of the differential equation goes to $\frac{l}{a}$ while $x \to +\infty$,

i.e. if $\phi$ is any solution of the differential equation, show that $\lim_{x \to +\infty} \phi(x)=\frac{l}{a}$.

That's what I have tried:

The solution of the differential equation will be of the form:

$\phi(x)=ce^{-ax}+e^{-ax} \int_0^x e^{at}b(t) dt$

$\lim_{x \to +\infty} c e^{-ax}=0$

So, $\lim_{x \to +\infty} \phi(x)=\lim_{x \to +\infty} e^{-ax} \int_0^x e^{at}b(t)dt$How can we calculate the limit $\lim_{x \to +\infty} e^{-ax} \int_0^x e^{at}b(t)dt$ ? (Thinking)

The problem is easily solved using the Laplace Transform... in term of the s variable the solution is...

$\displaystyle Y(s) = \frac{B(s) - y(0)}{s + a}\ (1)$

... and applying the final value theorem is...

$\displaystyle \lim_{s \rightarrow 0} s\ B(s) = l\ (2)$

$\displaystyle \lim_{ x \rightarrow \infty} y(x) = \lim_{s \rightarrow 0} \frac{s\ [B(s) - y(0)]}{ s + a}\ (3)$

... so that is...

$\displaystyle \lim_{ x \rightarrow \infty} y(x) = \frac{l}{a}\ (4)$

Kind regards

$\chi$ $\sigma$

 

[GJA]

Hi evinda,

First consider $$l\neq 0.$$ Note that this implies

$$\lim_{x\rightarrow\infty}\int_{0}^{x}e^{at}b(t)dt$$

diverges (to $$\pm\infty,$$ depending on the sign of $$l$$). Hence, we can compute the limit for $$l\neq 0$$ using L'Hopital's rule and the Fundamental Theorem of Calculus.Next consider $$l=0.$$ Note that we have the inequality

$$\left|\int_{0}^{x}e^{at}b(t)dt \right|\leq \int_{0}^{x}e^{at}|b(t)|dt\qquad\forall x$$

Now, either the integral on the right either converges to a number $$M$$ that is finite or it diverges to $$+\infty$$ as $$x\rightarrow\infty.$$ If the integral converges as $$x\rightarrow\infty$$, then the above inequality shows that

$$\lim_{x\rightarrow\infty}e^{-ax}\int_{0}^{x}e^{at}b(t)dt=0,$$

as desired. If the integral

$$\int_{0}^{x}e^{at}|b(t)|dt$$

diverges as $$x\rightarrow\infty$$, then we can once again use L'Hopital's rule and the Fundamental Theorem of Calculus to get that the limit is zero. Let me know if anything is unclear/not quite right.

 

[evinda]

Hi evinda,

First consider $$l\neq 0.$$ Note that this implies

$$\lim_{x\rightarrow\infty}\int_{0}^{x}e^{at}b(t)dt$$

diverges (to $$\pm\infty,$$ depending on the sign of $$l$$). Hence, we can compute the limit for $$l\neq 0$$ using L'Hopital's rule and the Fundamental Theorem of Calculus.

How do we deduce that $$\lim_{x\rightarrow\infty}\int_{0}^{x}e^{at}b(t)dt$$ diverges to $$\pm\infty$$ ? (Thinking)

 

[GJA]

Suppose $$l>0.$$ Since $$\lim_{x\rightarrow\infty}b(x)=l,$$ there is $$N>0$$ such that

$$\frac{l}{2}<b(x)\qquad\forall x\geq N$$

It then follows that

$$\infty = \frac{l}{2}\int_{N}^{\infty}e^{at}dt\leq \int_{N}^{\infty}e^{at}b(t)dt$$

I would try writing the proof for $$l<0$$ using the above as a guide. Let me know if anything is unclear/not quite right.

 

[evinda]

If the integral

$$\int_{0}^{x}e^{at}|b(t)|dt$$

diverges as $$x\rightarrow\infty$$, then we can once again use L'Hopital's rule and the Fundamental Theorem of Calculus to get that the limit is zero. Let me know if anything is unclear/not quite right.

If $$\int_{0}^{x}e^{at}|b(t)|dt$$ diverges, we cannot deduce that $\int_0^x e^{at} b(t) dt$ diverges, or can we? (Thinking)

 

[GJA]

If $$\int_{0}^{x}e^{at}|b(t)|dt$$ diverges, we cannot deduce that $\int_0^x e^{at} b(t) dt$ diverges, or can we? (Thinking)

You're 100% correct, we cannot conclude $\int_0^x e^{at} b(t) dt$ diverges as well - this was not part of the claim in the previous posts. Fortunately, we don't need to conclude this for the method outlined earlier.

In assuming $\int_{0}^{x}e^{at}|b(t)|dt$ diverges as $x\rightarrow\infty$, we can calculate the limit you want in your original post (using $\int_{0}^{x}e^{at}|b(t)|dt$ instead of $\int_{0}^{x}e^{at}b(t)dt$) via the Fundamental Theorem of Calculus and L'Hopital's Rule. If you haven't done this computation yet, give it a shot and see what you get. Then look at the inequality again. As before, I'm certain you can do this, so I am doing my best to avoid writing out all the details. Let me know, though, if you get stuck and I will help some more. Let me know if anything is unclear/not quite right.

 

[evinda]

You're 100% correct, we cannot conclude $\int_0^x e^{at} b(t) dt$ diverges as well - this was not part of the claim in the previous posts. Fortunately, we don't need to conclude this for the method outlined earlier.

In assuming $\int_{0}^{x}e^{at}|b(t)|dt$ diverges as $x\rightarrow\infty$, we can calculate the limit you want in your original post (using $\int_{0}^{x}e^{at}|b(t)|dt$ instead of $\int_{0}^{x}e^{at}b(t)dt$) via the Fundamental Theorem of Calculus and L'Hopital's Rule. If you haven't done this computation yet, give it a shot and see what you get. Then look at the inequality again. As before, I'm certain you can do this, so I am doing my best to avoid writing out all the details. Let me know, though, if you get stuck and I will help some more. Let me know if anything is unclear/not quite right.

Then it will be like that:

$$\lim_{x \to +\infty} e^{-ax} \int_0^x e^{at} |b(t)| dt= \lim_{x \to +\infty} \frac{\int_0^x e^{at} |b(t)| dt}{e^{ax}}= \lim_{x \to +\infty} \frac{e^{ax} |b(x)|}{a e^{ax}} dt \to \frac{|l|}{a}$$

Right?

However, having found this limit can we say something about the limit $\lim_{x \to +\infty} e^{-ax} \int_0^x e^{at} b(t) dt$? From the inequality don't we get that $-\frac{|l|}{a} \leq \lim_{x \to +\infty} e^{-ax} \int_0^x e^{at} b(t) dt \leq \frac{|l|}{a}$?
But this doen't help, or does it?

 

[GJA]

We need to recall what $l$ was in this case to finish the problem...

 

[evinda]

We need to recall what $l$ was in this case to finish the problem...

You mean that we have to use the fact that $\lim_{x \to +\infty} b(x)=l$? How does this help? (Thinking)

 

[GJA]

We knew specifically what $l$ was in the context of the argument you're quoting. Take a look at my previous post and you'll find the answer.