1.3 First Order Partial Differential Equations

Chapter1-03

1.3 First Order Partial Differential Equations

First order PDEs occur often in physicochemical processes. Let consider the unsteady one dimensional heat transfer when temperature T(x, t) depends on x and t then

dT = dt + dx  = +

The partial derivative denotes the change in temperature with respect to time at a fixed location x. The total derivative denotes the change in temperature with respect to time at a location moving with velocity c = . Therefore when = 0 or

+ c = 01.3-1

temperature T is a constant at the location moving in the x direction with velocity c. Consider the equation

+ = 01.3-2

where u(x,t) is the unknown function. Let u(x,t) = (x – t)2, then

= 2(x – t) and = – 2(x – t)

Therefore u(x,t) = (x – t)2 is a solution of the PDE (1.3-2). Other functions: , 3sin(x – t), 2cos(x – t), … are also solution of (1.3-2). In general the solution of u(x,t) has the form f(x – t) or

u(x,t) = f(x – t)

To obtain a unique solution to the PDE (1.3-2), a boundary or initial condition is required.

Example 1. Find solution of + = 0 that satisfies the initial condition u(x,t) = .

Since u(x,t) = f(x – t) is the solution of the PDE, this solution must satisfy the initial condition u(x,0) = . Therefore x must be replaced by x – t for the solution to have the form f(x – t) that satisfies the initial condition.

u(x,t) =

For any fixed value of t, the graph of u(x,t) = , as a function of x, represents a snapshot of a waveform as shown in Figure 1

Figure 1.a. at t = 0Figure 1.b. at t = 4

The general solution of + = 0, equation (1.3-2), can be obtained by a linear change of independent variables so that u(x, t) will become u(, ) where

 = ax + bt, = cx + dt.

The arbitrary constants a, b, c, d will be chosen so that the PDE can be reduced to an ordinary differential equation. The new independent variables are substituted into equation (1.3-2) by applying the chain rule

= + = a + c

= + = b + d

Substituting into (1.3-2) and simplifying we obtain

+ = (a + b)+ (c + d)= 0

The partial derivative can be eliminated by a proper choice of constants

a = 1, b = 0,c = 1, d = 1.

With this choice,  = a,  = x  t, and the equation becomes

= 0.

This equation can be integrated with respect to so that u = f() = f(x  t).

The partial derivative can also be eliminated by taking

a = 1, b = 1,c = 1, d = 0

With this choice,  = x  t,  =b, and the equation becomes

= 0.

The solution is then u = f() = f(x  t) which is the same as in the previous example.

1