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Chapter 1: Problem 8

If Laplace's equation is satisfied in three dimensions, show that $$ \oiint \nabla u \cdot \hat{\mathbf{n}} d S=0 $$ for any closed surface. (Hint: Use the divergence theorem.) Give a physical interpretation of this result (in the context of heat flow).

Short Answer

Expert verified
The surface integral of the gradient of a scalar function \( u \) over a closed surface \( S \) is zero when Laplace's equation is satisfied. This implies that for a steady-state heat conduction problem without any internal heat source or sink, the total heat flow across any closed surface in the domain is zero.

Step by step solution

01

Write Down Laplace鈥檚 Equation and Simplify

When Laplace's equation is satisfied in a three dimensional situation, we can write the equation as:\[\nabla^2 u = 0\]Laplace鈥檚 equation implies that the divergence of the gradient of \( u \) is zero. That is,\[\nabla \cdot \nabla u = \nabla \cdot \vec{F} = 0\]where \( \vec{F} = \nabla u \). It is important to note that \( \vec{F} = \nabla u = (u_x, u_y, u_z) \).
02

Apply the Divergence Theorem

We apply the divergence theorem which is stated for a solid volume \( V \) with boundary surface \( S \) as:\[\iiint_V \nabla \cdot \vec{F} dV = \oiint_S \vec{F} \cdot \hat{n} dS\]Here, \( \vec{F} = \nabla u \) and \( \hat{n} \) is the outward pointing unit normal vector on the surface \( S \). By putting the divergence of \( \vec{F} = 0 \) in our divergence theorem, we get:\[0 = \oiint_S \vec{F} \cdot \hat{n} dS = \oiint_S \nabla u \cdot \hat{n} dS\]which completes the first part of the problem.
03

Physical interpretation of the result

The above result has a physical interpretation in the context of heat flow. The scalar function \( u \) here could represent temperature in a steady-state heat conduction problem where there is no heat source or sink. The vector \( \nabla u \) would then represent the heat flux vector, and the surface integral would represent the net heat flow out of the volume enclosed by the surface \( S \). Since the Laplacian \( \nabla^2 u = 0 \), it implies that there is no heat source or sink within the volume \( V \), and the net heat flow across the boundary \( S \) has to be zero. This means that in a steady-state heat conduction problem without any internal heat source or sink, the total heat flow across any closed surface in the domain is always zero.

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Key Concepts

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

Divergence Theorem and Its Role
The divergence theorem is a powerful mathematical tool that connects the flow of a vector field through a closed surface to the behavior of the vector field within the enclosed volume. Simply put, it allows us to convert a surface integral into a volume integral and vice versa. This theorem can be expressed as:
  • For a vector field \( \vec{F} \), the volume integral over a region \( V \) is equivalent to the surface integral over the boundary surface \( S \).
  • This can be written as \( \iiint_V abla \cdot \vec{F} \, dV = \oiint_S \vec{F} \cdot \hat{n} \, dS \).
  • Here, \( \hat{n} \) is the unit normal vector pointing outward from the surface \( S \).
Laplace's equation, \( abla^2 u = 0 \), indicates that the divergence of \( abla u \) is zero. When applied to this context, the divergence theorem shows that for any closed surface, the integral of \( abla u \) over that surface is zero.
This indicates that there is no net flow of the vector field out of any closed surface. Understanding this helps clarify many physical phenomena, particularly when discussing flow and conservation principles.
Understanding Steady-State Heat Conduction
Steady-state heat conduction describes a scenario where the temperature distribution within a material is constant over time. In simpler terms, despite continuous heat transfer, the temperature at any given point within the material does not change with time. This condition is represented mathematically by Laplace's equation: \( abla^2 u = 0 \). Here, \( u \) indicates temperature, and the equation implies that there are no internal sources or sinks of heat.
Key points about steady-state heat conduction include:
  • The temperature remains steady, meaning it does not vary regardless of the continuous transfer of heat.
  • There's no accumulation of heat inside the material. Every part of the heat that enters a section of the material exits as well.
  • This principle is crucial when evaluating processes or systems designed to manage heat without undergoing over time changes.
This equation is helpful in solving real-world heat conduction problems where systems achieve a balance, preventing any further temperature changes inside the material.
Exploring the Heat Flux Vector
In the context of heat transfer, the heat flux vector \( \mathbf{q} \) is instrumental in describing the flow of heat through a material. It quantitatively conveys how heat energy moves through a point in space and is typically expressed as the gradient of the temperature \( u \), such that \( \mathbf{q} = -k abla u \), where \( k \) is the thermal conductivity of the material.
The heat flux vector provides crucial insights, such as:
  • The direction of heat flow is given by the direction of the vector: heat flows from higher to lower temperature areas.
  • The magnitude of \( \mathbf{q} \) reflects the rate of heat transfer per unit area.
  • In steady-state conditions, the heat flux vector is constant over time, embodying the wave of heat passing consistently through the material.
By incorporating this concept, we can understand how heat flows in various applications, ensuring systems are engineered to handle the heat transfer efficiently. Understanding \( abla u \) as the heat flux vector in a steady-state system ties together the mathematical frameworks with physical heat-related interpretations, supporting optimal design and analysis.

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Most popular questions from this chapter

Determine the equilibrium temperature distribution inside a circular annulus \(\left(r_{1} \leqslant r \leqslant r_{2}\right):\) (a) if the outer radius is at temperature \(T_{2}\) and the inner at \(T_{1}\). (b) if the outer radius is insulated and the inner radius is at temperature \(T_{1}\).

Derive the heat equation in two dimensions by using Green's theorem, (1.5.16), the two-dimensional form of the divergence theorem.

For conduction of thermal energy, the heat flux vector is \(\phi=-K_{0} \nabla u\). If in addition the molecules move at an average velocity \(\mathbf{V}\), a process called convection, then briefly explain why \(\boldsymbol{\phi}=-\boldsymbol{K}_{0} \nabla \boldsymbol{\nabla}+c \rho u \mathbf{V}\). Derive the corresponding equation for heat flow, including both conduction and convection of thermal energy (assuming constant thermal properties with no sources).

Determine the cquilibrium temperature distribution for a onc-dimensional rod with constant thermal properties with the following sources and boundary conditions: \(\begin{array}{lll}\text { (a) } Q=0, & u(0)=0, & u(L)=T \\ \text { (b) } Q=0, & u(0)=T, & u(L)=0 \\ \text { (c) } Q=0, & \frac{\partial u}{\partial x}(0)=0, & u(L)=T \\ \text { (d) } Q=0, & u(0)=T, & \frac{\partial u}{\partial x}(L)=\alpha \\ \text { (e) } \frac{Q}{K_{0}}=1, & u(0)=T_{1}, & u(L)=T_{2} \\ \text { (f) } \frac{Q}{K_{0}}=x^{2}, & u(0)-T, & \frac{\partial u}{\partial x}(L)=0 \\ \text { (g) } Q=0, & u(0)=T, & \frac{\partial u}{\partial x}(L)+u(L)=0 \\ \text { (h) } Q=0, & \frac{\partial u}{\partial x}(0)-[u(0)-T]=0, & \frac{\partial u}{\partial x}(L)=\alpha\end{array}\) In these you may assume that \(u(x, 0)=f(x)\).

For the following problems, determine an equilibrium temperature distribution (if one exists). For what values of \(\beta\) are there solutions? Explain physically. (a) \(\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}}+1\) \(u(x, 0)=f(x) \quad \frac{\partial u}{\partial x}(0, t)=1 \quad \frac{\partial u}{\partial x}(L, t)=\beta\) (b) \(\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}}\) \(u(x, 0)=f(x) \quad \frac{\partial u}{\partial x}(0, t)=1 \quad \frac{\partial u}{\partial x}(L, t)=\beta\) (c) \(\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}}+x-\beta\) \(u(x, 0)=f(x) \quad \frac{\partial u}{\partial x}(0, t)=0 \quad \frac{\partial u}{\partial x}(L, t)=0\)
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