ENTORNO COMPETITIVO DEL SECTOR

In order to see what problems could be solved by tracing characteristics it is plausible to assume, in view of the importance of the second derivative in the classification, that, when the conditions are such that

(10.8.1)

can be resolved, a similar problem will be reasonably posed for (10.6.1) and thereby for the general hyperbolic partial differential equation. Some examples will now be given where resolution is possible, together with the corresponding interpretation for the general equation.

(a) The two-characteristic problem

Suppose that u is given on the portion of the characteristic 0≤ξ≤a, η=0, say u=h(ξ), and on the piece of characteristic ξ=0, 0≤η≤b by u=k(η) (Figure 10.8.1). From (10.7.3)

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FIGURE 10.8.1: The two-characteristic problem.

By putting η=0 we obtain

(10.8.2) for 0≤ξ≤a and by placing ξ=0

(10.8.3) for 0≤η≤b. Hence

on making ξ=0 in (10.8.2). The same result is obtained from (10.8.3) since h(0)=k(0) for the data on u to be consistent at the origin.

This solution will hold everywhere in the rectangle Oacb, which is the region common to the characteristics that intersect the data lines.

Translating this back to the (x, y)-plane and general partial differential equation, we see that giving u on the two characteristics O′A, O′B (Figure 10.8.2) determines u in the shaded region O′ACB where AC and BC are the other characteristics through A and B, respectively.

(b) The mixed problem

In the mixed problem, u=h(ξ) on 0≤ξ≤a, η=0 as before but either u or a linear combination of its first partial derivatives is given on an are Oc in the first quadrant, the are being met at most once by any characteristic. Suppose that the equation of Oc is η=θ(ξ). Take a new variable χ=θ(ξ). Then Oc becomes the straight line η=χ and the form of (10.8.1) is unaltered so that the characteristics remain parallel to the coordinate axes. There is

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FIGURE 10.8.2: The two-characteristic problem in the general case.

therefore no loss of generality in selecting Oc as the straight line η=ξ in the first place. Suppose that u=k(ξ) on Oc with k(0)=h(0). Then we have

(10.8.5) for 0≤ξ≤a. Therefore

for 0≤ξ≤a, 0≤η≤a. Consequently, u has been determined everywhere within the square with side Oa (Figure 10.8.3).

Instead of u being specified on Oc, the linear combination might be given on Oc. In that case (10.8.5) will be replaced by

Since f is known from (10.8.4), g can be calculated by integration and again u has been discovered in the square.

Going back to the (x, y)-plane we can say that when u is given on the characteristic are O′A (Figure 10.8.4) and either u or a linear combination of its first partial derivatives is specified on the are O′C, which lies between the characteristics O′A, O′B and is met at most once by any characteristic, then u is determined in the shaded region O′ ACB where AC and BC are characteristics.

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FIGURE 10.8.3: The mixed problem.

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(c) Cauchy’s problem

Cauchy’s problem is to determine u when u, ∂u/∂ξ and ∂u/∂η are given on Oc, where Oc has the same

properties as in the previous section, i.e., it is not a characteristic and is met at most once by a characteristic. As before, there is no loss of generality in taking Oc to be η=ξ.

Suppose that on Oc

There is a connection between h, k and l because

(10.8.6) since the equation of Oc is η=ξ.

The imposition of the given conditions supplies

The three are consistent on account of (10.8.6). With f found from the second equation and g from the first, u is obtained in the same square as in the mixed problem.

For the general equation, data on u and its first partial derivative on O′C in Figure 10.8.4 will fix u in the characteristic domain O′ACB.

A return to the reflection problem for the wave equation in Section 10.7 is pertinent here, to examine it from the point of view of this section. AB of Figure 10.7.2 is not a characteristic and u, ∂u/∂t are given on it. Therefore there is a Cauchy problem and, from the above, a solution is available in APB. From this solution u is known on AP, which is a characteristic are, and u=0 on AQ, which is not a characteristic but lies between the two characteristics through A. Hence there is a mixed problem and u can be found in APQ. Similarly, a mixed problem gives u in BPR. Now u is known on PQ and PR, which are both characteristic arcs, and so by solving a two-characteristic problem we determine u in PRTQ. In this way, a solution to a reflection problem for a general hyperbolic equation can be built up by means of characteristics.

It is possible to find u when u is given on two arcs Oc, Od in the first quadrant when they are not characteristics. If Oc is in the first quadrant and Od is in the fourth there is insufficient information for a unique solution. For if we give u any values on the ξ-axis we have a mixed problem in the first quadrant and another one in the fourth, so that a solution can be found.

When elliptic partial differential equations were discussed it was pointed out that Dirichlet’s problem was an appropriate one. It will now be shown why it is not suitable for hyperbolic equations in general.

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FIGURE 10.8.5: The Dirichlet problem for the hyperbolic equation.

Let u be prescribed on the boundary ABCD of Figure 10.8.5. The aim is to find u inside; so assume some values for u on OB. Then the form of u in OAB is known by solving a mixed problem and similarly in OBC. From the discovered values of u on OC, u is determined in OCD via a mixed problem. Similarly, values in OAD can be found. In order that the solution be correct, the values of u on OD from OCD and OAD must agree. This places a condition on the assumed values on OB. There is no guarantee that this condition can be fulfilled. If it cannot be met there is no solution to the Dirichlet problem. If the condition on OB can be satisfied there will be at least one solution and there may be several since there may be more than one way of complying with the condition.

We conclude that the Dirichlet problem for the hyperbolic differential equation must be treated with great caution.