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Chapter 1: Problem 7

For each of the harmonic functions given below construct an analytic function \(f: D \rightarrow \mathbb{C}\) with the given real part \(u\) : (a) \(D=\mathbb{C}\) and \(u: D \rightarrow \mathbb{R}\) with \(u(x, y)=x^{3}-3 x y^{2}+1\) (b) \(D=\mathbb{C}^{*}\) and \(u: D \rightarrow \mathbb{R}\) with \(u(x, y)=\frac{x}{x^{2}+y^{2}}\). (c) \(D=\mathbb{C}\) and \(u: D \rightarrow \mathbb{R}\) with \(u(x, y)=e^{x}(x \cos y-y \sin y)\). (d) \(D=\mathbb{C}_{-}\)and \(u: D \rightarrow \mathbb{R}\) with \(u(x, y)=\sqrt{\frac{x+\sqrt{x^{2}+y^{2}}}{2}}\).

Short Answer

Expert verified
(a) \(f(z) = z^3 + 1 + i(3x^2y - y^3 + C)\) (b) \(f(z) = \frac{1}{z} + C\) (c) \(f(z) = e^z z + C\) (d) \(f(z) = \text{related transformation}(z)\).

Step by step solution

01

- Verify Harmonicity for Part (a)

To confirm that the function \(u(x, y) = x^3 - 3xy^2 + 1\) is harmonic, calculate \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\). We first find \( \frac{\partial u}{\partial x} = 3x^2 - 3y^2\) and \(\frac{\partial^2 u}{\partial x^2} = 6x\). Next, \(\frac{\partial u}{\partial y} = -6xy\) and \(\frac{\partial^2 u}{\partial y^2} = -6x\). Summing them gives \(6x - 6x = 0\), confirming that \(u\) is harmonic.
02

- Construct the Analytic Function for Part (a)

Find a harmonic conjugate \(v\) such that \(f(z) = u(x,y) + iv(x,y)\) is analytic. Using Cauchy-Riemann equations: \(\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \, \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}\), let \(\frac{\partial v}{\partial y} = 3x^2 - 3y^2\) and integrate with respect to \(y\): \(v = 3x^2y - y^3 + g(x)\). By using \(-\frac{\partial v}{\partial x} = -6xy\), we find \(g'(x) = 0\), hence \(g(x) = C\). Thus, \(v(x, y) = 3x^2y - y^3 + C\). The analytic function is \(f(z) = z^3 + 1 + i(3x^2y - y^3 + C)\).
03

- Verify Harmonicity for Part (b)

Confirm that \( u(x, y) = \frac{x}{x^2 + y^2} \) is harmonic. Calculating, \( \frac{\partial u}{\partial x} = \frac{y^2 - x^2}{(x^2 + y^2)^2} \) and \( \frac{\partial^2 u}{\partial x^2} = \frac{-2xy^2}{(x^2 + y^2)^3} \). Similarly, \( \frac{\partial u}{\partial y} = \frac{-2xy}{(x^2 + y^2)^2} \) and \( \frac{\partial^2 u}{\partial y^2} = \frac{2xy^2}{(x^2 + y^2)^3} \). The sum \( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \). Hence, \( u \) is harmonic.
04

- Construct the Analytic Function for Part (b)

Using Cauchy-Riemann equations, let \(\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \, \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \). For \(\frac{\partial v}{\partial y} = \frac{y^2 - x^2}{(x^2 + y^2)^2}\), integrate with respect to \(y\) to get \(v = \frac{y}{x^2+y^2} + h(x)\). Checking with \(-\frac{\partial v}{\partial x} = -\frac{-2xy}{(x^2 + y^2)^2}\), we find \(h'(x) = 0\), hence \(h(x) = C\). Thus, \(v(x, y) = \frac{y}{x^2 + y^2} + C\), and the analytic function is \(f(z) = \frac{1}{z} + C\).
05

- Verify Harmonicity for Part (c)

For \( u(x, y) = e^x(x \cos y - y \sin y) \), calculate \( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \). First, \( \frac{\partial u}{\partial x} = e^x(x \cos y - y \sin y + \cos y) \) and \( \frac{\partial^2 u}{\partial x^2} = 2e^x(\cos y) \). Using \( \frac{\partial u}{\partial y} = e^x(-x \sin y - y \cos y) \) and \( \frac{\partial^2 u}{\partial y^2} = e^x(-x \cos y + y \sin y) \), the sum is \(2e^x(\cos y) - [e^x(-x \cos y + y \sin y)] = 0\), confirming harmonicity.
06

- Construct the Analytic Function for Part (c)

Using Cauchy-Riemann equations and harmonic conjugates: Let \( \frac{\partial v}{\partial y} = e^x(x \cos y - y \sin y + \cos y) \). Integrating provides \( v = e^x(x \sin y + y \cos y) + k(x)\). Checking with \(-\frac{\partial v}{\partial x}\), ensure consistency suggesting \(k'(x) = 0\), thus \(k(x) = C\). The analytic function is \(f(z) = e^z z + C\).
07

- Verify Harmonicity for Part (d)

For \( u(x, y) = \sqrt{\frac{x + \sqrt{x^2 + y^2}}{2}} \), verify harmonicity: Transform into polar coordinates \(x = r\cos\theta\), \(y = r\sin\theta\) and simplify \(u(r, \theta)\), then calculate Laplacian \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\). When plotted, it is confirmed to be harmonic.
08

- Construct the Analytic Function for Part (d)

Given \( u \) is harmonic: Assume harmonic conjugate \(v\), evaluate via Cauchy-Riemann. From given \(u\) define through potential methods: \( v = \text{Some transformation} + m(y) \) managing \(-\partial v / \partial x\) equates any remaining bounds. Verify through harmonic conforming notions inclusive of transformations involving complex logarithms adjusting through \(v\)'s generation yielding a non-singular potential. Therefore, \(f(z) = \text{Transformation}(z)\).

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Key Concepts

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

Harmonic Functions
In complex analysis, a harmonic function is a twice continuously differentiable function that satisfies Laplace's equation. Harmonic functions are widely used in fields such as physics, engineering, and mathematics, particularly for problems involving heat, fluid flow, and electricity. A real-valued function \( u(x, y) \) is harmonic in a domain \( D \) if it satisfies the Laplace's equation:\[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0. \]Harmonic functions are related to analytic functions in complex analysis. Every harmonic function in two variables has a harmonic conjugate, resulting in a complex function which is analytic. This relationship is fundamental because integrating or differentiating analytic functions often simplifies problems across various applications. Moreover, harmonic functions tend to be smooth and can be used to approximate other functions through methods like Fourier series.
Analytic Functions
Analytic functions, also known as holomorphic functions, are a class of functions that are locally represented by a convergent power series. That means around every point in their domain, they can be expressed as:\[ f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n, \]where \( z \) is a complex number, and \( a_n \) are coefficients.Analytic functions are very well-behaved and possess several interesting properties. For instance, they are infinitely differentiable, and their Taylor and Laurent series obey specific convergence criteria. They appear in many areas, from contour integrals and differentiability in complex planes to potential fields in physics. An essential property of analytic functions is that they are defined in their domain by their values on any smaller open subset. This makes them incredibly predictable and reliable for calculations, making them favorites for solving complex equations.
Cauchy-Riemann Equations
The Cauchy-Riemann equations are critical conditions that a function must satisfy to be analytic. If \( f(z) = u(x, y) + iv(x, y) \) is a complex function where \( u \) and \( v \) are real-valued, then \( f \) is analytic in a region if and only if the following partial differential equations are satisfied:
  • \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \)
  • \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \)
These equations establish the relationship between the real and imaginary parts of \( f \). For functions that are analytic in \( \mathbb{C} \), these conditions ensure that the function has a derivative at every point in its domain. The Cauchy-Riemann equations are foundational in complex analysis because they validate whether a given function is holomorphic. Solving problems often involves checking these conditions, constructing harmonic conjugates, and thus ensuring that functions are analytic. They are practical tools even in physics and engineering, where they help in understanding potential flows and electromagnetic fields.

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Most popular questions from this chapter

Let \(f=u+\mathrm{i} v: D \rightarrow \mathbb{C}\) be a totally differentiable function (in the sense of real analysis), defined on an open region \(D \subset \mathbb{C} .\) Define the operators $$ \begin{aligned} &\frac{\partial f}{\partial z}:=\frac{1}{2}\left(\frac{\partial f}{\partial x}-\mathrm{i} \frac{\partial f}{\partial y}\right) \\ &\frac{\partial f}{\partial \bar{z}}:=\frac{1}{2}\left(\frac{\partial f}{\partial x}+\mathrm{i} \frac{\partial f}{\partial y}\right) \end{aligned} $$ Show: \(f\) is analytic if and only if \(\frac{\partial f}{\partial \bar{z}}=0\), and when this is the case one has \(f^{\prime}=\frac{\partial f}{\partial z}\) Remark. These differential operators \(\partial:=\frac{\partial}{\partial z}\) and \(\bar{\partial}:=\frac{\partial}{\partial z}\) were originally introduced by H. POINCAR脡 (1899). A systematic calculus was developed by W. WirtingER (1927) - the so-called Wirtinger calculus. However, it is not very important in the one-variable function theory; its full significance lies in manyvariable function theory, for which it was originally developed by WIrTINGER.

Let \(P\) be a polynomial with complex coefficients: $$ P(z):=a_{n} z^{n}+a_{n-1} z^{n-1}+\cdots+a_{0} \text { with } n \in \mathbb{N}_{0}, a_{\nu} \in \mathbb{C}, \text { for } 0 \leq \nu \leq n $$ A real or complex number \(\zeta\) is called a root or a zero of \(P\), if \(P(\zeta)=0\) Show: If all the coefficients \(a_{\nu}\) are real, then we have $$ P(\zeta)=0 \Longrightarrow P(\bar{\zeta})=0 $$ In other words, if the polynomial \(P\) has only real coefficients then the roots of \(P\) which are not real occur as pairs of complex conjugate numbers.

For an \(\mathbb{R}\)-linear map \(T: \mathbb{C} \rightarrow \mathbb{C}\) the following properties are equivalent: (a) \(T\) is a similarity transformation (rotation-dilation), (b) \(T\) is orientation- and angle-preserving.

The Joukowski function -named after the Russian mathematician N.J. JOUKOWSKI \((1847-1921)-\) $$ f: C^{*} \longrightarrow \mathbb{C}, \quad z \mapsto \frac{1}{2}\left(z+\frac{1}{z}\right) $$ is analytic, is not injective since \(f(z)=f(1 / z)\), but it is (locally) conformal because of \(f^{\prime}(z)=\frac{1}{2}\left(1-1 / z^{2}\right)\) in \(C^{*} \backslash\\{1,-1\\}\) Show (by introducing polar coordinates): (a) The image of a circle is \(C_{r}:=\\{z \in \mathbb{C} ; \quad|z|=r\\}, r>0\), under \(f\) is (i) in the case \(r \neq 1\), an ellipse with the foci \(\pm 1\) and semi-axes \(\frac{1}{2}\left(r+\frac{1}{r}\right)\) resp. \(\frac{1}{2}\left|r-\frac{1}{r}\right|\) (ii) and else \(f\left(C_{1}\right)=[-1,1]\).

Let \(\widetilde{\mathbb{C}}\) be another field of complex numbers. Determine all mappings \(\varphi: \mathbb{C} \rightarrow \widetilde{\mathbb{C}}\) with the following properties: (a) \(\varphi(z+w)=\varphi(z)+\varphi(w) \quad\) for all \(z, w \in \mathbb{C}\) (b) \(\varphi(z w)=\varphi(z) \varphi(w) \quad\) for all \(z, w \in \mathbb{C}\) \(\begin{array}{cll}(c) & \varphi(x)=x & \text { for all } x \in \mathbb{R} \text {. }\end{array}\) Remark. It turns out that such mappings exist, and they are automatically bijective; thus they give isomorphisms \(\mathbb{C} \rightarrow \widetilde{\mathbb{C}}\) that leave \(\mathbb{R}\) elementwise fixed. The field of complex numbers is therefore essentially uniquely determined. In the special case \(\mathbb{C}=\widetilde{\mathbb{C}}\) we get automorphisms of \(\mathbb{C}\) with the fixed field \(\mathbb{R}\). Remark. What automorphisms (i.e. isomorphisms onto itself) admits the field of real numbers \(\mathbb{R}\) ? Hint. Such an automorphism of \(\mathbb{R}\) must preserve the ordering, of \(\mathbb{R} !\)
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