# Homework Help: Characteristic equation in terms of the Laplace operator

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1. May 13, 2018

### TimeRip496

1. The problem statement, all variables and given/known data
The stability of a spinning body may be explored by using equation (3.40), with no
torque components present. It will be assumed here that the spin is about the z -axis and
has a rate ωZ = S.

2. Relevant equations
$$I_{xx}\dot{ω} - (I_{yy}-I_{zz})Sω_y = 0$$
$$I_{yy}\dot{y} - (I_{zz}-I_{xx})Sω_x = 0$$
These are linear equations whose characteristic equation in terms of the Laplace
operator, s, is
$$s^2 + (1-\frac{I_{zz}}{I_{xx}}) (1-\frac{I_{zz}}{I_{yy}}) S^2 = 0$$

3. The attempt at a solution
I have no idea what the variable is and thus I can't really apply the Laplace operator. Laplace operator is double differentiating the function but in this case I don't know what the function is and thus I have no way to solve this.

Source: https://books.google.com.sg/books?i...tion in terms of the Laplace operator&f=false

2. May 16, 2018

### pasmith

These should be $$I_{xx} \dot{\omega}_x - (I_{yy}-I_{zz})S\omega_y = 0 \\ I_{yy} \dot{\omega}_y - (I_{zz}-I_{xx})S\omega_x = 0$$

The context suggests the Laplace transform $$L(f) : s \mapsto \hat f(s) = \int_0^\infty f(t)e^{-st}\,dt$$ rather than the Laplacian operator $\nabla^2$.

However the linear constant-coefficient ODE $$\dot{\mathbf{x}} = \mathbf{A} \mathbf{x}$$
has a "characteristic equation" which is the eigenvalue equation of the matrix $\mathbf{A}$,
$$\det(\mathbf{A} - s\mathbf{I}) = 0$$
which in this case does (after some manipulation) yield $$s^2 + (1-\frac{I_{zz}}{I_{xx}}) (1-\frac{I_{zz}}{I_{yy}}) S^2 = 0.$$ But to call $s$ a "Laplace operator" in this context is not something I've seen anywhere else. What do you find if you look up "Laplace operator" in the index of your book? Does it lead you to a definition?

Last edited: May 16, 2018
3. May 24, 2018

### TimeRip496

There is no definition regarding it on the index section.
Thanks for the response. The index did not define the stated Laplace operator.

I still don't quite understand this part. Where did you get the matrix A and what did you use for it? Because to get these,
$$I_{xx} \dot{\omega}_x - (I_{yy}-I_{zz})S\omega_y = 0 \\ I_{yy} \dot{\omega}_y - (I_{zz}-I_{xx})S\omega_x = 0$$
the author combine $$\frac{d(H_c)}{dt}=\frac{d}{dt_{compts}}[H_c]+(\Omega X H_c)=T$$ and Hc which is a vector based on the principal axis.

Is A then $$\frac{d[I_c]}{dt}=\frac{d}{dt_{compts}}[I_c]+(\Omega X [I_c])$$? But how do I cross product a vector and a matrix?
where X is cross product, [Ic] is the moment of inertia tensor

4. May 24, 2018

### Ray Vickson

So you have a set of coupled DEs
$$\begin{array}{rcl} I_{xx}\: d \omega_x(t)/dt -(I_{yy}- I_{zz}) S \omega_y(t) &=& 0 \\ I_{yy}\: d \omega_y(t)/dt -(I_{zz}-I_{xx}) S \omega_x(t) &=& 0 \end{array}$$
If we let the Laplace transforms be
$$W_x(s) = \int_0^{\infty} e^{-st} \omega_x (t)\, dt, \; W_y(s) = \int_0^{\infty} e^{-st} \omega_y (t) \, dt,$$
we can get a set of coupled algebraic linear equations in $(W_x(s), W_y(s))$. To do that, you need to know the relationship between the Laplace transform of a function $\omega_x (t)$ and the transform of its time-derivative $\dot{\omega}_x (t)$. Those are standard Laplace transform properties that you can look up in a textbook or on line.

Once you know $W_x(s)$ and $W_y(s)$ you can "invert" them to determine $\omega_x (t)$ and $\omega_y (t)$. There are extensive tables of Laplace transforms and their inverses that can be helpful in doing that.

Last edited: May 24, 2018
5. May 27, 2018

### TimeRip496

I try and I get
$$W_x(s)= \mathcal{L} (ω_x(t))=\frac{1}{s} \mathcal{L} (\dotω_x(t))-\frac{ω_x}{s}$$
where $$\mathcal{L} (\dotω_x(t))=∫^∞_oe^{-st}\dotω_x(t)dt$$
However I can't integrate ω dot as it is an unknown function and thus I cannot find anything from the Laplace table to inverse it. Did I make a mistake somewhere?

6. May 27, 2018

### Ray Vickson

Yes: you did not use a standard property of Laplace transforms, that relate $\mathcal{L}(\dot{\omega})$ to $\mathcal{L}(\omega)$. It is widely available in textbooks and on-line, and it is something that every student of differential equations should know. It is the reason why Laplace-transform methods "work" at all when solving constant-coefficient linear DEs.

Last edited: May 27, 2018
7. May 28, 2018

### TimeRip496

The property you are talking about is this right?
$$\mathcal{L}(F'(t))= −F(0) + s f (s)$$

Then apply to above$$\mathcal{L}(\dotω_x(t))= −F(0) + s\mathcal{L}(ω_x(t))$$

where $$\mathcal{L}(\dotω_x(t))=∫^∞_oe^{-st}\dotω_x(t)dt$$
Thats the property I can only find that is related to unknown 1st-order derivative function

8. May 28, 2018

### Ray Vickson

That is exactly what you need. Carry on!

9. May 29, 2018

### TimeRip496

I get it but just to confirm,

$$\mathcal{L}(\dotω_x(t))= −F(0) + s\mathcal{L}(ω_x(t))$$
$$\mathcal{L}^{-1}(W_x(t))= \mathcal{L}^{-1}(\frac{1}{s}F(0)) + \mathcal{L}^{-1}(\frac{1}{s}\mathcal{L}(\dotω_x(t))$$
$$\omega_x=\frac{1}{s}\dotω_x(t)+0$$

where $$\mathcal{L}(\dotω_x(t))=∫^∞_oe^{-st}\dotω_x(t)dt$$

To solve it, am I right to take F(0) as zero or otherwise I will get infinity?

10. May 29, 2018

### Ray Vickson

No, obviously not! Lots of functions $F(t)$ have $F(0) \neq 0$ but have perfectly well-defined Laplace transforms. For example, the function $F(t) \equiv 1$ has transform $1/s$, and the function $F(t) = e^{-at}$ has transform $1/(a+s)$. You can even have $F(0+) = \infty$ but still have a perfectly good Laplace transform; for example, the function $F(t) = 1/\sqrt{t}$ has transform $\sqrt{\pi/s}.$ (Interestingly, though, the fact that for this function we have $F(0+) = \infty$ means that $F'(t)$ cannot have a Laplace transform (because the formula $s \tilde{F}(s) - F(0+)$ does not give a finite answer. The function $F'(t) = -(1/2)/t^{3/2}$ goes to $-\infty$ too fast as $t \to 0+$ to yield a convergent Laplace integral for any value of $s$.)

Anyway, what is $F(0)$ in the above? There is no function $F(t)$ anywhere in your differential equation.

Also: the "equation"
$$\omega_x=\frac{1}{s}\dotω_x(t)+0$$
is meaningless. It makes no sense to have both $s$ and $t$ in the same equation. You either have $t$ on both sides or else you have $s$ on both sides; you will never, ever, have both at the same time.

You still do not seem to have realized that you get two coupled linear equations for the transforms $W_x(s)$ and $W_y(s)$. When you solve that $2 \times 2$ linear system symbolically, the solutions will be some functions of $s$, and those are the things you need to "untransform" to get the solutions for $\omega_x(t)$ and $\omega_y(t)$.

Hint: If $\omega_x(t)$ and $\omega_y(t)$ have transforms $W_x(s)$ and $W_y(s)$, one of the equations you get is
$$s W_x(s) - \omega_x(0) = a W_y(s),$$
where
$$a = \frac{I_{yy}-I_{zz}}{I_{xx}} S.$$
There is a similar equation for $W_y(s)$, so you have all the equations you need to find $W_x(s)$ and $W_y(s)$.

Last edited: May 30, 2018
11. Jun 5, 2018

### TimeRip496

Okay thanks and I get
$$s^2-S^2(\frac{I_{yy}-I_{zz}}{I_{xx}})(\frac{I_{zz}-I_{xx}}{I_{yy}})=sW_y(0)+\frac{I_{zz}-I_{xx}}{I_{yy}}SW_x(0)$$

Thus, I can only assume that Wy(0) & Wx(0) are zero so I can finally solve it. Besides thanks a lot for your patience and help!!!

12. Jun 5, 2018

### Ray Vickson

This looks wrong to me, but I cannot tell where your errors lie because you do not show your work. Usually, "characteristic equations" do not involve initial conditions, so usually the constants $W_x(0), W_y(0)$ would not be involved. BTW: your original DEs involved $\omega_x$ and $\omega_y$, so any constants would need to be in terms of $\omega_x(0)$ and $\omega_y(0)$. Was your writing $W_x(0), W_y(0)$ instead just a typographical error?

13. Jun 6, 2018

### TimeRip496

$$\dot{W_x}(s)=-\omega_x(0)+sW_x(s)$$
$$\dot{W_y}(s)=-\omega_y(0)+sW_y(s)$$
$$\dot{W_x}-\frac{I_{yy}-I_{zz}}{I_{xx}}SW_y=0\ \ \longrightarrow\ sW_x(s)-\omega_x(0)=\frac{I_{yy}-I_{zz}}{I_{xx}}SW_y$$
Rearrange, $$W_x(s)=\frac{1}{s}(\frac{I_{yy}-I_{zz}}{I_{xx}}SW_y+\omega_x(0))$$
$$\dot{W_y}-\frac{I_{zz}-I_{xx}}{I_{yy}}SW_x=0\ \ \longrightarrow\ sW_y(s)-\omega_y(0)=\frac{I_{zz}-I_{xx}}{I_{yy}}SW_x$$
Then combine the above two by subbing in the Wx(s),
$$sW_y(s)-\omega_y(0)=S^2(\frac{I_{yy}-I_{zz}}{I_{xx}})(\frac{I_{zz}-I_{xx}}{I_{yy}})W_y*\frac{1}{s}+\frac{1}{s}\frac{I_{zz}-I_{xx}}{I_{yy}}S\omega_x(0)$$

And then rearrange,
$$s^2-S^2(\frac{I_{yy}-I_{zz}}{I_{xx}})(\frac{I_{zz}-I_{xx}}{I_{yy}})=s\omega_y(0)+\frac{I_{zz}-I_{xx}}{I_{yy}}S\omega_x(0)$$

I am left with the RHS which I can't get rid of

14. Jun 6, 2018

### Ray Vickson

(1) It is a complete waste of time and effort to keep writing/typing things like $S(I_{xx}-I_{zz})/I_{yy}$ over and over again; that is why I chose to write $a = S(I_{yy}-I_{zz})/I_{xx}$ and similarly, $b = S(I_{xx}-I_{zz})/I_{yy}$, so that the transformed equations are
$$s W_x - k_x = a W_y \; \hspace{3ex}(1)\\ s W_y - k_y = -b W_x \; \hspace{3ex} (2)$$
where $k_x =\omega_x(0)$ and $k_y = \omega_y(0)$.

(2) After that I have no idea what you are doing: you get an equation for $W_y$ of the form $s W_y - k_y =-ab W_y/s + a k_x/s.$ Then, you somehow "rearrange" the equation to have $W_y$ disappear completely! If I were doing it I would just go ahead and solve the equation to find $W_y$, because, after all, that is the object I want to know.

15. Jun 7, 2018

### TimeRip496

So I rearrange the equation to get
$$W_y=\frac{sk_y+akx}{s^2+ab}$$
Then I solve it be using inverse laplace transform to get ωy right?
But I can't find anything that represents the above equation from the Laplace table. the closest I get is for cos(at) where its
$$F(s)=\frac{s}{s^2+a^2}$$
Is this the correct one?

And I realised I did somehow removed the Wy unknowingly from my previous answer. Sorry that was my bad

16. Jun 7, 2018

### Ray Vickson

You can write
$$W_y(s) = k_y \frac{s}{s^2+c^2} + a k_x \frac{1}{s^2+c^2} \hspace{4ex}(1)$$
where $c^2 = a b$. Because of linearity of the Laplace transform ${\cal L}$ we have ${\cal L}\; [ \alpha f(t) + \beta g(t) ]= \alpha {\cal L} f(t) + \beta {\cal L} g(t)$ for constants $\alpha, \beta$. That means we can invert each term of (1) separately in terms of inverses of $s/(s^2+c^2)$ and $1/(s^2+c^2)$.

Last edited: Jun 7, 2018
17. Jun 8, 2018

### TimeRip496

Oh I see. So continuing on that,
$$\omega_y=k_ycos(ct)+\frac{ak_x}{c}sin(ct)$$
And then I continue by finding ωx by differentiating and subbing ωy into the given equation,
$$I_{yy}\dot\omega_y-(I_{zz}-I_{xx})S\omega_x=0$$
to get
$$\omega_x=\frac{k_yc}{b}sin(ct)-\frac{ak_x}{b}cos(ct)$$
Differentiating ωx and then subbing it, together with ωy into the given equation
$$I_{xx}\dot\omega_x-(I_{yy}-I_{zz})S\omega_y=0$$
to get $$(\frac{k_yc^2}{b}-ak_y)cos(ct)=(\frac{a^2k_x}{c}+\frac{ak_xc}{b})sin(ct)$$
But I am stuck at here. I don't know how to proceed as the trigos remain there and I can't get rid of them to get the final answer.

18. Jun 8, 2018

### Ray Vickson

If I had been doing it I would have solved for both $W_x(s)$ and $W_y(s)$ simultaneously, and then found both $\omega_x(t)$ and $\omega_y(t)$ by inverting both of the transforms. Of course, your method--of finding $\omega_y(t)$ and then solving a DE to get $\omega_x(t)$ -- will also work, but takes longer and requires more effort.

If all you want is an answer to your original question, namely, to find the "characteristic equation" of the DE system, you must first know what a characteristic equation really IS. If the word "Laplace" had never been used in the problem statement, what would you have done? How would you have found the characteristic equation then?

However, if you do want to use Laplace transforms, how would you get the characteristic equation? Here is a hint: all you need to do is examine the formula for $W_x(s)$ or $W_y(s)$; you do not need to invert them to find the $\omega_x(t)$ and/or $\omega_y(t).$

19. Jun 9, 2018

### TimeRip496

I am not very sure about what characteristic equation means but what I know is
1) For linear algebra, we can use it to find the eigenvalues using this formula det(A-λI)=0
2) For DE, it can be used to solve the n-th order homogeneous DE by converting it into a polynomial and finding its roots. But if the DE is non-homogeneous, we need to use method of undetermined coefficients or variation of parameters method to complement it.

But I don't know how can I use it to apply to the above question.

As for Wy, I am stuck at here
$$W_y=\frac{sk_y+akx}{s^2+ab}$$
I try rearranging it but I can't get rid of the RHS
$$s^2+ab=\frac{sk_y+ak_x}{W_y}$$

20. Jun 9, 2018

### Ray Vickson

What you wrote above is not a characteristic equation. As I have said already, the characteristic equation does not involve the initial conditions $k_x$ and $k_y$.

Basically, for a differential equation system, the characteristic equation is the equation whose roots $r$ are the ones that appear in solutions of the form $e^{rt}$. When you substitute the forms $\omega_x(t) = A e^{rt}$ and $\omega_y(t) = B e^{rt}$ into your DEs, you get a system of equations for $A, B$ that will have a nonzero solution only for certain values of $r$. Those critical $r$-values can be determined from some equation, called the characteristic equation.

So, if you use Laplace transforms instead, how would you find the characteristic equation? Where do you see the $r$-values in the formula for $W_x(s)?$ I suggest you get a book on Laplace transforms, or do an on-line search for differential equations and Laplace transforms; PF rules do not permit me to say much more, and in fact, I may have said too much already.

Last edited: Jun 9, 2018
21. Jun 12, 2018

### TimeRip496

Is it something like this,
$$\dot{W}_x(s)=-k_x+sW_x(s)$$
$$r=s\ \rightarrow \ (r-s)=0$$
Thus, $$W_x=Ae^{st}$$
However I am not sure about this because so far I had only learn how to use characteristic equation for 2nd order or higher order DE which will give me more than one value of r. The above is a first order DE and using characteristic equation doesn't seem right.

Or I should turn it into second order derivative so I can use the characteristic equation?
$$\ddot{W}_x(s)=-\dot{k}_x+s\dot{W}_x(s)$$
$$r^2-sr=r(r-s)=0$$
Thus,
$$W_x=A+Be^{st}$$

I am sorry about that and hopefully you can still tell me whether I am on the right track!

22. Jun 12, 2018

### Ray Vickson

You are on the wrong track. You have already found complete, explicit expressions for $W_x(s)$ and $W_y(s)$, so there is no need to re-do the work--incorrectly at that. Just looking at the formulas for $W_x(s), W_y(s)$ is enough for you to be able to write down the characteristic equation, without doing any more work. You just have to figure out where to look, and for that I have suggested you either get an appropriate book, or look on-line at material about differential equations and Laplace transforms.

Anyway, if you wanted to find the characteristic equation without using Laplace transforms, what would you do? Try that first. (And, when you do it, there should be no $W_x(s)$ or $W_y(s)$ anywhere, since these are Laplace transforms and you would NOT be using them; instead, there should be $\omega_x(t), \omega_y(t)$ and their time-derivatives.)

23. Jun 14, 2018

### TimeRip496

I think I get it already!!!
$$\omega_y=k_ycos(ct)+\frac{ak_x}{c}sin(ct)$$
$$C_1=k_y \ \ \ C_2=\frac{ak_x}{c}$$
$$\alpha=0 \ \ \ \ \beta=c \ \ \rightarrow \ \ \ \lambda=+/-ic$$
$$\lambda^2+c^2=0 \ \ [Solved]$$

As for doing it without Laplace transform,
$$\dot\omega_x-a\omega_y=0$$
$$\dot\omega_y-a\omega_x=0 \ \ \ \rightarrow \ \ \ \omega_y=-b\int\omega_xdt$$
$$\dot\omega_x+ab\int\omega_xdt=0$$
$$\lambda^2+ab=0 \ \ \ \ [Solved]$$

24. Jun 14, 2018

### Ray Vickson

OK, good: you have the solution. Now I can pull back the curtain to reveal the solution using Laplace transforms.

We have found the Laplace transforms $W_x(s)$ and $W_y(s)$ as
$$W_x(s) = \frac{s\, \omega_x(0) + a\, \omega_y(0)}{s^2 + ab} \\ W_y(s) = \frac{s\, \omega_y(0) -b\, \omega_x(0)}{s^2+ab}.$$
The characteristic equation is found by setting the denominators to zero: that is, $s^2 + ab = 0$.

Note that the initial conditions $\omega_x(0), \omega_y(0)$ appear only in the numerators, not the denominators. That is why they do not appear in the characteristic equation!

In general, if we have a system state-variable $y(t)$ whose Laplace transform $Y(s)$ has the form of a ratio of two polynomials, that is,
$$Y(s) = \frac{N(s)}{D(s)}$$
for some polynomials $N(s)$ and $D(s)$, the characteristic equation is just $D(s) = 0.$

The roots of the characteristic equation correspond to the "fundamental modes" of the system, so that if $r$ is a root of multiplicity $p$ -- that is, one factor of $D(s)$ is $(s-r)^p$ -- then $y(t)$ will contain terms that are constant multiples of $t^{p-1} e^{rt}/(p-1)!$ and some of its time-derivatives. If all the roots are simple (all $p = 1$) then $y(t)$ is a linear combination of the functions $e^{rt}$. (Note that time-derivatives of $e^{rt}$ are just constant multiples of $e^{rt}$ itself, so taking time-derivatives does not introduce new types of terms.)

In your case the roots are $\pm i \sqrt{ab}$ so you have $e^{\pm i \sqrt{ab} t},$ which are equivalent to $\sin( \sqrt{ab} t)$ and $\cos(\sqrt{ab} t).$

Last edited: Jun 14, 2018
25. Jun 15, 2018

### TimeRip496

Thanks a lot for your help and patience!!!

I am still new to Laplace transform and this is my first time seeing the Laplace in terms of ratio of 2 polynomials. The ratio you are referring to is the zeroes and poles of Laplace transform right? Is there any reason for such formulation? I did some online search and they mostly talk about using it for determining stability.