Part 1: Partial Differential Equations

Much of modern science, engineering, and mathematics is based on the study of partial differential equations, where a partial differential equation is an equation involving partial derivatives which implicitly defines a function of 2 or more variables.

For example, if u( x,t) is the temperature of a metal bar at a distance x from the initial end of the bar, then under suitable conditions u(x,t) is a solution to the heat equation

¶u ¶t = k  ¶2u ¶x2

where k is a constant. As another example, consider that if u(x,t) is the displacement of a string a time t, then the vibration of the string is likely to satisfy the one dimensional wave equation for a constant, which is

¶2u ¶t2 = a2  ¶2u ¶x2

(1)

When a partial differential equation occurs in an application, our goal is usually that of solving the equation, where a given function is a solution of a partial differential equation if it is implicitly defined by that equation. That is, a solution is a function that satisfies the equation.      

EXAMPLE 1    Show that if a is a constant, then u(x,y) = sin( at) cos( x) is a solution to

¶2u ¶t2 = a2  ¶2u ¶x2

(2)

Solution: Since a is constant, the partials with respect to t are

¶u ¶t = acos( at) cos( x),         ¶2u ¶t2 = -a2sin(at) sin( x)

(3)

Moreover, u_x = -sin( at) sin( x) and u_xx = -sin( at) cos( x) , so that we have

a2  ¶2u ¶x2 = -a2sin( at) cos( x)

(4)

Since (3) and (4) are the same, u( x,t) = sin( at) cos( x) is a solution to (2).       

EXAMPLE 2    Show that u( x,t) = e^ysin(x) is a solution to Laplace's Equation,

¶2u ¶x2  +   ¶2u ¶y2 = 0

Solution: To begin with, u_x = e^ycos( x) and u_xx = -e^ysin( x) . Moreover, u_y = e^ysin(x) and u_yy = e^ysin( x) , so that

¶2u ¶x2 +  ¶2u ¶y2 = -eysin( x) +eysin( x) = 0