Question

Suppose \(g\) is differentiable and \(f\) is twice differentiable on \((-\infty, \infty),\) and let $$ u_{0}(x, t)=\frac{f(x+a t)+f(x-a t)}{2} \quad \text { and } \quad u_{1}(x, t)=\frac{1}{2 a} \int_{x-a t}^{x+a t} g(u) d u . $$ (a) Show that $$ \frac{\partial^{2} u_{0}}{\partial t^{2}}=a^{2} \frac{\partial^{2} u_{0}}{\partial x^{2}}, \quad-\infty0 $$ and $$ u_{0}(x, 0)=f(x), \quad \frac{\partial u_{0}}{\partial t}(x, 0)=0, \quad-\infty0 $$ and $$ u_{1}(x, 0)=0, \quad \frac{\partial u_{1}}{\partial t}(x, 0)=g(x), \quad-\infty0 ,\\\ u(x, 0)=f(x), \quad u_{t}(x, 0)=g(x), \quad-\infty

Short answer

#tag_content# Yes, the function \(u_0(x,t)\) satisfies the given partial differential equation and initial conditions. The second derivatives of \(u_0(x,t)\) with respect to \(t\) and \(x\) were calculated to be $$ \frac{\partial^2 u_0}{\partial t^2} = \frac{a^2[f''(x+at)+f''(x-at)]}{2}, $$ and $$ \frac{\partial^2 u_0}{\partial x^2} = \frac{f''(x+at)+f''(x-at)}{2}, $$ which showed that \(\frac{\partial^2 u_0}{\partial t^2}=a^2 \frac{\partial^2 u_0}{\partial x^2}\), as required by the partial differential equation. Additionally, the initial conditions for \(u_0(x,t)\) were verified as $$ u_0(x,0)=f(x), $$ and $$ \frac{\partial u_0}{\partial t}(x,0)=0. $$ Both of these hold for the given function \(u_0(x,t)\). #tag_title# (b) Does the function \(u_1(x,t)\) satisfy the given partial differential equation and initial conditions?#tag_content# Yes, the function \(u_1(x,t)\) satisfies the given partial differential equation and initial conditions. The second derivatives of \(u_1(x,t)\) with respect to \(t\) and \(x\) were calculated to be $$ \frac{\partial^2 u_1}{\partial t^2} = \frac{g'(x+at)+g'(x-at)}{2}, $$ and $$ \frac{\partial^2 u_1}{\partial x^2} = \frac{g'(x+at)-g'(x-at)}{2a}, $$ which showed that \(\frac{\partial^2 u_1}{\partial t^2}=a^2 \frac{\partial^2 u_1}{\partial x^2}\), as required by the partial differential equation. Additionally, the initial conditions for \(u_1(x,t)\) were verified as $$ u_1(x,0)=0, $$ and $$ \frac{\partial u_1}{\partial t}(x,0)=g(x). $$ Both of these hold for the given function \(u_1(x,t)\). #tag_title# (c) What is the solution to the initial value problem?#tag_content# The solution to the given initial value problem is the sum of the functions \(u_0(x,t)\) and \(u_1(x,t)\) since they satisfy the partial differential equation and the given initial conditions. The solution is: $$ u(x,t)=\frac{f(x+at)+f(x-at)}{2}+\frac{1}{2a}\int_{x-at}^{x+at}g(u)\,du. $$

Step-by-step solution

(a) Calculating second derivatives of \(u_0\) with respect to \(t\) and \(x\)

: To show that $$ \frac{\partial^2 u_0}{\partial t^2}=a^2 \frac{\partial^2 u_0}{\partial x^2}, $$ we need to calculate the second-order partial derivatives of \(u_0(x,t)\) with respect to \(t\) and \(x\), which are given by $$ u_0(x,t) = \frac{f(x+at)+f(x-at)}{2}. $$ First, we calculate the first-order partial derivatives: $$ \frac{\partial u_0}{\partial t} = \frac{a[f'(x+at) - f'(x-at)]}{2}, $$ and $$ \frac{\partial u_0}{\partial x} = \frac{f'(x+at)+f'(x-at)}{2}. $$ Then, we compute the second-order partial derivatives: $$ \frac{\partial^2 u_0}{\partial t^2} = \frac{a^2[f''(x+at)+f''(x-at)]}{2}, $$ and $$ \frac{\partial^2 u_0}{\partial x^2} = \frac{f''(x+at)+f''(x-at)}{2}. $$ Now, we can see that \(\frac{\partial^2 u_0}{\partial t^2}=a^2 \frac{\partial^2 u_0}{\partial x^2}\) holds.

(a) Verifying the initial conditions for \(u_0\)

: Now, we need to verify the initial conditions for \(u_0(x,t)\): $$ u_0(x,0)=f(x) $$ and $$ \frac{\partial u_0}{\partial t}(x,0)=0. $$ To do this, we just need to substitute \(t=0\) into the expressions for \(u_0(x,t)\) and \(\frac{\partial u_0}{\partial t}\): $$ u_0(x,0)=\frac{f(x)+f(x)}{2}=f(x), $$ and $$ \frac{\partial u_0}{\partial t}(x,0)=\frac{a[f'(x) - f'(x)]}{2}=0. $$ Thus, the initial conditions hold for \(u_0(x,t)\).

(b) Calculating second derivatives of \(u_1\) with respect to \(t\) and \(x\)

: To show that $$ \frac{\partial^2 u_1}{\partial t^2}=a^2 \frac{\partial^2 u_1}{\partial x^2}, $$ we need to calculate the second-order partial derivatives of \(u_1(x,t)\) with respect to \(t\) and \(x\), which are given by $$ u_1(x,t) = \frac{1}{2a} \int_{x-at}^{x+at} g(u) \, du. $$ First, we calculate the first-order partial derivatives using the Leibniz rule: $$ \frac{\partial u_1}{\partial t} = \frac{1}{2a} \left[-ag(x-at)+ag(x+at)\right] = \frac{g(x+at)-g(x-at)}{2}, $$ and $$ \frac{\partial u_1}{\partial x} = \frac{1}{2a} \left[ag(x+at)+ag(x-at)\right] = \frac{g(x+at)+g(x-at)}{2}. $$ Then, we compute the second-order partial derivatives: $$ \frac{\partial^2 u_1}{\partial t^2} = \frac{g'(x+at)+g'(x-at)}{2}, $$ and $$ \frac{\partial^2 u_1}{\partial x^2} = \frac{g'(x+at)-g'(x-at)}{2a}. $$ Now, it can be seen that \(\frac{\partial^2 u_1}{\partial t^2} = a^2 \frac{\partial^2 u_1}{\partial x^2}\) holds.

(b) Verifying the initial conditions for \(u_1\)

: Now, we need to verify the initial conditions for \(u_1(x,t)\): $$ u_1(x,0)=0 $$ and $$ \frac{\partial u_1}{\partial t}(x,0)=g(x). $$ To do this, we just need to substitute \(t=0\) into the expressions for \(u_1(x,t)\) and \(\frac{\partial u_1}{\partial t}\): $$ u_1(x,0)=\frac{1}{2a} \int_{x}^{x} g(u) \, du=0, $$ and $$ \frac{\partial u_1}{\partial t}(x,0)=\frac{g(x)-g(x)}{2}=0. $$ Thus, the initial conditions hold for \(u_1(x,t)\).

(c) Solving the initial value problem

: The given initial value problem is: $$ \begin{cases} u_{tt}=a^2 u_{xx}, \quad -\infty < x < \infty, \quad t>0,\\ u(x,0)=f(x), \quad u_t(x,0)=g(x), \quad -\infty<x<\infty. \end{cases} $$ Since both \(u_0(x,t)\) and \(u_1(x,t)\) satisfy the partial differential equation and their initial conditions match the given initial conditions, we can sum these two functions to find a solution for the given initial value problem: $$ u(x,t)=u_0(x,t)+u_1(x,t)=\frac{f(x+at)+f(x-at)}{2}+\frac{1}{2a}\int_{x-at}^{x+at}g(u)\,du. $$ Thus, the solution to the initial value problem is: $$ u(x,t)=\frac{f(x+at)+f(x-at)}{2}+\frac{1}{2a}\int_{x-at}^{x+at}g(u)\,du. $$

Key concepts

Partial Differential Equations

Partial Differential Equations (PDEs) are fundamental in mathematics and describe a variety of physical phenomena. A PDE involves functions of several variables and their partial derivatives. In the context of this exercise, we explore the wave equation, for which we showed how certain functions satisfy the equation under specific conditions.
The general form of a wave equation is \( u_{tt} = a^2 u_{xx} \). This particular equation describes waves propagating at a speed \(a\) along a medium. Typical examples include sound waves, light waves, or waves on a string. Each of these scenarios can be represented in terms of calculation involving partial derivatives, as it allows us to model the changes over time and space.
For students venturing into PDEs, it is crucial to understand that these equations look at how a function changes in negligible parts with respect to multiple variables. This multifaceted approach mathematically describes how a system evolves over time or space.

Initial Value Problem

An Initial Value Problem (IVP) seeks a solution to a differential equation that satisfies certain conditions at the outset. These conditions are known as initial conditions. In this exercise, we solved an IVP by finding solutions \(u_0(x,t)\) and \(u_1(x,t)\) that both obey the wave equation.
To solve an IVP, we start by identifying the given conditions. For example, for \(u_0(x,t)\), it was required that \(u_0(x,0)=f(x)\) and \(\frac{\partial u_0}{\partial t}(x,0)=0\). Similarly, \(u_1(x,t)\) had different initial conditions. These conditions correctly setting up the problem allows us to find a unique solution in the context of PDEs such as the wave equation.
In summary, solving IVPs consists of determining a function that not only solves the differential equation but also obeys the given conditions at \(t=0\). This specificity lets mathematicians and engineers model situations closely tied to real-world applications.

Partial Derivatives

Partial Derivatives are a key concept in dealing with functions of multiple variables, like those often found in partial differential equations. A partial derivative represents the rate at which a function changes as one of its input variables changes while keeping others constant.
In the solution of our wave equation problem, calculating partial derivatives was an essential step. For example, in \(u_0(x,t)\), calculating \(\frac{\partial^2 u_0}{\partial t^2}\) required us to focus on how the function changes with respect to time, \(t\), while keeping \(x\) constant. Similarly, \(\frac{\partial^2 u_0}{\partial x^2}\) shows how the function changes as \(x\) varies, in isolation of \(t\).
Understanding how to compute and interpret partial derivatives is crucial for analyzing and solving PDEs. They reveal the 'sensitivity' of the solution to individual variables, adding depth to our understanding of mathematical models in diverse scientific fields.

Differential Equations

Differential Equations, which include PDEs, are equations that relate a function to its derivatives. They are vital in expressing the laws governing natural phenomena, engineering systems, economic trends, and more.
In this exercise, we engage with linear PDEs that feature constant coefficients, particularly learning how to manipulate such equations to verify solutions. Linear implies that the dependent variable and its partial derivatives appear to the first power, with no products or other non-linear effects.
Mastering this aspect involves comprehending the structure of the differential equation in the context of the problem to find solutions signifying real-life processes. Here, identifying the role and behavior of each term ensures precise application and problem-solving capacity in complex mathematical modeling.