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Chapter 15: Problem 1

Show that the solution of \(w^{\prime}+w / z^{2}=0\) has an essential singularity at \(z=0\).

Short Answer

Expert verified
The solution to the differential equation \(w^{\prime}+w / z^{2}=0\) is given as \(w = C exp (1/z)\), has an essential singularity at z=0.

Step by step solution

01

Solve the Differential Equation

The given differential equation can be written in the form \(dw/dz + w/z^2 = 0\). This is a first order linear differential equation, for which the integrating factor is \(exp (\int z^{-2} dz)\), so that gives us \(exp (-1/z)\). Multiply throughout by this factor to get \(exp (-1/z) dw/dz + w/z^2 exp(-1/z) = 0\). This simplifies to \(d/dz [w exp(-1/z)] = 0\), which can be integrated to obtain \(w exp(-1/z) = C\), where C is a constant.
02

Find the General Solution

From the above-equation, the general solution for this differential equation is \(w = C exp (1/z)\).
03

Identify the Singularity

A point \(z=a\) is an essential singularity of a function if neither the function nor its derivative is defined at that point. Let's put z=0 in our solution, \(w = C exp (1/z)\), we find that both the function and its derivative are undefined at z=0. This means that z=0 is an essential singularity for the function w.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equation
Understanding a differential equation is critical for solving various physics and engineering problems. In essence, a differential equation is a mathematical equation that relates some function with its derivatives. In the context of our exercise, we have a first-order linear differential equation \( w^{\prime} + \frac{w}{z^2} = 0 \).

This type of equation involves terms that are proportional to the function itself (here, \(w\)) and to its first derivative (here, \(w^{\prime}\)). Solving a differential equation often involves finding a function that meets the requirements specified by the equation鈥攊n our case, expressing \(w\) as a function of \(z\).
Integrating Factor
An integrating factor is a valuable function used when solving linear differential equations to simplify the expression down to an easily integrable form. Generally, for an equation like \(dy/dx + P(x)y = Q(x)\), the integrating factor is given by \(e^{\int P(x)dx}\).

In our exercise, the integrating factor is found by taking \(e^{\int z^{-2} dz}\), which results in \(e^{-1/z}\). When we multiply the entire differential equation by this integrating factor, it rearranges into a form where the left side becomes the derivative of a product. This way, we can integrate both sides to find the general solution. The integrating factor method is incredibly useful for turning otherwise complex differential equation problems into simpler ones that involve basic integration.
Singularity in Complex Analysis
In the realm of complex analysis, a singularity is a point at which a mathematical object is not defined or does not behave well, such as a point where a function approaches infinity. An essential singularity is a particularly intriguing type of singularity where the behavior of the function near the point is wildly unpredictable and there's no limit to what values the function can take on. In other words, near this point, the function can get arbitrarily large or small in every possible direction of approach.

In our exercise, the essential singularity at \(z=0\) for the function \(w\) is demonstrated by the fact that neither the function \(w = C e^{1/z}\) nor its derivative is defined at that point. This behavior is precisely characteristic of an essential singularity, according to the definition provided by complex analysis. The function's response around \(z=0\) is not merely undefined but also peculiarly erratic as \(z\) gets infinitely close to 0, which can have interesting implications in complex dynamics and other fields where non-linear behavior is studied.

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Most popular questions from this chapter

Let \(u=J_{v}(\lambda z)\) and \(v=J_{v}(\mu z)\). Multiply the Bessel DE for \(u\) by \(v / z\) and that of \(v\) by \(u / z\). Subtract the two equations to obtain $$ \left(\lambda^{2}-\mu^{2}\right) z u v=\frac{d}{d z}\left[z\left(u \frac{d v}{d z}-v \frac{d u}{d z}\right)\right] . $$ (a) Write the above equation in terms of \(J_{v}(\lambda z)\) and \(J_{v}(\mu z)\) and integrate both sides with respect to \(z\). (b) Now divide both sides by \(\lambda^{2}-\mu^{2}\) and take the limit as \(\mu \rightarrow \lambda\). You will need to use L'H么pital's rule. (c) Substitute for \(J_{v}^{\prime \prime}(\lambda z)\) from the Bessel DE and simplify to get $$ \int z\left[J_{v}(\lambda z)\right]^{2} d z=\frac{z^{2}}{2}\left\\{\left[J_{v}^{\prime}(\lambda z)\right]^{2}+\left(1-\frac{v^{2}}{\lambda^{2} z^{2}}\right)\left[J_{v}(\lambda z)\right]^{2}\right\\} $$ (d) Finally, let \(\lambda=x_{v n} / a\), where \(x_{v n}\) is the \(n\) th root of \(J_{v}\), and use Eq. (15.47) to arrive at $$ \int_{0}^{a} z J_{v}^{2}\left(\frac{x_{v n}}{a} z\right) d z=\frac{a^{2}}{2} J_{v+1}^{2}\left(x_{v n}\right) . $$

Use the recurrence relation for the Bessel function to show that \(J_{1}(z)=-J_{0}^{\prime}(z)\).

Transform \(d w / d z+w^{2}+z^{m}=0\) by making the substitution \(w=\) \((d / d z) \ln v .\) Now make the further substitutions $$ v=u \sqrt{z} \text { and } t=\frac{2}{m+2} z^{1+(1 / 2) m} $$ to show that the new DE can be transformed into a Bessel equation of order \(1 /(m+2)\).

Consider the function \(v(z) \equiv z^{r}(1-z)^{s} F\left(\alpha_{1}, \beta_{1} ; \gamma_{1} ; 1 / z\right)\) and assume that it is a solution of HGDE. Find a relation among \(r, s, \alpha_{1}, \beta_{1}\), and \(\gamma_{1}\) such that \(v(z)\) is written in terms of three parameters rather than five. In particular, show that one possibility is $$ v(z)=z^{\alpha-\gamma}(1-z)^{\gamma-\alpha-\beta} F(\gamma-\alpha, 1-\alpha ; 1+\beta-\alpha ; 1 / z) . $$ Find all such possibilities.

The linear combination $$ \begin{aligned} \Psi(\alpha, \gamma ; z) \equiv & \frac{\Gamma(1-\gamma)}{\Gamma(\alpha-\gamma+1)} \Phi(\alpha, \gamma ; z) \\ &+\frac{\Gamma(\gamma-1)}{\Gamma(\alpha)} z^{1-\gamma} \Phi(\alpha-\gamma+1,2-\gamma ; z) \end{aligned} $$ is also a solution of the CHGDE. Show that the Hermite polynomials can be Written as $$ H_{n}\left(\frac{z}{\sqrt{2}}\right)=2^{n} \Psi\left(-\frac{n}{2}, \frac{1}{2} ; \frac{z^{2}}{2}\right) $$.
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