The Integration of the Equation of Evolution

Abstract

The ordinary exponential function solves the initial value problem

$$dy/dx = \alpha y,\quad y(0) = 1.$$

We consider the Diffusion equation

$$\partial u/\partial t = \Delta u,where\Delta = \sum\limits_{j = 1}^m {{\partial ^2}} /\partial x_j^2$$

is the Laplacian in R^m We wish to find a solution u = u(x, t), t > 0, of this equation satisfying the initial condition u(x, 0) = f(x), where f(x) = f (x₁,..., x _m ) is a given function of x. We shall also study the wave equation

$${\partial ^2}u/\partial {t^2} = \Delta u, - \infty < t < \infty$$

With the initial data

$$u(x,0) = f(x)and{(\partial u/\partial t)_{t = 0}} = g(x)$$

f and g being given functions. This may be written in vector form as follows:

$$\frac{\partial }{{\partial t}}(\mathop u\limits_v ) = (\mathop 0\limits_\Delta \mathop I\limits_0 )(\mathop u\limits_v ),v = \frac{{\partial u}}{{\partial t}}$$

With the initial condition

$$\left( {\begin{array}{*{20}{c}} {u\left( {x,0} \right)} \\ {v\left( {x,0} \right)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}{c}} {f(x)} \\ {g(x)} \\ \end{array} } \right)$$

So in a suitable function space, the wave equation is of the same from as the diffusion (or heat) equation—differentiation with respect to the time parameter on the left and another operator on the right—or again similar to the equation dy/dt = αy.