Question

In each exercise, a Cauchy problem is given, with initial data specified on a curve \(\gamma\). (a) Sketch the curve \(\gamma\). (b) Determine the values of the parameter \(\tau\), if any, where the transversality condition fails to hold. (c) Assume that \(\omega(\tau)\) is continuously differentiable on the given interval \(\alpha \leq \tau \leq \beta .\) Are all the hypotheses of Theorem \(10.1\) satisfied? If not, which hypotheses do not hold? \(u_{x}-u_{t}=0, u(\cos \tau, \sin \tau)=\omega(\tau), \quad 0 \leq \tau \leq \pi / 2\)

Short answer

Question: Sketch the curve γ, determine the values of τ where the transversality condition fails to hold, if any, and check if all the hypotheses of Theorem 10.1 are satisfied or which do not hold, assuming that ω(τ) is continuously differentiable. Answer: The curve γ is a quarter-circle centered at the origin and ranging from the positive x-axis to the positive y-axis. The transversality condition holds for all values of τ in the interval [0, π/2] as their ratio can never equal -1. All the hypotheses of Theorem 10.1 are satisfied in this case, as the given PDE is a first-order linear quasi-homogeneous PDE, the initial condition is continuously differentiable, and the transversality condition holds for all values of τ.

Step-by-step solution

Find the equation and sketch the curve \(\gamma\)

The curve \(\gamma\) is defined by the initial condition \(u(\cos \tau, \sin \tau) = \omega(\tau)\) for \(0 \leq \tau \leq \pi/2\). To sketch this curve, we can rewrite it in parametric form as follows: $$ x(\tau) = \cos \tau, \quad y(\tau) = \sin \tau $$ This is the parametric equation of a quarter-circle with radius \(1\), centered at the origin \((0,0)\), and ranging from the positive x-axis to the positive y-axis. Hence, we can now sketch the curve \(\gamma\) as a quarter-circle.

Find the equation for characteristic curves

The given partial differential equation (PDE) is: $$ u_x - u_t = 0 $$ To find the equation for the characteristic curves, we can use the method of characteristics. The system of characteristic equations is given by: $$ \frac{dx}{ds} = 1, \quad \frac{dy}{ds} = -1, \quad \frac{du}{ds} = u_x - u_t $$ Solve the first two equations to get the general solution for \(x(s)\) and \(y(s)\): $$ x(s) = s + C_1, \quad y(s) = -s + C_2 $$ Now, we can eliminate the parameter \(s\) to obtain the characteristic curves. We can do this by solving \(s\) from the first equation: $$ s = x - C_1 $$ and then substituting it into the second equation: $$ y = - (x - C_1) + C_2 \Rightarrow y = x - C $$ where \(C = C_1 - C_2\) is an arbitrary constant.

Evaluate the transversality condition

The transversality condition states that at any point on the curve \(\gamma\), the gradients of \(\gamma\) and the characteristic curves must not be equal. The gradient of the curve \(\gamma\) is given by: $$ \frac{dy}{dx}\bigg|_{\gamma} = \frac{dy/d\tau}{dx/d\tau} = \frac{\sin\tau}{\cos\tau} $$ The gradient of the characteristic curve is: $$ \frac{dy}{dx}\bigg|_{Characteristic} = \frac{dy/ds}{dx/ds} = -1 $$ The transversality condition fails if: $$ \frac{\sin\tau}{\cos\tau} = -1 $$ However, since \(\tau\) varies from \(0\) to \(\pi/2\), both \(\sin\tau\) and \(\cos\tau\) are non-negative, and their ratio can never equal \(-1\). Therefore, the transversality condition holds for all values of \(\tau\) in the given interval.

Check the hypotheses of Theorem 10.1

Finally, we need to check if all the hypotheses of Theorem 10.1 are satisfied. Theorem 10.1 concerns the existence and uniqueness of solutions to Cauchy problems under certain conditions. The main hypotheses are: 1. The given PDE must be a first-order linear quasi-homogeneous PDE. 2. The initial condition must be defined on a curve \(\gamma\) and be continuously differentiable. 3. The transversality condition must hold. Our given PDE is \(u_x - u_t = 0\), which is a first-order linear quasi-homogeneous PDE. Thus, the first hypothesis is satisfied. We are given that \(\omega(\tau)\), which represents the initial condition, is continuously differentiable on the interval \(0 \leq \tau \leq \pi/2\). Thus, the second hypothesis is satisfied. In Step 3, we showed that the transversality condition holds for all values of \(\tau\) in the given interval. Thus, the third hypothesis is satisfied. As all the hypotheses of Theorem 10.1 are satisfied, we can conclude that the given Cauchy problem has a unique solution in the domain of \(\tau\).

Key concepts

Transversality Condition

The transversality condition is a crucial concept in solving partial differential equations (PDEs), especially when dealing with Cauchy problems. In the context of a Cauchy problem, we often have a curve \(\gamma\) which defines the initial conditions for the PDE.
For the PDE to be well-posed, and specifically for the method of characteristics to be applicable, we require that the characteristic curves of the PDE must intersect the curve \(\gamma\) transversally, that is, not tangent to it. However, how do we check this mathematically? We must look at the gradients, or slopes, of the curves at their points of intersection.
In the step by step solution provided, we calculated the gradients of both the initial curve \(\gamma\) and the characteristic curves. Given \(\gamma\) as a parameterized curve defined by \(\omega(\tau)\), the gradient is a ratio of derivatives with respect to the parameter \(\tau\). On the other hand, the characteristic curves have a constant gradient because they're straight lines with a slope of \( -1 \).
The condition fails if these gradients are equal, which would mean the curves are tangent—that does not happen in our case since the gradients are never equal within the specified domain. Hence, the transversality condition does indeed hold for all values of \(\tau\), ensuring that the characteristics properly intersect the initial data curve.

Parametric Equations

Parametric equations provide a method to express a curve in terms of an independent parameter, usually denoted \(\tau\) or \(s\). In our given problem, the curve \(\gamma\) is represented using parametric equations, allowing us to describe its shape without limiting ourselves to a direct relationship between \(x\) and \(y\).
In more practical terms, parametric equations help us describe the positioning of a point as it moves and are particularly useful in our scenario where a curve (the quarter-circle) defines the initial conditions of the PDE. Parametric equations offer a dynamic way to account for the motion of points and are essential in converting between different forms of complex curves.
Using parametric equations allows us to apply the method of characteristics more effectively and with greater precision, as we can closely follow how a point that starts out on the curve \(\gamma\) moves according to the dynamics dictated by the PDE.

Method of Characteristics

The method of characteristics is a powerful technique to solve first-order PDEs. It operates by transforming the PDE into a set of ordinary differential equations (ODEs) which describe the paths or 'characteristics' along which the solution to the PDE is constant.
These characteristics are the curves in the given exercise that allowed us to understand how a solution propagates in space and time from the initial data. By solving the associated characteristic ODEs, we derive relationships between the spatial variable \(x\), the time variable \(t\), and the solution \(u\).
Once we have these relationships, it becomes straightforward to thread the initial conditions along these characteristic lines, thereby constructing the solution to the entire Cauchy problem. Understanding the method of characteristics is key to visualizing and analyzing how the information travels across the domain of the PDE.

First-order Linear Quasi-Homogeneous PDE

Our Cauchy problem involves a first-order linear quasi-homogeneous partial differential equation (PDE). First-order indicates that the highest derivatives in the equation are of the first degree. Linear means that the equation involves these derivatives to the first power and without products of them. Quasi-homogeneous relates to the PDE having the same degree of derivatives for each term when considering the weights assigned to each variable.
Being quasi-homogeneous, the PDE preserves the balance of influence between the time and spatial derivatives. In our specific case, \(u_x - u_t = 0\), the derivatives of the solution \(u\) with respect to space (\(x\)) and time (\(t\)) equally contribute to the behavior of \(u\).
The solution to such an equation can be clearly determined as it adheres to the rules set by the theorem referenced, which includes the transversality condition holding. Understanding the nature of quasi-homogeneity helps us recognize patterns in the possible solutions and ensures that we can count on the uniqueness and existence of a solution under the proper conditions.