Second Order PDE
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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.
