Laplace Transform of...

JJacquelin

If the Laplace transform of u(x) is U(s), what is the Laplace transform of u''(x) ?

maistral

Oh, the question wasn't clear. My bad.

The problem is what constants should I use. Say for example, y(0) = 1.

the equation is in terms of u, u" - 0.625u = 0

The laplace transform is s²U - su(0) - u'(0) - 0.625U = 0.

What would be my u(0) and u'(0)? Thanks. :D

JJacquelin

If u(0) and u'(0) are not specified in the wording of the problem, then consider them as paramaters. For example, let u(0)=C1 and u'(0)=C2. They will appear into the general solution of the ODE. Since it is a second order ODE, the general solution must include two arbitrary constants anyway.

maistral

Oh. alright.

I have one last question about Laplace.

There's this convolution integral that I read, so it's integral from 0 to t of f(t-v)g(v)dv.

Solving this with some elementary algebraic maneuvers:

(2s^2 - 16) / (s^3 - 16s) = s^2 - 16 / s(s^2 - 16) + s^2 / s(s^2 - 16).

= 1 / s + s / (s^2 - 16) = 1 + cosh4t.

Using that integral; transforming the equation into:

= 2s^2 / s(s^2 - 16) - 16/s(s^2 - 16)
= 2s / (s^2 - 16) - 16/s * 1/(s^2 - 16)
= 2cosh4t - 16/4s * 4/(s^2 - 16)

= 2cosh4t - 4/s * 4/(s^2 - 16)

Using the second term;
F(s) = 4/s
G(s) = sinh4t
thus;
f(t-v) = 4
g(v) = 1/4 cosh4t

-int(0,t,4*cosh4t dt) = 4/4 cosh 4t = -cosh4t

= 2cosh4t - cosh4t = cosh4t.

The algebraic maneuver had 1 + cosh4t while this integral gave me just the cosh4t.

Where did the 1 go? Is it a constant generated from integrating? Isn't it supposed to be accounted already when the DE was transformed to the S domain (the y(0) and y'(0))?

Thanks. :D

maistral

Bump for my last question :(

chiro

For (2s^2 - 16) / (s^3 - 16s) = s^2 - 16 / s(s^2 - 16) + s^2 / s(s^2 - 16).
= 1 / s + s / (s^2 - 16) = 1 + cosh4t, I think you have made a mistake.

I get (2s^2 - 16) / s(s^2 - 16)
= (s^2 + s^2 - 16) / s(s^2 - 16)
= s^2 / s(s^2-16) + 1/s

The above will give you a different answer than your expansion.

maistral

But i think we got the same :\

I mean, taking yours:

= s^2 / s(s^2 - 16) + 1/s ;cancelling extra s
= s / (s^ - 16) + 1 / s = cosh4t + 1

:(

chiro

Haha yeah you're spot on! I guess if thats right the next thing to look at is the convolution.

So in the convolution you wrote:

Your integral should be int(0,t,4*sinh4tdt). Your inverse of 4/(s^2 - 16) is sinh4t not cosh4t. Calculating this we get

4*1/4(cosh4t - 1] = cosh4t - 1

Now 2cosh4t - (cosh4t - 1) = cosh4t + 1.

maistral

Oh, that was a typo. I was sleepy when typing that lol, sorry.

That's where i'm having issues. Where did the -1 come from? :(

i mean, f(t-v) = 4, g(v) = sinh4v

-int(0,t, 4 sinh4v dV) = -4/4 sinh 4t, right? where did the -1 come from? :S

maistral

argh, I'm thinking too much. I see now where it came from. Thanks for answering.