Laplace transform -- By parts?

[tomwilliam2]

Homework Statement

$A\dot{x} + By = 0$
$C\dot{y} + Dx = 0$

Homework Equations

$\int u'v = uv - \int uv'$

The Attempt at a Solution

This is a system of linear DE:
$A\dot{x} + By = 0$
$C\dot{y} + Dx = 0$
Where the constants A-D are non-zero and x and y are functions of time.

This is simple to solve by taking the derivative wrt time of the first:
$A\ddot{x}+ B\dot{y}=0$
Then rearranging for $\dot{y}$, removing it for the other expression, producing the characteristic equation:
$\lambda^2 - \frac{BD}{CA}=0$
Then finding the real, distinct roots and producing a family of solutions.
However, in my textbook it says "the characteristic equation (of these linear equations) in terms of the Laplace operator, s, is:
$s^2 - \frac{BD}{CA}=0$
Obviously that's the same as my solution...but I'm intrigued by the use of the "Laplace operator". I'm guessing the author doesn't mean the differential operator $\Delta$ - I can't see how you would use that to solve these equations. Does he mean a Laplace transform?
I understand that Laplace transforms can be used to solve these systems of linear equations, so I've been trying to do just that with this simple example. However, I get stuck trying to integrate:
$\int_{0}^{\infty}\dot{x}e^{-st} dt$
If I do integration by parts, I just get back to the same equation after I've integrated twice. I can't see how to solve it by substitution. Is there any other way to get to the characteristic equation which might be what the book is referring to?

 

[Ray Vickson]

Homework Statement

$A\dot{x} + By = 0$
$C\dot{y} + Dx = 0$

Homework Equations

$\int u'v = uv - \int uv'$

The Attempt at a Solution

This is a system of linear DE:
$A\dot{x} + By = 0$
$C\dot{y} + Dx = 0$
Where the constants A-D are non-zero and x and y are functions of time.

This is simple to solve by taking the derivative wrt time of the first:
$A\ddot{x}+ B\dot{y}=0$
Then rearranging for $\dot{y}$, removing it for the other expression, producing the characteristic equation:
$\lambda^2 - \frac{BD}{CA}=0$
Then finding the real, distinct roots and producing a family of solutions.
However, in my textbook it says "the characteristic equation (of these linear equations) in terms of the Laplace operator, s, is:
$s^2 - \frac{BD}{CA}=0$
Obviously that's the same as my solution...but I'm intrigued by the use of the "Laplace operator". I'm guessing the author doesn't mean the differential operator $\Delta$ - I can't see how you would use that to solve these equations. Does he mean a Laplace transform?
I understand that Laplace transforms can be used to solve these systems of linear equations, so I've been trying to do just that with this simple example. However, I get stuck trying to integrate:
$\int_{0}^{\infty}\dot{x}e^{-st} dt$
If I do integration by parts, I just get back to the same equation after I've integrated twice. I can't see how to solve it by substitution. Is there any other way to get to the characteristic equation which might be what the book is referring to?

Start over: one of the very standard (and very important) facts about Laplace Transforms is that if $x(t) \longleftrightarrow X(s)$ (meaning that $X(s)$ is the transform of $x(t)$) then
$$ x'(t) \longleftrightarrow s X(s) - x(0+) $$
That comes from integration by parts, if you do it properly.

 

[tomwilliam2]

Thanks for your help - I must be missing something then. I was aware of the identity

x′(t)⟷sX(s)−x(0+)x′(t)⟷sX(s)−x(0+)

but wasn't sure how to use it, given that I don't have any boundary condition to evaluate $x(0)$.
I'll try to approach it carefully:
Let $f(t) = A\dot{x} + By$ and let $X(s) \longleftrightarrow f(t)$
Then:
$X(s)= A \int_{0}^{\infty}\dot{x}e^{-st} dt + B \int_{0}^{\infty}ye^{-st}dt$
Using the t-derivative rule:
$\int_{0}^{\infty}\dot{x}e^{-st} dt = \left(s\int_{0}^{\infty}xe^{-st}dt - x(0)\right)$
So:
$X(s)= A\left(s\int_{0}^{\infty}xe^{-st}dt - x(0)\right) + B \int_{0}^{\infty}ye^{-st}dt$
Now, integration by parts:
Let $x = u$, so that $\dot{x} = u'$
Let $-\frac{1}{s}e^{-st}=v$, so that $e^{-st}=v'$
$\int uv' = uv - \int u'v$

Which means that:
$X(s) = A\left(s\left(-\frac{x}{s}e^{-st}+\int_{0}^{\infty}\frac{\dot{x}}{s}e^{-st}dt\right)-x(0)\right) + B \int_{0}^{\infty}ye^{-st}dt$
But then I apply the limits to the first t term in the RHS, from 0 to infinity, and I get:
$X(s) = A\left(x(0)+\int_{0}^{\infty}\dot{x}e^{-st}dt-x(0)\right) + B \int_{0}^{\infty}ye^{-st}dt$
Which simplifies to:
$X(s) = A\int_{0}^{\infty}\dot{x}e^{-st}dt + B \int_{0}^{\infty}ye^{-st}dt$
I'm not making any progress at all.
I can apply the first bit to the other equation as well, giving me:
$X(s)= A\left(s\int_{0}^{\infty}xe^{-st}dt - x(0)\right) + B \int_{0}^{\infty}ye^{-st}dt$
$Y(s) = C\left(s\int_{0}^{\infty}ye^{-st}dt - y(0)\right) + D \int_{0}^{\infty}xe^{-st}dt$
But I'm not sure if I'm getting anywhere.
Any thoughts?

 

[Ray Vickson]

Thanks for your help - I must be missing something then. I was aware of the identity

but wasn't sure how to use it, given that I don't have any boundary condition to evaluate $x(0)$.
I'll try to approach it carefully:
Let $f(t) = A\dot{x} + By$ and let $X(s) \longleftrightarrow f(t)$
Then:
$X(s)= A \int_{0}^{\infty}\dot{x}e^{-st} dt + B \int_{0}^{\infty}ye^{-st}dt$
Using the t-derivative rule:
$\int_{0}^{\infty}\dot{x}e^{-st} dt = \left(s\int_{0}^{\infty}xe^{-st}dt - x(0)\right)$
So:
$X(s)= A\left(s\int_{0}^{\infty}xe^{-st}dt - x(0)\right) + B \int_{0}^{\infty}ye^{-st}dt$
Now, integration by parts:
Let $x = u$, so that $\dot{x} = u'$
Let $-\frac{1}{s}e^{-st}=v$, so that $e^{-st}=v'$
$\int uv' = uv - \int u'v$

Which means that:
$X(s) = A\left(s\left(-\frac{x}{s}e^{-st}+\int_{0}^{\infty}\frac{\dot{x}}{s}e^{-st}dt\right)-x(0)\right) + B \int_{0}^{\infty}ye^{-st}dt$
But then I apply the limits to the first t term in the RHS, from 0 to infinity, and I get:
$X(s) = A\left(x(0)+\int_{0}^{\infty}\dot{x}e^{-st}dt-x(0)\right) + B \int_{0}^{\infty}ye^{-st}dt$
Which simplifies to:
$X(s) = A\int_{0}^{\infty}\dot{x}e^{-st}dt + B \int_{0}^{\infty}ye^{-st}dt$
I'm not making any progress at all.
I can apply the first bit to the other equation as well, giving me:
$X(s)= A\left(s\int_{0}^{\infty}xe^{-st}dt - x(0)\right) + B \int_{0}^{\infty}ye^{-st}dt$
$Y(s) = C\left(s\int_{0}^{\infty}ye^{-st}dt - y(0)\right) + D \int_{0}^{\infty}xe^{-st}dt$
But I'm not sure if I'm getting anywhere.
Any thoughts?

I don't know why you are making it so complicated; let $X(s), Y(s)$ be the transforms of $x(t),y(t)$. Your first DE becomes $A[s X(s) - x(0+)] + B Y(s) = 0$, or $AsX+BY=Ax(0+)$. The second DE can be expressed similarly. Now you will have a 2x2 linear system for $X,Y$.

 

[vela]

Thanks for your help - I must be missing something then. I was aware of the identity

but wasn't sure how to use it, given that I don't have any boundary condition to evaluate $x(0)$.
I'll try to approach it carefully:
Let $f(t) = A\dot{x} + By$ and let $X(s) \longleftrightarrow f(t)$

I'd suggest using F(s) to denote the Laplace transform of f(t) to avoid confusion.

Then:
$X(s)= A \int_{0}^{\infty}\dot{x}e^{-st} dt + B \int_{0}^{\infty}ye^{-st}dt$
Using the t-derivative rule:
$\int_{0}^{\infty}\dot{x}e^{-st} dt = \left(s\int_{0}^{\infty}xe^{-st}dt - x(0)\right)$
So:
$X(s)= A\left(s\int_{0}^{\infty}xe^{-st}dt - x(0)\right) + B \int_{0}^{\infty}ye^{-st}dt$

At this point, you can identify the integrals as the Laplace transforms of x and y, giving you
$$F(s) = A[sX(s)-x(0)] + B Y(s).$$ Note that you essentially rederived the identity Ray referred to.

Now, integration by parts:
Let $x = u$, so that $\dot{x} = u'$
Let $-\frac{1}{s}e^{-st}=v$, so that $e^{-st}=v'$
$\int uv' = uv - \int u'v$

Which means that:
$X(s) = A\left(s\left(-\frac{x}{s}e^{-st}+\int_{0}^{\infty}\frac{\dot{x}}{s}e^{-st}dt\right)-x(0)\right) + B \int_{0}^{\infty}ye^{-st}dt$
But then I apply the limits to the first t term in the RHS, from 0 to infinity, and I get:
$X(s) = A\left(x(0)+\int_{0}^{\infty}\dot{x}e^{-st}dt-x(0)\right) + B \int_{0}^{\infty}ye^{-st}dt$
Which simplifies to:
$X(s) = A\int_{0}^{\infty}\dot{x}e^{-st}dt + B \int_{0}^{\infty}ye^{-st}dt$
I'm not making any progress at all.
I can apply the first bit to the other equation as well, giving me:
$X(s)= A\left(s\int_{0}^{\infty}xe^{-st}dt - x(0)\right) + B \int_{0}^{\infty}ye^{-st}dt$
$Y(s) = C\left(s\int_{0}^{\infty}ye^{-st}dt - y(0)\right) + D \int_{0}^{\infty}xe^{-st}dt$
But I'm not sure if I'm getting anywhere.
Any thoughts?

All of this work simply undoes the first integration by parts, so it should be no surprise you're not getting anywhere following this tack.

You want to do as Ray suggests and convert the two equations and solve for X(s) and Y(s). Then you can invert each to find x(t) and y(t).

 

[tomwilliam2]

You want to do as Ray suggests and convert the two equations and solve for X(s) and Y(s). Then you can invert each to find x(t) and y(t).

Thanks for your responses.
Using Ray's suggestion, I can get to:
$AsX + BY = Ax(0)$ and
$CsY + DX = Cy(0)$
But I'm not sure how to proceed with the inversion. I can't find an integral table for inverse Laplace transforms which contains an s-term like this, (it's usually some negative power), and I don't know how to get rid of the $x(0)$ and $y(0)$ terms. Sorry if I'm missing something obvious.
In any case, although it would be nice to get all the way through to the solution, I'm really trying to see how to get to the middle step, i.e. the characteristic equation:
$s^2 - \frac{BD}{CA}=0$
I know this might be unnecessarily complicating things, but I'm just trying to get a grip on how it works. Thanks for your patience.

 

[Ray Vickson]

Thanks for your responses.
Using Ray's suggestion, I can get to:
$AsX + BY = Ax(0)$ and
$CsY + DX = Cy(0)$
But I'm not sure how to proceed with the inversion. I can't find an integral table for inverse Laplace transforms which contains an s-term like this, (it's usually some negative power), and I don't know how to get rid of the $x(0)$ and $y(0)$ terms. Sorry if I'm missing something obvious.
In any case, although it would be nice to get all the way through to the solution, I'm really trying to see how to get to the middle step, i.e. the characteristic equation:
$s^2 - \frac{BD}{CA}=0$
I know this might be unnecessarily complicating things, but I'm just trying to get a grip on how it works. Thanks for your patience.

What is stopping you from just solving your two equations for $X$ and $Y$? The fact that $X,Y$ happen to be Laplace transforms of something does not affect the simple algebraic problem of solving two linear equations in the two unknowns. You have been solving such problems ever since you first looked at algebra.

 

[tomwilliam2]

What is stopping you from just solving your two equations for $X$ and $Y$? The fact that $X,Y$ happen to be Laplace transforms of something does not affect the simple algebraic problem of solving two linear equations in the two unknowns. You have been solving such problems ever since you first looked at algebra.

Well, I can solve for X or Y, but I'm still missing a step. I get the result:
$X\left(s^2 - \frac{BD}{CA}\right) = \frac{Ax(0)-By(0)}{A}$
Which obviously contains the expression mentioned in the textbook. I'm not sure how to show that the part in brackets is equal to zero though. Presumably the step I'm missing is that $x(0) = y(0)$ or something.

 

[Ray Vickson]

Well, I can solve for X or Y, but I'm still missing a step. I get the result:
$X\left(s^2 - \frac{BD}{CA}\right) = \frac{Ax(0)-By(0)}{A}$
Which obviously contains the expression mentioned in the textbook. I'm not sure how to show that the part in brackets is equal to zero though. Presumably the step I'm missing is that $x(0) = y(0)$ or something.

For some values of $s$ the Laplace transforms $X(s)$ and $Y(s)$ can fail to exist. Certainly if $s^2 - (BD/AC) = 0$ your $X(s)$ (and your $Y(s)$) will fail to exist. What is so wrong with that? It is entirely normal behavior, as you can ascertain by reading some more about Laplace transforms and their properties.

Anyway, you are not finished; you need to complete the job by actually solving for $X(s)$. Then you can "invert" $X(s)$ to get $x(t)$. Believe it or not, that is one of the fastest and easiest ways to solve constant-coefficient linear differential equations or systems of equations, especially if they are non-homogeneous (that is, have non-zero right-hand-sides)

 

[tomwilliam2]

For some values of $s$ the Laplace transforms $X(s)$ and $Y(s)$ can fail to exist. Certainly if $s^2 - (BD/AC) = 0$ your $X(s)$ (and your $Y(s)$) will fail to exist. What is so wrong with that? It is entirely normal behavior, as you can ascertain by reading some more about Laplace transforms and their properties.

Anyway, you are not finished; you need to complete the job by actually solving for $X(s)$. Then you can "invert" $X(s)$ to get $x(t)$. Believe it or not, that is one of the fastest and easiest ways to solve constant-coefficient linear differential equations or systems of equations, especially if they are non-homogeneous (that is, have non-zero right-hand-sides)

Thanks again. I can produce an expression for X(s):
$X(s) = \frac{x(0)-By(0)/A}{s^2 - \frac{BD}{CA}}$
Then I got a bit lost applying partial fractions to sort this expression out, but I can see how once it's straightened out, I can invert it to get x(t).
My real question is why the textbook talked about the "characteristic equation in terms of the Laplace operator, s" and how that is an important stage in solving this. I'm still failing to understand what the relevance of that characteristic equation is for Laplace - although of course solving it the normal way I would have got to the characteristic equation immediately.
Just for the record, I can't find an entry in the table of inverse transforms that corresponds to:
$X(s) = \frac{x(0)-By(0)/A}{s^2 - \frac{BD}{CA}}$
And if I factorise it to:
$X(s) = \frac{x(0)-By(0)/A}{(s - \sqrt{\frac{BD}{CA}})(s+\sqrt{\frac{BD}{CA}})}$
I get a bit lost deciphering it with partial fractions:
Let $\alpha = \sqrt{\frac{BD}{CA}}$ for ease of notation
Then:
$x(0) - By(0)/A = P(s+\alpha) + Q(s-\alpha)$
Then as there is no s-term on the LHS, $P=-Q$, so:
$x(0) - By(0)/A = P(s+\alpha) -P(s-\alpha)$
Rearranging:
$P = \frac{x(0) - By(0)/A}{2\alpha}$
So
$X(s) = \frac{x(0)-By(0)/A}{s^2 - \frac{BD}{CA}} = \frac{x(0) - By(0)/A}{2\alpha(s-\alpha)}- \frac{x(0) - By(0)/A}{2\alpha(s+\alpha)}$

And I also can't find a solution of this format in the inverse laplace table. I've obviously gone wrong somewhere...

 

[tomwilliam2]

I may have solved it now:
$X(s) = \frac{x(0)-By(0)/A}{s^2 - \frac{BD}{CA}} = \frac{x(0) - By(0)/A}{2\alpha(s-\alpha)}- \frac{x(0) - By(0)/A}{2\alpha(s+\alpha)}$
$X(s) =\frac{x(0) - By(0)/A}{2\alpha}\frac{1}{(s-\alpha)}-\frac{x(0) - By(0)/A}{2\alpha}\frac{1}{(s+\alpha)}$
Using the table of inverse Laplace transforms:
$x(t) =\frac{x(0) - By(0)/A}{2\alpha}e^{\alpha t} - \frac{x(0)-By(0)/A}{2\alpha}e^{-\alpha t}$
Is this correct?

 

[Ray Vickson]

I may have solved it now:
$X(s) = \frac{x(0)-By(0)/A}{s^2 - \frac{BD}{CA}} = \frac{x(0) - By(0)/A}{2\alpha(s-\alpha)}- \frac{x(0) - By(0)/A}{2\alpha(s+\alpha)}$
$X(s) =\frac{x(0) - By(0)/A}{2\alpha}\frac{1}{(s-\alpha)}-\frac{x(0) - By(0)/A}{2\alpha}\frac{1}{(s+\alpha)}$
Using the table of inverse Laplace transforms:
$x(t) =\frac{x(0) - By(0)/A}{2\alpha}e^{\alpha t} - \frac{x(0)-By(0)/A}{2\alpha}e^{-\alpha t}$
Is this correct?

I have not checked the details, but you can most easily get results in such problems by using the simple facts that for $|a| \neq 0$ we have
$$ \displaystyle \frac{1}{s^2+a^2} \rightarrow \frac{1}{a} \sin(at), \;\; \frac{s}{s^2+a^2} \rightarrow \cos(at) \\
\displaystyle \frac{1}{s^2-a^2} \rightarrow \frac{1}{a} \sinh(at), \;\; \frac{s}{s^2-a^2} \rightarrow \cosh(at) $$

As a general rule, I personally prefer to simplify the work by introducing auxiliary symbols before starting, so I would have written $X(s) = b/(s^2-r^2)$, where $b =x(0) - B y(0) /A$ and $r = \sqrt{BD/AC}$. Then, after getting the answer in terms of $b$, $r$ and $t$, I could take the un-needed step of putting back things in terms of $A,B,C,D,x(0),y(0),t$. (I say this is un-needed because if I clearly state what $b$ and $r$ are, an answer in terms of $b$, $r$ and $t$ ought to suffice.)

 

[tomwilliam2]

Thanks. I think I've satisfied myself with how the Laplace transform works now, although my textbook's reference to the characteristic equation still seems mysterious.

 

[Ray Vickson]

Thanks. I think I've satisfied myself with how the Laplace transform works now, although my textbook's reference to the characteristic equation still seems mysterious.

The characteristic equation is the polynomial equation $P(r) = 0$ that you get when you substitute a trial solution of the form $x(t) = e^{rt}$ into the DE. It turns out that for a linear, constant-coefficient DE, the denominator in the expression for $X(s)$ is just $P(s)$; that is, $X(s) = Q(s)/P(s)$, where $Q$ is a polynomial of degree less than that of $P$ (at least for homogeneous DEs with 0 right-hand-sides). So, the singularities in the Laplace transform (the places where you would be dividing by 0) are just the roots of the characteristic equation.

For example, look at the DE $x'''(t) - 3 x''(t) +4x(t) =0$. If we write $D = \frac{d}{dt}$ this becomes $D^3 x - 3 D^2 x + 4x = 0$, and the characteristic equation is obtained just by replacing $D$ with some numerical parameter $r$: $r^3 - 3 r^4 + 4 = 0$. Now look what happens if we apply Laplace transforms: $x(t) \to X(s)$, $x'(t) \to sX(s)-x(0)$, $x''(t) \to s^2 X(s) - x'(0) - s x(0)$ and $x'''(t) \to s^3 X(s) - x''(0) - s x'(0) - s^2 x(0)$, so the DE becomes
$$ s^3 X - x''(0)-s x'(0) - s^2 x(0) -3 s^2 X + 3 x(0) + 3s x'(0) + 4X = 0,$$
so
$$X(s) =\displaystyle \frac{s^2 x(0) + s x'(0) - 3 s x(0) + x''(0) - 3 x'(0)}{s^3 - 3 s^2 + 4}, $$
exactly as stated before.

 

[tomwilliam2]

Thanks! I think I've got it now.