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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 = [pic]dt + [pic]dx ( [pic] = [pic] + [pic][pic]

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

[pic] + c[pic] = 0 1.3-1

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

[pic] + [pic] = 0 1.3-2

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

[pic] = 2(x – t) and [pic]= – 2(x – t)

Therefore u(x,t) = (x – t)2 is a solution of the PDE (1.3-2). Other functions: [pic], 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 [pic] + [pic] = 0 that satisfies the initial condition u(x,t) = [pic].

Since u(x,t) = f(x – t) is the solution of the PDE, this solution must satisfy the initial condition u(x,0) = [pic]. 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) = [pic]

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

[pic][pic]

Figure 1.a. [pic] at t = 0 Figure 1.b. [pic] at t = 4

The general solution of [pic] + [pic] = 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

[pic]= [pic][pic] + [pic][pic] = a[pic] + c[pic]

[pic] = [pic][pic] + [pic][pic] = b[pic] + d[pic]

Substituting into (1.3-2) and simplifying we obtain

[pic] + [pic] = (a + b)[pic] + (c + d)[pic]= 0

The partial derivative [pic]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

[pic] = 0.

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

The partial derivative [pic] can also be eliminated by taking

a = 1, b = (1, c = 1, d = 0

With this choice, ( = x ( t, ( =b, and the equation becomes

[pic] = 0.

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

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