Second Order PDE
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Table of Contents
Problem: 5
Find the characterictic curves of the PDE \[\frac{\partial^{2} z}{\partial y^{2}}-y\frac{\partial^{2} z}{\partial x^{2}} \]
Problem Restatement
We need to find the characteristic curves of the second-order PDE \[ \frac{\partial^{2} z}{\partial y^{2}} – y\,\frac{\partial^{2} z}{\partial x^{2}} = 0. \] Characteristic curves are special curves in the \((x,y)\)–plane along which the PDE may simplify.
Background
For a second-order linear PDE written in the form
\[ \begin{align} A(x,y) \, z_{xx} + 2B(x,y) \, z_{xy} + C(x,y) \, z_{yy} + \text{lower order terms} = 0, \end{align} \]the characteristic curves satisfy the equation
\[ \begin{align} A\,\left(\frac{dy}{dx}\right)^2 – 2B\,\frac{dy}{dx} + C = 0. \tag{1} \end{align} \]In our PDE, we can identify:
\[ \begin{align} A(x,y) &= -y, \quad B(x,y)=0, \quad C(x,y)=1. \end{align} \]Step 1. Form the Characteristic Equation
Substitute the coefficients into (1):
\[ \begin{align} -y\,\left(\frac{dy}{dx}\right)^2 + 1 &= 0. \tag{2} \end{align} \]Solve (2) for \(\frac{dy}{dx}\):
\[ \begin{align} -y\,\left(\frac{dy}{dx}\right)^2 + 1 &= 0 \quad \Longrightarrow \quad y\,\left(\frac{dy}{dx}\right)^2 = 1, \tag{3} \end{align} \] \[ \begin{align} \left(\frac{dy}{dx}\right)^2 &= \frac{1}{y}. \tag{4} \end{align} \]Thus,
\[ \begin{align} \frac{dy}{dx} &= \pm \frac{1}{\sqrt{y}}, \quad y>0. \tag{5} \end{align} \]Step 2. Solve the Differential Equation
Separate variables in (5):
\[ \begin{align} \sqrt{y}\,dy &= \pm\,dx. \tag{6} \end{align} \]Integrate both sides:
\[ \begin{align} \int \sqrt{y}\,dy &= \pm\int dx. \tag{7} \end{align} \]Compute the integrals:
\[ \begin{align} \frac{2}{3}y^{\frac{3}{2}} &= \pm\,x + C, \tag{8} \end{align} \]where \(C\) is an arbitrary constant.
Step 3. Express the Characteristic Curves
We can rewrite (8) as
\[ \begin{align} \frac{2}{3}y^{\frac{3}{2}} \mp x &= C. \tag{9} \end{align} \]Equation (9) represents a family of curves in the \((x,y)\)–plane. These are the characteristic curves for the given PDE.
Summary
Starting with the PDE \(\displaystyle z_{yy} – y\,z_{xx} = 0\), we identified the coefficients \(A=-y\), \(B=0\), and \(C=1\) and set up the characteristic equation \(\displaystyle -y\left(\frac{dy}{dx}\right)^2 + 1 = 0\). Solving this differential equation yielded \(\displaystyle \frac{2}{3}y^{3/2} \mp x = C\), which describes the family of characteristic curves.
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FAQs
Partial Differential Equations
- What is a partial differential equation (PDE)?
A PDE is an equation that involves unknown multivariable functions and their partial derivatives. It describes how the function changes with respect to multiple independent variables.
- How do PDEs differ from ordinary differential equations (ODEs)?
Unlike ODEs, which involve derivatives with respect to a single variable, PDEs involve partial derivatives with respect to two or more independent variables.
- What are the common types of PDEs?
PDEs are generally classified into three types based on their characteristics:
- Elliptic: e.g., Laplace’s equation
- Parabolic: e.g., the heat equation
- Hyperbolic: e.g., the wave equation
- What role do boundary and initial conditions play?
- Boundary conditions specify the behavior of the solution along the edges of the domain.
- Initial conditions are used in time-dependent problems to define the state of the system at the start.
- What methods are commonly used to solve PDEs?
There are several techniques, including:
- Analytical methods like separation of variables, Fourier and Laplace transforms, and the method of characteristics
- Numerical methods such as finite difference, finite element, and spectral methods
- What is the method of separation of variables?
This method assumes that the solution can be written as a product of functions, each depending on only one of the independent variables. This assumption reduces the PDE to a set of simpler ODEs that can be solved individually.
- In which fields are PDEs applied?
PDEs model a wide range of phenomena across various fields including physics (heat transfer, fluid dynamics), engineering (stress analysis, electromagnetics), finance (option pricing models), and more.
- What distinguishes linear from nonlinear PDEs?
- Linear PDEs have terms that are linear with respect to the unknown function and its derivatives, making them more tractable analytically.
- Nonlinear PDEs include terms that are nonlinear, often leading to complex behaviors and requiring specialized methods for solution.
- How do you determine the order of a PDE?
The order of a PDE is defined by the highest derivative (partial derivative) present in the equation. For example, if the highest derivative is a second derivative, the PDE is second order.
- What are some common challenges in solving PDEs?
Challenges include finding closed-form analytical solutions, handling complex geometries and boundary conditions, and the significant computational effort required for accurate numerical solutions.
