From the book First course on DE, there's the problem $2y''+ty'-2y=10, y(0)=0, y'(0)=0$. Implying laplace transform to both sides give a linear homogenuous DE $Y'(s)-(2s-\frac{3}{s})Y(s)=-\frac{10}{s^2}$ and it gives $Y=\frac{5}{s^3}+C\frac{e^{s^2}}{s^3}$, but I'm stuck here. The answer is $\frac{5}{2}t^2$, which means that $c=0$. Now I can't find the reason why $c=0$. Can anyone help???
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Can you apply the initial conditions? – gt6989b Dec 02 '19 at 14:31
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How? Y(s) is not very related to the conditions of y. – Hypernova Dec 02 '19 at 14:32
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did you tried to back apply the inverse laplace transform and prug the initial conditions, you might end finding the value of C this way – cand Dec 02 '19 at 14:34
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That was the first approach I tried, but had no idea of the inverse laplace transform of $\frac{e^{s^2}}{s^3}$. How can I do this? – Hypernova Dec 02 '19 at 14:36
3 Answers
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Apply the exponential order property of fuctions.
Definition: A function $f$ is said to be of exponential order $c$ if there exist constants $c,M>0,T>0$ such that $|f(t)|≤Me^{ct}$ for all $t>T$.
In order for $f(t)$ to have a Laplace Transform then in a race between $|f(t)|$ and $e^{ct}$ as $t\to\infty$ then $e^{ct}$ must approach its limit first, i.e. $\lim_{t\to\infty}\frac{f(t)}{e^{ct}}=0.$
Theorem: If $f$ is piecewise continuous on $[0,\infty)$ and of exponential order $c$, then $F(s)=L[f(t)]$ exists for $s>c$ and $\lim_{s\to\infty}F(s)=0.$
Therefore
$$\lim_{s\to\infty}Y(s)=0$$
implies
$$\lim_{s\to\infty}\frac{5}{s^3}+C\frac{e^{s^2}}{s^3}=0$$
which is true if and only if $C=0$. Thus
$$Y(s)=\frac{5}{s^3}$$
Axion004
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Wait isn't that a if(necessary) statement? It doesn't have to be exponential order to have a laplace transform. The converse holds but does the original statement hold? I've heard about such functions. – Hypernova Dec 02 '19 at 16:40
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For an exponential order function, we have existence and uniqueness of the Laplace transform. As you point out, the theorem is a sufficient condition for the existence of the Laplace transform. There are functions that have a Laplace transform which are not of exponential order. – Axion004 Dec 03 '19 at 05:58
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We have that any polynomial is of exponential order since
$$e^{at}=\sum_{n=0}^{\infty}\frac{t^na^n}{n!}\ge \frac{t^na^n}{n!} \implies t^n\le \frac{n!}{a^n}e^{at}$$
while $f(t)=e^{t^2}$ isn't of exponential order since given $c>0$
$$\lim_{t\to\infty}\frac{e^{t^2}}{e^{ct}}=\lim_{t\to\infty}e^{t(t-c)}=\infty$$
and for $c \le 0$ we have
$$\lim_{t\to\infty}{e^{t^2+ct}}=\infty$$
– Axion004 Dec 03 '19 at 05:58 -
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We have
$~\mathcal L\{y(t)\}=\bar y(p)~$
$\mathcal L\{y'(t)\}=p~\bar y(p)-y(0)=p~\bar y(p)~~~~$ (as it is given that $~y(0)=0~$)
$\mathcal L\{y''(t)\}=p^2~\bar y(p)-p~y(0)-y'(0)=p^2~\bar y(p)~~~~$(as it is given that $~y(0)=0,~y'(0)=0$)
$\mathcal L\{ty'(t)\}=-\dfrac{d}{dp}\{p~\bar y(p)\}=-\bar y(p)-p~\dfrac{d\bar y}{dp}~$
Now given equation is $$2y''+ty'-2y=10$$ Taking Laplace transform both side we have $$2~\mathcal L\{y''\}+\mathcal L\{ty'\}-2\mathcal L\{y\}=\mathcal L\{10\}$$ $$\implies 2p^2\bar y(p)-\bar y(p)-p~\dfrac{d\bar y}{dp}-2\bar y(p)=\dfrac{10}{p}$$ $$\implies \dfrac{d\bar y}{dp}+\left(\dfrac{3}{p}-2p\right)\bar y(p)=-\dfrac{10}{p^2}\tag1$$
Which is a first order linear differential equation.
Integrating factor (I.F.) $$~=e^{\int\left(\frac{3}{p}-2p\right)dp}=e^{3\ln p-p^2}=p^3e^{-p^2}~$$
Multiplying both side of $(1)$ by the I.F. we have
$$p^3e^{-p^2}\dfrac{d\bar y}{dp}+e^{-p^2}(3p^2-2p^4)\bar y(p)=-10pe^{-p^2}$$
$$\implies \dfrac{d}{dp}\left\{p^3e^{-p^2}\bar y\right\}=-10pe^{-p^2}$$
Integrating we have,
$$p^3e^{-p^2}\bar y=-10\int pe^{-p^2}$$
$$\implies p^3e^{-p^2}\bar y=5e^{-p^2}+c$$
$$\implies \bar y=\dfrac{1}{p^3}\left(c~e^{p^2}+5\right)\tag2$$where $~c~$ is integrating constant.
Now by the definition $~\bar y(p)=\int^{\infty}_0 y(t)e^{-pt}dt~,~$which$~~\to 0~$ if $~p\to\infty~$. So from $(2)$, $~c=0~$.
Therefore $$ \bar y=\dfrac{5}{p^3}$$
So
$~y(t)=\mathcal L^{-1}\{\bar y(p)\}~$
$~~~~~~=5~\mathcal L^{-1}\left\{\dfrac{1}{p^3}\right\}$
$~~~~~~=5\dfrac{t^2}{2}$
nmasanta
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If $p\to \infty$ then $e^{-pt}\to 0$, and therefore the integral and finally $\bar y(p)$ – nmasanta Dec 02 '19 at 17:08
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I think this link may help you "https://math.stackexchange.com/questions/859407/laplace-transform-non-exponential-order" – nmasanta Dec 02 '19 at 17:15
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Umm I still don't get it. So I know that there exists non-exponential order functions which have a laplace transform. Why can't $y(t)$ be one of those?? – Hypernova Dec 02 '19 at 17:19
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Look there is no $~f(t)~$ for which $~\mathcal L{f(t)}=F(p)=\dfrac{e^{p^2}}{p^3}~$. – nmasanta Dec 02 '19 at 17:32
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However I reached the answer using that this is the IVP! Note that the lVP uniqueness condition satifies to this problem. Now if we plug in the initial condition, then whatever $L^{-1} (\frac{e^{s^2}}{s^3})$ is, if it exists then $C=0$ since $\frac{5}{2}t^2$ is a solution for every single t. I think the approach of calculating $L^{-1} (\frac{e^{s^2}}{s^3})$ is harder.
Hypernova
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Why does the initial condition imply that $C=0$ for all values of $t$? The other answers assume that the function is of exponential order (which then directly implies that $C=0$). I don't see why the two initial conditions imply that $C=0$ for every value of $t$. – Axion004 Dec 03 '19 at 16:36
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@Axion004 IVP's solution should be unique and $\frac{5}{2}t^2$ is already a solution for the given IVP for all t. Now we can check that $C=0$ or $L^{-1} (\frac{e^{s^2}}{s^3})=0$ but the second one is absurd. – Hypernova Dec 04 '19 at 05:04