Question

(a) Solve the Cauchy problem consisting of the given nonhomogeneous equation together with the supplementary condition \(u(\tau, 0)=e^{-\tau^{2}},-\infty<\tau<\infty\). (b) Consider the upper half of the \(x t\)-plane, \(R=\\{(x, t):-\infty

Short answer

(a) To find the solution for the given Cauchy problem, we followed the method of characteristics, resulting in the following expression for u(x, t): \(u(x, t) = e^{-\frac{1}{2}s^2 + (x + \ln(t))s} \int(-s + x + \ln(t)) e^{\frac{1}{2}s^2 - (x + \ln(t))s} ds + k_3(\tau)\). Using the initial condition \(u(\tau, 0) = e^{-\tau^2}\), we find the value of \(k_3(\tau)\) and the final solution for \(u(x, t)\). (b) The region in the upper half of the \(x t\)-plane where the solution \(u(x, t)\) exists is given by \(R = \{(x, t): -\infty < x < \infty, 0 \leq t < \infty\}\), which is the entire upper half of the \(x t\)-plane.

Step-by-step solution

Derive the characteristic equations

We are given the PDE: \(t u_{x} - u_{t} = x\). To derive the characteristic equations, we'll first write down the parametric equations for the characteristic curves: \[ \frac{dt}{ds} = t, \quad \frac{dx}{ds} = -1, \quad \frac{du}{ds} = xu. \] Now we can solve each of these ODEs individually.

Solve the first ODE for t(s)

The first ODE is: \[ \frac{dt}{ds} = t. \] This is a first-order linear ODE, and its general solution can be found by separation of variables: \[ \frac{dt}{t} = ds \quad \Rightarrow \quad \ln t = s + \ln k_1 \quad \Rightarrow \quad t(s) = k_1 e^{s}. \]

Solve the second ODE for x(s)

The second ODE is: \[ \frac{dx}{s}= -1. \] This is a first-order linear ODE as well: \[ dx = -ds \quad \Rightarrow \quad x = -s + k_2. \]

Solve the third ODE for u(x, t)

The third ODE is: \[ \frac{du}{ds} = xu. \] This is a first-order linear ODE in terms of \(u\) and \(s\), with an integrating factor of \(e^{-\int x ds}\). After multiplying both sides by the integrating factor and integrating, we obtain: \[ u(s) = e^{\int x ds} \int x e^{-\int x ds} ds + k_3. \] Now we express \(x\) in terms of \(s\) using the result from step 3: \(x = -s + k_2\), and thus: \[ u(s) = e^{-\frac{1}{2}s^2 + k_2 s} \int(-s + k_2) e^{\frac{1}{2}s^2 - k_2 s} ds + k_3. \]

Eliminate the parameter s

We want to eliminate the parameter \(s\) from the equations obtained in steps 2, 3, and 4. Using step 2, we find \(s = \ln(\frac{t}{k_1})\). Plug this expression into the equation from step 3: \[ x = -\ln\left(\frac{t}{k_1}\right) + k_2. \] Now, solve for \(k_1\) and \(k_2\): \[ k_1 = \frac{t}{e^{-x + k_2}}, \] \[ k_2 = x + \ln(t). \]

Apply the initial condition and solve for u(x, t)

We are given the initial condition \(u(\tau, 0) = e^{-\tau^2}\). When \(t = 0\), we have \(k_1 = 0\). Thus, we find that \(k_2 = \tau\). Plug this into the equation for \(u(s)\) obtained in step 4: \[ u(x, t) = e^{-\frac{1}{2}s^2 + (x + \ln(t))s} \int(-s + x + \ln(t)) e^{\frac{1}{2}s^2 - (x + \ln(t))s} ds + k_3(\tau). \] Using the initial condition \(u(\tau, 0) = e^{-\tau^2}\), we find the value of \(k_3(\tau)\) and the final solution for \(u(x, t)\).

Determine the region of existence of the solution

We want to determine the region in the upper half of the \(x t\)-plane where the solution \(u(x, t)\) exists. From step 5, we found that \(x = -\ln\left(\frac{t}{k_1}\right) + k_2\). We also know that \(t > 0\) and \(k_2 = x + \ln(t)\). Consequently, the solution exists for any \((x, t)\) such that \(t > 0\) and \(x \in (-\infty, \infty)\). Thus, the solution exists on the entire upper half of the \(x t\)-plane, \(R = \{(x, t): -\infty < x < \infty, 0 \leq t < \infty\}\).

Key concepts

Partial Differential Equations

Partial Differential Equations (PDEs) are equations that involve rates of change with respect to more than one variable. They are fundamental in describing various phenomena found in engineering, physics, and finance. Different from ordinary differential equations, which deal with functions of a single variable, PDEs involve multiple independent variables.

• Order: The order of the highest derivative in the equation determines the order of a PDE.
• Linearity: A PDE may be linear or nonlinear, influencing solution methods and behavior.

PDEs are used to formulate problems involving functions of several variables and are often employed in simulations and computations.
A common application is modeling waves or heat distributions, like in our given problem: \[ t u_{x} - u_{t} = x.\] Here, it describes the relationship between changes in time and space, with an added term representing an external influence or source.

Characteristic Equations

Characteristic equations are used to find solutions to PDEs, particularly first-order linear PDEs. They help identify characteristic curves along which the PDE simplifies, turning into ordinary differential equations (ODEs).When solving a PDE, we derive parametric equations that describe these curves. For example, in our problem, the characteristic equations are derived as follows:

• \[ \frac{dt}{ds} = t \]
• \[ \frac{dx}{ds} = -1 \]
• \[ \frac{du}{ds} = xu \]

Solving these, each characteristic equation provides insights into how the values evolve along these special curves. This approach reduces a complex PDE into simpler ODEs, which are often easier to handle and solve.Understanding characteristic equations is key to solving Cauchy problems because they chart the path of the solution across the domain defined by the PDE.

Initial Conditions

Initial conditions specify the value of the solution at a particular point or curve, crucially influencing the solution to a differential equation. For PDEs, initial conditions like \( u(\tau, 0)=e^{-\tau^{2}} \) provide starting values necessary for integrating the solution over a given domain.Initial conditions relate our abstract solutions back to real-world problems by anchoring them at known points. They are essential in making sure the solutions to PDEs are unique and suitable for specific engineering or physical contexts. In our specific case, the condition given helps to determine the arbitrary constants introduced during the integration process of the PDE into ODEs. Without these conditions, solutions can be indeterminate or non-unique, rendering them practically useless.

Nonhomogeneous Equation

A nonhomogeneous equation has terms that are not solely dependent on the function and its derivatives. It typically includes external forces or sources, and is distinct from a homogeneous equation which equals zero throughout.The term \( x \) in our PDE \( t u_{x} - u_{t} = x \) makes it nonhomogeneous, introducing an additional challenge in solving it due to the presence of external influences.Such equations model systems that have external inputs or inhomogeneities. Solving these requires techniques like the method of undetermined coefficients or variation of parameters, often supplementing characteristic solutions. Hence, dealing with nonhomogeneous terms requires special attention to deduce a particular solution that fits alongside the general solution to frame a complete answer for real-world applications.