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Chapter 17: Problem 57

Stream function Recall that if the vector field \(\mathbf{F}=\langle f, g\rangle\) is source free (zero divergence), then a stream function \(\psi\) exists such that \(f=\psi_{y}\) and \(g=-\psi_{x}\) a. Verify that the given vector field has zero divergence. b. Integrate the relations \(f=\psi_{y}\) and \(g=-\psi_{x}\) to find a stream function for the ficled. $$\mathbf{F}=\left\langle-e^{-x} \sin y, e^{-x} \cos y\right\rangle$$

Short Answer

Expert verified
Question: Find the stream function for the given vector field \(\mathbf{F}=\langle -e^{-x} \sin y, e^{-x} \cos y\rangle\). Answer: The stream function for the given vector field is \(\psi(x, y) = e^{-x} \cos y + C\), where C is a constant.

Step by step solution

01

Verify zero divergence

To check if the given vector field has zero divergence, we need to compute the following: $$\frac{\partial f}{\partial x}+\frac{\partial g}{\partial y}$$ So let's find these partial derivatives: $$\frac{\partial f}{\partial x} = \frac{\partial (-e^{-x} \sin y)}{\partial x} = e^{-x}\sin y$$ $$\frac{\partial g}{\partial y} = \frac{\partial (e^{-x} \cos y)}{\partial y} = -e^{-x}\sin y$$ Now, let's compute the divergence: $$\text{div}(\mathbf{F}) = \frac{\partial f}{\partial x}+\frac{\partial g}{\partial y} = e^{-x}\sin y - e^{-x}\sin y = 0$$ Since the divergence is zero, we can continue to find the stream function.
02

Integrate f with respect to y

To find the stream function, we'll first integrate the relation \(f=\psi_{y}\) with respect to y: $$\psi_{y} = -e^{-x} \sin y$$ Integrate with respect to y: $$\psi(x, y) = \int (-e^{-x} \sin y) dy = e^{-x} \cos y + h(x)$$ where \(h(x)\) is an integration constant that may vary with x.
03

Integrate g with respect to x

Next, we need to integrate the relation \(g=-\psi_{x}\) with respect to x: $$-\psi_{x} = e^{-x} \cos y$$ Integrate with respect to x: $$-\psi(x, y) = \int (e^{-x} \cos y) dx = -e^{-x} \cos y + k(y)$$ where \(k(y)\) is an integration constant that may vary with y.
04

Combine the results

Now we can equate the two expressions for \(\psi(x,y)\) obtained in steps 2 and 3: $$e^{-x} \cos y + h(x) = -e^{-x} \cos y + k(y)$$ To determine \(h(x)\) and \(k(y)\), let's move all the terms involving x to one side, and all the terms involving y to the other side: $$2e^{-x} \cos y = k(y) - h(x)$$ Since the left side of the equation is equal to the difference of the two functions, one depending on y and one depending on x, we obtain that both sides are equal to a constant. Let's denote this constant as C: $$2e^{-x} \cos y = C$$ Then, the stream function \(\psi(x, y)\) can be expressed as $$\psi(x, y) = e^{-x} \cos y + C$$

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Key Concepts

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

Vector Field
A vector field is a mathematical construct that assigns a vector to every point in a space. Think of it like a weather map showing wind direction and intensity at different locations. Mathematically, in two dimensions, a vector field is represented as \(\mathbf{F}=\langle f, g\rangle\), where \(f\) and \(g\) are the components of the vector field in the x and y directions, respectively.

In the context of a stream function, which represents the flow of a fluid, the vector field describes the velocity of the fluid at any given point. It captures how fast the fluid is moving and in what direction. The components \(f\) and \(g\) of the vector field are derived from the stream function \(\psi\), pointing perpendicular to the flow lines.
Zero Divergence
Zero divergence is a property of a vector field that indicates there are no sources (where the field is created) or sinks (where it vanishes) within the field. Divergence basically measures the net flow of the field out of an infinitesimally small volume.

In our exercise, calculating the divergence of the vector field \(\mathbf{F}\) involves taking the partial derivative of \(f\) with respect to \(x\) and of \(g\) with respect to \(y\), and then adding these two derivatives together. If the sum is zero, this implies that the fluid's density remains constant, which can be characteristic of incompressible fluids. The result from the exercise shows that the divergence is zero, confirming the source-free nature of the vector field in question.
Partial Derivatives
Partial derivatives play a crucial role in the field of multivariable calculus, and they measure how a function changes as each variable is varied while keeping the other variables constant. It鈥檚 like examining the slope of the terrain in onedirection鈥攅ast to west or north to south鈥攔egardless of the sloping in the other direction.

When calculating the divergence of our vector field \(\mathbf{F}=\langle f, g\rangle\), we computed the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial g}{\partial y}\), which tell us how the components of the vector field change in space. Partial derivatives are foundational when dealing with functions of several variables and are essential in finding the stream function in this exercise.
Integration in Calculus
Integration is a fundamental operation in calculus that is essentially the inverse of differentiation. While differentiation measures rates of change, integration accumulates quantities. Think of it as adding up an infinite number of infinitesimally small pieces to find the whole.

In stream function calculus, we use integration to construct the stream function \(\psi\) from the components of the vector field. By integrating \(f\) with respect to \(y\) and \(g\) with respect to \(x\), as shown in the exercise, we find the stream function that represents the flow pattern of the fluid. The indefinite integration introduces a constant of integration, which can be a function of the other variable, as is in this case where \(h(x)\) and \(k(y)\) appear as results of the integration process.

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Most popular questions from this chapter

Alternative construction of potential functions Use the procedure in Exercise 71 to construct potential functions for the following fields. $$\quad \mathbf{F}=\langle-y,-x\rangle$$

Flux across concentric spheres Consider the radial fields \(\mathbf{F}=\frac{\langle x, y, z\rangle}{\left(x^{2}+y^{2}+z^{2}\right)^{p / 2}}=\frac{\mathbf{r}}{|\mathbf{r}|^{p}},\) where \(p\) is a real number. Let \(S\) consist of the spheres \(A\) and \(B\) centered at the origin with radii \(0

Outward normal to a sphere Show that \(\left|\mathbf{t}_{u} \times \mathbf{t}_{\mathbf{v}}\right|=a^{2} \sin u\) for a sphere of radius \(a\) defined parametrically by \(\mathbf{r}(u, v)=\langle a \sin u \cos v, a \sin u \sin v, a \cos u\rangle,\) where \(0 \leq u \leq \pi\) and \(0 \leq v \leq 2 \pi\)

Conservation of energy Suppose an object with mass \(m\) moves in a region \(R\) in a conservative force field given by \(\mathbf{F}=-\nabla \varphi,\) where \(\varphi\) is a potential function in a region \(R .\) The motion of the object is governed by Newton's Second Law of Motion, \(\mathbf{F}=m \mathbf{a},\) where a is the acceleration. Suppose the object moves from point \(A\) to point \(B\) in \(R\) a. Show that the equation of motion is \(m \frac{d \mathbf{v}}{d t}=-\nabla \varphi\) b. Show that \(\frac{d \mathbf{v}}{d t} \cdot \mathbf{v}=\frac{1}{2} \frac{d}{d t}(\mathbf{v} \cdot \mathbf{v})\) c. Take the dot product of both sides of the equation in part (a) with \(\mathbf{v}(t)=\mathbf{r}^{\prime}(t)\) and integrate along a curve between \(A\) and B. Use part (b) and the fact that \(\mathbf{F}\) is conservative to show that the total energy (kinetic plus potential) \(\frac{1}{2} m|\mathbf{v}|^{2}+\varphi\) is the same at \(A\) and \(B\). Conclude that because \(A\) and \(B\) are arbitrary, energy is conserved in \(R\).

What's wrong? Consider the rotation field \(\mathbf{F}=\frac{\langle-y, x\rangle}{x^{2}+y^{2}}\) a. Verify that the two-dimensional curl of \(F\) is zero, which suggests that the double integral in the circulation form of Green's Theorem is zero. b. Use a line integral to verify that the circulation on the unit circle of the vector field is \(2 \pi\) c. Explain why the results of parts (a) and (b) do not agree.
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