Problem 12 - Advanced Engineering Mathematics

Question

The electrostatic potential \(u\) between two concentric spheres of radius \(r=1\) and \(r=4\) is determined from $$ \frac{d^{2} u}{d r^{2}}+\frac{2}{r} \frac{d u}{d r}=0, \quad u(1)=50, u(4)=100 $$ Use the method of this section with \(n=6\) to approximate the solution of this boundary-value problem.

Step-by-Step Solution

Answer

The approximate potential function is \(u(r) = \frac{200}{r} + 250\). Apply this to each segment point to get specific values.

Step 1: Understand the Differential Equation

The equation given is a second-order ordinary differential equation (ODE): \[ \frac{d^{2} u}{d r^{2}}+\frac{2}{r} \frac{d u}{d r}=0 \]This is radial Laplace's equation in spherical coordinates for a potential function with spherical symmetry. We are asked to find an approximate solution using numerical methods with \(n=6\). The boundary conditions are \(u(1)=50\) and \(u(4)=100\).

Step 2: Transform the Differential Equation

Rewrite the equation to solve for a first-order equation. Let \(v = \frac{d u}{d r}\). Then the transformed equation will be:\[ \frac{d v}{d r} + \frac{2}{r} v = 0 \]Integrating factor method might be employed to solve for \(v\). The boundary condition will be applied at the end to solve for constants.

Step 3: Solve the First-Order ODE for v

Multiply through by the integrating factor \(r^2\), which gives:\[ r^2 \frac{d v}{d r} + 2rv = 0 \]This can be rewritten as:\[ \frac{d}{d r}(r^2 v) = 0 \]Integrating, we have:\[ r^2 v = C \]Solving for \(v\), we get:\[ v = \frac{C}{r^2} \]

Step 4: Integrate to find u(r)

Recall \(v = \frac{d u}{d r} = \frac{C}{r^2}\), so integrate with respect to \(r\):\[ u(r) = -\frac{C}{r} + D \]where \(C\) and \(D\) are constants to be determined using the boundary conditions.

Step 5: Apply Boundary Conditions

Apply the boundary conditions to determine \(C\) and \(D\):1. \(u(1) = 50: -\frac{C}{1} + D = 50\)2. \(u(4) = 100: -\frac{C}{4} + D = 100\)Solve these equations simultaneously to find \(C = -200\) and \(D = 250\).

Step 6: Equation of u(r)

Substitute \(C\) and \(D\) back into the expression for \(u(r)\):\[ u(r) = -\frac{(-200)}{r} + 250 \]which simplifies to:\[ u(r) = \frac{200}{r} + 250 \]

Step 7: Calculate approximate values with n=6

To approximate \(u(r)\) at specific points, divide the interval \([1, 4]\) into 6 equal segments: \(r_i = 1, 1.5, 2, 2.5, 3, 3.5, 4\). Calculate \(u(r)\) at these points:- \(u(1.5) = \frac{200}{1.5} + 250\)- \(u(2) = \frac{200}{2} + 250\)- \(u(2.5) = \frac{200}{2.5} + 250\)- \(u(3) = \frac{200}{3} + 250\)- \(u(3.5) = \frac{200}{3.5} + 250\)- \(u(4) = \frac{200}{4} + 250\)Calculate these values to provide an approximation.

Key Concepts

Ordinary Differential Equation

An ordinary differential equation (ODE) is a mathematical expression involving derivatives of a function with respect to one variable. In this exercise, the function is the electrostatic potential, denoted by \( u(r) \), and the variable is the radius \( r \). ODEs often describe physical phenomena where changes occur with respect to a single variable.

For instance, in our problem, we are dealing with a second-order ODE:

• \( \frac{d^{2} u}{d r^{2}}+\frac{2}{r} \frac{d u}{d r}=0 \)

This equation arises when analyzing radial Laplace's equation, which we encounter when solving problems with spherical symmetry. Understanding how to transform and solve these equations through various methods is crucial in mathematical physics.

The presence of derivatives indicates how the electrostatic potential changes with radius, while the boundary conditions \( u(1)=50 \) and \( u(4)=100 \) specify the potential values at specific points.

Numerical Methods

Numerical methods are techniques used to approximate solutions to mathematical problems that might not have a straightforward analytical solution. These methods utilize mathematical algorithms to provide approximate values for functions or equations.

In our exercise, we need to approximate the solution of the boundary-value problem by using a numerical approach with \(n=6\). This means subdividing the radius interval, \([1, 4]\), into 6 points: \(1, 1.5, 2, 2.5, 3, 3.5, 4\). By calculating the value of \( u(r) \) at these points, we gain an understanding of how the potential behaves across the region.

Such methods are particularly invaluable in physics and engineering when dealing with complex systems where exact solutions are hard to come by or when computational resources allow for more flexible analyses and solutions.

Integrating Factor Method

The integrating factor method is a technique for solving first-order linear ordinary differential equations. It is used when a direct integration of the equation is not straightforward. In our case, the integrating factor is used to handle the transformed equation for \( v \):

By rewriting the original second-order ODE, we introduce \( v = \frac{d u}{d r} \), simplifying our problem to:

• \( \frac{d v}{d r} + \frac{2}{r} v = 0 \)

The solution involves multiplying the equation by an integrating factor. Here, \( r^2 \) is chosen because it turns our left side into a simple derivative:

• \( \frac{d}{d r}(r^2 v) = 0 \)

Integrating this straightforward expression helps isolate \( v \), revealing terms that are convenient for further integration to find \( u(r) \).

This method highlights the power of assuming strategic forms for known systems to simplify and find solutions—especially when faced with boundary conditions.

Spherical Coordinates

Spherical coordinates are a system of coordinates that are naturally suited to problems where symmetry is about a point, such as spheres and circles. In spherical coordinates, positions are defined with distance from a central point (radius \( r \)), and angles in different planes.

Our differential equation:

• \( \frac{d^{2} u}{d r^{2}}+\frac{2}{r} \frac{d u}{d r}=0 \)

characterizes this radial component, focusing on how potentials like electrostatics behave radially from the center of a sphere.

By using spherical coordinates, we elegantly handle the complexities that arise from spherical symmetry, like naturally occurring variations due to radial distance. This makes spherical systems indispensable in physics, where potential problems often feature heavily in fields such as electromagnetism and fluid dynamics.

Laplace's Equation

Laplace's equation emerges frequently when examining phenomena of potential functions such as temperature distribution or electrostatic potentials. It is a second-order partial differential equation:

• \( abla^2 u = 0 \)

This equation represents situations where we seek equilibrium states, meaning no local "source" effects exist within the field.

In spherical coordinates, it takes a radial form as seen in our exercise:

• \( \frac{d^{2} u}{d r^{2}}+\frac{2}{r} \frac{d u}{d r}=0 \)

Given the boundary conditions formed by the concentric spheres, solving this equation allows us to find the potential that results purely from this setup.

Laplace's equation is foundational in various scientific fields. It describes how influences like temperature or electrostatic potentials evolve in a region, especially when they arise from or act through continuous mediums.