Question

Solve $$ \sin x \frac{\partial u}{\partial x}+\cos x \frac{\partial u}{\partial y}=\cos x $$ subject to (a) \(u(\pi / 2, y)=0\), (b) \(u(\pi / 2, y)=y(y+1)\).

Short answer

Characteristic method and boundary conditions yield \(u=g(-\ln|\cot(x/2)| - \ln|\sec(y)|)\)

Step-by-step solution

Identify the type of PDE

Recognize that the given equation is a first-order linear partial differential equation (PDE).

Use the method of characteristics

To solve the PDE, use the method of characteristics. This involves defining characteristic equations: \[ \frac{dx}{\text{sin}(x)} = \frac{dy}{\text{cos}(x)} = \frac{du}{\text{cos}(x)} \]}, {

Solve the characteristic equations

Integrate each portion of the characteristic equations: \[ \int \frac{dx}{\text{sin}(x)} = \int \frac{dy}{\text{cos}(x)} \], which results in \[ -\ln|\cot(x/2)| = \ln|\sec(y)| \]

Combine these integrals

Combine the integrals to find a relationship between x and y: \[ -\ln|\cot(x/2)| - \ln|\sec(y)| = C \].

Transform into solution form

Convert the relation to return u as: \[ u = g(C) \], where g is some arbitrary function, leading to: \[ u = g(-\ln|\cot(x/2)| - \ln|\sec(y)|) \]

Solve for specific solutions with boundary conditions

Apply the boundary conditions: (a) \(u(\pi / 2, y)=0\): Since \(\sin(\pi/2)=1\) and \(\cos(\pi/2)=0\): Substitute to find function g\[ g(-\ln|\cot((\pi/2)/2)| - \ln|\sec(y)|) = 0 \] which simplifies to the functional form. Repeat for (b) boundary: \(u(\pi / 2, y)=y(y+1)\)

Key concepts

First-Order Partial Differential Equations

First-order partial differential equations (PDEs) involve the first derivatives of the unknown function. In the given exercise, the PDE is \(\rxjsin x \frac{\partial u}{\partial x} + \cos x \frac{\partial u}{\partial y} = \cos x\). These PDEs are crucial in modeling various physical phenomena, such as heat diffusion and wave propagation. Understanding them helps in solving complex systems.
Since the PDE is first-order, it involves derivatives of the dependent variable \(u\) with respect to the independent variables \(x\) and \(y\). The term 'first-order' indicates that the highest degree of differentiation in the equation is one.
For any first-order PDE, solutions typically look for functions that satisfy the given differential relationship between the changing variables.

Boundary Conditions

Boundary conditions are additional constraints that a solution to a PDE must satisfy. In the given exercise, two boundary conditions are provided: (a) \(u(\frac{\pi}{2}, y)=0\) and (b) \(u(\frac{\pi}{2}, y)=y(y+1)\).

• Boundary conditions allow multiple potential solutions to be narrowed down to the one that correctly describes the physical situation.
• They define the value of the solution or its derivatives at specific points or regions, crucial for solving real-world problems.
• For instance, in thermal analysis, boundary conditions could represent the fixed temperature at the surface of an object.

Applying boundary conditions to our solution ensures that it satisfies the constraints derived from the problem's context.

Characteristic Equations

The method of characteristics transforms a PDE into a set of ordinary differential equations (ODEs) known as characteristic equations. For the given PDE, the characteristic equations are derived as follows:
\( \frac{dx}{\sin(x)} = \frac{dy}{\cos(x)} = \frac{du}{\cos(x)}\)

• These equations help us trace paths along which the PDE reduces to an ODE.
• Characteristic equations simplify solving PDEs by converting them into easier-to-handle ODEs.
• Solving these ODEs provides a relationship among the variables, forming the basis of the solution.

In this case, integrating the characteristic equations helps us find the link between \(x\) and \(y\).
This link becomes significant when we apply it to find solutions that satisfy the initial boundary conditions.

Linear PDE

A PDE is termed linear if it can be expressed as a linear combination of the unknown function and its derivatives. The term 'linear' implies that the unknown function and its partial derivatives appear in a linear form, without being multiplied or divided by each other.
In the given problem, the PDE \(\rxjsin x \frac{\partial u}{\partial x} + \cos x \frac{\partial u}{\partial y} = \cos x\) is linear because \(u\) and its partial derivatives \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\) appear in a linear manner.

• Linear PDEs are generally easier to solve compared to nonlinear PDEs.
• Solutions to linear PDEs can often be superimposed to form more complex solutions.
• Understanding linear PDEs builds a foundation for tackling more complicated non-linear scenarios.

Recognizing the linearity of the given PDE allows us to employ standard methods like the method of characteristics for finding solutions.