F X y G X h Y Where G and H Are Continuously Differentiable Functions of One Variable

Results from Elementary Analysis

R.F. Hoskins Research Professor , in Delta Functions (Second Edition), 2011

1.4.3

For differentiable functions in general the following results hold:

(i)

If u and v are differentiable functions, and a and b are constants, then w  = au  + bv is differentiable and

$\frac{d}{\mathit{dt}}\left(\mathit{au}+\mathit{bv}\right)=a\left(\frac{\mathit{du}}{\mathit{dt}}\right)+b\left(\frac{\mathit{dv}}{\mathit{dt}}\right)\text{.}$

(ii)

If u and v are differentiable then so also is the product function uv and

$\frac{d}{\mathit{dt}}\left(\mathit{uv}\right)=v\frac{\mathit{du}}{\mathit{dt}}+u\frac{\mathit{dv}}{\mathit{dt}}\text{.}$

Similarly

$\frac{d}{\mathit{dt}}\left(\frac{u}{v}\right)=\frac{1}{v^2}\left\{v\frac{\mathit{du}}{\mathit{dt}}-u\frac{\mathit{dv}}{\mathit{dt}}\right\}\text{.}$

(iii)

If y  = f(x) where x  = g(t) then, provided the derivatives concerned do exist, we have

$\frac{\mathit{dy}}{\mathit{dt}}=\frac{\mathit{dy}}{\mathit{dx}}\frac{\mathit{dx}}{\mathit{dt}}\text{.}$

(iv)

If y  = f(t) is a differentiable function with a well-defined inverse function, t  = f ^–1(y), then

$\frac{\mathit{dy}}{\mathit{dt}}={\left(\frac{\mathit{dt}}{\mathit{dy}}\right)}^{-1}\text{.}$

(v)

Rolle's Theorem. Let f be continuous on the closed interval [a, b] and differentiable at least in the open interval (a, b). If f(a)   = f(b)   =   0, then there exists a point t ₀ in (a, b) such that f′(t ₀)   =   0.

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Differentiation—Applications

Colin McGregor , ... Wilson Stothers , in Fundamentals of University Mathematics (Third Edition), 2010

9.5 Rates of Change

If u is a differentiable function of s, we sometimes refer to $\frac{\mathit{du}}{\mathit{ds}}$ as the rate of change of u with respect to s. For example, if A is the area of a circle of radius r, so that A  =   πr ², then $\frac{\mathit{dA}}{\mathit{dr}}$ = 2 πr is the rate of change of the area with respect to the radius. If v is a differentiable function of time t then $\frac{\mathit{dv}}{\mathit{dt}}$ , which is often written $\dot{v}$ , is simply called the rate of change of v.

The reason for the terminology can be understood by considering a particle moving on the x-axis. The rate of change of the position of the particle is its velocity. Suppose the x-coordinate of the particle at time t is x(i) where x is a differentiable function of t. The average velocity between times t and t  + h (h  ≠   0) is

$\frac{\mathrm{displacement}}{\mathrm{time}\mathrm{taken}}=\frac{x\left(t+h\right)-x\left(t\right)}{h}.$

See Figure 9.5.1. The limit of this quotient, as h approaches 0, is the velocity of the particle at time t. It is also $\frac{\mathit{dx}}{\mathit{dt}}$ . Thus

Figure 9.5.1.

$\mathrm{velocity}\mathrm{at}\mathrm{time}t=\frac{\mathit{dx}}{\mathit{dt}}=\dot{x}.$

If x is a twice differentiable function of t then, taking the rate of change of the velocity of the particle to be its acceleration and arguing as above, we get

$\mathrm{acceleration}\mathrm{at}\mathrm{time}t=\frac{d}{\mathit{dt}}\left(\frac{\mathit{dx}}{\mathit{dt}}\right)=\frac{d^{2}x}{d{t}^2}=\ddot{x}.$

Note the extended use of the 'dot notation' for differentiation with respect to t.

Example 9.5.1 At time t the x-coordinate of a particle moving on the x-axis is given by

$x\left(t\right)=t^3–6t^2+9t+2.$

Find (a) the position and velocity of the particle when its acceleration is 0, (b) the acceleration when the velocity is 0.

Solution Here,

$\begin{array}{l}\dot{x}\left(t\right)=3t^2–12t+9=3\left(t–1\right)(t–3),\\ \ddot{x}\left(t\right)=6t–12=6\left(t-2\right).\end{array}$

(a) The acceleration is 0 when t   =  2. Then

$\begin{array}{l}\mathrm{position}=x\left(2\right)=4,\\ \mathrm{velocity}=\dot{x}\left(2\right)=–3.\end{array}$

(b) The velocity is 0 when t  =   1 or t  =   3. When t  =   1,

$\mathrm{acceleration}=\ddot{x}\left(\mathrm{l}\right)=–6,$

and when t  =   3,

□ $\mathrm{acceleration}=\ddot{x}\left(3\right)=6.$

Example 9.5.2 An object is propelled, from ground level, vertically upwards with an initial velocity u. The deceleration of the object, due to gravity, is g (constant). Disregarding any other factors, when will the object return to earth?

Solution Let s, v and a be, respectively, the height, velocity and acceleration of the object at time t. Then

$\dot{v}\left(t\right)=a\left(t\right)=–g.$

Hence, using Corollary 9.4.8,

$v\left(t\right)=–\mathit{gt}+K\left(K\mathrm{a}\mathrm{constant}\right).$

But v(0)   = u. So K   =   u giving

$\dot{s}\left(t\right)=v\left(t\right)=–\mathit{gt}+u.$

Hence, using Corollary 9.4.8,

$s\left(t\right)=–\frac{1}{2}{\mathrm{gt}}^2+\mathit{ut}+C\left(C\mathrm{a}\mathrm{constant}\right).$

But s(0)   =   0. So C  =   0 giving

$\begin{array}{l}s\left(t\right)=\mathit{ut}–{\mathrm{gt}}^2=t\left(\mathrm{u}–\mathrm{gt}\right)\\ =0\mathrm{when}t=0\mathrm{or}2u/g.\end{array}$

Thus the object returns to earth at time t   =  2u/g.     □

In problems on related rates, one rate of change, $\frac{\mathit{du}}{\mathit{dt}}$ (say), is given and another, $\frac{\mathit{dv}}{\mathit{dt}}$ (say), has to be found. The first step is to relate u and v. This might be possible in one of several ways. For example, u = $\sqrt{1-v^2}$ or v = $\sqrt{1-u^2}$ or u²  +   v²   =   1 or u   =  cos w, v  =   sin w where w is a third function of t. Then differentiating with respect to t relates $\frac{\mathit{du}}{\mathit{dt}}$ and $\frac{\mathit{dv}}{\mathit{dt}}$ .

Example 9.5.3 Water is running out of a conical funnel at the rate of 0∙002 cubic metres per second. The radius of the top of the funnel is 0∙25 metres and its height is 0∙5 metres. At what rate is the water level changing when it is 0∙3 metres from the top?

Solution Let r and h be, respectively, the radius and height of the surface of the water at time t. See Figure 9.5.2. Let V be the volume of water left in the tank at time t.

Figure 9.5.2.

Here we are given $\frac{\mathit{dV}}{\mathit{dt}}$ = –0∙002 and have to find $\frac{\mathit{dh}}{\mathit{dt}}$ .

By similar triangles,

$\frac{r}{h}=\frac{0.25}{0.5}=\frac{1}{2},i.e.r=\frac{h}{2}.$

Hence,

$V=\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}\mathrm{\pi }\frac{h^2}{4}h=\frac{\mathrm{\pi }}{12}{h}^3$

so that, differentiating with respect to t,

$\frac{\mathit{dV}}{\mathit{dt}}=\frac{\mathrm{\pi }}{12}3h^2\frac{\mathit{dh}}{\mathit{dt}}=\frac{\mathrm{\pi }}{4}{h}^2\frac{\mathit{dh}}{\mathit{dt}}.$

So, when h   =  (0∙5 – 0∙3)   =   0∙2,

$\frac{\mathit{dh}}{\mathit{dt}}=\frac{\mathit{dV}}{\mathit{dt}}\frac{4}{\mathrm{\pi }}\frac{1}{h^2}=\left(-0\cdot 002\right)\frac{4}{\mathrm{\pi }}\frac{1}{0\cdot 04}=-\frac{0\cdot 2}{\mathrm{\pi }},$

i.e. the water level is dropping at 0∙2/π metres per second.     □

Example 9.5.4 A weight W is being raised vertically by a rope through a block B, 30 metres above a point A at ground level. The rope is 70 metres long and is held by a man M who walks away from A at 0∙75 metres per second. How fast is the weight rising when it is 10 metres above A?

Solution We ignore the dimensions of the weight and the block (pulley), the thickness of the rope and the height of the man. Let x and y be the distances from A to M and W, respectively. See Figure 9.5.3.

Figure 9.5.3.

Here we are given $\frac{\mathit{dx}}{\mathit{dt}}$ = 0∙75 and have to find $\frac{\mathit{dy}}{\mathit{dt}}$ .

From triangle ABM, we get

${\left(40+y\right)}^2=x^2+30^2.$

Differentiating with respect to time t gives.

$2\left(40+y\right)\frac{\mathit{dy}}{\mathit{dt}}=2x\frac{\mathit{dx}}{\mathit{dt}}\cdot$

Hence, when y  =   10 so that x  =   40 (see Figure 9.5.3),

$\frac{\mathit{dy}}{\mathit{dt}}=\frac{x}{40+y}\frac{\mathit{dx}}{\mathit{dt}}=\frac{40}{50}\frac{3}{4}=\frac{3}{5},$

i.e. the weight is rising at 0∙6 metres per second.     □

Example 9.5.5 The volume of a spherical balloon is increasing at the rate of 0∙05 cubic metres per second. How fast is the surface area increasing when the radius is 0∙4 metres?

Solution Let V, S and r be the volume, surface area and radius, respectively, at time t.

Here we are given $\frac{\mathit{dV}}{\mathit{dt}}$ = 0∙05 and have to find $\frac{\mathit{dS}}{\mathit{dt}}$ .

So, differentiating with respect to t,

$\frac{\mathit{dV}}{\mathit{dt}}=4\pi r^2\frac{\mathit{dr}}{\mathit{dt}}\mathrm{and}\frac{\mathit{dS}}{\mathit{dt}}=8\mathit{\pi r}\frac{\mathit{dr}}{\mathit{dt}}.$

Hence, when r  =   0∙4,

$\frac{\mathit{dS}}{\mathit{dt}}=8\mathrm{\pi }r\frac{1}{4\mathrm{\pi }{r}^2}\frac{\mathit{dV}}{\mathit{dt}}=\frac{2}{r}\frac{\mathit{dV}}{\mathit{dt}}=\frac{2}{0\cdot 4}0\cdot 05=0\cdot 25,$

i.e. the surface area is increasing at 0∙25 square metres per second.     □

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Random Matrices

In Pure and Applied Mathematics, 2004

Lemma 2.6.2

If three continuous and differentiable functions f _k (x), k = 1, 2, 3, satisfy the equation

(2.6.1) $f_1(xy)=f_2(x)+f_3(y),$

then they are necessarily of the form a ln x + b_k (k = 1, 2, 3), with b ₁ = b ₂ + b ₃.

Proof. Differentiating (2.6.1) with respect to x, we have

${f^{\prime }}_1(xy)=\frac{1}{y}{f^{\prime }}_2(x),$

which, on integration with respect to y, gives

(2.6.2) $\frac{1}{x}{f}_1(xy)={f^{\prime }}_2(x)\ln y+\frac{1}{x}g(x),$

where g(x) is still arbitrary. Substituting f ₁(xy) from (2.6.2) into (2.6.1),

(2.6.3) $x{f^{\prime }}_2(x)\ln y+g(x)-f_2(x)=f_3(y).$

Therefore the left-hand side of (2.6.3) must be independent of x; this is possible only if

$\begin{array}{ccc}x{f^{\prime }}_2(x)=a& \text{and}& g(x)-f_2\end{array}(x)=b_3,$

that is, only if

$f_2(x)=a\ln x+b_2=g(x)-b_3,$

where a, b ₂ and b ₃ are arbitrary constants.

Now (2.6.3) gives

$f_3(y)=a\ln y+b_3,$

and finally (2.6.1) gives

$f_1(xy)=a\ln (xy)+(b_2+b_3).$

Let us now examine the consequences of the statistical independence of the various components of H. Consider the particular transformation

(2.6.4) $H=U^{-1}{H}^{\prime }U,$

where

(2.6.5) $U=\left[\begin{array}{ccccc}\cos \theta & \sin \theta & 0& \dots & 0\\ -\sin \theta & \cos \theta & 0& \dots & 0\\ 0& 0& 1& \dots & 0\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& 0& \dots & 1\end{array}\right],$

or, in quaternion notation (provided N is even),

(2.6.6) $U=\left[\begin{array}{cccccc}\cos \theta +e_2\sin \theta & 0& \dots & 0& \dots & 0\\ 0& 1& \dots & 0& \dots & 0\\ \dots & \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & 0& \dots & 1\end{array}\right].$

This U is, at the same time, orthogonal, symplectic, and unitary.

Differentiation of (2.6.4) with respect to θ gives

(2.6.7) $\frac{\partial H}{\partial \theta }=\frac{\partial U^T}{\partial \theta }{H}^{\prime }U+U^T{H}^{\prime }\frac{\partial U}{\partial \theta }=\frac{\partial U^T}{\partial \theta }UH+H{U}^T\frac{\partial U}{\partial \theta },$

and by substituting for U, U^T , ∂U/∂θ and ∂U^T /∂θ from (2.6.5) or (2.6.6) we get

(2.6.8) $\frac{\partial H}{\partial \theta }=AH+H{A}^T,$

where

(2.6.9) $A=\frac{\partial U^T}{\partial \theta }U=\left[\begin{array}{cccc}0& -1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 0\\ \dots & \dots & \dots & \dots \\ 0& 0& 0& 0\end{array}\right],$

or, in quaternion notation, A is diagonal.

(2.6.10) $A=\left[\begin{array}{cccc}-e_2& 0& \dots & 0\\ 0& 0& \dots & 0\\ \dots & \dots & \dots & \dots \\ 0& 0& \dots & 0\end{array}\right].$

If the probability density function

(2.6.11) $P(H)={\displaystyle \prod _{(\alpha )}{\displaystyle \prod _{j⩽k}{f}_{kj}^{(\alpha )}\left(H_{kj}^{(\alpha )}\right)}}$

is invariant under the transformation U, its derivative with respect to θ must vanish; that

(2.6.12) ${\displaystyle \sum \frac{1}{f_{kj}^{(\alpha )}}\frac{\partial f_{kj}^{(\alpha )}}{\partial H_{kj}^{(\alpha )}}}\frac{\partial H_{kj}^{(\alpha )}}{\partial \theta }=0.$

Let us write this equation explicitly, say, for the unitary case. Equations (2.6.8) and (2.6.12) give

(2.6.13) $\begin{array}{l}\left[\left(-\frac{1}{f_{11}^{(0)}}\frac{\partial f_{11}^{(0)}}{\partial H_{11}^{(0)}}+\frac{1}{f_{22}^{(0)}}\frac{\partial f_{22}^{(0)}}{\partial H_{22}^{(0)}}\right)\left(2H_{12}^{(0)}\right)+\frac{1}{f_{12}^{(0)}}\frac{\partial f_{12}^{(0)}}{\partial H_{12}^{(0)}}\left(H_{11}^{(0)}-H_{22}^{(0)}\right)\right]\\ +{\displaystyle \sum _{k=3}^N\left(-\frac{1}{f_{1k}^{(0)}}\frac{\partial f_{1k}^{(0)}}{\partial H_{1k}^{(0)}}{H}_{2k}^{(0)}+\frac{1}{f_{2k}^{(0)}}\frac{\partial f_{1k}^{(0)}}{\partial H_{2k}^{(0)}}{H}_{1k}^{(0)}\right)}\\ +{\displaystyle \sum _{k=3}^N\left(-\frac{1}{f_{1k}^{(1)}}\frac{\partial f_{1k}^{(1)}}{\partial H_{1k}^{(1)}}{H}_{2k}^{(1)}+\frac{1}{f_{2k}^{(1)}}\frac{\partial f_{2k}^{(1)}}{\partial H_{2k}^{(1)}}{H}_{1k}^{(1)}\right)}=0.\end{array}$

The braces at the left-hand side of this equation depend on mutually exclusive sets of variables and their sum is zero. Therefore each must be a constant; for example,

(2.6.14) $-\frac{H_{2k}^{(0)}}{f_{1k}^{(0)}}\frac{\partial f_{1k}^{(0)}}{\partial H_{1k}^{(0)}}+\frac{H_{1k}^{(0)}}{f_{2k}^{(0)}}\frac{\partial f_{2k}^{(0)}}{\partial H_{2k}^{(0)}}=C_k^{(0)}.$

On dividing both side of (2.6.14) by H ⁽⁰⁾ _1k H ⁽⁰⁾ _2k and applying the Lemma 2.6.2, we conclude that the constant $C_k^{(0)}$ must be zero, that is,

(2.6.15) $\frac{1}{H_{1k}^{(0)}}\frac{1}{f_{1k}^{(0)}}\frac{\partial f_{1k}^{(0)}}{\partial H_{1k}^{(0)}}=\frac{1}{H_{2k}^{(0)}}\frac{1}{f_{2k}^{(0)}}\frac{\partial f_{2k}^{(0)}}{\partial H_{2k}^{(0)}}=\text{constant}=-2a,\text{say,}$

which on integration gives

(2.6.16) $f_{1k}^{(0)}(H_{1k}^{(0)})=\exp \left[-a{\left(H_{1k}^{(0)}\right)}^2\right].$

In the other two cases we also derive a similar equation. Now because the off-diagonal elements come on as squares in the exponential and all invariants are expressible in terms of the traces of powers of H, the function P H) is an exponential that contains traces of at most the second power of H.

Because P(H) is required to be invariant under more general transformations than we have here considered, one might think that the form of P(H) is further restricted. This, however, is not so, for

(2.6.17) $\begin{array}{ccc}P(H)& =& \exp \left(-a\text{tr}{H}^2+b\text{tr}H+c\right)\hfill \\ & =& e^c{\displaystyle \prod _j\exp }\left(b{H}_{jj}^{(0)}\right){\displaystyle \prod _{k⩽j}\exp \left[-a{\left(H_{kj}^{(0)}\right)}^2\right]}{\displaystyle \prod _{\lambda }{\displaystyle \prod _{k<j}\exp \left[-1{\left(H_{kj}^{(\lambda )}\right)}^2\right]}}\end{array}$

is already a product of functions, each of which depends on a separate variable. Moreover, because we require P(H) to be normalizable and real, a must be real and positive and b and c must be real.

Therefore we have proved the following theorem (Porter and Rosenzweig, 1960a).

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Differential Calculus with Several Independent Variables

Robert G. Mortimer , in Mathematics for Physical Chemistry (Fourth Edition), 2013

8.7 Maximum and Minimum Values of Functions of Several Variables

A point at which either a maximum or a minimum value in a function occurs is called an extremum. For example, Figure 8.3 shows a perspective view of a graph of the function $f=e^{-x^2-y^2}$ . The surface representing the function has a "peak" at the origin, representing a maximum value of the function.

Figure 8.3. The surface representing the function $y=e^{-x^2-y^2}$ with the absolute maximum and a constrained maximum shown.

The figure also shows a curve at which the surface intersects with a plane representing the equation $y=1-x$ . On this curve there is also a maximum, which has a smaller value than the maximum at the peak. We call this value a constrained maximum subject to the constraint that $y=1-x$ . We discuss the constrained maximum later.

The maximum at the origin in Figure 8.3 is called a local maximum or a relative maximum, because the value of the function at such a peak is larger than at any other point in the immediate vicinity. However, a complicated function can have more than one local maximum. Also, if we consider a finite region, the function might have a larger value somewhere on the boundary of the region that is larger than the value at a local maximum. To find the absolute maximum of the function for a given region, we must consider all local maxima and any points on the boundary of the region that might have greater values. The peak at the origin in Figure 8.3 is the absolute maximum of this function. Points of minimum value are completely analogous to points of maximum value.Local minima are located at the bottom of depressions or valleys in the surface representing the function. To find an absolute minimum for a given region, you must consider all local minima and any points on the boundary of the region that might have smaller values.

To locate a local maximum or minimum, we use the fact that the plane that is tangent to the surface will be horizontal at any local maximum or minimum. Therefore, the curve representing the intersection of any vertical plane with the surface will have a local maximum or a local minimum at the same place. The partial derivative with respect to one independent variable gives the slope of the tangent to the curve in the plane corresponding to a constant value of the other independent variable, so we can find a local maximum or minimum by finding the places where all the partial derivatives of the function vanish simultaneously.

Our method for a differentiable function of two variables is therefore to

1.

Solve the simultaneous equations

(8.46) ${\left(\frac{\partial f}{\partial x}\right)}_y=0\text{,}$

${\left(\frac{\partial f}{\partial y}\right)}_x=0\text{.}$

2.

Calculate the value of the function at all points satisfying these equations, and at the boundaries of the region being considered and any cusps or discontinuities. The maximum or minimum value in the region being considered must be in this set of values.

Example 8.14

Find the maximum value of the function shown in Figure 8.3, $f=e^{-x^2-y^2}$ .

At a local maximum or minimum

${\left(\frac{\partial f}{\partial x}\right)}_y=e^{-x^2-y^2}\left(-2x\right)=-2{\mathit{xe}}^{-x^2-y^2}\text{,}$

${\left(\frac{\partial f}{\partial y}\right)}_x=e^{-x^2-y^2}\left(-2y\right)=0=-2{\mathit{ye}}^{-x^2-y^2}\text{.}$

The only solution for finite values of x and y is $x=0$ , $y=0$ . Since no restricted region was specified, we consider all values of x and y. For very large magnitudes of x or y, the function vanishes, so we have found the desired absolute maximum, at which $f(0\text{,}0)=1$ .

In the case of one independent variable, a local maximum could be distinguished from a local minimum or an inflection point by determining the sign of the second derivative. With more than one variable, the situation is more complicated. In addition to inflection points, we can have points corresponding to a maximum with respect to one variable and a minimum with respect to another. Such a point is called a saddle point, and at such a point, the surface representing the function resembles a mountain pass or the surface of a saddle. Such points are important in the transition-state theory of chemical reaction rates.

For two independent variables, we calculate the following quantity:

(8.47) $D=\left(\frac{{\partial }^{2}f}{\partial x^2}\right)\left(\frac{{\partial }^{2}f}{\partial y^2}\right)-{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^2\text{.}$

The different cases are as follows:

1.

If $D>0$ and $({\partial }^{2}f/\partial x^2)>0$ , then we have a local minimum.

2.

If $D>0$ and $({\partial }^{2}f/\partial x^2)<0$ , then we have a local maximum.

3.

If $D<0$ , then we have neither a local maximum nor a local minimum.

4.

If $D=0$ , the test fails, and we cannot tell what we have.

Exercise 8.12

Evaluate D at the point (0,   0) for the function of the previous example and establish that the point is a local maximum.

For more than two independent variables, the method is similar, except that there is one equation for each independent variable.

8.7.1 Constrained Maximum/Minimum Problems

Sometimes we must find a maximum or a minimum value of a function subject to some condition, which is called a constraint. Such an extremum is called a constrained maximum or a constrained minimum. Generally, a constrained maximum is smaller than the unconstrained maximum of the function, and a constrained minimum is larger than the unconstrained minimum of the function. Consider the following example:

Example 8.15

Find the maximum value of the function in the previous example subject to the constraint $x+y=1$ .

The constraint corresponds to the specification of y as a function of x by

$y=1-x\text{.}$

This function is given by the line in the x–y plane of Figure 8.3. We are now looking for the place along this curve at which the function has a larger value than at any other place on the curve.

Since y is a function of x on the curve, the direct way to proceed is to replace y by $1-x$ :

$f=(x\text{,}1-x)=f(x)=e^{-x^2-(1-x{)}^2}=e^{-2x^2+2x-1}\text{.}$

Since f is now a function only of x, the local maximum is now at the point where $\mathrm{d}f/\mathrm{d}x$ vanishes:

$\frac{\mathrm{d}f}{\mathrm{d}x}=e^{-2x^2-(1-x{)}^2}(-4x+2)=0\text{.}$

This equation is satisfied by $x=\frac{1}{2}$ and by $|x|\to \infty$ . The constrained maximum corresponds to $x=\frac{1}{2}$ and the minimum corresponds to $|x|\to \infty$ . At the constrained maximum $y=1-\frac{1}{2}=\frac{1}{2}$ and the value of the function at the constrained maximum is

$\begin{array}{cc}\hfill f\left(\frac{1}{2}\text{,}\frac{1}{2}\right)=& \exp \left[-\left({\frac{1}{2}}^2\right)-{\left(\frac{1}{2}\right)}^2\right]\hfill \\ \hfill =& e^{-1/2}=0.6065\cdots \hfill \end{array}$

As expected, this value is smaller than the unconstrained maximum, at which $f=1$ .

8.7.2 Lagrange's Method of Undetermined Multipliers ³

If we have a constrained maximum or minimum problem with more than two variables, the direct method of substituting the constraint relation into the function might not be practical. Lagrange's methodfinds a constrained maximum or minimum without substituting the constraint relation into the function. If the constraint is written in the form $g(x\text{,}y)=0$ , the method for finding the constrained maximum or minimum in $f(x\text{,}y)$ is as follows:

1.

Form the new function

(8.48) $u(x\text{,}y)=f(x\text{,}y)+\lambda g(x\text{,}y)\text{,}$

where $\lambda$ is a constant called an undetermined multiplier. Its value is unknown at this point of the analysis.

2.

Form the partial derivatives of u, and set them equal to zero,

(8.49) ${\left(\frac{\partial u}{\partial x}\right)}_y={\left(\frac{\partial f}{\partial x}\right)}_y+\lambda {\left(\frac{\partial g}{\partial x}\right)}_y=0\text{,}$

(8.50) ${\left(\frac{\partial u}{\partial x}\right)}_x={\left(\frac{\partial f}{\partial x}\right)}_x+\lambda {\left(\frac{\partial g}{\partial x}\right)}_x=0\text{.}$

3.

Solve the set of equations consisting of $g=0$ and these two equations as a set of simultaneous equations for the value of x, the value of y, and the value of $\lambda$ that correspond to the local maximum or minimum.

Example 8.16

Find the constrained maximum of the previous example by the method of Lagrange.

The constraint equation is written

$g(x\text{,}y)=x+y-1=0\text{.}$

The function u is

$u(x\text{,}y)=e^{-x^2-y^2}+\lambda (x+y-1)\text{.}$

The three equations to be solved are

${\left(\frac{\partial u}{\partial x}\right)}_y=\left(-2x\right)e^{-x^2-y^2}+\lambda =0\text{,}$

${\left(\frac{\partial u}{\partial y}\right)}_x=\left(-2y\right)e^{-x^2-y^2}+\lambda =0\text{,}$

$g(x\text{,}y)=x+y-1=0\text{.}$

We begin by solving for $\lambda$ in terms of x and y. We multiply the first equation by y and the second equation by x and add the two equations. The result can be solved to give

$\lambda =\frac{4\mathit{xy}}{x+y}{e}^{-x^2-y^2}\text{.}$

Substitute this into the first equation to obtain

$\left(-2x\right)e^{-x^2-y^2}+\frac{4\mathit{xy}}{x+y}{e}^{-x^2-y^2}=0\text{.}$

The exponential factor is not zero for any finite values of x and y, so

$-2x+\frac{4\mathit{xy}}{x+y}=0\text{.}$

When the expression for $\lambda$ is substituted into the second equation in the same way, the result is

$-2y+\frac{4\mathit{xy}}{x+y}=0\text{.}$

The difference of these two equation is

$-2x+2y=0\text{,}$

which is solved for y in terms of x to obtain

$y=x\text{.}$

This is substituted into the third simultaneous equation to obtain

$x+x-1=0\text{,}$

which gives

$x=\frac{1}{2}\text{,}y=\frac{1}{2}\text{.}$

This is the same result as in the previous example. In this case, the method of Lagrange was more work than the direct method. In problems involving more variables, the method of Lagrange will usually be easier.

Exercise 8.13

(a)

Find the local minimum of the function

$f(x\text{,}y)=x^2+y^2+2x\text{.}$

(b)

Without using the method of Lagrange, find the constrained minimum subject to the constraint,

$x=-y\text{.}$

(c)

Find the constrained minimum using the method of Lagrange.

The method of Lagrange also works if there is more than one constraint. Assume that we desire the local maximum or minimum of the function

(8.51) $f=f(x\text{,}y\text{,}z)$

subject to the two constraints

(8.52) $g_1(x\text{,}y\text{,}z)=0$

and

(8.53) $g_2(x\text{,}y\text{,}z)=0\text{.}$

The procedure is similar, except that two undetermined multipliers are used. One forms the function

(8.54) $u=u(x\text{,}y\text{,}z)=f(x\text{,}y\text{,}z)+{\lambda }_1{g}_1(x\text{,}y\text{,}z)+{\lambda }_2{g}_2(x\text{,}y\text{,}z)$

and solves the set of simultaneous equations consisting of Eqs. (8.52) and (8.53), and the three equations:

(8.55) ${\left(\frac{\partial u}{\partial x}\right)}_{y\text{,}z}=0\text{,}$

(8.56) ${\left(\frac{\partial u}{\partial y}\right)}_{x\text{,}z}=0\text{,}$

(8.57) ${\left(\frac{\partial u}{\partial z}\right)}_{x\text{,}y}=0\text{.}$

The result is a value for ${\lambda }_1$ , a value for ${\lambda }_2$ , and values for $x\text{,}y$ , and z which locate the constrained local maximum or minimum.

Example 8.17

Find the minimum in the function

$f=x^2+y^2+z^2$

subject to the constraints

$\begin{array}{cc}\hfill y-1=& 0\text{,}\hfill \\ \hfill z-2=& 0\text{.}\hfill \end{array}$

We form the function

$u=x^2+y^2+z^2+{\lambda }_1(y-1)+{\lambda }_2\left(z-2\right)\text{.}$

We need to solve the equations

${\left(\frac{\partial u}{\partial x}\right)}_{y\text{,}z}=2x=0\text{,}$

${\left(\frac{\partial u}{\partial y}\right)}_{x\text{,}z}=2y+{\lambda }_1\text{,}$

${\left(\frac{\partial u}{\partial z}\right)}_{x\text{,}y}=2z+{\lambda }_2=0\text{,}$

plus the two constraint equations. This is a simple case, since the first equation gives

$x=0$

and the constraint equations give the values of y and z. The constrained minimum occurs at

$x=0\text{,}y=1\text{,}z=2\text{.}$

The value of the function at this point is $f=5$ .

Exercise 8.14

Find the minimum of the previous example without using the method of Lagrange.

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Differentiation

Agamirza Bashirov , in Mathematical Analysis Fundamentals, 2014

7.2 Differentiation and Continuity

The following establishes that the differentiable functions are more delicate than the continuous functions.

Theorem 7.4

If the function $f:E\to \mathbb{R}$ is differentiable at $c\in E$ , then it is continuous at $c$ .

Proof

Take any $\epsilon >0$ . Then there exists $\sigma >0$ such that $x\in E$ and $\mid x-c\mid <\sigma$ imply

$\mid f(x)-f(c)-f^{\prime }(c)(x-c)\mid <\epsilon \mid x-c\mid \text{.}$

Let $\delta =\min \{\sigma \text{,}\epsilon /(\epsilon +\mid f^{\prime }(c)\mid )\}$ . Then $x\in E$ and $\mid x-c\mid <\delta$ imply

$\begin{array}{cc}\hfill \mid f(x)-f(c)\mid \le & \mid f(x)-f(c)-f^{\prime }(c)(x-c)\mid +\mid f^{\prime }(c)(x-c)\mid \hfill \\ \hfill \le & (\epsilon +\mid f^{\prime }(c)\mid )\mid x-c\mid <\epsilon \text{.}\hfill \end{array}$

Thus, ${\lim }_{x\to c}f(x)=f(c)$ . This proves the theorem. $\blacksquare$

Example 7.5

The converse of Theorem 7.4 fails. Indeed, the function $f(x)=\mid x\mid \text{,}x\in \mathbb{R}$ is continuous at $c=0$ (see Example 5.5). But $f^{\prime }(0)$ does not exist, because for the function

$g(x)=\frac{f(x)-f(c)}{x-c}=\frac{\mid x\mid }{x}\text{,}x\in \mathbb{R}⧹\{0\}\text{,}$

we have $g(0-)=-1$ and $g(0+)=1$ , implying the nonexistence of ${\lim }_{x\to 0}g(x)$ . The graph of the function $f(x)=\mid x\mid \text{,}x\in \mathbb{R}$ is given in Figure 7.2. One can observe that at the point $(0\text{,}0)$ the graph of this function has a "corner," which is typical for being continuous and nondifferentiable.

In 1872 Weierstrass shocked the mathematical world by constructing an example of a function that is continuous at every $x\in \mathbb{R}$ , being differentiable nowhere. Another such pathological example was known by Bolzano earlier. Since that time, several such functions have been constructed. Next we present one of them due to Van der Waerden. ⁶⁴

Example 7.6

The idea of the continuous nowhere differentiable function that we are going to construct consists of increasing the number of "corner points" on the graph of the function from Example 7.5. For this, define the function

$f_0(x)=\mid x\mid \text{for}\mid x\mid \le 1/2$

and extend it to $\mathbb{R}$ by $f_0(x+k)=f_0(x)$ , where $\mid x\mid \le 1/2$ and $k\in \mathbb{Z}$ . Then $f_0$ is a continuous periodic function with the period $1$ that is not differentiable at the points $x=k/2$ , where $k\in \mathbb{Z}$ . Moreover, $f_0$ satisfies $0\le f_0(x)\le 1/2$ for every $x\in \mathbb{R}$ . Let

$f_n(x)=\frac{1}{4^n}{f}_0(4^{n}x)\text{,}x\in \mathbb{R}\text{,}$

and define

(7.3) $f(x)={\displaystyle \sum _{n=0}^{\infty }}{f}_n(x)={\displaystyle \sum _{n=0}^{\infty }}\frac{1}{4^n}{f}_0(4^{n}x)\text{,}x\in \mathbb{R}\text{.}$

By Theorems 3.30 and 3.34 Theorem 3.30 Theorem 3.34 , the series in Eq. (7.3) converges for every $x\in \mathbb{R}$ since

$\mid f_n(x)\mid =\left|\frac{1}{4^n}{f}_0(4^{n}x)\right|\le \frac{1}{2·4^n}\le \frac{1}{4^n}\text{.}$

Therefore, the function $f$ is well defined for every $x\in \mathbb{R}$ . Figure 7.3 illustrates how the number of "corner points" on the graphs of the functions $f_0\text{,}{f}_1\text{,}{f}_2$ increases and how this effects the graph of $f_0+f_1+f_2$ . We assert that the function $f$ is uniformly continuous and nowhere differentiable on $\mathbb{R}$ .

For uniform continuity, note that each of the functions $f_n$ is uniformly continuous on $\mathbb{R}$ . Indeed, for $\epsilon >0$ , we can take $\delta =\epsilon$ . Then from $\mid x-y\mid <\delta$ , we deduce

$\mid f_n(x)-f_n(y)\mid =\left|\frac{1}{4^n}{f}_0(4^{n}x)-\frac{1}{4^n}{f}_0(4^{n}y)\right|\le \frac{1}{4^n}\mid 4^{n}x-4^{n}y\mid =\mid x-y\mid <\epsilon \text{.}$

This proves the uniform continuity of $f_n$ . Furthermore, for $x\text{,}y\in \mathbb{R}$ , we have

$\begin{array}{cc}\hfill \mid f(x)-f(y)\mid \le & \left|{\displaystyle \sum _{k=n+1}^{\infty }}{f}_k(x)\right|+\left|{\displaystyle \sum _{k=n+1}^{\infty }}{f}_k(y)\right|+\left|{\displaystyle \sum _{k=0}^n}(f_k(x)-f_k(y))\right|\hfill \\ \hfill \le & 2{\displaystyle \sum _{k=n+1}^{\infty }}\frac{1}{2·4^k}+{\displaystyle \sum _{k=0}^n}\mid f_k(x)-f_k(y)\mid \hfill \\ \hfill =& \frac{1}{3·4^n}+{\displaystyle \sum _{k=0}^n}\mid f_k(x)-f_k(y)\mid \text{.}\hfill \end{array}$

For $\epsilon >0$ , we can take sufficiently large $n\in \mathbb{N}$ such that $1/4^n<3\epsilon /2$ and let $\delta =\epsilon /2(n+1)$ . Then from $\mid x-y\mid <\delta$ , we obtain

$\mid f(x)-f(y)\mid <\frac{\epsilon }{2}+{\displaystyle \sum _{k=0}^n}\frac{\epsilon }{2(n+1)}=\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \text{.}$

This proves the uniform continuity of $f$ .

To prove that $f$ is nowhere differentiable on $\mathbb{R}$ , assume the contrary: there exists $x\in \mathbb{R}$ at which $f$ is differentiable. Then for every sequence $\{a_m\}$ with the terms $\pm 1$ ,

${\displaystyle \underset{m\to \infty }{\lim }}{4}^m{a}_m\left(f\left(x+\frac{1}{4^m{a}_m}\right)-f(x)\right)=f^{\prime }(x)\text{.}$

We will deduce a contradiction by construction of a sequence $\{a_m\}$ for which the preceding limit does not exist. Since $f_0$ has the period 1, for every $n\in \mathbb{N}$ and $k\in \mathbb{Z}$ ,

$f_n\left(x+\frac{k}{4^n}\right)-f_n(x)=\frac{1}{4^n}(f_0(4^{n}x+k)-f_0(4^{n}x))=0\text{.}$

Fix $m\in \mathbb{N}$ . Then for $n\ge m$ ,

$f_n\left(x+\frac{1}{4^m{a}_m}\right)-f_n(x)=f_n\left(x+\frac{4^{n-m}{a}_m}{4^n}\right)-f_n(x)=0\text{,}$

where $4^{n-m}{a}_m$ is a possible value of $k$ . Therefore,

$\begin{array}{c}\hfill 4^m{a}_m\left(f\left(x+\frac{1}{4^m{a}_m}\right)-f(x)\right)=4^m{a}_m{\displaystyle \sum _{n=0}^{\infty }}\left(f_n\left(x+\frac{1}{4^m{a}_m}\right)-f_n(x)\right)\\ \hfill =4^m{a}_m{\displaystyle \sum _{n=0}^{m-1}}\left(f_n\left(x+\frac{1}{4^m{a}_m}\right)-f_n(x)\right)\text{.}\end{array}$

On the other hand, $n\le m-1$ implies

$\frac{1}{4^{m-n}}\le \frac{1}{4}\text{.}$

Therefore, at least one of the equalities

$\left|f_0\left(4^{n}x+\frac{1}{4^{m-n}}\right)-f_0(4^{n}x)\right|=\frac{1}{4^{m-n}}$

and

$\left|f_0\left(4^{n}x-\frac{1}{4^{m-n}}\right)-f_0(4^{n}x)\right|=\frac{1}{4^{m-n}}$

holds for every $n=0\text{,}1\text{,}\dots \text{,}m-1$ . Let $a_m=1$ if the first of the preceding equalities hold. Otherwise, let $a_m=-1$ . Then

$\begin{array}{cc}\hfill & 4^m{a}_m\left(f\left(x+\frac{1}{4^m{a}_m}\right)-f(x)\right)\hfill \\ \hfill & =4^m{a}_m{\displaystyle \sum _{n=0}^{m-1}}\left(f_n\left(x+\frac{1}{4^m{a}_m}\right)-f_n(x)\right)\hfill \\ \hfill & =4^m{a}_m{\displaystyle \sum _{n=0}^{m-1}}\frac{1}{4^n}\left(f_0\left(4^{n}x+\frac{1}{4^{m-n}{a}_m}\right)-f_0(4^{n}x)\right)\hfill \\ \hfill & =4^m{a}_m{\displaystyle \sum _{n=0}^{m-1}}\frac{1}{4^n}·\frac{\pm 1}{4^{m-n}}={\displaystyle \sum _{n=0}^{m-1}}(\pm a_m)={\displaystyle \sum _{n=0}^{m-1}}(\pm 1)\text{.}\hfill \end{array}$

Thus, if $f^{\prime }(x)$ exists, then $f^{\prime }(x)={\sum }_{n=0}^{\infty }(\pm 1)$ , where the terms of the series are either $1$ or $-1$ . By Corollary 3.27, this series diverges, contradicting the existence of $f^{\prime }(x)$ .

Figure 7.2. Function $f(x)=\mid x\mid \text{,}x\in \mathbb{R}$ .

Figure 7.3. Example 7.6.

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Processing, Analyzing and Learning of Images, Shapes, and Forms: Part 2

Christos Sakaridis , ... Petros Maragos , in Handbook of Numerical Analysis, 2019

3.1 Geometric gradient approximation on graphs

The first term of level set-based active contour evolution models to be approximated is the gradient of the bivariate embedding function. A general method is thus developed in Sakaridis et al. (2017) for calculating the gradient of a real-valued, bivariate function that is implicitly defined on a continuous domain, although its values are known only at a finite set of 2D points, which constitute the vertices of the graph. The context is the same as in the preceding work of Drakopoulos and Maragos (2012).

3.1.1 Formulation

Compared to the gradient approximations proposed in Drakopoulos and Maragos (2012), the geometric gradient approximation of Sakaridis et al. (2017) leverages the local spatial configuration of vertices in the formulation of the approximation. More specifically, it introduces the concept of the neighbour angle, i.e., the angle around a vertex which is "occupied" by each of its neighbours. The motivation for this approach comes from the following lemma in bivariate calculus.

Lemma 1

The gradient of a differentiable function $u:{\mathbb{R}}^2\to \mathbb{R}$ at point x is

(3) $\begin{array}{l}\hfill \nabla u(\mathit{x})=\frac{{\int }_0^{2\pi }{D}_{\varphi }u(\mathit{x}){\mathit{e}}_{\varphi }d\varphi }{\pi },\end{array}$

where e _ϕ is the unit vector in direction ϕ and D _ϕ u(x) is the directional derivative of u at x in this direction, defined by

$\begin{array}{l}\hfill D_{\varphi }u(\mathit{x})=\underset{h\to 0}{lim}\frac{u(\mathit{x}+h{\mathit{e}}_{\varphi })-u(\mathit{x})}{h}.\end{array}$

Based on Lemma 1, the goal is to approximate the gradient at a vertex of the graph by substituting the integral

(4) $\begin{array}{l}\hfill \mathcal{I}={\int }_0^{2\pi }{D}_{\varphi }u(\mathbf{x}){\mathbf{e}}_{\varphi }d\varphi \end{array}$

with a sum over all the neighbours of the vertex. To this end, let us first introduce several key concepts.

The Euclidean distance between vertices v and w of a graph $\mathcal{G}$ is denoted by d(v, w) and the unit vector in the direction of the edge vw starting at v is denoted by e _vw . We define ϕ(w) ∈ [0, 2π) as the angle between the vector e _vw and the horizontal axis, as in Fig. 1. A vertex v will be alternatively denoted by v to declare its position vector. Moreover, we denote by $\mathcal{N}(v)$ the set of neighbours of v in $\mathcal{G}$ , with cardinality N(v). For the sake of brevity in notation, this cardinality will be written simply as N. We write $\mathcal{N}(v)=\left\{w_1,w_2,\dots ,w_N\right\}$ so that the angles ϕ(w _i ) are in ascending order. Based on this ordering, we define the angle around v "occupied" by w _i , which is called neighbour angle, as

Fig. 1. Angles ϕ(w _i ), Δϕ(w _i ) and ω(w _i ).

From Sakaridis, C., Drakopoulos, K., Maragos, P. 2017. Theoretical analysis of active contours on graphs. SIAM J. Imaging Sci. 10 (3), 1475–1510.

(5) $\begin{array}{l}\hfill \mathrm{\Delta }\varphi (w_i)=\left\{\begin{array}{ll}\frac{\varphi \left(w_{i+1}\right)-\left(\varphi \left(w_N\right)-2\pi \right)}{2}& \text{if}i=1,\\ \frac{\varphi \left(w_1\right)+2\pi -\varphi \left(w_{i-1}\right)}{2}& \text{if}i=N,\\ \frac{\varphi \left(w_{i+1}\right)-\varphi \left(w_{i-1}\right)}{2}& \text{otherwise.}\end{array}\right.\end{array}$

In a similar fashion, we define the angle corresponding to the bisector between two consecutive neighbours as

(6) $\begin{array}{l}\hfill \omega (w_i)=\left\{\begin{array}{ll}\frac{\varphi \left(w_i\right)+\varphi \left(w_N\right)-2\pi }{2}& \text{if}i=1,\\ \frac{\varphi \left(w_i\right)+\varphi \left(w_{i-1}\right)}{2}& \text{otherwise.}\end{array}\right.\end{array}$

A visual representation of the neighbour angle is provided in Fig. 1. Using the above notation, the geometric gradient approximation at v is given by the following formula:

(7) $\begin{array}{l}\hfill \nabla u(v)\approx \frac{\sum _{i=1}^N\frac{u(w_i)-u(v)}{d(v,w_i)}{\mathbf{e}}_{v{w}_i}\mathrm{\Delta }\varphi (w_i)}{\pi }.\end{array}$

The directional derivative term in (3) is approximated by the difference quotient of the function along each edge. On the other hand, the angle differential is handled through the neighbour angles, which effectively constitute a Voronoi tessellation of the circle around v, created from its neighbours. The reasoning behind this approach is to use information about the change of u along each particular direction that comes from the neighbour which is closest to this direction.

If the neighbour angles were not taken into account, we would place equal importance on all neighbours of v and return to an approximation similar to the weighted sum that was introduced in Drakopoulos and Maragos (2012):

(8) $\begin{array}{l}\hfill \nabla u(v)\approx \frac{\sum _{i=1}^N\frac{u(w_i)-u(v)}{d(v,w_i)}{\mathbf{e}}_{v{w}_i}}{N}.\end{array}$

3.1.2 Convergence for random geometric graphs

In the remaining theoretical analysis of this section, we mainly focus on a certain type of graphs, namely random geometric graphs (Penrose, 2003).

Definition 1

A random geometric graph (RGG) $\mathcal{G}(n,\rho (n))$ is comprised of a set $\mathcal{V}$ of vertices and a set $\mathcal{E}$ of edges. The set $\mathcal{V}$ consists of n points distributed uniformly at random and independently in a bounded region $D\subset {\mathbb{R}}^2$ . The set $\mathcal{E}$ of edges is defined through the radius ρ(n) of the graph: an edge connects two vertices v and w if and only if their distance is at most ρ(n), i.e., d(v, w) ≤ ρ(n).

An instance of an RGG is given in Fig. 2. For RGGs, the approximation of (7) converges in probability to the true value of the gradient as the number of vertices increases, under some conditions on the radius which constrain the density of the graph. Before stating the related theorem, we remind the reader of some definitions for the asymptotic notations which are used in the analysis.

Fig. 2. A random geometric graph embedded in D = [0, 1]², with n = 80 vertices and radius ρ = 0.25.

From Sakaridis, C., Drakopoulos, K., Maragos, P. 2017. Theoretical analysis of active contours on graphs. SIAM J. Imaging Sci. 10 (3), 1475–1510.

Definition 2

Let f and g be two nonnegative functions. Then,

$\begin{array}{ll}\hfill f(n)\in O(g(n))& \iff \exists k>0\exists n_0\forall n\ge n_0:f(n)\le kg(n),\hfill \\ \hfill f(n)\in \mathrm{\Theta }(g(n))& \iff f(n)\in O(g(n))\wedge g(n)\in O(f(n)),\hfill \\ \hfill f(n)\in o(g(n))& \iff \forall k>0\exists n_0\forall n\ge n_0:f(n)<kg(n),\hfill \\ \hfill f(n)\in \omega (g(n))& \iff \forall k>0\exists n_0\forall n\ge n_0:f(n)>kg(n).\hfill \end{array}$

Theorem 1

Let $u:{\mathbb{R}}^2\to \mathbb{R}$ be a differentiable function and $\mathcal{G}(n,\rho (n))$ an RGG embedded in D = [0, 1]², with $\rho (n)\in \omega \left(n^{-1/2}\right)\cap o\left(1\right)$ . For every vertex v of $\mathcal{G}$ , the gradient approximation of (7) converges in probability to ∇u(v).

The proof of Theorem 1 is provided in Sakaridis et al. (2017).

3.1.3 Asymptotic analysis of approximation error

Note that Drakopoulos and Maragos (2012) also proved that two of their gradient approximations converge in probability to the true value of the gradient. Going one step further, Sakaridis et al. (2017) obtain an asymptotic bound on the rate of convergence to the true gradient as the size of the RGG grows large. Since the framework in RGGs is stochastic, their result involves the expectation of the approximation error.

Let us denote the error in approximating $\mathcal{I}$ with

(9) $\begin{array}{l}\hfill \mathcal{S}=\sum _{i=1}^N\frac{u(w_i)-u(v)}{d(v,w_i)}{\mathbf{e}}_{v{w}_i}\mathrm{\Delta }\varphi (w_i)\end{array}$

by $\mathcal{E}=\mathcal{S}-\mathcal{I}$ . Comparing the two expressions, we deduce that the approximation with $\mathcal{S}$ is threefold:

1.

Directional derivatives along edges are approximated with difference quotients.

2.

The approximate value for the directional derivative along each edge is used as a constant estimate for all the directions "falling into" the respective neighbour angle.

3.

The unit vector in the direction of each edge is also used for all the directions corresponding to the respective neighbour angle.

The calculation of the error that is performed in the proof of the following theorem involves construction of intermediate expressions between $\mathcal{S}$ and $\mathcal{I}$ , bounding the magnitudes of the resulting differences and combining these individual bounds using the triangle inequality.

Theorem 2

Let $u:{\mathbb{R}}^2\to \mathbb{R}$ be a differentiable function and $\mathcal{G}(n,\rho (n))$ an RGG embedded in D = [0, 1]², with $\rho (n)\in \omega \left(n^{-1/2}\right)\cap o\left(1\right)$ . For the gradient approximation error at every vertex v of $\mathcal{G}$ , it holds that

(10) $\begin{array}{l}\hfill \mathrm{E}[‖\mathcal{E}‖]\in O\left(\rho (n)+\frac{1}{n{\rho }^2(n)}\right).\end{array}$

The full proof of Theorem 2 is given in Sakaridis et al. (2017). The radius of the graph is effectively the factor that determines the strictness of the asymptotic bound. To provide better intuition, we study the case when $\rho (n)\in \mathrm{\Theta }\left(n^{-a}\right),a\in (0,1/2)$ . Substituting in (10), we obtain

(11) $\begin{array}{l}\hfill \mathrm{E}[‖\mathcal{E}‖]\in O\left(n^b\right),b=\left\{\begin{array}{ll}-a& \text{if}a\in \left(0,\frac{1}{3}\right],\\ -1+2a& \text{if}a\in \left(\frac{1}{3},\frac{1}{2}\right).\end{array}\right.\end{array}$

The strictest upper bound is $O\left(n^{-1/3}\right)$ , it is achieved for a = 1/3 and it constitutes a trade-off between minimizing the first error term, which calls for small radii, and the other two terms, which require more neighbours and consequently larger radii. On the other hand, when a∉(0, 1/2), the conditions of Theorems 1 and 2 for ρ(n) are not met and convergence to the true value of the gradient is not guaranteed in general. For instance, for a = 0, we get ρ(n) ∈ Θ(1), which means that the radius does not approach 0 in the limit. In turn, this implies that the difference quotients are not guaranteed to converge to the respective directional derivatives, since the distance d(v, w) ≤ ρ(n) does not go to 0 in the limit. In addition, for a = 1/2, it holds that ρ(n) ∈Θ(n ^−1/2) and hence ρ ²(n) ∈ Θ(1/n). As a result, the expected number of neighbours is E[N] = (n − 1)πρ ²(n) ∈ Θ(1). In other words, the sum $\mathcal{S}$ is finite in the limit, which forbids convergence to the integral $\mathcal{I}$ .

In Theorems 1 and 2, the domain D of the RGG is assumed to be the unit square. It is, however, straightforward to generalize the results of both theorems to arbitrary rectangular regions, since their proofs in Sakaridis et al. (2017) only use this assumption for calculating the area of D. This generalization involves applying a uniform scaling to both coordinates by $1/\sqrt{|D|}$ , so that the transformed region has unitary area. The values of u are also scaled by the same factor. These steps ensure that all terms in (7) remain unaffected by the transformation. The two theorems are then applicable to the transformed input. In order to transfer the results back to the original input, one just needs to scale the radius ρ by the constant factor $\sqrt{|D|}$ , which leaves the asymptotic bounds in both theorems unaffected.

3.1.4 Practical application

Despite convergence of the geometric gradient approximation to the true value of the gradient in the case of RGGs, in practice there is a nonnegligible error for graphs with finite number of vertices. This error is propagated to the embedding function of the active contour after each update and may be accumulated after several iterations. To mitigate this, Sakaridis et al. (2017) apply smoothing filtering on the approximate gradient values at a local, neighbourhood level as an empirical means to eliminate potential outliers by taking into account the values at neighbouring vertices. This smoothing is also applicable to curvature, as we discuss in Section 3.2. The smoothing filter can be either an average or a median filter, receiving as input the set of function values at the vertex itself and all its neighbours. In the case of curvature this is straightforward, while for gradient, each of the two vector components are filtered separately. Application of smoothing filtering is statistically motivated for the case of smooth (differentiable) functions by showing that using the neighbours of a vertex to form an ensemble of estimators of the approximated quantity at that vertex reduces the variance of the estimation compared to the basic approximation while not changing the bias.

Experimental validation of the gradient and curvature approximations as well as the aforementioned smoothing filtering is performed by using closed-form functions defined on RGGs. In the experiments that follow in the rest of Sections 3 and 5, the radius of an RGG is chosen as ρ(n) = 0.6n ^−1/3 unless otherwise mentioned, so as to achieve the strictest asymptotic bound for gradient approximation error according to the results of Section 3.1.3. Besides, in Sections 3.1.4 and 3.2.2, all RGGs are embedded in [0, 1]². For each graph, the function's gradient and the curvature of its level sets are approximated at each vertex and afterwards the results are filtered with an average or median filter. The analytical expressions of the function's gradient and curvature are then compared to the estimates. Performance is measured using a global, graph-level error metric which is called relative error and denoted by e _r . The error at each individual vertex of the graph is defined as the difference between the approximate value and the true analytical value, and the relative error is simply the ratio of the energy of the error signal to the energy of the true signal on the entire graph:

(12) $\begin{array}{l}\hfill e_r=\frac{E_{\text{error}}}{E_{\text{analytical}}}.\end{array}$

The relative error of the geometric gradient approximation is evaluated for an isotropic Gaussian on RGGs whose size ranges from 1000 to 10,000 vertices. The analytical form of the Gaussian is

(13) $\begin{array}{l}\hfill exp\left\{-\left[{(x-x_0)}^2+{(y-y_0)}^2\right]/2{\sigma }^2\right\},\end{array}$

with σ = 0.25 and x ₀ = y ₀ = 0.5. Fig. 3A shows average values of e _r over 10 different graphs for each size to reduce variance in the reported performance. Using either an average or a median filter reduces the relative error substantially irrespective of size. Based on this result, smoothing filtering is generally applied for gradient in practice when performing active contour evolution.

Fig. 3. (A) Relative error of gradient approximations for a Gaussian function defined on RGGs of increasing size. The geometric approximation with no filtering is compared to its filtered versions with an average or median filter. (B) Relative error of curvature approximations for a conic function defined on RGGs of increasing size. All four combinations of type of approximation (geometric or gradient-based) and smoothing filter (average or median) are compared.

Modified from Sakaridis, C., Drakopoulos, K., Maragos, P. 2017. Theoretical analysis of active contours on graphs. SIAM J. Imaging Sci. 10 (3), 1475–1510.

Another practical consideration about the geometric gradient approximation (7) is that its absolute error increases as the magnitude of the true gradient grows large, i.e., when the function exhibits abrupt variations. This does not pose a problem for the calculation of gradient direction (which is relevant as input for approximating the curvature), since the latter does not depend on the range of the function's variation around the examined vertex. Utilizing all incident edges in the weighted sum of (7) ensures that all available information in the neighbourhood of the vertex is used to estimate which direction the gradient points to, as emphasized in Drakopoulos and Maragos (2012). However, the estimated gradient magnitude with the geometric gradient approximation (7) is prone to greater error, as it depends on the range of the function's variation. The use of difference quotients in (7) accentuates this effect for dense graphs, where distances between neighbouring vertices that appear in the denominator of the quotients approach zero. Thus, the preferable approximation for gradient magnitude in practice is the maximum absolute difference of values of the function along edges that are incident on v, introduced in Drakopoulos and Maragos (2012):

(14) $\begin{array}{l}\hfill ‖\nabla u(v)‖\approx \underset{w\in \mathcal{N}(v)}{max}\{|u(w)-u(v)|\}.\end{array}$

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Stochastic Dynamics

Don Kulasiri , Wynand Verwoerd , in North-Holland Series in Applied Mathematics and Mechanics, 2002

2.3 What is Stochastic Calculus?

In standard calculus we deal with differentiable functions which are continuous except perhaps in certain locations of the domain under consideration. To understand the continuity of the functions better we make use of the definitions of the limits. We call a function f, a continuous function at the point t = t₀ if

$\underset{t\to t_0}{lim}f\left(t\right)=f\left(t_0\right)$

regardless of the direction t approaches t₀ . A right-continuous function at t₀ has a limiting value only when t approaches t₀ from the right direction, i.e. t is larger than t₀ in the vicinity of t ₀. We will denote this as

$f\left(t+\right)=\underset{t\downarrow t_0}{lim}f\left(t\right)=f\left(t_0\right)\text{.}$

Similarly a left-continuous function at t₀ can be represented as

$f\left(t-\right)=\underset{t\uparrow t_0}{lim}f\left(t\right)=f\left(t_0\right)\text{.}$

These statements imply that a continuous function in both right-continuous and left-continuous at a given point of t. Often we encounter functions having discontinuities; hence the need for the above definitions. To measure the size of a discontinuity, we define the term "jump" at any point t to be a discontinuity where the both f(t   +) and f(t-) exist and the size of the jump be ∆f (t)=f(t   +)− f(t   −). The jumps are the discontinuities of the first kind and any other discontinuity is called a discontinuity of the second kind. Obviously a function can only have countable number of jumps in a given range. From the mean value theorem in calculus it can be shown that we can differentiate a function in a given interval only if the function is either continuous or has a discontinuity of the second kind during the interval. Stochastic calculus is the calculus dealing with often non-differentiable functions having jumps without discontinuities of the second kind. One such example of a function is the Wiener process (Brownian motion). One realization of the standard Wiener process is given in Figure 2.1.

Figure 2.1. An example of a function dealt in stochastic calculus. This function is continuous but not differentiable at any point.

Without going into details of how we computed this function- we will do that in Chapter 3 – we can see that the increments are irregular and we can not define a derivative according to the mean value theorem. This is because of the fact that the function changes erratically within small intervals, however small that interval may be, and we can not define a derivative at a given point in the conventional sense. Therefore we have to devise new mathematical tools that would be useful in dealing with these irregular non-differentiable functions.

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Calculus

Huw Fox , Bill Bolton , in Mathematics for Engineers and Technologists, 2002

Sums of functions

Consider a function which can be considered to be a sum of a number of other functions, e.g. y = f(x) + g(x):

[11] $\begin{array}{ll}\frac{\text{d}}{\text{d}x}\left[f\left(x\right)+g\left(x\right)\right]\hfill & =\underset{\delta x\to 0}{\lim }\frac{\left[f\left(x+\delta x\right)+g\left(x+\delta x\right)\right]-\left[f\left(x\right)+g\left(x\right)\right]}{\delta x}\hfill \\ \hfill & =\underset{\delta x\to 0}{\lim }\left[\frac{f\left(x+\delta x\right)-f\left(x\right)}{\delta x}+\frac{g\left(x+\delta x\right)-g\left(x\right)}{\delta x}\right]\hfill \\ \hfill & =\underset{\delta x\to 0}{\lim }\left[\frac{f\left(x+\delta x\right)-f\left(x\right)}{\delta x}\right]+\underset{\delta x\to 0}{\lim }\left[\frac{g\left(x+\delta x\right)-g\left(x\right)}{\delta x}\right]\hfill \\ \hfill & =\frac{\text{d}}{\text{d}x}f\left(x\right)+\frac{\text{d}}{\text{d}x}g\left(x\right)\hfill \end{array}$

The derivative of the sum of two differentiable functions is the sum of their derivatives.

Key point

The derivative of the sum of two differentiable functions is the sum of their derivatives.

As an illustration, consider the differentiation of the hyperbolic function y = sinh x. This function (see Section 1.8) can be written as ½(e ^x − e^−x ). Thus:

[12] $\frac{\text{d}}{\text{d}x}\left(\sinh x\right)=\frac{\text{d}}{\text{d}x}\frac{1}{2}\left({\text{e}}^x-{\text{e}}^{-x}\right)=\frac{1}{2}\left({\text{e}}^x+{\text{e}}^x\right)=\cosh x$

In a similar way we can differentiate sinh ax and cosh ax, obtaining:

Key point

$\begin{array}{l}\frac{\text{d}}{\text{d}x}\left(\sinh ax\right)=a\cosh ax\\ \frac{\text{d}}{\text{d}x}\left(\cosh ax\right)=a\text{sinh}ax\end{array}$

The hyperbolic function tanh ax can be differentiated using the quotient rule (see later in this Section).

$\frac{\text{d}}{\text{d}x}\left(\tanh ax\right)=a\frac{1}{{\cosh }^{2}ax}$

[13] $\frac{\text{d}}{\text{d}x}\left(\sinh ax\right)=a\cosh ax$

[14] $\frac{\text{d}}{\text{d}x}\left(\cosh ax\right)=a\sinh ax$

Example

Determine the derivatives of:

(a)

y = 2x ³ + x ²,

(b)

y = sin x + cos 2x,

(c)

y = e^4x + x

(a)

$\frac{\text{d}y}{\text{d}x}=6x^2+2x$

(b)

$\frac{\text{d}y}{\text{d}x}=\cos x-2\mathrm{\hspace{0.17em}sin}2x$

(c)

$\frac{\text{d}y}{\text{d}x}=4{\text{e}}^{4x}+1$

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Deterministic Partial Differential Equations

Jinqiao Duan , Wei WANG , in Effective Dynamics of Stochastic Partial Differential Equations, 2014

2.5 Sobolev Inequalities

We now review some inequalities for weakly differentiable functions [264, Ch. 7], [287, Ch. II] or [144, Ch. 7]. Let $D$ be a domain in ${\mathbb{R}}^n$ . The Lebesgue spaces $L^p=L^p(D)\text{,}p\ge 1$ , are the spaces of measurable functions that are $p$ th order Lebesgue integrable on a domain $D$ in ${\mathbb{R}}^n$ . The norm for $f$ in $L^p$ is defined by

$\parallel f{\parallel }_{L^p}\triangleq {\left({\int }_D\mid f(x){\mid }^p\mathit{dx}\right)}^{\frac{1}{p}}\text{.}$

In particular, $L^2(D)$ is a Hilbert space with the following scalar product $〈·\text{,}·〉$ and norm $\parallel ·\parallel$ :

$〈f\text{,}g〉≔{\int }_{D}f\overline{g}\mathit{dx}\text{,}\parallel f\parallel ≔\sqrt{〈f\text{,}f〉}=\sqrt{{\int }_D\mid f(x){\mid }^2\mathit{dx}}\text{,}$

for $f\text{,}g$ in $L^2(D)$ . Now we introduce some common Sobolev spaces. For $k=1\text{,}2\text{,}\dots \text{,}$ define

$H^k(D)≔\{f:{\partial }^{\alpha }f\in L^2(D)\text{,}\mid \alpha \mid \le k\}\text{.}$

Here $\alpha =({\alpha }_1\text{,}\dots \text{,}{\alpha }_n)$ with ${\alpha }_i$ 's being nonnegative integers, $\mid \alpha \mid ={\alpha }_1+\cdots +{\alpha }_n$ , and ${\partial }^{\alpha }\triangleq {\partial }_{x_1}^{{\alpha }_1}\cdots {\partial }_{x_n}^{{\alpha }_n}$ . Each of these is a Hilbert space with scalar product

$〈u\text{,}v{〉}_k={\int }_D{\displaystyle \sum _{\mid \alpha \mid \le k}}{\partial }^{\alpha }u{\partial }^{\alpha }\overline{v}\mathit{dx}$

and the norm

$\parallel u{\parallel }_k=\sqrt{〈u\text{,}u{〉}_k}=\sqrt{{\int }_D{\displaystyle \sum _{\mid \alpha \mid \le k}}\mid {\partial }^{\alpha }u{\mid }^2\mathit{dx}}\text{.}$

For $k=1\text{,}2\text{,}\dots$ and $p\ge 1$ , we further define another class of Sobolev spaces,

$W^{k\text{,}p}(D)=\{u:D^{\alpha }u\in L^p(D)\text{,}\mid \alpha \mid \le k\}\text{,}$

with norm

$\parallel u{\parallel }_{k\text{,}p}={\left({\displaystyle \sum _{\mid \alpha \mid \le k}}\parallel {\partial }^{\alpha }u{\parallel }_{L^p}^p\right)}^{\frac{1}{p}}\text{.}$

Recall that $C_c^{\infty }(D)$ is the space of infinitely differentiable functions with compact support in the domain $D$ . Then $H_0^k(D)$ is the closure of $C_c^{\infty }(D)$ in Hilbert space $H^k(D)$ (under the norm $\parallel ·{\parallel }_k$ ). It is a Hilbert space contained in $H^k(D)$ . Similarly, $W_0^{k\text{,}p}(D)$ is the closure of $C_c^{\infty }(D)$ in Banach space $W^{k\text{,}p}(D)$ (under the norm $\parallel ·{\parallel }_{k\text{,}p}$ ). It is a Banach space contained in $W^{k\text{,}p}(D)$ .

Standard abbreviations $L^2=L^2(D)\text{,}{H}_0^k=H_0^k(D)\text{,}k=1\text{,}2\text{,}\dots$ are used for these common Sobolev spaces.

Let us list several useful inequalities.

Cauchy–Schwarz inequality

For $f\text{,}g\in L^2(D)$ ,

$\left|{\int }_{D}f(x)g(x)\mathit{dx}\right|\le \sqrt{{\int }_D\mid f(x){\mid }^2\mathit{dx}}\sqrt{{\int }_D\mid g(x){\mid }^2\mathit{dx}}\text{.}$

Hölder inequality

For $f\in L^p(D)$ and $g\in L^q(D)$ with $\frac{1}{p}+\frac{1}{q}=1\text{,}p>1$ , and $q>1$ ,

$\left|{\int }_{D}f(x)g(x)\mathit{dx}\right|\le {\left({\int }_D\mid f(x){\mid }^p\mathit{dx}\right)}^{\frac{1}{p}}{\left({\int }_D\mid g(x){\mid }^q\mathit{dx}\right)}^{\frac{1}{q}}\text{.}$

Minkowski inequality

For $f\text{,}g\in L^p(D)$ ,

${\left({\int }_D\mid f(x)\pm g(x){\mid }^p\mathit{dx}\right)}^{\frac{1}{p}}\le {\left({\int }_D\mid f(x){\mid }^p\mathit{dx}\right)}^{\frac{1}{p}}+{\left({\int }_D\mid g(x){\mid }^p\mathit{dx}\right)}^{\frac{1}{p}}\text{.}$

Poincaré inequality

For $g\in H_0^1(D)$ ,

$\parallel g{\parallel }^2={\int }_D\mid g(x){\mid }^2\mathit{dx}\le {\left(\frac{\mid D\mid }{{\omega }_n}\right)}^{\frac{1}{n}}{\int }_D\mid \nabla g{\mid }^2\mathit{dx}\text{,}$

where $\mid D\mid$ is the Lebesgue measure of the domain $D$ , and ${\omega }_n$ is the volume of the unit ball in ${\mathbb{R}}^n$ in terms of the Gamma function $\Gamma$ :

${\omega }_n=\frac{{\pi }^{n2}}{\Gamma (\frac{n}{2}+1)}\text{.}$

It is clear that ${\omega }_1=2\text{,}{\omega }_2=\pi$ and ${\omega }_3=\frac{4}{3}\pi$ .

Similarly, for $u\in W_0^{1\text{,}p}(D)\text{,}1\le p<\infty$ , and $D\subset {\mathbb{R}}^n$ a bounded domain,

$\parallel u{\parallel }_p\le {\left(\frac{\mid D\mid }{{\omega }_n}\right)}^{\frac{1}{n}}\parallel \nabla u{\parallel }_p\text{.}$

Let $u\in W^{1\text{,}p}(D)\text{,}1\le p<\infty$ , and $D\subset {\mathbb{R}}^n$ a bounded convex domain. Take $S\subset D$ , be any measurable subset, and define the spatial average of $u$ over $S$ by $u_S=\frac{1}{\mid S\mid }{\int }_{S}u\mathit{dx}$ (with $\mid S\mid$ being the Lebesgue measure of $S$ ). Then

$\parallel u-u_S{\parallel }_p\le {\left(\frac{{\omega }_n}{\mid D\mid }\right)}^{1-\frac{1}{n}}{d}^n\parallel \nabla u{\parallel }_p\text{,}$

where $d$ is the diameter of $D$ .

Agmon inequality

Let $D\subset {\mathbb{R}}^n$ be an open domain with piecewise smooth boundary. There exists a positive constant $C$ , depending only on domain $D$ , such that

$\begin{array}{cc}\hfill & \parallel u{\parallel }_{L^{\infty }(D)}\le C\parallel u{\parallel }_{H^{\frac{n-1}{2}}(D)}^{\frac{1}{2}}\parallel u{\parallel }_{H^{\frac{n+1}{2}}(D)}^{\frac{1}{2}}\text{,}\mathrm{for}n\mathrm{odd}\text{,}\hfill \\ \hfill & \parallel u{\parallel }_{L^{\infty }(D)}\le C\parallel u{\parallel }_{H^{\frac{n-2}{2}}(D)}^{\frac{1}{2}}\parallel u{\parallel }_{H^{\frac{n+2}{2}}(D)}^{\frac{1}{2}}\text{,}\mathrm{for}n\mathrm{even}\text{.}\hfill \end{array}$

In particular, for $n=1$ and $u\in H^1(0\text{,}l)$ ,

$\parallel u{\parallel }_{L^{\infty }(0\text{,}l)}\le C\parallel u{\parallel }_{L^2(0\text{,}l)}^{\frac{1}{2}}\parallel u{\parallel }_{H^1(0\text{,}l)}^{\frac{1}{2}}\text{.}$

Moreover, for $n=1$ and $u\in H_0^1(0\text{,}l)$ ,

$\parallel u{\parallel }_{L^{\infty }(0\text{,}l)}\le C\parallel u{\parallel }_{L^2(0\text{,}l)}^{\frac{1}{2}}\parallel u_x{\parallel }_{L^2(0\text{,}l)}^{\frac{1}{2}}\text{.}$

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Half-Linear Differential Equations

In North-Holland Mathematics Studies, 2005

5.6.1 Asymptotic formula for distribution of zeros

Here we present an asymptotic formula for number of zeros of oscillatory solutions of the equation

(5.6.1) ${\left(\text{Φ}\left(x^{\prime }\right)\right)}^{\prime }+\left(p-1\right)c\left(t\right)\text{Φ}\left(x\right)=0.$

It is supposed that c(t) > 0 for large t and the results are based on the generalized Prüfer transformation from Section 1.2. In this transformation, a nontrivial solution and its derivative are expressed via the generalized half-linear sine and cosine functions. Recall that the half-linear sine function, denoted by sin _p t, is the solution of the equation

${\left(\text{Φ}\left(x^{\prime }\right)\right)}^{\prime }+\left(p-1\right)\text{Φ}\left(x\right)=0$

satisfying the initial condition x(0) = 0, x'(0) = 1 and the half-linear cosine function is denned by cos _p t = sin' _p t. Generalized π, denoted by π _p , is introduced in Subsection 1.1.2.

Theorem 5.6.1

Suppose that c is a differentiable function such that c(t) > 0 on an interval [T, ∞), and

(5.6.2) $\underset{t\to \infty }{\lim }{c}^{\prime }\left(t\right){\left[c\left(t\right)\right]}^{-\frac{p+1}{p}}=0$

holds. Then (5.6.1) is oscillatory. Moreover, if N[x; T] denotes the number of zeros of a solution x of (5.6.1) in the interval [a, T], then

(5.6.3) $N\left[x;T\right]=P\left[x;T\right]+R\left[x;T\right],$

where P[x; T] is the principal term given by

$P\left[x;T\right]=\frac{1}{{\pi }_p}{\displaystyle {\int }_a^T{\left[c\left(s\right)\right]}^{\frac{1}{p}}ds}$

and R[x; T] is the remainder which is of smaller order than P[x; T] as T → ∞ and satisfies

$\left|R\left[x;T\right]\right|\le \frac{1}{p{\pi }_p}{\displaystyle {\int }_a^T\frac{\left|c^{\prime }\left(s\right)\right|}{c^{\prime }\left(s\right)}ds+O\left(1\right).}$

Proof. Set $C\left(t\right):=c^{\prime }\left(t\right){\left[c\left(t\right)\right]}^{-\frac{p+1}{p}}$ and define

(5.6.4) $C^{*}\left(t\right)=\sup \left\{\left|C\left(s\right)\right|:s\ge t\right\},t\ge a.$

Then C ^*(t) is nonincreasing and satisfies lim _t→∞ C ^*(t) = 0 by (5.6.2). We have

$\left|{\left[c\left(t+h\right)\right]}^{-\frac{1}{p}}-{\left[c\left(t\right)\right]}^{-\frac{1}{p}}\right|=\frac{1}{p}\left|{\displaystyle {\int }_t^{t+h}C\left(s\right)ds}\right|\le \frac{\left|h\right|}{p}{C}^{*}\left(t\right),$

which implies that

$\underset{h\to \infty }{\lim \sup }\frac{{\left[c\left(t+h\right)\right]}^{-\frac{1}{p}}}{t+h}\le \frac{C^{*}\left(t\right)}{p}.$

It follows that lim _t→∞ t ⁻¹[c(t)]^−1/p = 0, or equivalently, lim _t→∞ t ^p c(t) = ∞. This implies, by Theorem 1.4.5, that (5.6.1) is oscillatory.

Now we turn our attention to the proof of the asymptotic formulas for numbers of zeros. By the Sturmian comparison theorem (Theorem 1.2.4) we have that N[x ₁; T] and N[x ₂; T] differ at most by one for any solutions x ₁ and x ₂ of (5.6.1), so we may restrict our attention to the solution x ₀ of (5.6.1) determined by the initial conditions x ₀(a) = 0, x'₀(a) = 1. This solution is oscillatory by the first part of the our theorem.

We introduce the polar coordinates ρ(t), φ(t) for x ₀(t) by setting

(5.6.5) ${\left[c\left(t\right)\right]}^{\frac{1}{p}}{x}_0\left(t\right)=\rho \left(t\right){\sin }_p\varphi \left(t\right),{x^{\prime }}_0\left(t\right)=\rho \left(t\right){\cos }_p\varphi \left(t\right).$

It can be shown without difficulty that ρ(t) and φ(t) are continuously differentiable on [a, ∞) and satisfy the differential equations

(5.6.6) $\begin{array}{l}\frac{{\rho }^{\prime }}{\rho }=\frac{c^{\prime }\left(t\right)}{pc\left(t\right)}{\left|{\sin }_p\varphi \right|}^p,\\ {\varphi }^{\prime }={\left[c\left(t\right)\right]}^{\frac{1}{p}}+\frac{c^{\prime }\left(t\right)}{pc\left(t\right)}{\sin }_p\varphi \text{Φ}\left({\cos }_p\varphi \right).\end{array}$

We use the notation

$g\left(\varphi \right)={\sin }_p\varphi \text{Φ}\left({\cos }_p\varphi \right),$

in terms of which (5.6.6) is written as

(5.6.7) ${\varphi }^{\prime }={\left[c\left(t\right)\right]}^{\frac{1}{p}}+\frac{c^{\prime }\left(t\right)}{pc\left(t\right)}g\left(\varphi \right).$

From the first equation in (5.6.5) we see that x ₀(t) = 0 if and only if φ(t) = jπ _p , j ∈ ℤ. We may suppose that φ(a) = 0. In view of (5.6.2) there is no loss of generality in assuming that

$C^{*}\left(t\right)<p\text{for}t\ge a,$

where C ^*(t) is defined by (5.6.4). Since

(5.6.8) $\left|g\left(\varphi \right)\right|\le 1\text{all}\varphi ,$

we have

${\left[c\left(t\right)\right]}^{\frac{1}{p}}+\frac{c^{\prime }\left(t\right)}{pc\left(t\right)}g\left(\varphi \left(t\right)\right)\ge {\left[c\left(t\right)\right]}^{\frac{1}{p}}\left(1-\frac{1}{p}{C}^{*}\left(t\right)\right)>0,$

which implies that φ'(t) > 0, so that φ(t) is increasing for t ≥ a.

We now integrate (5.6.7) over [a, T], obtaining

(5.6.9) $\varphi \left(T\right)={\displaystyle {\int }_a^T{\left[c\left(s\right)\right]}^{\frac{1}{p}}ds}+\frac{1}{p}{\displaystyle {\int }_a^T\frac{c^{\prime }\left(s\right)}{c\left(s\right)}g\left(\varphi \left(s\right)\right)ds}=F\left(T\right)+G\left(T\right),$

where

$F\left(T\right)≔{\displaystyle {\int }_a^T{\left[c\left(s\right)\right]}^{\frac{1}{p}}ds},G\left(T\right)≔\frac{1}{p}{\displaystyle {\int }_a^T\frac{c^{\prime }\left(s\right)}{c\left(s\right)}g\left(\varphi \left(s\right)\right)ds}$

From (5.6.8) it is clear that

(5.6.10) $|G\left(T\right)|\le \frac{1}{p}{\displaystyle {\int }_a^T\frac{|c^{\prime }\left(s\right)|}{c\left(s\right)}ds}$

Noting that the number of zeros of x ₀(t) in [a, T] is given by

$N\left[x_0;T\right]=\left[\frac{\varphi \left(T\right)}{{\pi }_p}\right]+1,$

where [u] denotes the greatest integer not exceeding u, we see from (5.6.9) and (5.6.10) that the conclusion of the theorem holds with the choice

$P\left[x_0;T\right]=\frac{1}{{\pi }_p}F\left(T\right)=\frac{1}{{\pi }_p}{\displaystyle {\int }_a^T{\left[c\left(s\right)\right]}^{\frac{1}{p}}ds.}$

That the term R[x ₀; T] = N[x ₀; T] − P[x ₀; T] is of smaller order than P[x ₀; T] follows from the observation that

$\begin{array}{lll}{\displaystyle {\int }_a^T\frac{c^{\prime }\left(s\right)}{c\left(s\right)}ds}\hfill & =\hfill & {\displaystyle {\int }_a^T\left|C\left(s\right)\right|{\left[c\left(s\right)\right]}^{\frac{1}{p}}ds}\hfill \\ \hfill & \le \hfill & {\displaystyle {\int }_a^T{C}^{*}\left(s\right){\left[c\left(s\right)\right]}^{\frac{1}{p}}ds}=o\left({\displaystyle {\int }_a^T{\left[c\left(s\right)\right]}^{\frac{1}{p}}ds}\right)\text{as}T\to \infty .\hfill \end{array}$

This completes the proof.

Example 5.6.1

Consider the equation

(5.6.11) ${\left(\text{Φ}\left(x^{\prime }\right)\right)}^{\prime }+\left(p-1\right)t^{\beta }\text{Φ}\left(x\right)=0,t\ge 1,$

where β is a constant with p + β > 0. The function c(t) = t ^β satisfies

$\begin{array}{l}{\displaystyle {\int }_1^T{\left[c\left(s\right)\right]}^{\frac{1}{p}}ds}=\frac{p}{p+\beta }\left(T^{\frac{p+\beta }{p}}-1\right),\\ {\displaystyle {\int }_1^T\frac{\left|c^{\prime }\left(s\right)\right|}{c\left(s\right)}ds}=|\beta |\log T,\end{array}$

and so we conclude from Theorem 5.6.1 that the quantity P[x; T] can be taken to be

$P\left[x;T\right]=\frac{p}{\left(p+\beta \right){\pi }_p}{T}^{\frac{p+\beta }{p}}$

and (5.6.3) holds with this P[x; T] and R[x; T] satisfying

$R\left[x;T\right]=\frac{|\beta |}{p{\pi }_p}\log T+O\left(1\right).$

Remark 5.6.1

(i)

The results of this subsection cannot be applied to the generalized Euler equation 1.4.20

${\left(\text{Φ}\left(x^{\prime }\right)\right)}^{\prime }+\text{λ}\left(p-1\right)t^{-p}\text{Φ}\left(x\right)=0$

shows that both of them are of the same logarithmic order as T → ∞.

(ii)

In [317], M. Piros has investigated a similar problem under a more stringent restriction on c(t), namely he supposed that c ^v (t) is a concave function of t for some v > 0. Then he proved that the error term R[x; T] in (5.6.3) is O(1). Exactly, the differential equation (5.6.11) with β = 1/v plays the exceptional role in determining the precise value of R[x; T].

(iii)

It is not difficult to see how the results can be extended to the case of a general r in (1.1.1), which satisfies ƒ^∞ r ^1-q (s) ds = ∞, by using the transformation of independent variable, see Subsection 1.2.7.

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