HARD QUESTION RE Laplace transform and green's function and convolution?

Question

Hi,

Before reading anything, I shall just mention that '∗' is the convolution operator and 'δ(t)' is the delta dirac (or impulse) function. The question is to be solved using the Laplace Transform (it comes from a sheet of Laplace Transform questions).

The problem is as follow:
Show that y'' - 2y' - 8y = f(t), y(0) = 1, y'(0) = 0;
can be represented by:
y(t) = 1/6 · f(t) ∗ e^(4t) - 1/6 · f(t) ∗ e^(-2t) + 1/3 · e^(4t) + 2/3 · e^(-2t);

Okay so I managed to solve it; however, I was wondering if it were possible to solve the question by letting f(t) = δ(t) such that F(s) = 1.

I found that Green's function (by letting f(t) = δ(t)) is g(t) = 1/2 · e^(4t) + 1/2 · e^(-2t).

Then I tried to find y(t) by finding g(t) ∗ f(t); however, that didn't work out.

Personally, I think this question is quite a challanging one and I appreciate anyone who even has an attempt.

Thank you.

Answer 1

The reason for the difficulty is that the initial conditions are not homogeneous;

specifically the condition y(0) = 1. So the sum of 2 solutions is not a solution.



This is evident from the explicit representation of the solution S(f) in terms

of the function f(t):



S(f)(t) = y(t) = (f*h)(t) + [1/3 e^(4t) + (2/3) e(-2t)]



* = convolution



h(t) = 1/6 e^(4t) - 1/6 e^(-2t)



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Note that for any two functions f & g



f*g(t) (0) = 0 because (f*g)(t) = Int_(0,t) f(t-u) g(u) du



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Fred: does this answer your question?

Answer 2

Not a solution; just observations that I hope will help.



Ignoring LT's for the moment, and

solving the simpler DE without f(t)

y'' - 2y' - 8y = f(t), y(0) = 1, y'(0) = 0

auxiliary equation m^2 - 2m - 8 = 0

with roots 4, -2 leads to

y = (1/3)· e^4t + (2/3)·e^-2t (complementary function).....(a)



The function y(t) required by the question is

y(t) = 1/6 · f(t) ∗ e^(4t) - 1/6 · f(t) ∗ e^(-2t) + 1/3 · e^(4t) + 2/3 · e^(-2t)

Since solution = particular integral + complementary function we have

particular integral = 1/6 · f(t) ∗ e^(4t) - 1/6 · f(t) ∗ e^(-2t)

which can be can be rewritten as

PI = (1/2) · f(t) ∗ [(1/3)·e^(4t)] - 1/6 · f(t) ∗ (2/3)·4e^(-2t) ....(b)



The similarity of b) and a) suggest some simple integration probably to do

with the convolution operator. Is this the same or related to this?



Laplace transform of convolution

If L[f(t)] = F(s) and L[g(t)] = G(s) then

F(s)G(s) = L[(0 to t)∫f(u)g(t - u)du]



"The convolution of f and g, (written as f∗g following your convention) is defined

as the integral of the product of the two functions after one is reversed and shifted".



The proof of that is given (using different variables) at:-

http://planetmath.org/encyclopedia/LaplaceTransformOfConvolution.html



Useful image here: Overlap of integral areas with one function reversed & shifted.

http://en.wikipedia.org/wiki/Convolution_operator

~~~~~~~~~~~~~~~~~~~~~~~~~~~

Perhaps you can make something of these pieces

of a jigsaw and make the jump from a) to b)



Regards - Ian

Answer 3

The Laplace rework of a convolution is the fabricated from the convolutions of the two applications. it particularly is in many situations utilized in opposite. in case you realize the inverse transforms of F(s) and G(s) and that they are applications f(x) and g(x), then the inverse rework of F(s)G(s) is the convolution f*g(x)=int_0^x f(t)g(x-t)dt.