Question

In Exercises \(17-28\) define the formal solution of $$ u_{x x}+u_{u y}=0, \quad 0

Short answer

Based on the given partial differential equation (PDE) \(u_{xx} + u_{uy} = 0\) and boundary conditions, we found the formal solution to be \(u(x, y) = f(x - y)\). Furthermore, for the specified values of \(a = 1\), \(b = 2\), and \(f(x) = 3x^3 - 4x^2 + 1\), we obtained the solution of the boundary value problem as \(u(x, y) = 3(x-y)^3 - 4(x-y)^2 + 1\).

Step-by-step solution

Analyze the PDE

Our given PDE is: \(u_{xx} + u_{uy} = 0\). It is a second-order quasi-linear PDE. To solve this PDE, we will use the method of characteristics.

Find the characteristics equations

To apply the method of characteristics, we need to find characteristic equations. The characteristic equations for the given PDE are: $$ \frac{dx}{dt}=u_{y}, \quad \frac{dy}{dt}=u_{x}, \quad \frac{du}{dt}=0 $$

Solve the characteristic equations

Integrate the third equation to get: $$ u(x,y) = C_1(z) $$ where \(C_1(z)\) is an arbitrary function of \(z\). To determine \(z\), we integrate the first two equations as follows: $$ \frac{dy}{dx} = \frac{u_x}{u_y} $$ Now, plug in the derivatives of \(u(x,y)\) using \(u_{xx} + u_{uy} = 0\), we get: $$ \frac{dy}{dx} = -\frac{u_{xx}}{u_{uy}} $$ Let \(z = x + \int \frac{u_{xx}}{u_{uy}}dy\). We get \(z = x - \int \frac{u_{uy}}{u_{uy}}dy = x - y\). Thus, the general solution of the given PDE is: $$ u(x,y) = C_1(x-y) $$

Apply boundary conditions

Apply the boundary condition \(u(x, 0) = f(x)\): $$ C_1(x) = f(x) $$ So, \(u(x, y) = f(x - y)\). Next, apply the boundary condition \(u(x, b) = 0\): $$ 0 = f(x - b) $$ We know \(f(x) = 3x^3 - 4x^2 + 1\), therefore \(f(x - b) = 3(x-b)^3 -4(x-b)^2 +1\). Now, apply the boundary conditions \(u_x(0, y) = 0\) and \(u(a, y) = 0\) to obtain a particular solution that satisfies all the conditions. Notice that applying these boundary conditions won't change the form of \(u(x,y)=f(x-y)\), so the formal solution to the given PDE is \(u(x,y)=f(x-y)\).

Solve boundary value problem

Now, we will solve the boundary value problem for specified values of \(a\), \(b\), and \(f(x)\). Given \(a = 1\), \(b = 2\), and \(f(x) = 3x^3 - 4x^2 + 1\). Substitute these values in the solution \(u(x, y) = f(x - y)\): $$ u(x, y)=3(x-y)^3 - 4(x-y)^2 + 1 $$ This is the solution of the boundary value problem for the specified values of \(a\), \(b\), and \(f(x)\).

Key concepts

Quasi-linear Partial Differential Equations

Quasi-linear partial differential equations (PDEs) are equations that involve unknown functions and their derivatives. These equations are considered "quasi-linear" because they are linear in the highest order derivatives. It allows some nonlinearities in lower-order derivatives or the functions themselves. Quasi-linear PDEs often appear in applications like fluid dynamics, wave propagation, and many more fields.
A useful characteristic of quasi-linear PDEs is that they often transform into simpler or more familiar forms under certain methods, such as the Method of Characteristics. Understanding these properties helps to analyze complex systems where these equations arise.

Method of Characteristics

The Method of Characteristics is a powerful technique to solve certain types of partial differential equations, particularly first-order and some quasi-linear forms. This method transforms a PDE into a set of ordinary differential equations (ODEs) which are often easier to handle. Here's a basic idea of how it works:

• We express the PDE in a way that highlights characteristic curves.
• These curves, or paths, show how information propagates through the field or domain.
• By integrating along these paths, we reduce the PDE to a system of ODEs describing the behavior of the solution along these curves.

In the original exercise, characteristic equations were derived from the PDE using the method. This approach simplified finding the solution by integrating the ODEs along characteristics, ultimately leading to the solution $u(x,y) = f(x-y)$. The Method of Characteristics is extremely useful for handling boundary conditions and understanding the behavior of solutions in quasi-linear problems.

Second-order PDEs

Second-order partial differential equations involve second derivatives of the unknown function and play a crucial role in mathematical physics and engineering. These equations often model physical phenomena such as heat transfer, fluid flow, and elasticity.
In the given exercise, the PDE $u_{xx} + u_{uy} = 0$ is a second-order PDE. Such equations can be elliptic, parabolic, or hyperbolic classification based on the relationships between their coefficients, which heavily influences the method of solution.
The classification helps determine appropriate boundary and initial conditions. Depending on this, we use different mathematical techniques and approaches to solve them. Understanding second-order PDEs enables the solutions to effectively describe and predict physical and engineering problems.

Boundary Conditions Analysis

Boundary conditions are crucial in solving PDEs, as they help determine a unique solution for a given problem. In the context of the exercise, several boundary conditions were imposed, such as $u(x, 0) = f(x)$ and $u(a, y) = 0$.
Analyzing these conditions involves checking their compatibility with the PDE, ensuring the conditions do not contradict each other, and that they decisively "close" the problem such that a unique solution emerges.
These boundary conditions give crucial information to "shape" the solution space. When using the Method of Characteristics, boundary conditions can significantly impact the choice of characteristic curves and can refine the solution form. Understanding and correctly applying boundary conditions are vital to solving boundary value problems effectively.