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 \(\displaystyle l\neq 0.\) Note that this implies

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

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

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

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

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

as desired. If the integral

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

diverges as \(\displaystyle x\rightarrow\infty\), 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 \(\displaystyle l\neq 0.\) Note that this implies

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

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

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

 

[GJA]

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

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

It then follows that

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

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

 

[evinda]

If the integral

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

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

 

[GJA]

If \(\displaystyle \int_{0}^{x}e^{at}|b(t)|dt\) 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.