Verify Green's Formula for a Simple DE

[zenterix]

Homework Statement

1. (a) Compute $t*1$. More generally, compute $(q*1)(t)$ in terms of $q=q(t)$.

(b) Compute $1*t$. More generally, compute $(1*q)(t)$ in terms of $q=q(t)$.

2. What is the differential operator $p(D)$ whose unit impulse response is the unit step function $u=u(t)$?

In 1(b) you computed $1*q=u*q$. Is the assertion in the box in the beginning of this session true in this case?

Relevant Equations

Assertion: Suppose $w(t)$ is the unit impulse response for the operator $p(D)$. Let $q(t)$ be a (perhaps generalized) function. Then the solution to $p(D)x=q(t)$ with rest initial conditions is given (on $t>0$) by $w(t)*q(t)$.

These problems are from a practice problem set from MIT OCW's 18.03 "Differential Equations.

Computing the convolutions $t*1$ and $1*t$ is straightforward. They both equal $\frac{t^2}{2}$.

Then, $(q*1)(t)=\int_0^t q(\tau)d\tau$

and $(1*q)(t)=\int_0^t q(t-\tau)d\tau$ which after a change of variables is equal to $\int_0^t q(\tau)d\tau$.

What we see here is what is actually a general property: the convolution operation "$*$" is commutative.

My question is about problem 2.

Since $u'(t)=\delta(t)$ we see that for $p(D)=D$ we have $p(D)u(t)=\delta(t)$.

That is, the unit step function $u(t)$ is the unit impulse response for the differential equation $p(D)x(t)=\delta(t)$.

In problem 1, we computed $1*q$, and since this is an integral from $0$ to $t$, this is the same as the convolution $u*q$.

Suppose we have a differential equation $p(D)x(t)=q(t)$ with $p(D)=D$ and in which $q(t)$ may be a generalized function (ie, a regular function plus a linear combination of delta functions).

We know the unit impulse response function is $w(t)=u(t)$.

The assertion says that the solution to the differential equation with rest initial conditions is given on $t>0$ by $w(t)*q(t)$.

We are asked if this is true.

To show that it is true, we need to show that $x(t)=u(t)*q(t)$ solves the differential equation.

That is, $\dot{x}(t)=q(t)$.

As we showed in problem 1, $u(t)*q(t)=1*q(t)=\int_0^t q(\tau)d\tau$.

If $q(t)$ is just a regular function then by the FTC indeed we have $\dot{x}=q(t)$.

Now suppose $q(t)$ contains delta functions.

For example, for a point $a>0$ suppose $q(t)=\delta_a(t)$.

Then $x(t)=u(t)*q(t)=\int_{0^-}^{t^+} \delta_a(\tau)d\tau= u_a(t)$ and so $\dot{x}=\delta_a(t)=q(t)$.

Then there is the following comment

It also holds for $q=\delta_0=\delta$, but that case falls outside the domain over which we are checking the derivative.

Note that to show this we finally had to use the signs on the limits of integration that we had been carefully keeping track of thus far.

"Falls outside the domain over which we are checking the derivative".

What is this domain exactly?

I guess it is $t>0$.

Why is the integral done starting at $0^-$?

 

[pasmith]

By convention, $$\int_0^\infty \delta(x)\,dx = \frac12 \int_{-\infty}^\infty \delta(x)\,dx = \frac12.$$ Thus the limit $0^{-}$ is necessary to include the entire contribution of $\delta$ in the result.