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Differential Operators

The idea of a function as ``something'' that takes a value (real, complex, vector, etc.) as ``input'' and returns ``something else'' as ``output'' should be very familiar and useful.

This idea can be generalized to operators that take a function as an argument and return another function.

The derivative operator operates on a function and returns another function that describes how the function changes:

$$\begin{split}\mathcal{D}[f(x)] & = \ensuremath{\frac{d{f}}{d{... ...)]\\ \mathcal{D}[f(x) + g(x)] = & D[f(x)] + D[g(x)] \end{split}$$ (22-1)

The last two equations above indicate that the ``differential operator'' is a linear operator.

The integration operator is the right-inverse of $\mathcal{D}$

$$\mathcal{D} [\mathcal{I}[ f(x)]] = \mathcal{D}[ \int f(x) dx]$$ (22-2)

but is only the left-inverse up to an arbitrary constant.

Consider the differential operator that returns a constant multiplied by itself

$$\mathcal{D} f(x) = \lambda f(x)$$ (22-3)

which is another way to write the the homogenous linear first-order ODE and has the same form as an eigenvalue equation. In fact, $f(x) = \exp(\lambda x)$ , can be considered an eigenfunction of Eq. 22-3.

For the homogeneous second-order equation,

$$\left(\mathcal{D}^2 + \beta \mathcal{D} - \gamma \right)[ f(x)] = 0$$ (22-4)

It was determined that there were two eigensolutions that can be used to span the entire solution space:

$$f(x) = C_+ e^{\lambda_+ x} + C_- e^{\lambda_- x}$$ (22-5)

Operators can be used algebraically, consider the inhomogeneous second-order ODE

$$\left(a \mathcal{D}^2 + b \mathcal{D} + c\right)[y(x)] = x^3$$ (22-6)

By treating the operator as an algebraic quantity, a solution can be found¹

$$\begin{split}y(x) & = \left(\frac{1}{a \mathcal{D}^2 + b \mat... ...{6(b^2 - a c)x}{c^3} - \frac{6 b(b^2 - 2 a c)}{c^3} \end{split}$$ (22-7)

which solves Eq. 22-6.

The Fourier transform is also a linear operator:

$$\begin{split}\mathcal{F}[f(x)] =& g(k) = \frac{1}{\sqrt{2 \pi... ...i}} \int_{-\infty}^{\infty} g(k) e^{-\imath k x} dk \end{split}$$ (22-8)

Combining operators is another useful way to solve differential equations. Consider the Fourier transform, $\mathcal{F}$ , operating on the differential operator, $\mathcal{D}$ :

$$\mathcal{F}[\mathcal{D}[f]] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \ensuremath{\frac{d{f(x)}}{d{x}}} e^{i k x} dx$$ (22-9)

Integrating by parts,

$$= \frac{1}{\sqrt{2 \pi}}f(x)\left.\right\vert _{x=-\infty}^{x=\in... ...{2 \pi}} \int_{-\infty}^{\infty} \ensuremath{\frac{d{f(x)}}{d{x}}} e^{i k x} dx$$ (22-10)

If the Fourier transform of $f(x)$ exists, then typically² $\lim_{x\rightarrow\pm\infty} f(x) = 0$ . In this case,

$$\mathcal{F}[\mathcal{D}[f]] = -i k \mathcal{F}[f(x)]$$ (22-11)

and by extrapolation:

$$\begin{split}\mathcal{F}[\mathcal{D}^2[f]] & = -k^2 \mathcal{... ...{D}^n[f]] & = (-1)^n \imath^n k^n \mathcal{F}[f(x)] \end{split}$$ (22-12)

Operational Solutions to ODEs

Consider the heterogeneous second-order linear ODE which represent a forced, damped, harmonic oscillator that will be discussed later in this lecture.

$$M \ensuremath{\frac{d^2{y(t)}}{d{t}^2}} + V \ensuremath{\frac{d{y(t)}}{d{t}}} + K_s y(t) = \cos (\omega_o t)$$ (22-13)

Apply a Fourier transform (mapping from the time ($t$ ) domain to a frequency ($\omega$ ) domain) to both sides of 22-13:

$$\begin{split}\mathcal{F}[M \ensuremath{\frac{d^2{y(t)}}{d{t}^... ...ta(\omega-\omega_o) +\delta(\omega+\omega_o)\right] \end{split}$$ (22-14)

because the Dirac Delta functions result from taking the Fourier transform of $\cos(\omega_o t)$ .

Equation 22-14 can be solved for the Fourier transform:

$$\mathcal{F}[y] = \sqrt{\frac{-\pi}{2}} \frac{ \left[ \delta(\omeg... ...ega_o) +\delta(\omega+\omega_o)\right] } { M \omega^2 + \imath \omega V - K_s }$$ (22-15)

In other words, the particular solution Eq. 22-13 can be obtained by finding the function $y(t)$ that has a Fourier transform equal the the right-hand-side of Eq. 22-15-or, equivalently, operating with the inverse Fourier transform on the right-hand-side of Eq. 22-15.

MATHEMATICA$^{\text{\scriptsize {\textregistered }}}$ Example

(notebook Lecture-22)

(html Lecture-22)

(xml+mathml Lecture-22)

Operator Calculus and the Solution to the Damped-Forces Harmonic Oscillator Model

MATHEMATICA$^{\text{\scriptsize {\textregistered }}}$ does have built-in functions to take Fourier (and other kinds of) integral transforms. However, as will be seen below, using operational calculus to solve ODEs is not necessarily simple in MATHEMATICA$^{\text{\scriptsize {\textregistered }}}$ . Nevertheless, it may be instructive to force it--if only as an an example of using the good tool for the wrong purpose.

Operators to Functionals

Equally powerful is the concept of a functional which takes a function as an argument and returns a value. For example $\mathcal{S}[y(x)]$ , defined below, operates on a function $y(x)$ and returns its surface of revolution's area for $0 < x < L$ :

$$\mathcal{S}[y(x)] = 2 \pi \int_0^L y \sqrt{1 + \left(\ensuremath{\frac{d{y}}{d{x}}}\right)^2} dx$$ (22-16)

This is the functional to be minimized for the question, ``Of all surfaces of revolution that span from $y(x=0)$ to $y(x=L)$ , which is the $y(x)$ that has the smallest surface area?''

This idea of finding ``which function maximizes or minimizes something'' can be very powerful and practical.

Suppose you are asked to run an ``up-hill'' race from some starting point $(x=0,y=0)$ to some ending point $(x=1,y=1)$ and there is a ridge $h(x,y)=x^2$ . What is the most efficient running route $y(x)$ ?³

Figure: The terrain separating the starting point $(x=0,y=0)$ and ending point $(x=1,y=1)$ . Assuming a model for how much running speed slows with the steepness of the path--which route would be quicker, one ($y_1(x)$ ) that starts going up-hill at first or another ($y_2(x)$ ) that initially traverses a lot of ground quickly?

A reasonable model for running speed as a function of climbing-angle $\alpha$ is

$$v(s) = \cos(\alpha(s))$$ (22-17)

where $s$ is the arclength along the path. The maximum speed occurs on flat ground $\alpha=0$ and running speed monotonically falls to zero as $\alpha \rightarrow \pi/2$ . To calculate the time required to traverse any path $y(x)$ with endpoints $y(0)=0$ and $y(1)=1$ ,

$$\begin{split}\ensuremath{\frac{d{s}}{d{t}}} & = v(s) = \cos(\... ...{d{x}}}}^2 + {\ensuremath{\frac{d{h}}{d{x}}}}^2} dx \end{split}$$ (22-18)

So, with the hill $h(x)=x^2$ , the time as a functional of the path is:

$$\mathcal{T}[y(x)] = \int_0^1 \sqrt{1 + {\ensuremath{\frac{d{y}}{d{x}}}}^2 + 4 x^2} \; dx$$ (22-19)

MATHEMATICA$^{\text{\scriptsize {\textregistered }}}$ Example

(notebook Lecture-22)

(html Lecture-22)

(xml+mathml Lecture-22)

Functionals: Introduction to Variational Calculus by Variation of Parameters

1. Instead of trying to find the function $y(x)$ (if such a function exists) that minimizes the function in Eq. 22-19, consider the polynomial $y(x) = a + b x + c x^2$ as an ``approximating function'' and then find the parameters $a$ , $b$ , and $c$ , that minimize the functional.
2. Ensure that the cubic equation satisfies the boundary conditions and thereby fix two of the three free parameters
3. By integrating $y(x)$ in Eq. 22-19, the functional equation is transformed to a function of the remaining free variable. (It is much easier in MATHEMATICA$^{\text{\scriptsize {\textregistered }}}$ to integrate without limits in Eq. 22-19 and then evaluate the limits in a separate step.)
4. Find the parameter that minimizes the integral.
5. Visualize the quickest path.

There is a powerful and beautiful mathematical method for finding the extremal functions of functionals which is called Calculus of Variations.

By using the calculus of variations, the optimal path $y(x)$ for Eq. 22-19 can be determined:

$$y(x) = \frac{2 x \sqrt{1 + 4 x^2} + \sinh^{-1}(2 x)} {2 \sqrt{5} + \sinh^{-1}(2)}$$ (22-20)

The approximation determined in the MATHEMATICA$^{\text{\scriptsize {\textregistered }}}$ example above is pretty good.