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Chapter 12: Problem 2

Define the formal solution of $$ \begin{array}{r} u_{r r}+\frac{1}{r} u_{r}+\frac{1}{r^{2}} u_{\theta \theta}=0, \quad \rho_{0}

Short Answer

Expert verified
In this problem, we are given a partial differential equation (PDE) with boundary conditions: $$u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^2}u_{\theta\theta}=0$$ with boundary conditions: 1. \(u(\rho_0, \theta)=0\) 2. \(u(\rho, \theta)=f(\theta)\) 3. \(u(r, 0)=0\) 4. \(u(r, \gamma)=0\) Our goal is to find the function \(u(r, \theta)\) that satisfies both the PDE and the given boundary conditions. To do so, we used the separation of variables technique assuming that the solution is of the form: $$u(r, \theta)=R(r)\Theta(\theta)$$ By plugging this form into the PDE and dividing through by the product \(R(r)\Theta(\theta)\), we were able to separate the variables. This left us with two ordinary differential equations (ODEs) to solve; one for \(R(r)\) and another for \(\Theta(\theta)\). After solving these ODEs, we applied the boundary conditions to derive the specific form of the functions and obtain the final solution for the problem: $$u(r, \theta) = R(r)\Theta(\theta)$$ It is important to note that the specific details of solving the ODEs and applying the boundary conditions will depend on the functions and conditions given in each individual problem. However, this step-by-step method provides the general framework for solving such problems.

Step by step solution

01

Assume the solution is separable

We assume that the solution \(u(r, \theta)\) can be represented as the product of two functions, one depending on \(r\) and another depending on \(\theta\). We write: $$u(r, \theta)=R(r)\Theta(\theta)$$
02

Plug the assumed solution into the PDE

Substitute the assumed solution \(u(r, \theta)=R(r)\Theta(\theta)\) into the PDE equation. This yields: $$R''(r)\Theta(\theta)+\frac{1}{r}R'(r)\Theta(\theta)+\frac{1}{r^2}R(r)\Theta''(\theta)=0$$
03

Separate the variables

Divide the equation by the product \(R(r)\Theta(\theta)\) to separate the variables: $$\frac{R''(r)}{R(r)}+\frac{R'(r)}{rR(r)}=-\frac{\Theta''(\theta)}{\Theta(\theta)}$$ Since the left side of the equation depends only on \(r\) and the right side of the equation depends only on \(\theta\), both sides must be equal to a constant, say, \(-k^2\).
04

Solve the ODEs for \(R(r)\) and \(\Theta(\theta)\)

Now, we have two ODEs to solve: 1. For \(R(r)\): $$R''(r)+\frac{1}{r}R'(r)+k^2R(r)=0$$ 2. For \(\Theta(\theta)\): $$\Theta''(\theta)+k^2\Theta(\theta)=0$$ To solve these ODEs, we can use techniques like the Frobenius Method for the \(R(r)\) equation and a simple harmonic oscillator solution for the \(\Theta(\theta)\) equation.
05

Apply the boundary conditions

With the ODEs solved, we obtain the general solutions for \(R(r)\) and \(\Theta(\theta)\). Now we can apply the boundary conditions to determine the coefficients and specific function forms for those general solutions.
06

Combine \(R(r)\) and \(\Theta(\theta)\) for the final solution

Finally, we need to combine our derived solutions for \(R(r)\) and \(\Theta(\theta)\) to provide the formal solution to the original problem. The formal solution will be in the form: $$u(r, \theta) = R(r)\Theta(\theta)$$ This step-by-step method provides the general path to finding a formal solution to the given problem. Note that the specific details of solving ODEs and applying the boundary conditions will depend on the functions and conditions given in each individual problem.

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Key Concepts

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

Separable Solutions
In solving partial differential equations (PDEs), a common technique employed to find solutions is leveraging separable solutions. In our context, assuming the solution can be split into two distinct functions鈥攅ach dependent on a different variable鈥攇reatly simplifies the problem at hand.

By expressing the solution for our PDE as a product, namely, \( u(r, \theta) = R(r)\Theta(\theta) \), we are indicating that the changes in \( r \) affects only \( R(r) \) and the changes in \( \theta \) affects only \( \Theta(\theta) \). Afterward, the original complex PDE turns into a set of simpler ordinary differential equations (ODEs), for which established techniques of solution are available.
Boundary Value Problems
Boundary value problems are a type of differential problem where the function is defined within certain limits and the solution must satisfy the conditions given at the boundaries. They shed light on the behavior of physical systems at the edges of a domain.

Our problem presents a spatial domain bounded in radial and angular directions and the function \( u \) must satisfy certain values at these edges. This is critical because these conditions shape the unique solution of the function within the given area. Without the proper application of these boundary conditions, the solution to a PDE remains indeterminate and loses its ability to model real-life situations effectively.
Ordinary Differential Equations (ODEs)
After separating the variables in a PDE, we usually get a set of ordinary differential equations, which are easier to handle as they involve derivatives with respect to only one variable. These ODEs can be addressed by a variety of methods depending on their type and the boundary conditions.

The transformed ODEs from our problem are second-order linear differential equations, which are a well-studied type and have well-known solution strategies. For instance, the Frobenius Method is an advanced technique that can be used to solve ODEs, especially when they have a singularity at the origin, which is often the case in radial equations like \( R(r) \) in our problem.
Frobenius Method
The Frobenius Method is particularly useful for tackling second-order linear ODEs with variable coefficients that are regular singular points. This method assumes a power series solution and involves calculating coefficients in the series expansion.

It represents a formal approach to dealing with singularities by expanding around them. This involves writing the solution as an infinite series and then calculating each term's coefficient so that the series satisfies the equation. For radial equations like ours, this approach is beneficial for handling the \( \frac{1}{r} \) term, which complicates the equation.
Harmonic Oscillator
The harmonic oscillator is a fundamental concept used to solve second-order differential equations with constant coefficients, which frequently arise in physics. It characterizes systems where the restoring force is directly proportional to the displacement, leading to sinusoidal motion.

In our equation regarding \( \Theta(\theta) \), we are dealing with such a second-order ODE. The solutions to these types of equations are sinusoidal functions like sines and cosines. The natural connection to the harmonic oscillator model simplifies finding an analytical solution that satisfies the angular part of our boundary value problem.

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Most popular questions from this chapter

Use Exercise 17 to solve the initial-boundaryvalue problem. In some of these exercises Theorem 11.3.5(c) or Exercise 11.3.42(b) will simplify the computation of the coefficients in the mixed Fourier cosine series. $$ \begin{array}{l} u_{t t}=u_{x x}, \quad 00, \\ u_{x}(0, t)=0, \quad u(1, t)=0, \quad t>0, \\ u(x, 0)=x^{4}-4 x^{3}+6 x^{2}-3, \quad u_{t}(x, 0)=0, \quad 0 \leq x \leq 1. \end{array} $$

In Exercises \(17-28\) define the formal solution of $$ u_{x x}+u_{u y}=0, \quad 0

Justify defining the formal solution of the initial-boundary value problem $$ \begin{array}{c} u_{t t}=a^{2} u_{x x}, \quad 00 \\ u(0, t)=0, \quad u_{x}(L, t)=0, \quad t>0 \\ u(x, 0)=f(x), \quad u_{t}(x, 0)=g(x), \quad 0 \leq x \leq L \end{array} $$ to be $$ u(x, t)=\sum_{n=1}^{\infty}\left(\alpha_{n} \cos \frac{(2 n-1) \pi a t}{2 L}+\frac{2 L \beta_{n}}{(2 n-1) \pi a} \sin \frac{(2 n-1) \pi a t}{2 L}\right) \sin \frac{(2 n-1) \pi x}{2 L} $$ where $$ S_{M f}(x)=\sum_{n=1}^{\infty} \alpha_{n} \sin \frac{(2 n-1) \pi x}{2 L} \quad \text { and } \quad S_{M g}(x)=\sum_{n=1}^{\infty} \beta_{n} \sin \frac{(2 n-1) \pi x}{2 L} $$ are the mixed Fourier sine series of \(f\) and \(g\) on \([0, L] ;\) that is, $$ \alpha_{n}=\frac{2}{L} \int_{0}^{L} f(x) \sin \frac{(2 n-1) \pi x}{2 L} d x \quad \text { and } \quad \beta_{n}=\frac{2}{L} \int_{0}^{L} g(x) \sin \frac{(2 n-1) \pi x}{2 L} d x . $$

Use Exercise 17 to solve the initial-boundaryvalue problem. In some of these exercises Theorem 11.3.5(c) or Exercise 11.3.42(b) will simplify the computation of the coefficients in the mixed Fourier cosine series. $$ \begin{array}{l} u_{t t}=u_{x x}, \quad 00, \\ u_{x}(0, t)=0, \quad u(1, t)=0, \quad t>0 ,\\\ u(x, 0)=0, \quad u_{t}(x, 0)=x^{4}-4 x^{3}+6 x^{2}-3, \quad 0 \leq x \leq 1. \end{array} $$

Solve the nonhomogeneous initial-boundary value problem. $$ \begin{array}{l} u_{t}=u_{x x}-2, \quad 00 \\ u(0, t)=1, \quad u(1, t)=3, \quad t>0 \\ u(x, 0)=2 x^{2}+1, \quad 0 \leq x \leq 1 \end{array} $$
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