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Let \(w=f(u)+g(v),\) where \(u=x+i y\) \(v=x-i y,\) and \(i=\sqrt{-1} .\) Show that \(w\) satisfies the Laplace equation \(w_{x x}+w_{y y}=0\) if all the necessary functions are differentiable.

Short Answer

Expert verified
\(w_{xx} = -w_{yy}\) leading to \(w_{xx} + w_{yy} = 0\).

Step by step solution

01

Compute Partial Derivatives

To start, compute the partial derivatives of \(w\) with respect to \(x\) and \(y\). Remember that \(u=x+i y\) leads to \(\frac{\partial u}{\partial x} = 1\) and \(\frac{\partial u}{\partial y} = i\). Similarly, for \(v=x-i y\), we have \(\frac{\partial v}{\partial x} = 1\) and \(\frac{\partial v}{\partial y} = -i\). Use the chain rule to find the derivatives of \(w\).

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Key Concepts

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

Partial Derivatives
In mathematics, a partial derivative represents the rate of change of a function with respect to one of its variables, keeping all other variables constant. When dealing with functions of several variables, partial derivatives are essential tools. They help us understand how a change in one variable affects the entire function. For example, consider the function \(w = f(u) + g(v)\). The variables \(u\) and \(v\) are themselves functions of \(x\) and \(y\), defined as:
  • \(u = x + iy\)
  • \(v = x - iy\)
To find the partial derivative of \(w\) with respect to \(x\), we differentiate the entire expression with respect to \(x\), considering that \(u\) and \(v\) change as \(x\) changes while \(y\) remains constant. Similarly, we calculate \(\frac{\partial w}{\partial y}\) by differentiating while treating \(x\) as constant. These derivatives lay the groundwork for further calculations, including checking if a function satisfies certain differential equations, such as Laplace's equation.
The partial derivatives involve understanding small changes and are foundational in vector calculus and physics.
Chain Rule
The chain rule is a fundamental concept in calculus used for computing the derivative of a composite function. When a function is composed of other functions, like \(w = f(u) + g(v)\), where \(u\) and \(v\) are in turn dependent on \(x\) and \(y\), the chain rule helps us find derivatives with respect to \(x\) and \(y\).
To apply the chain rule, we note that:
  • For \(u = x + iy\), \( \frac{\partial u}{\partial x} = 1 \) and \( \frac{\partial u}{\partial y} = i \).
  • For \(v = x - iy\), \( \frac{\partial v}{\partial x} = 1 \) and \( \frac{\partial v}{\partial y} = -i \).
Now, suppose we want to find \( \frac{\partial w}{\partial x} \), the chain rule tells us: \[ \frac{\partial w}{\partial x} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial g}{\partial v} \cdot \frac{\partial v}{\partial x} \] Applying this rule allows us to understand and simplify the differentiation process of composite functions. It's a crucial technique for solving problems in mathematical analysis, physics, and engineering.
Complex Variables
Complex variables involve numbers of the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, satisfying \(i^2 = -1\). These variables can represent more complex equations and phenomena, crucial in fields like engineering, physics, and applied mathematics. When dealing with complex variables, the functions, such as \(f(u)\) and \(g(v)\), exhibit behavior with respect to both "real" and "imaginary" parts.
  • In our example, the complex variables are \(u = x + iy\) and \(v = x - iy\).
  • These variables simplify the manipulation of equations involving trigonometric and exponential functions.
Complex analysis is an area that extends calculus to the complex plane, providing tools like the Cauchy-Riemann equations and conformal mappings. Ensuring differentiability in the complex sense opens paths to more robust theorems and results, such as satisfying the Laplace equation. In the context of solving the Laplace equation, complex variables enable expressing solutions elegantly and efficiently.

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Most popular questions from this chapter

You plan to calculate the area of a long, thin rectangle from measurements of its length and width. Which dimension should you measure more carefully? Give reasons for your answer.

Value of a \(2 \times 2\) determinant If \(|a|\) is much greater than \(|b| .|c| .\) and \(|d| .\) to which of \(a, b, c,\) and \(d\) is the value of the determinant $$f(a, b, c, d)=\left|\begin{array}{ll}a & b \\\c & d \end{array}\right|$$ most sensitive? Give reasons for your answer.

Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$f(x, y)=x^{2}+y^{3}-3 x y, \quad-5 \leq x \leq 5, \quad-5 \leq y \leq 5$$

Suppose that the Celsius temperature at the point \((x, y, z)\) on the sphere \(x^{2}+y^{2}+z^{2}=1\) is \(T=400 x y z^{2} .\) Locate the highest and lowest temperatures on the sphere.

Just as you describe curves in the plane parametrically with a pair of equations \(x=f(t), y=g(t)\) defined on some parameter interval \(I\), you can sometimes describe surfaces in space with a triple of equations \(x=f(u, v), y=g(u, v), z=h(u, v)\) defined on some parameter rectangle \(a \leq u \leq b, c \leq v \leq d .\) Many computer algebra systems permit you to plot such surfaces in parametric mode. (Parametrized surfaces are discussed in detail in Section 16.5.) Use a CAS to plot the surfaces in Exercises \(77-80 .\) Also plot several level curves in the \(x y\) -plane. $$\begin{array}{l} x=u \cos v, \quad y=u \sin v, \quad z=u, \quad 0 \leq u \leq 2 \\ 0 \leq v \leq 2 \pi \end{array}$$

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