MHB How to Determine the Limit of a Differential Equation Solution?

[evinda]

Hello! (Wave)

I am looking at the following exercise:

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)

I am looking at the following exercise:

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 trivial using the Laplace Tranform... defining...

$\displaystyle \mathcal {L} \{ f(t) \} = F(s) = \int_{0}^{\infty} f(t)\ e^{- s t}\ d t\ (1)$

... is...

$\displaystyle \mathcal {L} \{ f^{\ '} (t)\} = s\ F(s) - f(0)\ (2)$

... and...

$\displaystyle \lim_{t \rightarrow \infty} f(t) = \lim_{ s \rightarrow 0} s\ F(s)\ (3)$

Now, writing the ODE in term of Laplace Transform You obtain...

$\displaystyle s\ Y(s) - y(0) + a\ Y(s) = B(s) \implies Y(s) = \frac{B(s) + y(0)}{s + a} \ (4)$

... so that is...

$\displaystyle \lim_{t \rightarrow \infty} y(t) = \lim_{s \rightarrow 0} \frac{s\ \{B(s) + y(0)\}}{s + a} = \frac{l}{a}\ (5)$

Kind regards

$\chi$ $\sigma$

 

[evinda]

How else could we calculate the limit? Because we haven't seen [m] Laplace [/m] yet... (Thinking)

 

[chisigma]

How else could we calculate the limit? Because we haven't seen [m] Laplace [/m] yet... (Thinking)

All right!...

a) take into account that if b(x) is continuous and $\displaystyle \lim_{x \rightarrow \infty} b(x) = l$...

b) suppose that for all its derivatives is $\displaystyle \lim_{x \rightarrow \infty} b^{(n)} (x) = 0$...

... now You consider that for the 'particular solution' is...

$\displaystyle \phi(x) = e^{- a\ x}\ \int_{0}^{x} e^{a\ t}\ b(t)\ d t = e^{- a\ x} |\frac{e ^{a\ t}}{a}\ b(t)|_{0}^{x} - \frac {e^{- a x}}{a}\ \int_{0}^{x} e^{a\ t}\ b^{\ '} (t)\ dt =$

$\displaystyle = \frac{b(x)}{a} - \frac{e ^{- a\ x}}{a}\ b(0) - \frac{b^{\ '} (x)}{a^{2}} + \frac{e^{- a\ x}}{a^{2}}\ b^{\ '}(0) + ...\ (1)$

... and if b(x) is 'regular enough' around x=0 the result is...

$\displaystyle \lim_{x \rightarrow \infty} \phi(x) = \frac{l}{a}\ (2)$

If b) isn't true however, honestly I don't have at the moment any idea on how to proceed...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

All right!...

a) take into account that if b(x) is continuous and $\displaystyle \lim_{x \rightarrow \infty} b(x) = l$...

b) suppose that for all its derivatives is $\displaystyle \lim_{x \rightarrow \infty} b^{(n)} (x) = 0$...

... now You consider that for the 'particular solution' is...

$\displaystyle \phi(x) = e^{- a\ x}\ \int_{0}^{x} e^{a\ t}\ b(t)\ d t = e^{- a\ x} |\frac{e ^{a\ t}}{a}\ b(t)|_{0}^{x} - \frac {e^{- a x}}{a}\ \int_{0}^{x} e^{a\ t}\ b^{\ '} (t)\ dt =$

$\displaystyle = \frac{b(x)}{a} - \frac{e ^{- a\ x}}{a}\ b(0) - \frac{b^{\ '} (x)}{a^{2}} + \frac{e^{- a\ x}}{a^{2}}\ b^{\ '}(0) + ...\ (1)$

... and if b(x) is 'regular enough' around x=0 the result is...

$\displaystyle \lim_{x \rightarrow \infty} \phi(x) = \frac{l}{a}\ (2)$

If b) isn't true however, honestly I don't have at the moment any idea on how to proceed...

It's my opinion that, when in doubt, the safest thing is to look for a few 'counterexample '... and what I found was this ...

$\displaystyle y^{\ '} + y = \frac{\sin x^{2}}{1 + x},\ y(0) = 0\ (1)$

... which, being $a=1$ and $\displaystyle b(x) = \frac{\sin x^{2}}{1 + x}$, is a particular case of the general equation proposed by evinda...

... here is $\displaystyle \lim_{x \rightarrow \infty} b(x) = 0$ but $\displaystyle \lim_{x \rightarrow \infty} b^{\ '} (x)$ doesn't exist...

The numeric solution of (1) given by 'Monster Wolfram' is here...

y''' '+' y '=' sin '('x'^'2')''/''('1'+' x')', y'('0')''='0 - Wolfram|Alpha

... and it seeems that $\displaystyle \lim_{x \rightarrow \infty} y(x)$ doesn't exist...

... the conclusion could be that the assumption that for all n> 0 is $\displaystyle \lim_{x \rightarrow \infty} b^{(n)} (x) = 0$ is required ...

Kind regards

$\chi$ $\sigma$