Question: How do we use the Laplace transform to find the inverse of a function?

[pierce15]

Homework Statement

Find

$$ L^{-1} \left[ \frac{1}{ (p^2 + a^2)^2} \right] $$

Homework Equations

$$ L [ x \cos ax ] = \frac{p^2 - a^2} { (p^2 + a^2) ^2 } $$

The Attempt at a Solution

I have no idea. Any thoughts?

 

[LCKurtz]

Homework Statement

Find

$$ L^{-1} \left[ \frac{1}{ (p^2 + a^2)^2} \right] $$

Homework Equations

$$ L [ x \cos ax ] = \frac{p^2 - a^2} { (p^2 + a^2) ^2 } $$

The Attempt at a Solution

I have no idea. Any thoughts?

Look at this thread:

https://www.physicsforums.com/showthread.php?t=725295

 

[berkeman]

Thread closed temporarily for Moderation...

@piercebeatz -- Please check your PMs. You must always show some effort before we can offer tutorial help.

Thread is re-opened.

 

[pierce15]

I'm terribly sorry for that. Here's where I got (I'm supposed to use the above identity in the problem, and convolutions is in the next chapter):

$$ L^ {-1} \left[ \frac{1}{(p^2 + a^2)^2} \right] = L^{-1} \left[ \frac{1}{p^2 - a^2} \frac{p^2 - a^2}{(p^2 + a^2)^2} \right] $$

By the theorem

$$ L^{-1} [ F(p) G(p) ] = \int_0^x f(x-t)g(t) \, dt $$

(where F(p) = L(f(t)) etc.), this becomes

$$ L^{-1} \left[ \frac{1}{p^2 - a^2} \frac{p^2 - a^2}{(p^2 + a^2)^2} \right] = \int_0^x \frac{1}{a} \sinh (a (x-t)) \cos(at) \, dt $$

By the substitution t = iy, this becomes

$$ i/a \int_0^{x/i} \sinh( a( x-iy)) \cosh ay \, dy $$

$$ = i/2a \int_0^{x/i} \left[ \sinh (a (x + y - iy)) + \sinh(a( x- y - iy) ) \right] \, dy $$

$$ = i/2a \left[ \frac{\cosh( a ( x+ y - iy)) }{a(1 - i)} + \frac{ \cosh(a (x - y -iy))}{a(-1-i)} \right]_0^{x/i} $$

$$ = i/2a \left[ \frac{\cosh( -aix)}{a(1-i)} + \frac{\cosh(aix)}{a(-1-i)} \right] $$

$$ = \frac{1}{a} \cos ax \cdot \frac{i}{2} \left[ \frac{1}{a(1-i)} - \frac{1}{a(1+i)} \right] $$

$$ = \frac{1}{a} \cos ax \cdot \frac{i}{2} \left[ \frac{ a( -1 - i + 1 - i )}{-2 a^2 } \right] $$

$$ = - \frac{ \cos ax}{2a^2} $$

That's the only way I could think of doing it

 

[LCKurtz]

I'm terribly sorry for that. Here's where I got (I'm supposed to use the above identity in the problem, and convolutions is in the next chapter):

$$ L^ {-1} \left[ \frac{1}{(p^2 + a^2)^2} \right] = L^{-1} \left[ \frac{1}{p^2 - a^2} \frac{p^2 - a^2}{(p^2 + a^2)^2} \right] $$

By the theorem

$$ L^{-1} [ F(p) G(p) ] = \int_0^x f(x-t)g(t) \, dt $$

That is the convolution theorem stated in inverse form.

(where F(p) = L(f(t)) etc.), this becomes

$$ L^{-1} \left[ \frac{1}{p^2 - a^2} \frac{p^2 - a^2}{(p^2 + a^2)^2} \right] = \int_0^x \frac{1}{a} \sinh (a (x-t)) \cos(at) \, dt $$

Why would you break it up that way? Why not $$
\mathcal L^{-1}\left(\frac {1}{p^2+a^2}\cdot \frac {1}{p^2+a^2}\right)$$I haven't checked the rest of your work, but, at least according to Maple, your answer isn't correct.

 

[vanhees71]

Why don't you use complex integration and the theorem of residues to invert the Laplace transform?

 

[pierce15]

I don't know that theorem. Does anyone see any error in the above calculation? I can't find one

 

[LCKurtz]

I don't know that theorem. Does anyone see any error in the above calculation? I can't find one

You don't know what theorem? Please quote the message to which you are replying.

 

[pierce15]

You don't know what theorem? Please quote the message to which you are replying.

Sorry, I meant the "theorem of residues"

 

[vela]

I don't know that theorem. Does anyone see any error in the above calculation? I can't find one

You didn't take the inverse Laplace transform of the two pieces correctly. According to Mathematica, the inverse Laplace transform of $\frac{p^2-a^2}{(p^2+a^2)^2}$ is $t \cos at$. You forgot the $t$ out front.

In any case, I suggest you take LCKurtz's suggestion for how to break up the original transform to use with the convolution theorem.

 

[pierce15]

Right, I forgot the t. Putting it in and doing integration by parts yields the answer, which I have obtained on paper. Thank you all for your time.

 

[vanhees71]

Sometimes, it's easier to directly use the back-transformation formula
f(t)=\frac{1}{2 \pi \mathrm{i}} \int_{C} \mathrm{d} p \exp(p t) F(p),
where C is a straight line parallel to the imaginary axis in the complex p line.

The integral can be evaluated with help of the theorem of residues very easily in this case. The image function is
F(p)=\frac{p^2-a^2}{(p^2+a^2)^2}.
For simplicity let's assume that a>0. Then the poles of F are on the imaginary axis, p_{1,2}=\pm \mathrm{i} a. For the real part of the integration path we can thus choose any positive value, and thus we can close this path by a semi-circle with infinite radius to the left p plane (since in the back-transformation formula above we always tacitly assume t>0.

Thus the integral is just given by the sum over the residues of the function F(p)\exp(p t). Since the poles are of second order, the residues are evaluated as
\text{res}_{\pm \mathrm{i} a}[F \exp(p t)]=\lim_{p \rightarrow \mathrm{i} a} \frac{\mathrm{d}}{\mathrm{d} p}[(p \mp \mathrm{i}a)^2 F(p) \exp(p t)] = \frac{t}{2} \exp(\pm \mathrm{i} a t).
Thus the original function is
f(t)=\text{res}_{\mathrm{i} a}[F \exp(p t)] + \text{res}_{-\mathrm{i} a}[F \exp(p t)]=t \cos(a t).