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 \(\omega(\tau)\). \(u(\tau, \tau)=\omega(\tau), \quad 2 \leq \tau \leq 4\). The solution is \(u(x, t)=\left(x e^{\alpha t}\right)^{3}\). \(x u_{x}+u_{t}=0\)

Short answer

In conclusion, we have solved the given Cauchy problem involving a partial differential equation. We found the constant \(\alpha = -\frac{1}{C_1'}\), sketched the curve \(\gamma\) as a straight line with slope 1 in the \(xt\)-plane, and determined the function \(\omega(\tau) = \left(\tau e^{-\frac{1}{C_1'} \tau}\right)^3\) for \(2 \leq \tau \leq 4\). Note that the exact form of \(\omega(\tau)\) depends on the value of \(\alpha\), which was not specified in the problem.

Step-by-step solution

Solve the PDE using the method of characteristic equations

For the given PDE, \(xu_x + u_t = 0\), we can use the method of characteristics to rewrite it as a system of ordinary differential equations (ODEs). The characteristic equations are: \begin{align*} \frac{dx}{dt} &= x, \\ \frac{du}{dt} &= -u. \end{align*} Solving the first ODE, we get: $$ \frac{dx}{x} = dt \implies \ln x = t + C_1 \implies x = e^{t + C_1} = e^{t}e^{C_1}, $$ where \(C_1\) is an arbitrary constant. Now, let \(C_1' = e^{C_1}\), then \(x = e^t C_1'\), and the characteristic equations become \(u = C_2(xe^{-t})^3\), where \(C_2\) is another arbitrary constant. The general solution of the PDE can thus be written as: $$ u(x, t) = \left(\frac{x}{C_1' e^t}\right)^3, $$ where \(C_1'\) is an arbitrary constant.

Determine the curve \(\gamma\) and find the values of \(\alpha\) and \(\omega(\tau)\)

Given that \(u(\tau, \tau) = \omega(\tau)\) for \(2 \leq \tau \leq 4\), the function \(u(x,t)\) along this curve is \(u(\tau,\tau) = \left(\frac{\tau}{C_1' e^\tau}\right)^3\) for \(2 \leq \tau \leq 4\). Comparing this solution to \(u(x, t) = \left(x e^{\alpha t}\right)^3\), we have: $$ \left(\frac{\tau}{C_1' e^\tau}\right)^3 = \left(\tau e^{\alpha \tau }\right)^3. $$ Equating the powers of \(\tau\) and \(e^{\tau}\) in the two expressions, we find that \(\alpha = -\frac{1}{C_1'}\). Using the given solution, we rewrite \(u(x, t)\) as: $$ u(x, t) = \left(x e^{-\frac{1}{C_1'} t}\right)^3. $$ Thus, we have determined the constant \(\alpha = -\frac{1}{C_1'}\). The curve \(\gamma\) is the straight line where \(x = t\), so \(\gamma\) needs to be sketched as a straight line with slope equal to 1 in the \(xt\)-plane, starting at \((2, 2)\) and ending at \((4, 4)\). Finally, to determine the function \(\omega(\tau)\), note that: $$ \omega(\tau) = u(\tau, \tau) = \left(\tau e^{-\frac{1}{C_1'} \tau}\right)^3, $$ for \(2 \leq \tau \leq 4\). Since \(\alpha\) is not specified, \(\omega(\tau)\) depends on the value of \(\alpha\).

Key concepts

Partial Differential Equations

Partial Differential Equations (PDEs) are a type of mathematical equation used to describe a wide range of phenomena such as heat conduction, wave propagation, and fluid dynamics. They involve rates of change with respect to multiple variables, which makes them more complex than Ordinary Differential Equations (ODEs) that deal with functions of a single variable.

PDEs are formulated based on known relationships between the variables involved and the rates at which they change. Solutions to PDEs can be quite challenging to find, and they often require sophisticated techniques and assumptions. The solution to a PDE generally depends on boundary conditions or initial conditions, such as the values of the function on a boundary or at a specific time, which are provided by the Cauchy problem.

When solving the Cauchy problem for a PDE, we seek a function that satisfies both the PDE and the initial conditions. The process usually entails determining the shape of the curve where the initial values are given, which can greatly influence the behavior of the solution over time.

Method of Characteristics

The Method of Characteristics is a powerful technique for solving certain types of PDEs, particularly first-order linear and quasi-linear PDEs. It transforms the original PDE into a set of ODEs along curves called characteristics, which can then be solved with standard methods for ODEs.

To apply this method, one must first identify these characteristic curves, which are essentially paths in the domain of the solution where information about the solution propagates. These curves are found by solving a system of ODEs derived from the original PDE. Each characteristic carries along it a constant value of some function, which simplifies the problem.

Once the characteristic curves are found, we can project the initial or boundary conditions onto these curves to find the solution of the PDE. This turns a complex problem in multiple variables into a simpler problem involving a single variable, along the curve.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations that involve functions of a single variable and their derivatives. They describe the relationship between a function and the rates at which it changes and are used to model the dynamics of physical systems.

ODEs are classified by their order, which is determined by the highest derivative present in the equation. A first-order ODE involves only the first derivative of the function, while a second-order ODE involves the second derivative, and so on. The solution to an ODE is a function that satisfies the equation when the function and its derivatives are substituted into it.

To solve an ODE, one can employ various methods, such as separation of variables, integrating factors, or advanced numerical techniques. In the context of PDEs and the Cauchy problem, solving ODEs along characteristic curves is an integral part of finding the overall solution to the PDE.