Is there a mistake in my calculation or in my reasoning?

[Jonas E]

Homework Statement

y'' + 3y' + 2y = r(t),

r(t) = u(t - 1) - u(t - 2),

y(0) = y'(0) = 0.

I need to solve this by convolution, which I know is commutative. The problem is that my calculation gives (f * g) =/= (g * f). Could someone please tell me where my mistake is?

Homework Equations

(f * g) = ∫f(τ)g(t-τ)dτ, from 0 to t

The Attempt at a Solution

I split the problem into 3 cases:

First (0 < t < 1): r(t) = 0, so y(t) = 0

Second (1 < t < 2): r(t) = 1 This is where it goes wrong.

I get f(t) = 1, g(t) = e^(-t) - e^(-2t), then y = (f * g) = ∫[ e^-(t-τ) - e^-2(t-τ) ]dτ, from 1 to t = (1/2) - e^-(t-1) - (1/2)e^-2(t-1), which is the correct answer according to my textbook.

But, y = (g * f) = ∫[ e^(-τ) - e^(-2τ) ]dτ, from 1 to t = -e^(-t) + (1/2)e^(-2t) + e^(-1) - (1/2)e^(-2) =/= (f * g)

I integrated from 1 instead of 0 since r=0 for t < 1.

 

[Orodruin]

I suggest you look over your integration boundaries for the second case. Think about what the convolution really means.

 

[Jonas E]

I suggest you look over your integration boundaries for the second case. Think about what the convolution really means.

I don't think I understand. I get the right answer for (g * f) if I integrate from 0 to (t - 1) instead, but I don't understand why. Could you please explain?

 

[Orodruin]

Try writing the integrals using the actual expressions for the functions instead. Where are the heaviside functions non-zero?

 

[Jonas E]

Try writing the integrals using the actual expressions for the functions instead. Where are the heaviside functions non-zero?

I have now tried that for both (f * g) and (g * f), but in both cases I get the correct answer for case 3 (t > 2), when I'm trying to solve for case 2:

(f * g) = ∫[ e^(-τ) - e^(-2τ) ] * [ u(t - τ - 1) - u(t - τ - 2)]dτ

I find that for case 2:

t - τ - 1 > 0 Λ t - τ - 2 < 0

which gives me:

t - 2 < τ < t - 1

So the integral becomes:

[ -e^(-τ) + (1/2)e^(-2τ) ] from (t - 2) to (t - 1) = e^-(t - 2) - (1/2)e^-2(t - 2) - (e^-(t - 1) - (1/2)e^-2(t - 2))

but this is correct only if t > 2, so I don't understand why I get that when I try to solve for 1 < t < 2

 

[Ray Vickson]

I have now tried that for both (f * g) and (g * f), but in both cases I get the correct answer for case 3 (t > 2), when I'm trying to solve for case 2:

(f * g) = ∫[ e^(-τ) - e^(-2τ) ] * [ u(t - τ - 1) - u(t - τ - 2)]dτ

I find that for case 2:

t - τ - 1 > 0 Λ t - τ - 2 < 0

which gives me:

t - 2 < τ < t - 1

So the integral becomes:

[ -e^(-τ) + (1/2)e^(-2τ) ] from (t - 2) to (t - 1) = e^-(t - 2) - (1/2)e^-2(t - 2) - (e^-(t - 1) - (1/2)e^-2(t - 2))

but this is correct only if t > 2, so I don't understand why I get that when I try to solve for 1 < t < 2

I don't see why you want to use convolutions, instead of just applying standard properties of Laplace transforms to get an easily-recognized form of inverse Laplace transform. However, if you insist on using convolutions you will have
y(t) = \int_0^t f(t - \tau) r(\tau) \, d \tau = \underbrace{\int_0^t f(t-\tau) u(\tau-1) \, d\tau}_{= I_1} -<br /> \underbrace{\int_0^t f(t-\tau) u(\tau-2) \, d\tau}_{= I_2}
For $t \leq 1$ we have $I_1 = 0$ because the $u$-factor vanishes. For $t > 1$ we have $I_1 = \int_1^t f(t-\tau) \, d \tau$, etc. The term $I_2$ is similar.