Second Order PDE
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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.
Related Docs
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.
