Question

Exercises 1-6: A Cauchy problem is given in each exercise. (a) Determine the unspecified constant \(\alpha\). (b) The values of \(u\) are prescribed along a curve \(\gamma\) in the \(x t\)-plane. Sketch the curve \(\gamma\). (c) Determine the function \(\omega(\tau)\). \(u_{x}+\alpha t u_{t}=0\) \(u(\tau, 1)=\omega(\tau), \quad 0 \leq \tau<\infty\). The solution is \(u(x, t)=x-2 \ln t, \quad 0

Short answer

Question: Determine the constant \(\alpha\), sketch the curve \(\gamma\), and find the function \(\omega(\tau)\) for the given Cauchy problem. Solution: The constant \(\alpha = \frac{1}{2}\), the curve \(\gamma\) is a horizontal line at \(t = 1\), and the function \(\omega(\tau) = \tau\).

Step-by-step solution

Determine the constant \(\alpha\)

To determine the constant \(\alpha\), we need to plug the solution \(u(x,t)=x-2\ln t\) into the PDE \(u_x+\alpha t u_t=0\). Firstly, we find the partial derivatives of \(u(x, t)\) with respect to \(x\) and \(t\): $$ u_x = \frac{\partial u(x, t)}{\partial x} = 1 $$ and $$ u_t = \frac{\partial u(x, t)}{\partial t} = -\frac{2}{t}. $$ Now, plug these derivatives into the PDE: $$ 1 + \alpha t\left(-\frac{2}{t}\right) = 0. $$ Simplifying, we get $$ 1 - 2\alpha = 0. $$ Solving for \(\alpha\), we have \(\alpha = \frac{1}{2}\).

Sketch the curve \(\gamma\)

The curve \(\gamma\) is where the initial condition is prescribed, which is \(u(\tau, 1) = \omega(\tau)\) for \(0 \leq \tau < \infty\). In the \(xt\)-plane, this means that the curve \(\gamma\) is a horizontal line at \(t = 1\). To sketch it, draw a horizontal line at \(t = 1\) intersecting the \(x\)-axis from \(x = 0\) to \(x = \infty\).

Determine the function \(\omega(\tau)\)

Now, we will find the function \(\omega(\tau)\) using the initial condition \(u(\tau, 1) = \omega(\tau)\) and the solution \(u(x, t) = x - 2 \ln t\). Plug the solution into the initial condition: $$ \omega(\tau) = u(\tau, 1) = \tau - 2 \ln 1. $$ Since \(\ln 1 = 0\), we have $$ \omega(\tau) = \tau. $$ To summarize, the constant \(\alpha = \frac{1}{2}\), the curve \(\gamma\) is a horizontal line at \(t = 1\), and the function \(\omega(\tau) = \tau\).

Key concepts

Partial Differential Equations

Partial differential equations (PDEs) are fundamental in the mathematical modeling of systems involving multiple variables and their rates of change. Unlike ordinary differential equations, which involve derivatives with respect to a single variable, PDEs involve derivatives with respect to more than one independent variable.

For instance, in the given exercise, we have a first-order PDE given by the equation
\( u_{x} + \alpha t u_{t} = 0 \)
where \( u_{x} \) and \( u_{t} \) are the partial derivatives of \( u \) with respect to \( x \) and \( t \), respectively. This particular equation models how the function \( u \), which can be a physical quantity such as temperature or concentration, changes with both position \( x \) and time \( t \), influenced by an unknown constant \( \alpha \).

To solve such an equation, one typically needs to determine the unknown constant and apply initial or boundary conditions to find a unique solution. The solving process often involves finding the characteristic curves along which the PDE reduces to an ordinary differential equation, making it easier to solve.

In the step-by-step solution, we saw how substituting the solution into the PDE and its partial derivatives leads to finding the value of the constant \( \alpha \), step 1. It is an essential part of solving the PDE and clarifies the intrinsically linked nature of these various mathematical components.

Initial Condition

Initial conditions play a crucial role in solving partial differential equations. They allow us to specify the value of the function or its derivatives at a specific point or curve in time and space. With adequate initial and boundary conditions, PDEs usually have unique solutions.

In the context of our example, the initial condition is given along a curve \( \gamma \):
\( u(\tau, 1) = \omega(\tau) \), for \( 0 \leq \tau < \infty \). This means that at time \( t = 1 \), we know the values of \( u \) at every position \( \tau \) along the curve. The initial condition injects a physical interpretation into the mathematical problem, transforming it from an abstraction to a model that can be used to predict real-world behavior.

The step-by-step solution demonstrated that by incorporating the initial condition in step 3, we can unwrap the function \( \omega \), which is the output of the function \( u \) along the initial curve \( \gamma \). This is essential to understanding the evolution of the system as defined by the PDE over time.

Characteristic Curve

Characteristic curves are a key mathematical tool used to solve partial differential equations. They are the paths along which the information from the initial conditions travel, thereby shaping the solution. For first-order PDEs, characteristic curves simplify the problem by reducing the PDE to an ordinary differential equation along these curves.

In our exercise's case, understanding the characteristic curves gives insight into how the prescription of initial conditions along a specific curve, such as \( \gamma \), affects the entire solution of the PDE. However, the characteristic curves are not explicitly covered in the steps provided. To fully understand their importance, one must delve into the theory of first-order PDEs, where the method of characteristics is a standard approach to tackling such equations.

Characteristic curves essentially provide the skeleton of the solution's structure. They often need to be sketched, like in step 2, to visualize how the solution propagates through the problem's domain. This helps students intuitively grasp the dynamic nature of solutions to PDEs, which can significantly aid in conceptualizing and solving more complex problems.