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Chapter 3: Problem 9

Determine in each case an entire function \(f: \mathbb{C} \rightarrow \mathbb{C}\), which satisfies (a) \(f(0)=1, f^{\prime}(z)=z f(z)\) for all \(z \in \mathbb{C}\), (b) \(f(0)=1, f^{\prime}(z)=z+2 f(z)\) for all \(z \in \mathbb{C}\).

Short Answer

Expert verified
(a) \( f(z) = e^{\frac{z^2}{2}} \). (b) \( f(z) = \frac{5}{4}e^{2z} + \frac{1}{2}z - \frac{1}{4} \).

Step by step solution

01

Analyze the differential equation for part (a)

We are given the differential equation \( f'(z) = z f(z) \) with the initial condition \( f(0) = 1 \). This is a first-order linear differential equation.
02

Solution of the differential equation for part (a)

Let's asume a solution of the form \( f(z) = e^{g(z)} \). Therefore, \( f'(z) = g'(z) e^{g(z)} = z e^{g(z)} \). This implies \( g'(z) = z \). Integrating both sides, we get \( g(z) = \frac{z^2}{2} + C \). Since \( f(0) = 1 \), we find \( C = 0 \). Hence, \( f(z) = e^{\frac{z^2}{2}} \).
03

Verify the solution for part (a)

We substitute \( f(z) = e^{\frac{z^2}{2}} \) in the differential equation and verify that it satisfies \( f'(z) = z e^{\frac{z^2}{2}} \). Therefore, it satisfies both the equation and the initial condition.
04

Analyze the differential equation for part (b)

For part (b), we have \( f'(z) = z + 2 f(z) \) with \( f(0) = 1 \). This is a linear inhomogeneous differential equation.
05

Solve the homogeneous part for part (b)

Consider the homogeneous equation \( f_h'(z) = 2f_h(z) \). The solution is \( f_h(z) = Ce^{2z} \).
06

Find a particular solution for the inhomogeneous equation

For the particular solution, assume \( f_p(z) = Az + B \). Substitute into \( f'(z) = z + 2f(z) \) to find suitable \( A \) and \( B \). We find \( A = \frac{1}{2} \), so \( f_p(z) = \frac{1}{2}z - \frac{1}{4} \).
07

Combine solutions to find the general solution for part (b)

The general solution is \( f(z) = f_h(z) + f_p(z) = Ce^{2z} + \frac{1}{2}z - \frac{1}{4} \).
08

Apply the initial condition for part (b)

Use \( f(0) = 1 \) to find \( C \): \( C - \frac{1}{4} = 1 \), giving \( C = \frac{5}{4} \). So, \( f(z) = \frac{5}{4}e^{2z} + \frac{1}{2}z - \frac{1}{4} \).
09

Verification of the solution for part (b)

Verify \( f(z) = \frac{5}{4}e^{2z} + \frac{1}{2}z - \frac{1}{4} \) satisfies \( f'(z) = z + 2f(z) \) and the initial condition.

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Key Concepts

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

Complex Differential Equations
Complex differential equations involve derivatives of complex-valued functions with respect to complex variables. These equations can be quite challenging because they deal with functions that map a complex number to another complex number, allowing them to behave differently than typical real-valued functions. In our exercise, we encountered complex differential equations in the form of:
  • Part (a): The equation \( f'(z) = z f(z) \) demonstrates a first-order linear differential equation.
  • Part (b): The equation \( f'(z) = z + 2 f(z) \) illustrates a linear inhomogeneous differential equation, where the "inhomogeneous" term is \( z \).
To solve such equations, we commonly use methods like separation of variables, integrating factors, or particular solutions. Mastering these techniques is essential in fields like quantum mechanics and signal processing, where these equations frequently occur.
Initial Value Problem
An initial value problem (IVP) is a type of differential equation paired with a specific value, called the initial condition, which allows us to find a unique solution to the equation. The initial condition is provided to determine the constant that arises from the integration process. In the exercise:
  • For part (a), the initial condition is \( f(0) = 1 \).
  • For part (b), the initial condition remains \( f(0) = 1 \).
This initial value is crucial as it enables us to find the constant terms in expressions like \( g(z) = \frac{z^2}{2} + C \) for part (a), ensuring the solution aligns with the given condition. Solving initial value problems ensures we find the appropriate function behavior at the specified initial point, which can model numerous real-world phenomena accurately.
Linear Inhomogeneous Differential Equation
A linear inhomogeneous differential equation is a linear differential equation that includes a non-zero term on the right-hand side. This term means the equation is not homogeneous and usually requires a slightly more complex solving strategy compared to homogeneous equations.For part (b) in this exercise, we deal with:
  • The equation \( f'(z) = z + 2f(z) \), where the term \( z \) makes it inhomogeneous.
To solve such equations, we follow these steps:
  • Solve the corresponding homogeneous equation (without the inhomogeneous term) to find the homogeneous solution \( f_h(z) = Ce^{2z} \).
  • Find a particular solution \( f_p(z) \) by assuming a simple form, such as \( Az + B \).
  • Combine the homogeneous and particular solutions to get the general solution.
  • Apply initial conditions to determine any constants involved.
Understanding linear inhomogeneous equations is essential because they frequently model systems influenced by external forces, such as electrical circuits with a constant power supply.

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Most popular questions from this chapter

Show: $$ \int_{0}^{\infty} \frac{d x}{1+x^{5}}=\frac{\pi \sqrt{10} \sqrt{5+\sqrt{5}}}{25} \approx 1.069896 \ldots $$ Hint. Let \(\zeta\) be a primitive root of unity of order \(5 .\) The integrand takes on the half-lines \(\\{t ; t \geq 0\\}\) and \(\\{t \zeta ; t \geq 0\\}\) the same values. Compare the integrals along these half-lines. Generalize the exponent 5 to an arbitrary odd natural exponent, and compute the integral.

Give examples of power series with finite radius of convergence \(r \neq 0\), which have respectively one of the following properties: (a) the power series converges on the full boundary of the convergence disk, (b) the power series diverges on the full boundary of the convergence disk, (c) there are at least two convergence points and at least two divergence points on the boundary of the convergence disk.

Let \(D \subset \mathbb{C}\) be a domain, \(\alpha:[0,1] \rightarrow D\) a smooth closed curve, and \(a \notin\) Image \(\alpha\) Show: The winding number $$ \chi(\alpha ; a)=\frac{1}{2 \pi \mathrm{i}} \int_{\alpha} \frac{1}{\zeta-a} d \zeta $$ is always an integer. Hint. Define for \(t \in[0,1]\) $$ G(t):=\int_{0}^{t} \frac{\alpha^{\prime}(s)}{\alpha(s)-a} d s \quad \text { and } \quad F(t):=(\alpha(t)-a) \exp (-G(t)) $$ then compute \(F^{\prime}(t)\), and finally show \(\alpha(t)-a=(\alpha(0)-a) \exp G(t)\) for all \(t \in[0,1]\)

Let us consider the recursively defined FIBONACCI sequence \(\left(f_{n}\right)\) with \(f_{0}=\) \(f_{1}=1\) and \(f_{n}:=f_{n-1}+f_{n-2}\) for \(n \geq 2\) Show: (a) The power series \(f(z):=\sum_{n=0}^{\infty} f_{n} z^{n}\) coincides with the rational function $$ z \mapsto \frac{1}{1-z-z^{2}} $$ (b) For all \(n \in \mathbb{N}_{0}\) we have the following formula of BINET for the FIBONACCI numbers $$ f_{n}=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}. $$

Let \(u\) be a non-constant harmonic function on a domain \(D \subset \mathbb{R}^{2}\). Show that \(u(D)\) is an open interval.
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