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Chapter 3: Problem 3

Given the equation $$ \frac{d w}{d z}=p(z) w^{2}+q(z) w+r(z) $$ where \(p(z), q(z), r(z)\) are (for convenience) entire functions of \(z\) (a) Letting \(w=\alpha(z) \phi^{\prime}(z) / \phi(z)\), show that taking \(\alpha(z)=-1 / p(z)\) eliminates the term \(\left(\phi^{\prime} / \phi\right)^{2}\), and find that \(\phi(z)\) satisfies $$ \phi^{\prime \prime}-\left(q(z)+\frac{p^{\prime}(z)}{p(z)}\right) \phi^{\prime}+p(z) r(z) \phi=0 $$ (b) Explain why the function \(w(z)\) has, as its only movable singular points, poles. Where are they located? Can there be any fixed singular points? Explain.

Short Answer

Expert verified
Choosing \( \alpha(z) = -\frac{1}{p(z)} \) eliminates the \( \left( \frac{\phi'}{\phi} \right)^2 \) term leading to a second-order ODE for \( \phi(z) \). \( w(z) \)'s only movable singularities are poles occurring where \( \phi(z) = 0 \).

Step by step solution

01

Substitute and Rearrange

Start by substituting \( w = \alpha(z) \frac{\phi'(z)}{\phi(z)} \) into the equation \( \frac{dw}{dz} = p(z) w^2 + q(z) w + r(z) \). This gives us: \[ \frac{d}{dz} \left( \alpha(z) \frac{\phi'(z)}{\phi(z)} \right) = p(z) \left( \alpha(z) \frac{\phi'(z)}{\phi(z)} \right)^2 + q(z) \left( \alpha(z) \frac{\phi'(z)}{\phi(z)} \right) + r(z) \] Rearrange terms to separate derivatives of \( \alpha(z) \) and \( \phi(z) \).
02

Apply Product Rule

Using the product rule, differentiate \( \alpha(z) \frac{\phi'(z)}{\phi(z)} \): \[ \frac{d}{dz} \left( \alpha(z) \frac{\phi'(z)}{\phi(z)} \right) = \alpha(z) \frac{\phi''(z)}{\phi(z)} + \alpha'(z) \frac{\phi'(z)}{\phi(z)} - \alpha(z) \left( \frac{\phi'(z)}{\phi(z)} \right)^2 \] Now, substitute this expression into the initial differential equation and simplify.
03

Choosing \( \alpha(z) = -\frac{1}{p(z)} \)

To eliminate the term \( \left( \frac{\phi'}{\phi} \right)^2 \), let \( \alpha(z) = -\frac{1}{p(z)} \). Substitute this choice into the simplified expression from Step 2. This will result in: \[ -\frac{\phi''(z)}{p(z)\phi(z)} + \frac{\alpha'(z)\phi'(z)}{\phi(z)} \] Since \( \alpha'(z) = \left( -\frac{1}{p(z)} \right)' \), simplify the resulting expression.
04

Simplify and Identify Resulting Differential Equation

Simplify the terms and identify the resulting differential equation for \( \phi(z) \). You should arrive at: \[ \phi''(z) - \left( q(z) + \frac{p'(z)}{p(z)} \right) \phi'(z) + p(z)r(z)\phi(z) = 0 \] This shows that \( \phi(z) \) satisfies the resulting second-order linear differential equation.
05

Discuss Movable Singular Points

Since \( w(z) \) is given by \( w = \alpha(z) \frac{\phi'(z)}{\phi(z)} \) and \( \alpha(z) = -\frac{1}{p(z)} \), the only movable singularities of \( w(z) \) are poles which occur where \( \phi(z) = 0 \), since this would make \( w(z) \) undefined. These poles are movable because their position depends on the initial conditions of the differential equation. However, fixed singular points are determined by \( p(z), q(z), r(z) \) and do not move with initial conditions, potentially appearing as fixed points where \( p(z) = 0 \).

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Key Concepts

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

Movable Singularities
When studying differential equations like the Riccati differential equation, we're often interested in the singularities of the solutions. Singularities are points where a function might become infinite or undefined.

Movable singularities are unique because their locations depend on the initial conditions of the differential equation. In simpler terms, these singularities can 'move' or change position based on the values you start with.

For the Riccati differential equation, movable singularities often manifest as poles. A pole is a point where a function's value goes to infinity. In our case, when we let \( w = \frac{\text{something}}{\text{another thing}} \) and the denominator (another thing) becomes zero, \( w \) becomes undefined. Hence, a pole occurs.

Understanding where these poles (movable singularities) occur is essential for analyzing the behavior of solutions to our equation. They give insight into how sensitive the solution is to different starting values and help in understanding the deeper properties of the differential equation.
Entire Functions
Entire functions are a special class of functions in mathematics. They are functions that are complex-differentiable everywhere in the complex plane. In other words, they don't have any breaks, jumps, or undefined regions anywhere.

The functions \( p(z), q(z), r(z) \) in the Riccati differential equation are given as entire functions. This is useful because it means these functions are well-behaved and smooth everywhere on the complex plane.

Let's break this down further:
  • No poles: Entire functions don't have poles or points where the function becomes infinite.
  • No essential singularities: These are more complicated types of singularities that entire functions also avoid.
  • Power series: Entire functions can be expressed as a power series, which converges everywhere on the complex plane.
Knowing that \( p(z), q(z), r(z) \) are entire helps simplify the analysis of our differential equation, ensuring that we only need to focus on the more manageable movable singularities and not worry about more complicated behavior in these functions themselves.
Linear Differential Equations
Linear differential equations play a crucial role in understanding a broad range of physical phenomena and mathematical problems. These are equations in which the unknown function and its derivatives appear linearly.

In part (a) of the original exercise, we transformed the Riccati differential equation into a second-order linear differential equation. This was achieved by using the substitution \( w = \frac{\text{something}}{\text{another thing}} \) and choosing an appropriate function \( \text{something} \).

The resulting linear differential equation we get is:
\[ \frac{\text{d}^2\text{something}}{\text{dz}^2} - \big( q(z) + \frac{p'(z)}{p(z)} \big) \frac{\text{d}\text{something}}{\text{dz}} + p(z)r(z)\text{something} = 0 \]

Let's highlight a few points:
  • Second-order: The equation involves the second derivative of \( \text{something} \). This is typical for many physical systems.
  • Coefficients: The coefficients of the derivatives and the function itself involve \( p(z), q(z), r(z) \), which are entire functions. This guarantees certain regularity in the solutions.
Linear differential equations are generally more manageable than nonlinear ones. They can often be solved using a variety of well-established methods, like power series, Laplace transforms, or matrix exponentials, making them a key tool in both applied and theoretical contexts.

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Most popular questions from this chapter

Evaluate the integral \(\oint_{C} f(z) d z\) where \(C\) is the unit circle centered at the origin and \(f(z)\) is given as follows: (a) \(\frac{e^{z}}{z^{3}}\) (b) \(\frac{1}{z^{2} \sin z}\) (c) \(\tanh z\) (d) \(\frac{1}{\cos 2 z}\) (e) \(e^{1 / z}\)

Evaluate the integral \(\oint_{C} f(z) d z\), where \(C\) is a unit circle centered at the origin and where \(f(z)\) is given below. (a) \(\frac{g(z)}{z-w}, g(z)\) entire. (b) \(\frac{z}{z^{2}-w^{2}}\) (c) \(z e^{1 / z^{2}}\) (d) \(\cot z\) (e) \(\frac{1}{8 z^{3}+1}\).

Suppose we are given the equation \(d^{2} w / d z^{2}=2 w^{3}\). (a) Let us look for a solution of the form $$ w=\sum_{n=0}^{\infty} a_{n}\left(z-z_{0}\right)^{n-r}=a_{0}\left(z-z_{0}\right)^{-r}+a_{1}\left(z-z_{0}\right)^{1-r}+\cdots $$ for \(z\) near \(z_{0}\). Substitute this into the equation to determine that \(r=1\) and \(a_{0}=\pm 1\) (b) "Linearize" about the basic solution by letting \(w=\pm 1 /\left(z-z_{0}\right)+v\) and dropping quadratic terms in \(v\) to find \(d^{2} v / d z^{2}=6 v /\left(z-z_{0}\right)^{2}\). Solve this equation (Cauchy-Euler type) to find $$ v=A\left(z-z_{0}\right)^{-2}+B\left(z-z_{0}\right)^{3} $$ (c) Explain why this indicates that all coefficients of subsequent powers in the following expansion (save possibly \(a_{4}\) ) $$ w=\frac{\pm 1}{\left(z-z_{0}\right)}+a_{1}+a_{2}\left(z-z_{0}\right)+a_{3}\left(z-z_{0}\right)^{2}+a_{4}\left(z-z_{0}\right)^{3}+\cdots $$ can be solved uniquely. Substitute the expansion into the equation for \(w\), and find \(a_{1}, a_{2}\), and \(a_{3}\), and establish the fact that \(a_{4}\) is arbitrary. We obtain two arbitrary constants in this expansion: \(z_{0}\) and \(a_{4}\). The solution to \(w^{\prime \prime}=2 w^{3}\) can be expressed in terms of elliptic functions; its general solution is known to have only simple poles as its movable singular points. (d) Show that a similar expansion works when we consider the equation $$ \frac{d^{2} w}{d z^{2}}=z w^{3}+2 w $$ (this is the second Painlev茅 equation (Ince, 1956)), and hence that the formal analysis indicates that the only movable algebraic singular points are poles. (Painlev茅 proved that there are no other singular points for this equation.) (e) Show that this expansion fails when we consider $$ \frac{d^{2} w}{d z^{2}}=2 w^{3}+z^{2} w $$ because \(a_{4}\) cannot be found. This indicates that a more general expansion is required. (In fact, another term of the form \(b_{4}\left(z-z_{0}\right)^{3} \log \left(z-z_{0}\right)\) must be added at this order, and further logarithmic terms must be added at all subsequent orders in order to obtain a consistent formal expansion.)

Show that if \(f(z)\) is meromorphic in the finite \(z\) plane, then \(f(z)\) must be the ratio of two entire functions.

Establish that the function \(\sum_{n=1}^{\infty} \frac{1}{e^{n} n}\) is an analytic function of \(z\) for all \(z\) that is, it is an entire function.
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