Uniqueness of Solutions for Laplace Initial Value Problems

[sssr]

Hi

I have to solve this equation using laplace transform

ty''+(t-1)y'+y=t^2 y(0)=0 y'(0)=0

I tried it using MATLAB and get to this point

(Int(-2/s^4*exp(1/2*s*(s+6)),s)+C1)*exp(-1/2*s*(s+6))

I think I've gone wrong somewhere

Any Help would be appreciated.

Thanks

 

[saltydog]

Hi

I have to solve this equation using laplace transform

ty''+(t-1)y'+y=t^2 y(0)=0 y'(0)=0

I tried it using MATLAB and get to this point

(Int(-2/s^4*exp(1/2*s*(s+6)),s)+C1)*exp(-1/2*s*(s+6))

I think I've gone wrong somewhere

Any Help would be appreciated.

Thanks

Hello sssr,

Suppose I've waited long enough for someone better than me to weigh in and they still may, but I don't think you can do this:

$$\mathcal{L}\{ty^{'}\}=\int_0^{\infty} te^{-st}y^{'}dt$$

Let alone the second derivative! Might be a convolution for it that I'm not aware of however. Please someone say so if so.

And in fact, I don't see how the power series method can work either.

Ergo: Unless someone else says otherwise, my only recourse if I were solving it would be old faithful: numerically.

 

[Justin Lazear]

Suppose I've waited long enough for someone better than me to weigh in and they still may, but I don't think you can do this:

$$\mathcal{L}\{ty^{'}\}=\int_0^{\infty} te^{-st}y^{'}dt$$

Let alone the second derivative! Might be a convolution for it that I'm not aware of however. Please someone say so if so.

It seems to me if you apply parts enough times, the only hitch is in evaluating

$$\int_0^{\infty} te^{-st}ydt$$

But both ty'' and ty' reduce to that and known transforms, if I'm not mistaken.

Then again, small comfort, since that's not so easy to evaluate anyway.

--J

 

[saltydog]

I don't think that integral can be evaluated Justin.

Also Sssr, The ODE is singular at t=0 thus the initial value problem you proposed is, as I see it anyway, ill-poised (no offense). However, good for t>0 you know.

So what do you think guys?

 

[sssr]

Hi thanks everyone for the replies
I actually made a mistake in my working and now
have come to this point

(2/s^2+2/3/s^3+2/s+C1)/(s+1)^3

So can anyone help in where do i go from here

Thanks

 

[Corneo]

Not sure if this is helpful or not.

if we have

$$F(s) = \int_0^\infty e^{-st} f(t) \ dt = \mathcal{L}\{f(t) \}$$

Then

$$F'(s) = \mathcal\{ -t f(t)\}$$

and in general

$$F^{(n)}(s) = \mathcal{L} \{ (-t)^n f(t)} \}$$

 

[Corneo]

Given that

$$ty''+(t-1)y'+y = t^2 \qquad y(0) = 0, y'(0) = 0$$

$$-\mathcal{L} \{- ty''\} + -\mathcal{L} \{-ty' \} - \mathcal{L}\{ y'\} + \mathcal{L}\{y\} = \frac {2}{s^3}$$

I force a negative sign in there so I can use those two formulas above.

$$- \frac {d}{ds}\bigg[ s^2Y(s) - sy(0) - y'(0) \bigg] - \frac {d}{ds} \bigg [ sY(s) - y(0) \bigg] - \bigg[ sY(s) - y(0)\bigg] + Y(s) = \frac {2}{s^3}$$

Where $$\mathcal{L}\{ y\} = Y(s)$$

You will have to compute those derivatives using the product rule.

 

[sssr]

Given that

$$ty''+(t-1)y'+y = t^2 \qquad y(0) = 0, y'(0) = 0$$

$$-\mathcal{L} \{- ty''\} + -\mathcal{L} \{-ty' \} - \mathcal{L}\{ y'\} + \mathcal{L}\{y\} = \frac {2}{s^3}$$

I force a negative sign in there so I can use those two formulas above.

$$- \frac {d}{ds}\bigg[ s^2Y(s) - sy(0) - y'(0) \bigg] - \frac {d}{ds} \bigg [ sY(s) - y(0) \bigg] - \bigg[ sY(s) - y(0)\bigg] + Y(s) = \frac {2}{s^3}$$

Where $$\mathcal{L}\{ y\} = Y(s)$$

You will have to compute those derivatives using the product rule.

Thanks Corneo
I've done that part and get

Y'(s)*(-s-s^2) + Y(s)*(-3s) = -2/s^3

and then simplify

Y'(s) + Y(s)*(3/(s^2+s) = -2/(s^5+s^4)

Then solving that i end up with

(-2/s^2-2/3/s^3-2/s+C1)/(s+1)^3

and this is where i am stuck - how do i solve for C1

Thanks

 

[saltydog]

$$F'(s) = \mathcal\{ -t f(t)\}$$

and in general

$$F^{(n)}(s) = \mathcal{L} \{ (-t)^n f(t)} \}$$

Very nice guys. I see I should have waited a bit longer and have learned a lot. I'll try and solve it too now. That's just a typo right Corneo. Should be:

$$F'(s) = \mathcal{L}\{ -t f(t)\}$$

Edit: Ok Corneo, I think you did that tutorial for my benefit as Sssr already knew that. Thanks a bunch.

 

[saltydog]

Y'(s)*(-s-s^2) + Y(s)*(-3s) = -2/s^3

Alright, even though I'm hesitant to say anything now since I was off from the get-go, I feel I have it now. I think you have an extra minus sign up there. With regards to the constant of integration, just invert the Laplace transform with the 'c' in place and see what you get.

I'm still unclear about what I think is a singularity of the ODE at t=0. However, the solution I obtained (and verified by back-substitution) does not have one.

Edit: Oh yea, thanks guys. Interesting problem and I learned a lot!

 

[Corneo]

I haven't had much luck with this problem. I don't know how to solve for Y(s). I thought it would be seperable, but I don't think so.

 

[saltydog]

Hello guys. Sssr, allow me to summarize what I've done with Corneo's kind help:

So we have:

$$ty''+(t-1)y'+y = t^2 \qquad y(0) = 0, y'(0) = 0$$

And we let $$\mathcal{L}\{ y\} = Y(s)$$

So I make the transforms as Corneo suggest, do the differentiation and integration and come up with:

$$Y(s)=\frac{2}{s(s+1)^3}+\frac{2}{s^2(s+1)^3}+\frac{2}{3s^3(s+1)^3}+\frac{c}{s(s+1)^3}$$

Note I kept the c. However I wish to point out that in Kreyszig, which goes over this type of problem for Laguerre's ODE (learning this only after Corneo's post), they set it equal to zero but I don't understand why.

The inverse Laplace transform of the above is:


$$\mathcal{L}^{-1}\{Y(s)\}=\frac{1}{6}e^{-t}(-2+3c+2e^t)t^2$$

That is:

$$y(t)=\frac{1}{6}e^{-t}(-2+3c+2e^t)t^2$$

This satisfies the ODE (and the initial conditions) for any value of c. However, I suspect it needs to be 0 but don't know why.

Also, I thought I would arrive at a "specific" solution without arbitrary constants. Can anyone explain this to me?

 

[sssr]

to
saltydog & Corneo just want to say thanks for all your help with this
problem

Thanks heaps
Cheers

 

[Justin Lazear]

In the brief intermission from me being overworked that I'm stealing from myself, did we figure out what the story with the c was? I noted that the numerical solution Mathematica found set c to zero, as saltydog said, but I've had little time to look into it more than that.

--J

 

[saltydog]

In the brief intermission from me being overworked that I'm stealing from myself, did we figure out what the story with the c was? I noted that the numerical solution Mathematica found set c to zero, as saltydog said, but I've had little time to look into it more than that.

--J

Hello Justin. I've been unavailable for a while and am only now catching up. You know, I hate it when I don't understand something in math and I don't understand the bit about the constant of integration up there. I'm a purist with things like that and will no doubt eventually trek 100 miles to the nearest university math department and just start asking questions. They're usually very nice about things like that you know. I'll try and research it a bit via the library first though. It's a pleasant ride.

 

[saltydog]

Some follow-up

I've been looking into the problem above concerning the constant of integration and the lack of uniqueness of solutions. Today I spoke with a professor I know concerning this equation and he suggested I look into uniqueness requirements for the following equation:

$$b_0(x)y^{''}+b_1(x)y^{'}+b_2(x)y=R(x)$$

His suggestion is that uniqueness will not be guaranteed at the singular points of this equation. That is, for the equation:

$$ty^{''}+(t-1)y^{'}+y=t^2;\quad y(0)=0,\quad y^{'}(0)=0$$

The solution obtained via Laplace transform:

$$y(t)=\frac{1}{6}e^{-t}(-2+3C+2e^t)t^2$$

May in fact be the non-unique solution for this equation at the singular point t=0.

I might add Mathematica's NDSolve will return a solution only if initial conditions at t$\neq$0 are supplied.

This is only a proposal and needs to be verified. Think I'll work on it.

 

[Justin Lazear]

*sighs* If only I didn't have to write an essay and do a few more finals, move out of my room, then drive for 30-some-odd hours, I could look into it more. But alas, I've been screwed by finals week.

So don't bust your butt on my account, salty, because I won't be able to do anything for a week, at least.

--J

 

[saltydog]

*sighs* If only I didn't have to write an essay and do a few more finals, move out of my room, then drive for 30-some-odd hours, I could look into it more. But alas, I've been screwed by finals week.

So don't bust your butt on my account, salty, because I won't be able to do anything for a week, at least.

--J

Hello Justin. I have a suggestion for you that works for me: that 30 hour trip, stop frequently and eat meals. It makes it a much more pleasant trip. Oh yea, the math up there, that's dessert for me, you know, like cheese cake with strawberries and I'm finding it a new challenge. Keep in mind that resolution of the problem is not my primary objective but rather the journey and I stop often.

 

[Justin Lazear]

Unfortunately, it's a 30 hour drive that must be completed in one weekend. Sometimes I wonder about the things I get myself into... But there's a bubble bath waiting for me at the end of it, so it'll all be worth it.

And while I'm driving for two days straight, you get to wander around in the playground. Oh well. At least I'm done with the essay now.

--J

 

[Hurkyl]

You sure that's right?

When I assume the solution is of the form:

$$y(t) := e^{-t} p(t) + q(t)$$

Where p and q are arbitrary polynomials, and plug into the equation, I get this as the only solultion:

$$y(t) = A t^2 e^{-t} + \frac{1}{3} t^2$$

And plugging this back into the original equation works.

Oh, I guess this is your solution -- you just didn't collapse the (1/6)(-2+3C) into a single constant A. Silly me!


By the way, where is the other family of solutions? I should have two arbitrary constants available to me! It's probably a rational function times an exponential... or maybe a different exponential?


Anyways, back to the problem at hand -- uniqueness of solution.

It helps to think geometrically, I think. A differential equation is nonsingular when you can solve for the value of the second derivative, if you're given the value and first derivative at that point.

So, if you have a solution curve that starts at your favorite point and with your favorite slope, there's a unique solution for its curvature near that point.

(This sort of thing is much easier to picture with a first order equation! Anyways...)

It's a powerful theorem that there exists at least one solution, but once you have that, it's intuitively clear that it must be locally unique, by looking at an appropriate "infinitessimal" neighborhood of the point.


Now, when you have a singular point, you cannot solve for the curvature. Either no solution is possible, or all values work. Since these solutions extend into areas where it is locally unique, you can look at it in the other direction... solutions "pinch" together as they approach this point, making you unable to distinguish them near the singular point.

 

[saltydog]

Thanks Hurkyl.

By the way, where is the other family of solutions? I should have two arbitrary constants available to me! It's probably a rational function times an exponential... or maybe a different exponential?

I'm working on it. I'm a slow-poke and have never up until now worked on an initial-value problem that did NOT produce a unique solution. This is an epiphany for me.

Now, when you have a singular point, you cannot solve for the curvature. Either no solution is possible, or all values work. Since these solutions extend into areas where it is locally unique, you can look at it in the other direction... solutions "pinch" together as they approach this point, making you unable to distinguish them near the singular point.

Why can't you solve for the curvature at a singular point?

Also, Schaums has:

$$ty^{''}+y^{'}+4ty=0;\quad y(0)=3\quad y^{'}(0)=0$$

This one is solved in terms of Bessel functions:

$$y(t)=3J_0(2t)$$

How come that one yields a unique solution?

Edit: You know, I'm not sure now (don't have the book), that $J_0$ above is a Bessel function. I think so.

 

[Hurkyl]

Why can't you solve for the curvature at a singular point?

Curvature involves the second derivative. If you tried solving for y''(t) in:

t y''(t) + (t - 1) y'(t) + y(t) = t^2

at t = 0, you get this after plugging in t = 0:

-y'(0) + y(0) = 0

y''(0) doesn't appear in the equation at all! That makes it a little hard to solve for it.

How come that one yields a unique solution?

Heuristic reasoning doesn't apply in all circumstances. Nothing I've said in my post was a "proof", just an explanation. (And remember, I'm nowhere near an expert at these things!)

In this case, I think the singularity arises because things like to go to infinity, rather than clustering together -- another way that things can happen.

In this case, the "other solutions" would yield inconsistencies, I suppose. I haven't worked out something that would be heuristic to make one think that, though. Maybe the fact that (y'' + 4y) must vary like (y' / t) near t = 0 is enough of a constraint to "solve" for y''?

(My perception is based on assuming the answer has a laurent series about x = 0)

 

[arildno]

Hi

I have to solve this equation using laplace transform

ty''+(t-1)y'+y=t^2 y(0)=0 y'(0)=0

I tried it using MATLAB and get to this point

(Int(-2/s^4*exp(1/2*s*(s+6)),s)+C1)*exp(-1/2*s*(s+6))

I think I've gone wrong somewhere

Any Help would be appreciated.

Thanks

This has probably been said already, but:
The easiest way to see why you might have a non-unique solution here, is to convert it to the first order system:
$$y_{1}'=y_{2}(t), y_{2}'=t-\frac{y_{1}}{t}+(\frac{1}{t}-1)y_{2}, y_{1}(0)=y_{2}(0)=0\vec{Y}=(y-{1},y_{2}), \vec{F}(t,\vec{Y})=(y_{2},t-\frac{y_{1}}{t}+(\frac{1}{t}-1)y_{2})$$

But since $$\frac{\partial\vec{F}}{\partial\vec{Y}}$$ does not exist at t=0, one of the conditions needed for existence of unique solutions (in at least one type of uniqueness proof) is violated.

 

[saltydog]

This has probably been said already, but:
The easiest way to see why you might have a non-unique solution here, is to convert it to the first order system:
$$y_{1}'=y_{2}(t), y_{2}'=t-\frac{y_{1}}{t}+(\frac{1}{t}-1)y_{2}, y_{1}(0)=y_{2}(0)=0\vec{Y}=(y-{1},y_{2}), \vec{F}(t,\vec{Y})=(y_{2},t-\frac{y_{1}}{t}+(\frac{1}{t}-1)y_{2})$$

But since $$\frac{\partial\vec{F}}{\partial\vec{Y}}$$ does not exist at t=0, one of the conditions needed for existence of unique solutions (in at least one type of uniqueness proof) is violated.

Thanks Arildno. I'll work on that. However, since your comments are not easy for me to follow, I must first approach it in simpler terms and then work up to your analysis (just my way of doing things). In this regards, I'm first studying:

$$ty^{''}+(t-1)y^{'}+y=0$$

via power series. Then I'll look at it a bit more via Laplace transforms, then look at existence and uniqueness proofs in general for ODEs, then systems, and then interpret your comments.

 

[saltydog]

I've calculated the two power series solutions for:

$$ty^{''}+(t-1)y^{'}+y=0$$

They are:

$$y_1(x)=-x^2+\sum_{n=3}^{\infty}\frac{(-1)^{n-1} x^n}{(n-2)!}$$

$$y_2(x)=1+x-x^2+y_1(x) ln(x)-\sum_{n=3}^{\infty}a_n (H_{n-2}-1) x^n$$

with:

$$a_n=\frac{(-1)^{n-1}}{(n-2)!}$$

$$H_n=\sum_{k=1}^n \frac{1}{k}$$


Thus the general solution is:

$$y(x)=Ay_1(x)+By_2(x)$$

I verify these by calculating the derivatives (for the first 100 terms of the series), and then back-substitution into the LHS of the ODE and plotting the results. The attached plots show this back-substition for both functions. As you can see, the results are no larger than 10^-14. Some reading this may wonder, "what does this have to do with Laplace Transforms?" . . . sometimes you have to move backwards in order to move forwards.

 

[Hurkyl]

Ah, that makes me happy! Two free variables, and the typical solution isn't twice differentiable at the origin.

 

[saltydog]

Interesting you note that Hurkly.

In the interest of being complete, I've calculated the initial-value problem:

$$ty^{''}+(t-1)y^{'}+y=0;\quad y(0.1)=3,\quad y'(0.1)=-1$$

using the series I calculated above (just take the derivative of y(x)) and obtained the system to solve for A and B:

$$3=Ay_1(0.1)+By_2(0.1)$$

$$-1=Ay_1^{'}(0.1)+By_2^{'}(0.1)$$

Doing this, I obtain the particular solution:

$$y(x)\approx 24.48y_1(x)+2.9y_2(x)$$

I next ploted this particular solution and compared it against Mathematica's NDSolve for this initial-value problem. The superposition of both results are below; they are identical. This along with the results above, gives me confidence the series calculated are correct.

 

[saltydog]

So now that the general solution has been determined for the homogeneous equation, one seeks to determine the general solution for the non-homogeneous case:

$$xy^{''}+(x-1)y^{'}+y=x^2$$

That's easy. We need only a particular solution. But we calculated an infinite number of them up there. One is:

$$x^2e^{-x}+\frac{1}{3}x^2$$

And so the general solution is:

$$y(x)=Ay_1(x)+By_2(x)+x^2e^{-x}+\frac{1}{2}x^2$$

With $y_1(x)$ and [itex]y_2(x)[/tex] defined above.

This is interesting in that I can think of no other way of solving the non-homogeneous equation analytically for the general solution. The Laplace Transform in conjunction with the power-series approach are thus shown to be effective in accomplishing this task. Note that the general solution is not defined for x=0. I am thus further led to suspect that uniqueness is lost at any point where the general solution ceases to exist.

 

[saltydog]

Hello guys. I wish to further investigate this problem from the perspective of the following theorem (in Rainville and Bedient but no proof):

Consider the equation:

$$y^{''}=f(x,y,y^{'})$$

If f and its partial derivatives with respect to y and $y^{'}$ are continuous functions in a region T defined by:

$$|x-x_0|\leq a,\quad |y-y_0|\leq b, \quad |y^{'}-y_0^{'}|\leq c,$$

then there exists an interval $|x-x_0|\leq h$ AND a unique function $\phi(x)$ such that $\phi(x)$ is a solution of the differential equation for all x in the interval $|x-x_0|\leq h$ such that:

$$\phi(x_0)=y_0\quad\text{and}\quad \phi^{'}(x_0)=y_0^{'}$$

Thus for:

$$y^{''}=t-\frac{t-1}{t}y^{'}-\frac{1}{t}y=f(t,y,y^{'})$$

we have:

$$\frac{\partial f}{\partial y}=-\frac{1}{t}$$

$$\frac{\partial f}{\partial y^{'}}=-\frac{t-1}{t}$$

Obviously f and the partials aren't continuous at t=0 and is consistent with Arildno's analysis above.

However, that's not good enough. The theorem makes no mention of "if and only if". Are there some equations with discontinuities which still allow unique solutions? I really need to determine for myself at what specific point in the proof does the absence of continuity affects uniqueness. I'll first work on it myself and if I have problems, E.L. Ince, "Ordinary Differential Equations" has a complete proof. Or I can ask you guys.

 

[arildno]

Great work, saltydog!

I did, however say, that the non-existence of the partial derivative MIGHT indicate the lack of uniqueness..

I do not know the precise necessary and sufficient conditions for uniqueness, I'm only familiar (or rather, was, several years ago) with one particular uniqueness proof (in Marsden's "Real Analysis") which did rely upon the existence of the partial derivative.