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Chapter 21: Problem 2

Find the characteristic curves for \(\mathbf{L}_{x}[u]=\partial u / \partial x\).

Short Answer

Expert verified
The characteristic curves for \(\mathbf{L}_{x}[u]=\partial u / \partial x\) are straight lines along the x-axis.

Step by step solution

01

Write down the characteristic equation

The characteristic equation for this partial differential equation is given by \(dx/dt = 1\). This indicates that the slope of the curve at any point is equal to 1.
02

Solve the characteristic equation

To solve the characteristic equation, we can integrate both sides. Integrating \(dx/dt = 1\) with respect to 't' gives \(x = t + C1\). This is an equation of a line where 'C1' is the integration constant and represents the y-intercept of the lines.
03

Deduce the characteristic curves

The solutions of the characteristic equation, namely \(x = t + C1\), indicate that the characteristic curves are straight lines along the x-axis with different y-intercepts. These are the characteristic curves of the given partial differential equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Characteristic Curves
Characteristic curves are a fundamental concept when dealing with first-order partial differential equations. They help transform complex partial equations into simpler ordinary differential equations.

When we have a partial differential equation like \( \mathbf{L}_x[u]=\partial u / \partial x \), finding the characteristic curves involves determining the path along which the rate of change is consistent.
  • They are determined by the characteristic equation.
  • By solving this, we identify paths in the domain where the solution behaves predictably.
In the given problem, the characteristic equation \( dx/dt = 1 \) tells us the slope of these curves is uniform. By finding these paths, we simplify the problem of solving partial differential equations by reducing them into lines along which the equation holds true.
Integration in Differential Equations
Integration plays a crucial role in solving differential equations. Here, integrating allows us to move from the rate of change to the function itself.

In our step-by-step solution, we start with the equation \( \frac{dx}{dt} = 1 \). Solving this requires integrating both sides with respect to \( t \). This gives:
  • \( \int \frac{dx}{dt} \, dt = \int 1 \, dt \)
  • After integration, we obtain \( x = t + C1 \).
This solution reveals that the characteristic curves are lines with different intercepts due to the constant \( C1 \). These integrals convert the dynamic nature of rates into a static solution, providing clarity on how our system evolves.
Partial Derivatives
Partial derivatives are central to understanding the behavior of functions of multiple variables. They quantify how a function changes with respect to one variable while keeping others constant.

For the equation \( \partial u / \partial x \), we're looking at how the function \( u \) changes as \( x \) changes.
  • Partial derivatives are like holding other dimensions steady while you peek into one specific direction.
  • They help in constructing linear approximations of functions in multiple dimensions.
In our exercise, focusing on \( \partial u / \partial x \) simplifies our journey to finding characteristic curves, directing us to examine changes with respect to \( x \). By understanding partial derivatives, we can break down complex systems into manageable parts.

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Most popular questions from this chapter

Use \(J \delta(\mathbf{x}-\mathbf{a})=\delta(\boldsymbol{\xi}-\boldsymbol{\alpha})\) and the coordinate transformation from the spherical coordinate system to Cartesian coordinates to express the \(3 \mathrm{D}\) Cartesian delta function in terms of the corresponding spherical delta function at a point \(P=\left(x_{0}, y_{0}, z_{0}\right)=\left(r_{0}, \theta_{0}, \varphi_{0}\right)\) where the Jacobian \(J\) is nonvanishing.

Show that for scattering problems \((E>0)\) (a) the integral form of the Schrödinger equation in one dimension is $$ \Psi(x)=e^{i k x}-\frac{i \mu}{\hbar^{2} k} \int_{-\infty}^{\infty} e^{i k|x-y|} V(y) \Psi(y) d y . $$ (b) Divide \((-\infty,+\infty)\) into three regions \(R_{1}=(-\infty,-a), R_{2}=(-a,+a)\) and \(R_{3}=(a, \infty)\). Let \(\psi_{i}(x)\) be \(\psi(x)\) in region \(R_{i} .\) Assume that the potential \(V(x)\) vanishes in \(R_{1}\) and \(R_{3}\). Show that $$ \psi_{1}(x)=e^{i k x}-\frac{i \mu}{\hbar^{2} k} e^{-i k x} \int_{-a}^{a} e^{i k y} V(y) \psi_{2}(y) d y $$ This shows that determining the wave function in regions where there is no potential requires the wave function in the region where the potential acts. (c) Let $$ V(x)=\left\\{\begin{array}{ll} V_{0} & \text { if }|x|a \end{array}\right. $$ and find \(\psi_{2}(x)\) by the method of successive approximations. Show that the \(n\) th term is less than \(\left(2 \mu V_{0} a / \hbar^{2} k\right)^{n-1}\) (so the Neumann series will converge) if \(\left(2 V_{0} a / \hbar v\right)<1\), where \(v\) is the velocity and \(\mu v=\) \(\hbar k\) is the momentum of the wave. Therefore, for large velocities, the Neumann series expansion is valid.

Solve the Cauchy problem for the two-dimensional Laplace equation subject to the Cauchy data \(u(0, y)=0,(\partial u / \partial x)(0, y)=\epsilon \sin k y\), where \(\epsilon\) and \(k\) are constants. Show that the solution does not vary continuously as the Cauchy data vary. In particular, show that for any \(\epsilon \neq 0\) and any preassigned \(x>0\), the solution \(u(x, y)\) can be made arbitrarily large by choosing \(k\) large enough.

Consider the operator \(\mathbf{L}_{\mathbf{x}}=\nabla^{2}+\mathbf{b} \cdot \boldsymbol{\nabla}+c\) for which \(\left\\{b_{i}\right\\}_{i=1}^{m}\) and \(c\) are functions of \(\left\\{x_{i}\right\\}_{i=1}^{m}\). (a) Show that \(\mathbf{L}_{\mathbf{x}}^{\dagger}[v]=\nabla^{2} v-\boldsymbol{\nabla} \cdot(\mathbf{b} v)+c v\), and $$ \mathbf{Q}\left[u, v^{*}\right]=\mathbf{Q}[u, v]=v \nabla u-u \nabla v+\mathbf{b} u v $$ (b) Show that a necessary condition for \(\mathbf{L}_{\mathbf{x}}\) to be self- adjoint is \(2 \mathbf{b} \cdot \nabla u+\) \(u(\boldsymbol{\nabla} \cdot \mathbf{b})=0\) for arbitrary \(u .\) (c) By choosing some \(u\) 's judiciously, show that (b) implies that \(b_{i}=0\). Conclude that \(\mathbf{L}_{\mathrm{x}}=\nabla^{2}+c(\mathbf{x})\) is formally self-adjoint.

Show that the integral equation associated with the damped harmonic oscillator DE \(\ddot{x}+2 \gamma \dot{x}+\omega_{0}^{2} x=0\), having the \(\mathrm{BCs} x(0)=x_{0}\) \((d x / d t)_{t=0}=0\), can be written in either of the following forms. (a) \(x(t)=x_{0}-\frac{\omega_{0}^{2}}{2 \gamma} \int_{0}^{t}\left[1-e^{-2 \gamma\left(t-t^{\prime}\right)}\right] x\left(t^{\prime}\right) d t^{\prime}\). (b) \(x(t)=x_{0} \cos \omega_{0} t+\frac{2 \gamma x_{0}}{\omega_{0}} \sin \omega_{0} t-2 \gamma \int_{0}^{t} \cos \left[\omega_{0}\left(t-t^{\prime}\right)\right] x\left(t^{\prime}\right) d t^{\prime}\). Hint: Take \(\omega_{0}^{2} x\) or \(2 \gamma \dot{x}\), respectively, as the inhomogeneous term.
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