PRODUCTO EFICIENTE INSTALACIÓN CORRECTA USO ADECUADO DEL PRODUCTO

In this section we set out a method of finding solutions of linear partial differential equations of the first order. We start by considering a particular example, namely

(10.1.1)

Associated with this are certain curves in the (x, y)-plane specified by obliging them to have the slope which satisfies

(10.1.2)

the right-hand side being the ratio of the coefficients of the two partial derivatives in (10.1.1). The general solution of (10.1.2) is where C is a constant and the curves are, in fact, straight lines. They are drawn for various values of C in Figure 10.1.1. These special curves are known as the characteristics of the partial differential equation (10.1.1).

The idea in solving (10.1.1) is to introduce a new set of coordinates, called ξ and η, in which the

characteristics can be expressed as ξ=constant. The other coordinate η must be selected so that to a given point (ξ, η) there corresponds only one point in the x- and y-coordinates. In other words, we want the curve η=constant to intersect the curve ξ=constant in the (x, y)-plane in one, and only one, point. Now, in Figure 10.1.1, a line parallel to the y-axis meets a characteristic in one, and only one, point. Therefore make the change of coordinates

(10.1.3)

Notice that (10.1.3) does supply just one (x, y) for a given (ξ, η); in fact x=η and . Other choices of η are acceptable providing that they meet the criterion that the curve η=constant intersects ξ=constant once and once only; for example, η=y and η=3y+2x are other possibilities.

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FIGURE 10.1.1: The characteristics of (10.1.1).

From (10.1.3), by the chain rule for partial derivatives,

since ∂ξ/∂x=−3/2, ∂ξ/∂y=1, ∂η/∂x=1, ∂η/∂y=0. Substitution in (10.1.1) gives (10.1.4)

Thus the change of coordinates (10.1.3) has simplified the partial differential equation and, in fact, one partial derivative ∂u/∂ξ has been removed altogether. It was with the object of causing this partial derivative to disappear that the special choice of ξ was made.

Equation (10.1.4) tells us that u is independent of η and so it must be a function of ξ only, i.e., (10.1.5)

on employing (10.1.3).

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The formula (10.1.5) shows that u is a constant on a characteristic since is constant on a characteristic. Hence, if u is known at one point of a characteristic it is known at every point of the

characteristic. The constant can be different on different characteristics, but we cannot say how it will change without additional information.

Suppose, indeed, that it is required that u=3y on x=0 for y≥0. Then, putting x=0 in (10.1.5), we are obliged to have

for y≥0. Using this expression for F we have and so, from (10.1.5),

u=3y−9x/2

for y≥3x/2. Thus u has been found at every point on and above the line y=3x/2. Nothing can be said about the behaviour of u below this line, other than that it is a function of y−3x/2, because there is insufficient information. In other words, the solution is restricted to that part of the (x, y)-plane that is covered by the characteristics that intersect the are on which the initial data are given.

We might have been asked to find u so that u=3y on the line y=3x/2 for y≥0. Then (10.1.5) would require F(0)=3y for y≥0, which no F will satisfy because the left-hand side is constant whereas the right-hand side is not. Indeed, the only problem which can be solved when u is given on y=3x/2 is for u to be specified as a

constant there. A similar situation arises for the characteristic y−3x/2=constant. Thus the initial data can be specified arbitrarily for u only when given on an arc that does not coincide with a characteristic. When initial data are prescribed on a characteristic they must take a special form if inconsistency is to be avoided.

Let us turn now to the more general linear partial differential equation

(10.1.6) We succeeded in solving (10.1.1) because we were able to discover a substitution which eliminated one partial derivative and led to (10.1.4). So we want to try to find a similar substitution for (10.1.6). Now, if a≡0 or b≡0 in the region under consideration, only one partial derivative occurs anyway so it is only when neither a nor b vanishes identically that further discussion is necessary. We then consider the curves whose slope satisfies

(10.1.7)

They are known as characteristics and have no multiple points provided that a and b do not vanish at the same point, a possibility which will be

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excluded here. The characteristics need not be straight lines as they were for (10.1.1). Let the equation for the characteristics be written in the form

(10.1.8)

where C is a constant which gives different characteristics for different values. A derivative of (10.1.8) with respect to x supplies

But, since the slope is forced to be the same as that of (10.1.7), we must have

(10.1.9) Suppose now that there is a set of curves with equation

C1 being constant, with the property that each curve meets each characteristic in one and only one point in the region of interest while ψ varies continuously along a characteristic. There are usually many possible ψ which are open to the solver and the one most convenient to the problem in hand should be chosen. Make the substitution

(10.1.10)

The conditions placed on and ψ ensure that a point may be fixed by either (x, y) or (ξ, η). Since (10.1.6) transforms to

(10.1.11)

on account of (10.1.9). Equation (10.1.11) is often called the characteristic form of the partial differential equation.

If values for x, y in terms of ξ, η are inserted in (10.1.11) from (10.1.10), the coefficients in (10.1.11) become known functions of ξ and η. For fixed ξ (10.1.11) is an ordinary differential equation of the first order for u in terms of η. Therefore u can be determined but contains an arbitrary constant (which may be different for different values of ξ). Thus, if u is specified at a point

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of ξ=constant it is known at all points of that curve. In other words, if u is given at one point of a

characteristic it can be found at all points of the characteristic. In particular, when u is designated to have certain values on an are that is met at most once by a characteristic it is determined in the region covered by the characteristics intersecting the are. Again, as for (10.1.1), data cannot be prescribed arbitrarily all along a characteristic; they must comply with (10.1.11) if there is to be a solution.

The existence theorems of Chapter 5 can be employed to check the existence and uniqueness of solutions to (10.1.11).

Example 10.1.1 Find the solution of

in x>0 such that u=2y when x=1.

The differential equation for the characteristics is, from (10.1.7), Therefore the characteristics are

(10.1.12)

The slope of the characteristics is positive in x>0. Also y→−∞ as x→0 and y→∞ as x→∞. The shape of the characteristics is, consequently, that shown in Figure 10.1.2, the larger C the closer the characteristic is to the y-axis.

A set of curves that intersects characteristics once and once only is furnished by lines parallel to the x-axis; another set is provided by lines parallel to the y-axis. Choose those parallel to the y-axis since the initial values are intimated on x=1. Then, according to (10.1.10),

(10.1.13)

Note that it is always worthwhile considering parallel straight lines first for the ψ curves since they lead to the simplest form for η. Now

and the given partial differential equation is converted to

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FIGURE 10.1.2: The characteristics of (10.1.12).

From (10.1.13), x=η and so that

has to be solved. From

(10.1.14) is deduced

It is required that u=2y when x=1, i.e., u=2ξ+η+ln η when η=1 or u=2ξ+1 when η=1, and this is to hold for all ξ. Therefore

whence

Replacing ξ, η by x, y via (10.1.13) we obtain

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This solution is valid everywhere in x>0, because all the characteristics intersect x=1 and every point in x>0 lies on some characteristic.

Example 10.1.2 Find the solution of

in 0<x<y such that u=2x on y=3x. The characteristics satisfy

whence their equation is ln y=−ln x+constant, which may be simplified to

Each characteristic in the first quadrant is therefore the branch of a hyperbola (Figure 10.1.3). The intersecting curves may again be chosen as straight lines and to illustrate the fact that they need not be parallel to the coordinate axes we select them to be parallel to y=x. (What would be the objection to making them parallel to y=−x?) Thus the substitution

(10.1.15) is made.

With this change of variable

and the partial differential equation is transformed to

FIGURE 10.1.3: The characteristic hyperbolae.

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Now, if y is eliminated from (10.1.15),

so that . However, x and ξ must both be positive in the first quadrant so that the upper

sign must be selected, i.e., Hence we are led to

It follows that

on inserting (10.1.15) and the formulae for x and y just found. Since u=2x when y=3x,

which implies that

when z is positive. Consequently

Since each characteristic intersects y=3x this solution is valid throughout the first quadrant. Nothing can be said about what happens outside the first quadrant because the initial data are given only on characteristics that are confined to the first quadrant.