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.
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.
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