Question

Find the steady-state temperature \(u(\rho, \phi)\) in a sphere of unit radius if the temperature is independent of \(\theta\) and satisfies the boundary condition $$ u(1, \phi)=f(\phi), \quad 0 \leq \phi \leq \pi $$ Hint: Refer to Problem 9 and to Problems 22 through 29 of Section \(5.3 .\) Use the fact that the only solutions of Legendre's equation that are finite at both \(\pm 1\) are the Legendre polynomials.

Short answer

#tag_title#Summary#tag_content#: In this problem, we were asked to find the steady-state temperature \(u(\rho, \phi)\) in a unit sphere with the given boundary condition \(u(1, \phi) = f(\phi)\) and independence of temperature on \(\theta\). We utilized Laplace's equation and the separation of variables to transform the problem into two differential equations for \(R(\rho)\) and \(\Phi(\phi)\). The solutions for these equations involved Legendre polynomials and the radial part contained constant coefficients. By applying the given boundary condition, we found that the steady-state temperature in the unit sphere can be expressed as a linear combination of Legendre polynomials with coefficients determined by the function \(f(\phi)\): $$ u(\rho, \phi) = (A \rho^n + B \rho^{-(n+1)})P_n(\cos \phi) $$ Where \(f(\phi) = \sum_{n=0}^{\infty} C_n P_n(\cos \phi)\), and \(C_n\) are the coefficients that can be determined using the given \(f(\phi)\) function.

Step-by-step solution

Revisit Laplace's equation and separation of variables approach

: As mentioned in the hint, we need to refer to Problem 9, which deals with finding solutions of Laplace's equation in spherical coordinates. Laplace's equation in spherical coordinates is given by: $$ \dfrac{1}{\rho^2} \dfrac{\partial}{\partial \rho}\left(\rho^2 \dfrac{\partial u}{\partial \rho}\right) + \dfrac{1}{\rho^2 \sin \phi}\dfrac{\partial}{\partial \phi}\left(\sin \phi \dfrac{\partial u}{\partial \phi}\right) + \dfrac{1}{\rho^2 \sin^2 \phi} \dfrac{\partial^2 u}{\partial \theta^2} = 0 $$ Since the temperature is independent of \(\theta\), the term containing \(\frac{\partial^2 u}{\partial \theta^2}\) can be eliminated: $$ \dfrac{1}{\rho^2} \dfrac{\partial}{\partial \rho}\left(\rho^2 \dfrac{\partial u}{\partial \rho}\right) + \dfrac{1}{\rho^2 \sin \phi}\dfrac{\partial}{\partial \phi}\left(\sin \phi \dfrac{\partial u}{\partial \phi}\right) = 0 $$ We will now proceed with the separation of variables, similar to Problem 9.

Apply separation of variables

: Assume the solution is of the form \(u(\rho, \phi) = R(\rho) \Phi (\phi)\). Substitute this into the simplified Laplace's equation, which simplifies as follows: $$ \dfrac{1}{\rho^2} \dfrac{\partial}{\partial \rho}\left(\rho^2 \dfrac{\partial (R(\rho) \Phi(\phi))}{\partial \rho}\right) + \dfrac{1}{\rho^2 \sin \phi}\dfrac{\partial}{\partial \phi}\left(\sin \phi \dfrac{\partial (R(\rho) \Phi(\phi))}{\partial \phi}\right) = 0 $$ $$ \dfrac{1}{R(\rho)}\dfrac{\partial}{\partial \rho}\left(\rho^2 \dfrac{\partial R(\rho)}{\partial \rho}\right) + \dfrac{1}{\Phi(\phi)\sin \phi}\dfrac{\partial}{\partial \phi}\left(\sin \phi \dfrac{\partial \Phi(\phi)}{\partial \phi}\right) = 0 $$ We can now separate the equation, setting both terms equal to the same negative constant: $$ \dfrac{1}{R(\rho)}\dfrac{\partial}{\partial \rho}\left(\rho^2 \dfrac{\partial R(\rho)}{\partial \rho}\right) = -\lambda $$ $$ \dfrac{1}{\Phi(\phi)\sin \phi}\dfrac{\partial}{\partial \phi}\left(\sin \phi \dfrac{\partial \Phi(\phi)}{\partial \phi}\right) = \lambda $$

Solve the differential equation for \(\Phi(\phi)\)

: First, we will find a general solution for the differential equation for \(\Phi(\phi)\): $$ \dfrac{1}{\Phi(\phi)\sin \phi}\dfrac{\partial}{\partial \phi}\left(\sin \phi \dfrac{\partial \Phi(\phi)}{\partial \phi}\right) = \lambda $$ This differential equation is Legendre's equation, and its general solution is: $$ \Phi(\phi) = P_n(\cos \phi) $$ Where \(P_n(x)\) is the Legendre polynomial of order \(n\) and \(\lambda=n(n+1)\).

Solve the differential equation for \(R(\rho)\)

: Now we will find a general solution for the differential equation for \(R(\rho)\): $$ \dfrac{1}{R(\rho)}\dfrac{\partial}{\partial \rho}\left(\rho^2 \dfrac{\partial R(\rho)}{\partial \rho}\right) = -n(n+1) $$ This differential equation has a general solution: $$ R(\rho) = A \rho^n + B \rho^{-(n+1)} $$ Where \(A\) and \(B\) are constants.

Determine the general solution for \(u(\rho, \phi)\)

: Combining the results from Step 3 and Step 4, we have the general solution for \(u(\rho, \phi)\): $$ u(\rho, \phi) = (A \rho^n + B \rho^{-(n+1)})P_n(\cos \phi) $$

Apply boundary condition and find the steady-state temperature

: Now we apply the given boundary condition, \(u(1, \phi) = f(\phi)\), to find the steady-state temperature: $$ u(1, \phi) = (A + B)P_n(\cos\phi) = f(\phi) $$ This equation implies that the steady-state temperature in the sphere is a linear combination of Legendre polynomials of \(\cos\phi\) with given coefficients that yield the given boundary condition function \(f(\phi)\). Specifically: $$ f(\phi) = \sum_{n=0}^{\infty} C_n P_n(\cos \phi) $$ Where \(C_n\) are the coefficients that can be determined using the given \(f(\phi)\) function. With this, we can find the desired steady-state temperature \(u(\rho,\phi)\) in the unit sphere using the previously found general solution.

Key concepts

Laplace's Equation

In the realm of physics and mathematics, Laplace's equation holds a special place, particularly when it comes to understanding how potential fields, like gravitational and electrostatic fields, behave. When we address the steady-state temperature distribution in a sphere, we are indeed dealing with a potential field, where the temperature is the potential function.

Laplace's Equation in its general form is \[abla^2u = 0\], where \(abla^2\) is the Laplacian operator and \(u\) represents the potential function—in our case, the temperature. The equation effectively says that in a region where there are no internal sources or sinks of heat, the temperature at any point is the average of the temperature in the surrounding points.

In spherical coordinates, this equation becomes particularly important as it neatly adapts to the symmetry of spherical objects, like our unit sphere. The goal of solving Laplace's equation, within this context, is to find a solution that satisfies the given boundary conditions, which essentially define the temperature at the surface of the sphere.

Separation of Variables

The method of separation of variables is a powerful mathematical technique used to solve partial differential equations like Laplace's equation. It involves assuming that the solution can be split into a product of functions, each depending only on a single coordinate. This allows us to break down a complicated multi-variable problem into simpler, single-variable problems.

In our sphere temperature problem, the assumed form of the solution is \(u(\rho, \phi) = R(\rho) \Phi (\phi)\). Through this technique, we separate the variables \(\rho\) and \(\phi\), which represent the radial and angular coordinates, respectively. By dividing the entire equation by the product \( R(\rho) \Phi (\phi) \) we're able to isolate terms depending on \(\rho\) from those depending on \(\phi\), resulting in two ordinary differential equations (ODEs) that are much more manageable.

Legendre Polynomials

Legendre polynomials are a sequence of orthogonal polynomials which arise naturally when solving Laplace's equation in spherical coordinates using the separation of variables. They are particularly useful because they satisfy a specific type of differential equation called Legendre's equation, which carries the same form as part of our separated Laplace's equation.

A key property of Legendre polynomials is that they are finite at \(\pm 1\), which corresponds to the physical requirement that our temperature function must be finite at the poles of the sphere. They form a complete set of functions over the interval \([-1,1]\), which means any reasonable function defined over that interval can be expressed as an infinite series of Legendre polynomials—just like a Fourier series. This property is crucial, as it allows us to express the angular part of our temperature distribution, \(f(\phi)\), as a series of these polynomials.

Spherical Coordinates

Spherical coordinates (\(\rho, \theta, \phi\)) offer a three-dimensional coordinate system that extends polar coordinates into 3D by adding an angle \(\phi\) for the third dimension. Unlike Cartesian coordinates, spherical coordinates are defined by the distance from a central point (\(\rho\)), like the radius of a sphere, and two angles (\(\theta\) and \(\phi\)).

In problems involving spheres, like our temperature example, spherical coordinates are natural and intuitive because they directly mirror the geometry of the sphere. For our steady-state temperature problem, we're only concerned with \(\rho\) and \(\phi\) because of the symmetry about the \(\theta\) angle. This simplification in spherical coordinates significantly eases the complexity of the problem and is a testament to the utility of choosing an appropriate coordinate system for a given problem.

Boundary Conditions

Boundary conditions are a set of constraints that a solution to a differential equation must satisfy at the boundaries of the domain. In physical terms, they are the conditions at the surface of an object or the edges of a region where we know what the state of the system is like. For our temperature distribution problem, the boundary condition is provided by the function \(f(\phi)\), which gives the temperature at the surface of the sphere (\(\rho=1\)).

These conditions are not just arbitrary; they are usually based on physical reality. In this case, the function \(f(\phi)\) could represent, for example, how sunlight heats the surface of a planet, with our steady-state solution representing the final temperature distribution without any further changes with time. When we find our final solution for the temperature inside the sphere, it must match these boundary conditions at the surface to be considered valid. This ensures our mathematical solution has real-world applicability.