# Preface

This section presents some properties of first order partial differential equations.

Introduction to Linear Algebra with Mathematica

# First order PDEs

A partial differential equation (PDE) for a function u(x1, x1, … , xn) of n variables is of the form
$F\left( x_1 , \ldots , x_n , u_{x_1} , \ldots , u_{x_n}, u_{x_1 x_1} , u_{x_1 x_2} , \ldots \right) = 0.$
The order of the above differential equation is the order of the highest derivative occuring in the equation, The equation is linear if it depends linearly on u and its derivatives. If all highest-order derivatives of u occur linearly with coefficients depending only on independent variables, u, and lower-order derivatives of u, then the equation is quasilinear. A simple example of a fist-order PDE is
$u_t + a(u)\,u_x = 0 ,$
where $$u_t \overset{\mathrm def}{=} \partial u/\partial t$$ and $$u_x \overset{\mathrm def}{=} \partial u/\partial x .$$ When $$a(u) \equiv a$$ is a constant, the linear equation ut + aux = 0 is called the transport equation. When $$a(u) \equiv u,$$ the quasilinear equation ut + uux = 0 is called the inviscid Burger's equation, which arises in the study of a one-dimensional stream of particles or fluid having zero viscosity.
ClearAll[u,x,t]; pde=D[u[x,t],t]+D[u[x,t],x]==0; bc=u[0,t]==0; ic=u[x,0]==x; sol=DSolve[{pde,ic,bc},u[x,t],{x,t}]
$u_x^2 + u_y^2 = c^2 ,$
is an example of a first order nonlinear PDE.

Example: Consider the transverse electromagnetic planar where the waves propagating in the z-direction having unboundedlossy medium is given by

$\begin{split} \frac{\partial E_x}{\partial z} + \mu\,\frac{\partial H_y}{\partial t} &= 0 , \\ \frac{\partial H_y}{\partial z} + \epsilon\,\frac{\partial E_x}{\partial t} + \sigma\, E_x &= 0 , \end{split}$
where μ, ε, and σ > 0 are the constants permeability, permittivity and the conductivity, respectively. The boundary conditions are given as follow
\begin{align*} E_x (\infty , t) &= \mbox{ finite}, \\ E_x (z,0) &= H_y (z,0) =0 , \\ \frac{\partial E_x (z,0)}{\partial z} &= \frac{\partial H_y (z,0)}{\partial z} =0, \\ \left. \frac{\partial E_x (z,t)}{\partial t} \right\vert_{t=0} &= \left. \frac{\partial H_y (z,t)}{\partial t} \right\vert_{t=0} =0 , \\ E(0,t) &= f(t)\, H(t) , \end{align*}
where H(t) is the Heaviside function.    ■

Example: Consider the nonlinear system of equations

$\begin{split} u_t + v\,u_x +u &= 1 , \\ v_t + u\,v_x -v &=1, \end{split}$
subject to the initial condition
$u(x,0) = e^x, \qquad v(x,0) = e^{-x} ,$
The true solution of the given initial value problem is
$u(x,t) = e^{x-t}, \qquad v(x,t) = e^{t-x} .$
In order to solve the given IVP using the power series method, the unknown functions u and v are considered as
\begin{align*} u(x,t) &= \sum_{r=0}^N \sum_{s=0}^N a_{r,s} x^r t^s , \\ v(x,t) &= \sum_{i=0}^N \sum_{j=0}^N b_{i,j} x^i t^j . \end{align*}
Substitution these series into the system of differential equations yields
\begin{align*} a_{i, j+1} &= \frac{1}{j+1} \left[ \delta_{i,j,0} - a_{i.j} + \sum_{r=0}^i \sum_{s=0}^j \left( s+1 \right) b_{i-r,j-s} a_{r+1,j} \right] , \\ b_{i,j+1} = \frac{1}{j+1} \left[ \delta_{i,j,0} + b_{i.j} + \sum_{r=0}^i \sum_{s=0}^j \left( s+1 \right) a_{i-r,j-s} b_{r+1,j} \right] , \end{align*}
where the Kroneker's delta function is
$\delta_{i,j,0} = \begin{cases} 1, & \ \mbox{ if } i=j=0, \\ 0, & \ \mbox{ otherwise.} \end{cases}$
Solving the above recurence relation, we obtain power series approximation
\begin{align*} u(x,t) &= 1-t + \frac{t^2}{2} - \frac{t^3}{3!} + x - xt + \frac{x t^2}{2} - \frac{x t^3}{3!} + \frac{x^2}{2} - \frac{x^2 t}{2} \\ & \quad + \frac{x^2 t^2}{4} - \frac{x^2 t^3}{12} + \frac{x^3}{6} - \frac{x^3 t}{6} + \frac{x^3 t^2}{12} - \frac{x^3 t^3}{36} + \frac{x^3 t^4}{144} + \cdots , \\ v(x,t) &= 1+t + \frac{t^2}{2} + \frac{t^3}{6} -x -xt - \frac{xt^2}{2} - \frac{xt^4}{24} + \frac{x^2}{2} + \frac{x^2 t}{2} + \frac{x^2 t^3}{12} \\ & \quad - \frac{x^3}{6} - \frac{x^3 t}{6} - \frac{x^3 t^2}{12} - \frac{x^3 t^3}{36} + \cdots . \end{align*}
■

Method of Characteristics

Before we discuss the method of characteristics, let us start with one typical example. Consider the initial value problem for the transport equation
$\begin{split} u_t + a\,u_x &= 0 , \\ u(x,0) &= h(x) , \end{split}$
where $$a$$ is a constant. We can reduce the transport equation to an exact ordinary differential equation by considering x-variable as a function of time t. So we consider the transport equation along some curve x(t). Then we will have
$\frac{\text d}{{\text d}t}\, u(x(t),t) = u_t + a\,u_x ,$
if we require $${\text d}x/{\text d} t = a.$$ In this case, x = 𝑎 t + x0, where x0 is the x-intercept of the curve. Along this curve we have $$\frac{\text d}{{\text d}t}\, u(x(t),t) = 0 ,$$ that is, u is a constant, which we know from the initial condition to be h(x0). Thus, u(x,t) = h(x0) = h(x - at). Indeed, ih h is C¹ (= continuously differentiable), then we can check that u(x,t) = h(x - at satisfies the transport equation and prescribed initial condition. This solution corresponds to "tranporting" the initial data h(x) without change along the x-axis at a speed $${\text d}x/{\text d} t = a.$$ The lines x = 𝑎 t + x0 are called the characteristic curves for u,sub>t + 𝑎 ux = 0. The reduction of a first order partial differential equation to an ordinary differential equation along its characteristics is called the method of characteristics.

Let us consider the quasilinear equation for a function of two variables x and y:
$a(x,y,u)\, u_x + b(x,y,u)\, u_y = c(x,y,u) ,$
where the coefficient functions 𝑎, b, and c are continuous in x, y, and u. If u(x,y) is a solution of the above partial differential equation, let us consider the graph z = u(x,y). This surface has normal vector N0 = ⟨ -ux(x0,y0), -uy(x0,y0) ⟩ at the point (x0,y0,u(x0,y0)). However, if we let z0 = u(x0,y0), then the above partial differential equation implies that that the vector $${\bf V} = \langle a\left( x_0 , y_0 , z_0 \right) , b\left( x_0 , y_0 , z_0 \right) , c\left( x_0 , y_0 , z_0 \right) \rangle$$ is perpendicular to this normal vector, and hence must lie in the tangent plane to the graph of z = u(x,y) at the point z0.

In other words, $${\bf V}(x,y,z) = \langle a\left( x , y , z \right) , b\left( x , y , z \right) , c\left( x , y , z \right) \rangle$$ defines a vector field in ℝ³, to which graphs of solutions must be tangent at each point. Surfaces that are tangent at each point to a vector field in ℝ³ are called integral surfaces of the vector field. Thus, in order to find a solution of the given partial differential equation $$a\,u_x + b\,u_y = c ,$$ we should try to find integral surfaces. Of course, there may be many integral surfaces of V, so we might try to be more specific and find the integral surface containing a given curve Γ ⊂ ℝ3. So we are led to formulating the following:

The Cauchy problem: : For a given curve Γ ⊂ ℝ³, can we find a solution u of the first order partial differential equation whose graph contains Γ?    ⧫

In special case when Γ is the graph ⟨ x, h(x) ⟩ in the xz-plane of a function h, the Cauchy problem is just an initial value problem with the obvious interpretation of the variable y as "yime".

We constract the integral surfaces by using characteristic curves that are the integral curves of the vector field V. These curves are called the characteristic curves is they satisfy the system of equations

$a(x,y,z) = \frac{{\text d}x}{{\text d}t} , \qquad b(x,y,z) = \frac{{\text d}y}{{\text d}t} , \qquad c(x,y,u) = \frac{{\text d}z}{{\text d}t} .$
We can solve this system of ODEs subject to the initial condition
$x\left( t_0 \right) = x_0 , \qquad y\left( t_0 \right) = y_0 , \qquad z\left( t_0 \right) = z_0 .$

Divergent form

Let u be a smooth function in some domain Ω ⊂ ℝd+1. A partial differential equation
\begin{equation} \label{Eq.First.3} \partial_t {\bf u} + \nabla_x \cdot {\bf f}({\bf u}) = 0 , \qquad {\bf u} \in \mathbb{R}^n , \quad {\bf x} \in \mathbb{R}^d , \end{equation}
is said to be in the divergent form. Here
$\partial_t = \frac{\partial}{\partial t} , \qquad \nabla_x = \left( \partial_{x_1} , \partial_{x_2} , \ldots , \partial_{x_n} \right) , \quad \partial_{x_i} = \frac{\partial}{\partial x_i} ,$
and f :  ℝn →ℝnd.
Every linear partial differential equation with constant coefficients can be written in the divergent form.
Let Ω be a region in ℝ² x1 < x < x2, t1 < t < t2, and suppose that in Ω ⊂
${\bf u} = {\bf u}(x,t) = \left( u_1 )x,t), \ldots , u_n (x,t) \right)$
solves the divergent form first order partial differential equation
\begin{equation} \label{Eq.First.1} \nabla \cdot {\bf v} = \frac{\partial}{\partial x} \,F({\bf u}) + \frac{\partial}{\partial t} \,G({\bf u}) = 0, \end{equation}
where
$\nabla = \left( \frac{\partial}{\partial x} , \frac{\partial}{\partial t} \right) , \qquad {\bf v} = {\bf v}({\bf u}) = \left( F({\bf u}) , G({\bf u}) \right) .$
An infinitely differentiable function φ = φ(x, t) that vanish in a neighborhood of the boundary ∂Ω of Ωis called test function on Ω. We say that u = u(x, t) is a weak solution of equation \eqref{Eq.First.1} if
\begin{equation} \label{Eq.First.2} \iint_{\Omega} \left[ \frac{\partial \varphi}{\partial x} \, F({\bf u}) + \frac{\partial \varphi}{\partial t} \, G({\bf u}) \right] {\text d}x {\text d} t = 0 \end{equation}
holds for all text functions φ = φ(x, t) on Ω.

Example: Consider the initial value problem

$u\,u_x + u_t = 0, \qquad u(x,0) = \begin{cases} 1, & \ 0 < x < 1, \\ 0, & \ x < 0 \quad\mbox{or} \quad x > 1. \end{cases}$
It has two weak solutions
$v = v(x,t) = \begin{cases} 0, & \ x < t/2 , \\ 1, & \ \frac{t}{2} < x < 1 + \frac{t}{2} , \\ 0, & \ x > 1 + \frac{t}{2} , \end{cases} \qquad w = w (x,t) = \begin{cases} 0, & \ x < 0 , \\ \frac{x}{t}, & \ 0 < x < t , \\ 1, & \ t < x < 1 + \frac{t}{2} , \\ 0, & \ x > 1 + \frac{t}{2} . \end{cases}$
The given differential equation cna be written in the divergent form:
$\frac{\partial}{\partial x} \left( u^2 \right) + \frac{\partial}{\partial t} \left( u \right) = 0 ,$
where F(u) = u² , G(u) = u.    ■

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