# Preface

In the following series of web pages, we discuss basic partial differential equations (PDEs for short) of hyperbolic type. The wave equation $$\Box_c u \overset{\mathrm def}{=} u_{tt} - c^2 \Delta u$$ is one of the most important representative of hyperbolic equations. The wave equation usually describes water waves, the vibrations of a string or a membrane, the propagation of electromagnetic and sound waves, or the transmission of electric signals in a cable.

Introduction to Linear Algebra with Mathematica

# Hyperbolic equations

A hyperbolic partial differential equation of order n is a partial differential equation (PDE) that, roughly speaking, has a well-posed initial value problem for the first n − 1 derivatives. More precisely, the Cauchy problem can be locally solved for arbitrary initial data along any non-characteristic hypersurface.

The wave equation is an important representative of a hyperbolic equation. The wave equation for a function u(x1, … , xn, t) = u(x, t) of n space variables x1, … , xn and the time t is given by

$\square u = \square_c u \equiv u_{tt} - c^2 \nabla^2 u = 0 , \qquad \nabla^2 = \Delta = \frac{\partial^2}{\partial x_1^2} + \cdots + \frac{\partial^2}{\partial x_n^2} ,$
with a positive constant c (having dimensions of speed). The operator □ defined above is known as the d'Alembertian or the d'Alembert operator.

Let us consider a partial differential equation with two independent variables. By a linear change of variables, any equation of the form

$A\,\frac{\partial^2 u}{\partial x^2} + 2B\,\frac{\partial^2 u}{\partial x \partial y} + C\, \frac{\partial^2 u}{\partial y^2} + (\mbox{lower order derivative terms}) = 0$
with
$B^2 - AC > 0$
can be transformed to the wave equation, apart from lower order terms which are inessential for the qualitative understanding of the equation.

Example: Consider the hyperbolic equation with the initial condition at the characteristic

$u_{xy} = 0 , \qquad u(x,y= \mbox{constant}) = f(x) \quad u_y (x,y= \mbox{constant}) = g(x) .$
This initial value problem has no solution. Its general solution is
$u(x,y) = \phi (x) + \psi (y)$
for some smooth functions φ and ψ. To satisfy the initial conditions, we should solve the system of equations
$\begin{cases} \phi (x) + \psi (c) &= f(x) , \\ \psi' (c) &= g(x) . \end{cases}$
Then function g(x) = C, a constant. If this is true, then function ψ(y) can be arbitrary. Hence,
$\phi (x) = f(x) - \psi (c) .$
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Example: Consider the initial value problem for one dimensional wave equation

$\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2} \quad (-\infty < x < \infty ), \qquad u(x,0) = d(x), \quad \left. \frac{\partial u}{\partial t} \right\vert_{t=0} = v(x) ,$
where d(x) and v(x) are given initial displacement and velocity functions, respectively. A constant c has dimensions of speed.

Application of the Fourier transform leads to the initial value problem for the ordinary differential equation

$\frac{{\text d}^2 u^F}{{\text d}t^2} + c^2 \xi^2 u^F = 0, \qquad u^F (0) = d^F , \quad \dot{u}^F (0) = v^F ,$
where
$u^F = ℱ\left[ u(x,t) \right] = \int_{-\infty}^{\infty} u(x,t)\, e^{{\bf j} x\xi}\, {\text d}x , \qquad d^F = ℱ\left[ d(x) \right] = \int_{-\infty}^{\infty} d(x)\, e^{{\bf j} x\xi}\, {\text d}x , \qquad v^F = ℱ\left[ d(x) \right] = \int_{-\infty}^{\infty} d(x)\, e^{{\bf j} x\xi}\, {\text d}x$
are Fourier transforms of the the functions u(x,t), d(x), and v(x) with respect to spacial variable x, respectively. Solution of the above initial value problem is
$u^F = d^F \cos\left( c\xi t \right) + v^F \,\frac{\sin\left( c\xi t \right)}{c\xi} .$
Upon introducing two auxiliary functions
$\Phi (\xi ) = \frac{\sin\left( c\xi t \right)}{c\xi} \qquad\mbox{and} \qquad \Psi (\xi ) = \cos\left( c\xi t \right)$
we can represent the Fourier transform of the solution as the sum
$u^F = d^F (\xi )\,\Psi (\xi ) + v^F (\xi )\,\Phi (\xi ) ,$
which allows us to use the convolution rule (with respect to variable x):
$u(x,t) = d(x) \underset{x}{*} \psi (x,t) + v(x) \underset{x}{*} \phi (x,t) ,$
where ψ and ϕ are inverse Fourier transforms of Ψ and Φ, respectively:
\begin{align*} \phi (x,t) &= ℱ^{-1}\left[ \Phi (\xi ) \right] (x) = \frac{1}{2\pi} \, \int_{-\infty}^{\infty} \frac{\sin\left( c\xi t \right)}{c\xi} \, e^{-{\bf j} x\xi} \,{\text d}\xi , \\ \psi (x,t) &= ℱ^{-1}\left[ \Psi (\xi ) \right] (x) = \frac{1}{2\pi} \, \int_{-\infty}^{\infty} \cos\left( c\xi t \right) e^{-{\bf j} x\xi} \,{\text d}\xi = \frac{\partial}{\partial t} \phi (x,t) . \end{align*}
Since function ψ is a derivative (in weak sense) of ϕ, it is sufficient to determine the latter. Taking the inverse Fourier transform, we obtain
$\phi (x) = \frac{1}{2\pi} \, \int_{-\infty}^{\infty} \frac{\sin\left( c\xi t \right)}{c\xi} \, e^{-{\bf j} x\xi} \,{\text d}\xi = \frac{1}{4\pi c}\,\int_{-\infty}^{\infty} \frac{1}{{\bf j}\xi} \left( e^{{\bf j} ct\xi} - e^{-{\bf j} ct\xi} \right) e^{-{\bf j} x\xi} \,{\text d}\xi$
because of Euler's formula $$\sin t = \frac{1}{2{\bf j}} \, e^{{\bf j} t} - \frac{1}{2{\bf j}} \, e^{-{\bf j} t} .$$ We know from Example D in section that
$ℱ \left[ \mbox{sign}(x) \right] (\xi ) = \frac{2}{{\bf j}\xi} \qquad ℱ^{-1} \left[ \frac{2}{{\bf j}\xi} \right] = \mbox{sign}(x) ,$
Therefore
$\phi (x,t) = \frac{1}{2c}\,\mbox{sign}(x+ ct) - \frac{1}{2c}\,\mbox{sign}(x- ct) .$
Its derivative (in weak sense) is
$\psi (x,t) = \frac{1}{2}\,\delta (x+ ct) + \frac{1}{2}\,\delta (x- ct) .$
Taking convolutions, we obtaine the explicit formula for the given initial value problem for the wave equation
$d(x) \underset{x}{*} \psi (x,t) + v(x) \underset{x}{*} \phi (x,t) = \frac{1}{2} \left[ d(x+ct) + d(x-ct) \right] + \frac{1}{2c} \int_{x-ct}^{x+ct} v(\xi ) \,{\text d}\xi ,$
lnown as d'Alembert's formula.    ■

Example: Consider the initial value problem for the wave equation

$u_{tt} = u_{xx} , \qquad u(x,0) = x^3 , \quad u_t (x,0) = x ,$
where uxx stands for the partial derivative $$u_{xx} = \partial^2 u/\partial x^2 ,$$ and so on. Assuming that the solution is represented by the power series
$u(x,t) = \sum_{i\ge 0} \sum_{j\ge 0} a_{i.j} x^i t^j ,$
and sybstituting into the wave equation, we recursively find the values of coefficients. So
$u(x,t) = xt + x^3 + 3c^2 x t^2$
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Example: Consider the initial value problem for the wave equation

$u_{tt} = u_{xx} +6, \qquad u(x,0) = x^2 , \quad u_t (x,0) = 4x ,$
where uxx stands for the partial derivative $$u_{xx} = \partial^2 u/\partial x^2 ,$$ and so on. Assuming that the solution is represented by the power series
$u(x,t) = \sum_{i\ge 0} \sum_{j\ge 0} a_{i.j} x^i t^j ,$
and sybstituting into the wave equation, we recursively find the values of coefficients. So
$u(x,t) = 4xt + x^2 + 4 t^2 .$
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Example: Consider the initial value problem for the nonhomogeneous wave equation

$u_{tt} = u_{xx} - x^2 u(x,t) + x, \qquad u(x,0) = 0 , \quad u_t (x,0) = 0 ,$
where uxx stands for the partial derivative $$u_{xx} = \partial^2 u/\partial x^2 ,$$ and so on. Assuming that the solution is represented by the power series
$u(x,t) = \sum_{i\ge 0} \sum_{j\ge 0} a_{i.j} x^i t^j ,$
and sybstituting into the wave equation, we recursively find the values of coefficients. So
$u(x,t) = \frac{x t^2}{2} + \frac{x^3 t^4}{24} + \frac{x t^6}{120} .$
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Example:

$u_{tt} = u_{xx} + u_{xxxx} + \left( u^2 \right)_{xx} , \qquad x\in \mathbb{R}$
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Example:

$u_{tt} = u_{xx} - u_{xxxx} + \left( u^2 \right)_{xx} , \qquad x\in \mathbb{R}$
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1. Grigoryan, V, Parial Differential Equations, 2010, University of California, Santa Barbara.
2. McOwen, R.C., Partial Differential Equations, Prentice Hall, New Jersey, (1996).
3. Abdul-Majid Wazwaz, Partial Differential Equations and Solitary Waves Theory, Nonlinear Physical Science. Springer, Berlin, Heidelberg, 2009. https://doi.org/10.1007/978-3-642-00251-9_5
4. Abdul-Majid Wazwaz, Partial Differential Equations: Methods and Applications, Balkema, Leiden, (2002).