Second Order PDE
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Table of Contents
Problem: 6
Solve \[\frac{\partial^{2} z}{\partial x^{2}}-2\frac{\partial^{2} z}{\partial x\partial y}+\frac{\partial^{2} z}{\partial y^{2}}=0 \]
Problem Restatement
We wish to solve the second-order PDE \[ \frac{\partial^2 z}{\partial x^2} – 2\,\frac{\partial^2 z}{\partial x\partial y} + \frac{\partial^2 z}{\partial y^2} = 0. \]
Background and Key Concept
Notice that the left-hand side can be factored as \(\displaystyle \left(\frac{\partial}{\partial x} – \frac{\partial}{\partial y}\right)^2 z\). That is,
\[ \begin{align} \left(\frac{\partial}{\partial x} – \frac{\partial}{\partial y}\right)^2 z &= z_{xx} – 2\,z_{xy} + z_{yy} = 0. \tag{1} \end{align} \]Recognizing this factorization allows us to simplify the problem considerably.
Step 1. Factorization and Change of Variables
Since (1) implies \(\displaystyle \left(\frac{\partial}{\partial x} – \frac{\partial}{\partial y}\right)^2 z = 0\), it follows that the derivative \(\displaystyle w(x,y)=\left(\frac{\partial}{\partial x} – \frac{\partial}{\partial y}\right) z\) must satisfy \(\displaystyle \left(\frac{\partial}{\partial x} – \frac{\partial}{\partial y}\right)w=0\). Equivalently, \(w\) is constant along the characteristics of the operator \(\partial_x-\partial_y\).
Alternatively, we introduce new variables: \(\displaystyle u=x-y\) and \(\displaystyle v=x+y\). Then, using the chain rule,
\[ \begin{align} \frac{\partial}{\partial x} &= \frac{\partial u}{\partial x}\frac{\partial}{\partial u} + \frac{\partial v}{\partial x}\frac{\partial}{\partial v} = 1\cdot \frac{\partial}{\partial u} + 1\cdot \frac{\partial}{\partial v}, \tag{2.1} \\ \frac{\partial}{\partial y} &= \frac{\partial u}{\partial y}\frac{\partial}{\partial u} + \frac{\partial v}{\partial y}\frac{\partial}{\partial v} = (-1)\cdot \frac{\partial}{\partial u} + 1\cdot \frac{\partial}{\partial v}. \tag{2.2} \end{align} \]Thus,
\[ \begin{align} \frac{\partial}{\partial x} – \frac{\partial}{\partial y} &= \left(\frac{\partial}{\partial u}+\frac{\partial}{\partial v}\right) – \left(-\frac{\partial}{\partial u}+\frac{\partial}{\partial v}\right) = 2\,\frac{\partial}{\partial u}. \tag{2.3} \end{align} \]Therefore, the operator in (1) becomes
\[ \begin{align} \left(2\,\frac{\partial}{\partial u}\right)^2 z = 4\,z_{uu} = 0. \tag{2.4} \end{align} \]Step 2. Solve the Transformed Equation
Equation (2.4) implies
\[ \begin{align} z_{uu} = 0. \tag{3.1} \end{align} \]The general solution of this ordinary differential equation with respect to \(u\) (with \(v\) as a parameter) is
\[ \begin{align} z(u,v) = A(v)\,u + B(v), \tag{3.2} \end{align} \]where \(A(v)\) and \(B(v)\) are arbitrary functions of \(v\).
Step 3. Express the General Solution in Original Variables
Recall the change of variables: \(\displaystyle u=x-y\) and \(\displaystyle v=x+y\). Thus, the solution (3.2) becomes
\[ \begin{align} z(x,y) = A(x+y)\,(x-y) + B(x+y). \tag{4.1} \end{align} \]This is the general solution of the given PDE.
Summary
By recognizing the factorization \(\displaystyle z_{xx}-2z_{xy}+z_{yy}=(\partial_x-\partial_y)^2z\), we introduced the variables \(u=x-y\) and \(v=x+y\) to simplify the PDE to \(z_{uu}=0\). Integrating twice with respect to \(u\) yielded \(\displaystyle z(u,v)=A(v)\,u+B(v)\), which translates back to the general solution \(\displaystyle z(x,y)=A(x+y)(x-y)+B(x+y)\).
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.
