Differentiation and integration of functions.
Theorem: Derivative of unit step function is dirac delta
$$\begin{eqnarray}
\frac{d}{dt}U(t) = \delta(t)
\end{eqnarray}$$
Proof: Applying the definition of derivative
$$\begin{eqnarray}
\frac{d}{dt}U(t) = \lim_{\varepsilon \rightarrow 0}\frac{U(t+\varepsilon)-U(t-\varepsilon)}{2\varepsilon} = \lim_{\varepsilon \rightarrow 0}\frac{\chi_{[-\varepsilon,\varepsilon]}}{2\varepsilon} =\delta(t)
\end{eqnarray}$$
More generally, differentiation of any discontinuity gives a dirac delta.
Theorem: Integration of dirac delta is unit step.
Proof:
\[
\int_{-\infty}^t \delta(s)ds =
\begin{cases}
0, & t \leq 0 \\ 1, & 0<t
\end{cases}
\]
or,
\begin{eqnarray}
\int_{-\infty}^t \delta(s)ds =U(t)
\end{eqnarray}
Example: Consider the function $f(t)$. at $t=0$ this function is continuous and its derivative is also continuous.
\begin{eqnarray}
f(t) = (at^3+bt^4+ct^5+dt^6)U(t)
\end{eqnarray}
at $t=0$ this function is continuous and its derivative $Df(t)$. is also continuous:
\begin{eqnarray}
\frac{d}{dt}\left[(at^3+bt^4+ct^5+dt^6)U(t)\right] &=& \frac{d}{dt} \left(at^3+bt^4+ct^5+dt^6 \right) U(t) + \left(at^3+bt^4+ct^5+dt^6 \right) \frac{d U(t)}{dt}\\
&=& (3at^2+4bt^3+5ct^4+6dt^5)U(t) + \left(at^3+bt^4+ct^5+dt^6 \right) \delta(t) \\
&=& (3at^2+4bt^3+5ct^4+6dt^5)U(t)
\end{eqnarray}
What about the second derivative? $D^2f(t)=(6at+12bt^2+20ct^3+30dt^4)U(t)$ is continuous $t=0$.
What about the third derivative? $D^3f(t)=(6a+24bt+60ct^2+120dt^3)U(t)$. .Therefore the third derivative is not continuous at $t=0$. It has a discontinuity of size $6a$.
Fourth derivative is $D^4f(t)=6a\delta(t)+(24b+120ct+3600dt^2)U(t)$. It has a dirac delta and a discontinuity at $t=0$.
If we take one more derivative what will happen? We will see in the next section.
Differentiation and integration of dirac deltas.
One may be surprised that a function as strange as dirac delta may have derivative. Contrary to the intiution, it has well defined derivatives of any order. We will denote $n$th order derivative of dirac delta as $\delta^{n}(t)$
\begin{eqnarray}
\frac{d^n}{dt^n}\delta(t) = D^n \delta(t) = \delta^{(n)}(t)
\end{eqnarray}
We can use integration by parts to write
\begin{eqnarray}
\int_{-\infty}^{\infty} \delta^{(n)}(t) f(t) dt = \left. \delta^{(n-1)}(t) f(t) \right\vert_{-\infty}^{\infty} – \int_{-\infty}^{\infty} \delta^{(n-1)}(t) Df(t) dt
\end{eqnarray}
Imposing the boundary conditions
\begin{eqnarray}
\left. \delta^{(n-1)}(t) f(t) \right\vert_{-\infty}^{\infty} = 0
\end{eqnarray}
will result in
\begin{eqnarray}
\int_{-\infty}^{\infty} \delta^{(n)}(t) f(t) dt = – \int_{-\infty}^{\infty} \delta^{(n-1)}(t) Df(t) dt
\end{eqnarray}
As $f(t)$ is completely general, the integrands must be equal
\begin{eqnarray}
\delta^{(n)}(t) f(t) = – \delta^{(n-1)}(t) Df(t)
\end{eqnarray}
by iteration, we can get
\begin{eqnarray}
\delta^{(n)}(t) f(t) = (-1)^n \delta(t) D^nf(t)
\end{eqnarray}
Summarizing our results:
\begin{equation}
\boxed{
\delta^{(n)}(t) = – \delta^{(n-1)}(t) D =(-1)^n \delta(t) D^n
}
\end{equation}
which is a generalization of theProperty 6, Multiplication of dirac deltas with functions, defined in the previous chapter.
Example: Find $(3t^2+4t+7)\delta(t)$
Answer: $7\delta(t)$
Example:Find $(3t^2+4t+7)\delta^{(2)}(t)$
Answer: $6\delta(t)$
Example:Find $(3t^2+4t+7)\delta^{(1)}(t)$
Answer: $-4\delta(t)$
Example: Consider the function $f(t)$
\[
f(t) =
\begin{cases}
0, & t \leq 0 \\ a+bt+ct^2+dt^3, & 0<t
\end{cases}
\]
Or, equivalently, $f(t)=(a+bt+ct^2+dt^3)U(t)$. Compute its derivatives. Note that there is a discontinuity at $t=0$.
Answer:
\begin{eqnarray}
Df(t) &=& a\delta(t) + ( b+2ct + 3dt^2) U(t). \\
D^2f(t) &=& a\delta^{(1)}(t) + b\delta(t) +(2c + 6dt) U(t). \\
D^3f(t) &=& a\delta^{(2)}(t) + b\delta^{(1)}(t) +2c\delta(t) + 6d U(t). \\
D^4f(t) &=& a\delta^{(3)}(t) + b\delta^{(2)}(t) +2c\delta^{(1)}(t) + 6d \delta(t)
\end{eqnarray}
Exercise: Integrate $D^4f(t)$ found above four times to get back $f(t)$.
Answer: Left to the student.
Exercise: Compute the derivatives of the function $g(t)=(4e^{-2t} + 3 e^{-3t} + e^{t})U(t)$.
Answer:
\begin{eqnarray}
Dg(t) &=& 8\delta(t) + (-8e^{-2t} -9e^{-3t} + e^{t})U(t). \\
D^2g(t) &=& 8\delta^{(1)}(t) -16 \delta(t)+ (16e^{-2t} + 27e^{-3t} + e^{t})U(t). \\
D^3g(t) &=& 8\delta^{(2)}(t) -16 \delta^{(1)}(t)+44 \delta(t)+(-32e^{-2t} – 84e^{-3t} + e^{t})U(t). \\
D^4g(t) &=& 8\delta^{(3)}(t) -16 \delta^{(2)}(t)+ 44 \delta^{(1)}(t) -115 \delta(t)+ (64e^{-2t} + 252e^{-3t} + e^{t})U(t)
\end{eqnarray}
Exercise: Integrate $Dg(t)$ to get back $g(t)$
Answer:
\begin{eqnarray}
\int_{s=-\infty}^t Dg(s)ds &=& \int_{s=-\infty}^t ( 8\delta(s) + (-8e^{-2s} -9e^{-3s} + e^{s})U(s))ds \\
\end{eqnarray}
- for $t<0$ this integral is 0
- for $t>0$ this integral is $\int_{s=0^-}^t Dg(s)ds = 8+ (4e^{-2t} + 3 e^{-3t} + e^{t} – 8) = 4e^{-2t} + 3 e^{-3t} + e^{t}$
Joining the two parts we get $g(t)=(4e^{-2t} + 3 e^{-3t} + e^{t})U(t)$.
General rule for differentiation at discontinuities
Consider the function
\begin{eqnarray}
y(t) = g(t)U(t)
\end{eqnarray}
where $g(t)$ is continuous at $t=0$ with finite and continuous derivatives, $\frac{d^k}{dt^k}g(t)=g^{(k)}(t)$, of all orders.
What is the derivatives of $y(t)$ at $t=0$, of all orders? Obviously, left and right derivatives will differ. Left derivatives of any order is zero. So, we only need to find the right derivatives at $t=0^+$.
\begin{eqnarray}
y(t) &=& g(t)U(t) \\
Dy(t) &=& g^{(1)}(t) U(t) + g(t)\delta(t) = g^{(1)}(t) U(t) + g(0)\delta(t)\\
D^2y(t) &=& g^{(2)}(t) U(t) + g^{(1)}(t)\delta(t)+g(0) \delta^{(1)}(t)= g^{(2)}(t) U(t) + g^{(1)}(t)\delta(0)+g(0) \delta^{(1)}(t)
\end{eqnarray}
And the general rule is:
\begin{eqnarray}
\boxed{
D^ky(t) = g^{(k)}(t) U(t) + \sum_{i=0}^{k-1} g^{(i)}(0)\delta^{(k-1-i)}(t)
}
\end{eqnarray}