First Order PDE
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Table of Contents
Problem: 4
Show that the following PDEs \(px-qy=x \), \(px^{2}+q=xz \) are compatible.
Problem Restatement
We wish to show that the two first‐order PDEs
\[ \begin{aligned} & (R)\quad p\,x – q\,y = x, \\ & (T)\quad p\,x^{2} + q = xz, \end{aligned} \]where \(p=z_{x}\) and \(q=z_{y}\), are compatible (i.e. the integrability condition \(z_{xy}=z_{yx}\) holds).
Step 1. Solve for p and q
We view equations (R) and (T) as a linear system in \(p\) and \(q\). From (R):
\[ \begin{align} p\,x – q\,y &= x \quad \Longrightarrow \quad q = \frac{x(p-1)}{y}, \tag{1.1} \end{align} \]Substitute \(q\) from (1.1) into (T):
\[ \begin{align} p\,x^{2} + \frac{x(p-1)}{y} &= xz. \end{align} \]Multiplying through by \(y\) (assuming \(y\neq 0\)) and then dividing by \(x\) (assuming \(x\neq 0\)), we obtain
\[ \begin{align} p(xy+1) = yz + 1 \quad \Longrightarrow \quad p = \frac{yz+1}{xy+1}. \tag{1.2} \end{align} \]Now substitute (1.2) into (1.1) to get \(q\):
\[ \begin{align} q &= \frac{x\left(\frac{yz+1}{xy+1} – 1\right)}{y} = \frac{x\left(\frac{yz+1 – (xy+1)}{xy+1}\right)}{y} \nonumber \\ &= \frac{x\,(yz – xy)}{y\,(xy+1)} = \frac{x(z-x)}{xy+1}. \tag{1.3} \end{align} \]Step 2. Write the PDEs in Evolution Form
Since \(p=z_x\) and \(q=z_y\), equations (1.2) and (1.3) give
\[ \begin{align} z_{x} &= \frac{yz+1}{xy+1}, \tag{2.1} \\ z_{y} &= \frac{x(z-x)}{xy+1}. \tag{2.2} \end{align} \]Step 3. Verify the Compatibility Condition
We now compute the mixed derivatives of \(z\) using (2.1) and (2.2).
Compute \(z_{xy}\)Differentiate (2.1) with respect to \(y\). Write \(\displaystyle A(x,y,z)=\frac{yz+1}{xy+1}\). Then by the chain rule,
\[ \begin{align} z_{xy} &= A_{y} = \frac{\partial A}{\partial y} + \frac{\partial A}{\partial z}\, z_{y}. \end{align} \]First, compute the partial derivatives:
\[ \begin{align} \frac{\partial A}{\partial y} &= \frac{z\,(xy+1) – (yz+1)x}{(xy+1)^2} = \frac{zxy+z – xzy – x}{(xy+1)^2} = \frac{z – x}{(xy+1)^2}, \tag{3.1} \\ \frac{\partial A}{\partial z} &= \frac{y}{xy+1}. \tag{3.2} \end{align} \]Substitute \(z_y\) from (2.2):
\[ \begin{align} z_{xy} &= \frac{z-x}{(xy+1)^2} + \frac{y}{xy+1} \cdot \frac{x(z-x)}{xy+1} \nonumber \\ &= \frac{z-x}{(xy+1)^2} + \frac{xy(z-x)}{(xy+1)^2} \nonumber \\ &= \frac{(z-x)(1+xy)}{(xy+1)^2} = \frac{z-x}{xy+1}. \tag{3.3} \end{align} \] Compute \(z_{yx}\)Differentiate (2.2) with respect to \(x\). Write \(\displaystyle B(x,y,z)=\frac{x(z-x)}{xy+1}\). Then
\[ \begin{align} z_{yx} &= B_{x} = \frac{\partial B}{\partial x} + \frac{\partial B}{\partial z}\, z_{x}. \end{align} \]Compute the partial derivatives (treating \(z\) as independent):
\[ \begin{align} \frac{\partial B}{\partial x} &= \frac{(z-x – x)(xy+1) – x(z-x)y}{(xy+1)^2} = \frac{(z-2x)(xy+1) – xy(z-x)}{(xy+1)^2}. \tag{4.1} \end{align} \]Expanding the numerator:
\[ \begin{align} (z-2x)(xy+1) – xy(z-x) &= zxy + z – 2x^2y – 2x – xzy + x^2y \nonumber \\ &= z – 2x^2y -2x + x^2y \nonumber \\ &= z – x^2y – 2x. \tag{4.2} \end{align} \]So,
\[ \begin{align} \frac{\partial B}{\partial x} = \frac{z – x^2y – 2x}{(xy+1)^2}. \tag{4.3} \end{align} \]Next,
\[ \begin{align} \frac{\partial B}{\partial z} = \frac{x}{xy+1}. \tag{4.4} \end{align} \]Substitute \(z_x\) from (2.1):
\[ \begin{align} z_{yx} &= \frac{z – x^2y – 2x}{(xy+1)^2} + \frac{x}{xy+1}\cdot \frac{yz+1}{xy+1} \nonumber \\ &= \frac{z – x^2y – 2x + x(yz+1)}{(xy+1)^2}. \tag{4.5} \end{align} \]Simplify the numerator:
\[ \begin{align} z – x^2y – 2x + x(yz+1) &= z – x^2y – 2x + xyz + x \nonumber \\ &= z + xyz – x^2y – x \nonumber \\ &= (xy+1)(z-x). \tag{4.6} \end{align} \]Thus,
\[ \begin{align} z_{yx} &= \frac{(xy+1)(z-x)}{(xy+1)^2} = \frac{z-x}{xy+1}. \tag{4.7} \end{align} \]Step 4. Conclude Compatibility
Since \(\displaystyle z_{xy}=\frac{z-x}{xy+1}\) [see (3.3)] and \(\displaystyle z_{yx}=\frac{z-x}{xy+1}\) [see (4.7)], we have \(\displaystyle z_{xy}=z_{yx}\).
Therefore, the two PDEs are compatible.
Summary
We solved the given system for \(p\) and \(q\) in terms of \(z\) and the independent variables. Then, by computing the mixed derivatives \(z_{xy}\) and \(z_{yx}\) using the obtained expressions, we verified that they are equal. This confirms that the integrability condition is satisfied and hence the PDE system is compatible.
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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.
