Problem 3 - Applied Partial Differential Equations

Question

Solve $$ \frac{\partial u}{\partial t}=k \frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right) $$ with \(u(r, 0)=f(r), u(0, t)\) bounded, and \(u(a, t)=0\). You may assume that the corresponding eigenfunctions, denoted \(\phi_{n}(r)\), are known and are complete. (Hint: See Sec. 5.2.2.)

Step-by-Step Solution

Answer

The solution to the partial differential equation is \(u(r,t)=\sum_{n=1}^{∞}B_{n}e^{-kλ_{n}t}\phi_{n}(r)\) where λn and \(\phi_{n}(r)\) are the eigenvalues and eigenfunctions from the radial part, and \(B_{n}\) is determined by the boundary conditions.

Step 1: Apply the method of separation of variables

Assume that the solution can be written as a product of two functions, one of r and one of t. We write \(u(r,t)=T(t)R(r)\). Substitute this into the PDE.

Step 2: Apply the chain rule and separate the variables

Using the chain rule, we write the derivatives and divide through by \(T(t)R(r)\) to get \[\frac{1}{k}\frac{1}{T}\frac{dT}{dt}=\frac{1}{R}\frac{1}{r}\frac{d}{dr}\left(r\frac{dR}{dr}\right)\].\ The left side of this equation only depends on t and the right side only on r. Hence, both sides should be equal to a separation constant. Let us denote this constant with -λ

Step 3: Solve the ODEs

Now we have to solve the ordinary differential equations (ODEs). From the left side we get the ODE for T: \(\frac{dT}{dt}=-kλT\). The general solution for this equation is \(T(t)=e^{-kλt}\).\ Likewise, from the right side we get an ODE for R: \(\frac{1}{r}\frac{d}{dr}\left(r\frac{dR}{dr}\right)=-λR\). From the problem, the eigenfunctions \(\phi_{n}(r)\) are known solutions to this equation.

Step 4: Apply the initial and boundary conditions and generalize solution

We use the initial condition \(u(r,0)=f(r)\). Since at time t=0, our solution must match the initial u(r,0), we have \(R(r)T(0)=f(r)\) which gives us R(r)=f(r). The general solution of the PDE is a superposition (i.e., sums and scalar multiples) of all solutions, hence, the solution of the PDE is:\ \(u(r,t)=\sum_{n=1}^{∞}B_{n}e^{-kλ_{n}t}\phi_{n}(r)\)\ where λn and \(\phi_{n}(r)\) are the eigenvalues and eigenfunctions from the radial part, and \(B_{n}\) is determined by applying the boundary condition \(u(0,t)\) is bounded and \(u(a,t)=0\).

Key Concepts

Separation of Variables

Separation of variables is a mathematical technique used to simplify complex differential equations. In this method, we assume that the solution to the partial differential equation (PDE) can be written as a product of two or more simpler functions, each depending on a single independent variable. For instance, in the heat equation given in the exercise, the solution is presented as a product of a time-dependent function, \( T(t) \), and a radius-dependent function, \( R(r) \). This turns the PDE into a format that allows us to understand and solve it more easily.

• Start by writing the solution as a product: \( u(r, t) = T(t)R(r) \).
• Substitute this form into the original PDE.
• Notice how the equation splits into parts, each dependent on one of the separate variables \( t \) and \( r \).

This approach essentially decouples the variables, enabling us to solve for each function individually and later combining them for the full solution.

Initial and Boundary Conditions

Initial and boundary conditions are critical in solving differential equations as they provide the necessary parameters to determine a unique solution. Initial conditions specify the state of the system at the beginning of the observation (often at \( t = 0 \)), while boundary conditions describe the behavior at the boundaries of the region under consideration. In this problem:

• We are given an initial condition \( u(r, 0) = f(r) \), meaning the temperature distribution at \( t = 0 \) as a function of radius \( r \).
• The boundary condition \( u(0, t) \) is bounded implies the temperature at the center remains finite for any \( t \).
• Another boundary condition \( u(a, t) = 0 \) suggests the temperature at the outer edge is zero for all \( t \).

These conditions help in determining the constants and functions in the solution, making it more applicable to real-world scenarios.

Ordinary Differential Equations (ODEs)

An ordinary differential equation (ODE) involves functions of a single variable and their derivatives. In separation of variables, converting a PDE into ODEs allows us to solve these simpler individual equations first. In this problem:

• The equation for time \( T(t) \) comes from part of the separated PDE: \( \frac{dT}{dt} = -k \lambda T \).
• The solution for this ODE is an exponential function: \( T(t) = e^{-k\lambda t} \).
• The spatial component results in another ODE: \( \frac{1}{r} \frac{d}{dr}\left(r \frac{dR}{dr}\right) = -\lambda R \).

Solving these ODEs individually, then combining the results gives us the overall solution to the original PDE.

Eigenfunctions and Eigenvalues

Eigenfunctions and eigenvalues are key concepts especially when dealing with boundary value problems in linear differential equations. They form the core solutions that satisfy specific conditions:

• Eigenfunctions, denoted as \( \phi_n(r) \) in this exercise, solve the spatial part of the differential equation under the boundary conditions provided (like \( u(a,t)=0 \)). Each eigenfunction corresponds to an eigenvalue \( \lambda_n \).
• These pairs help form a complete set of solutions to the problem.

In our solution:

• The solution is expressed as a series: \( u(r,t) = \sum_{n=1}^{\infty} B_n e^{-k \lambda_n t} \phi_n(r) \).
• The constants \( B_n \) are determined by initial and boundary conditions.

This method leverages the orthogonality of eigenfunctions, facilitating the construction of a general solution that spans the possible states of the system.