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# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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u0(x, y) = c0,
un (x, y) = cn cosh(nnx/b) cos(nny/b), n = 1, 2, 3,....
(b) By superposing the fundamental solutions of part (a), formally determine a function u also satisfying the nonhomogeneous boundary condition ux (a, y) = f (y). Note that when ux (a, y) is calculated, the constant term in u(x, y) is eliminated, and there is no condition
10.8 Laplace’s Equation
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from which to determine c0. Furthermore, it must be possible to express f by means of a Fourier cosine series of period 2b, which does not have a constant term. This means that
[ f (y) dy = 0
0
is a necessary condition for the given problem to be solvable. Finally, note that c0 remains arbitrary, and hence the solution is determined only up to this additive constant. This is a property of all Neumann problems.
11. Find a solution u(r, 6) of Laplace’s equation inside the circle r = a, also satisfying the boundary condition on the circle
ur(a, 6) = g(6), 0 < 6 < 2n.
Note that this is a Neumann problem, and that its solution is determined only up to an arbitrary additive constant. State a necessary condition on g(6) for this problem to be solvable by the method of separation of variables (see Problem 10).
> 12. (a) Find the solution u(x, y) ofLaplace’s equation in the rectangle 0 < x < a,0 < y < b,
also satisfying the boundary conditions
u(0, y) = 0, u(a, y) = 0, 0 < y < b,
uy(x, 0) = 0, u(x, b) = g(x), 0 < x < a.
Note that this is neither a Dirichlet nor a Neumann problem, but a mixed problem in which u is prescribed on part of the boundary and its normal derivative on the rest.
(b) Find the solution if
, , fx, 0 < x < a/2,
g(x) =1 nZZ
ya — x, a/2 < x < a.
(c) Let a = 3 and b = 1. By drawing suitable plots compare this solution with the solution of Problem 1.
> 13. (a) Find the solution u(x, y) ofLaplace’s equation in the rectangle 0 < x < a,0 < y < b,
also satisfying the boundary conditions
u(0, y) = 0, u(a, y) = f (y), 0 < y < b,
u(x, 0) = 0, uy(x, b) = 0, 0 < x < a.
Hint: Eventually it will be necessary to expand f (y ) in a series making use of the functions sin(ny/2b), sin(3ny/2b), sin(5ny/2b), ... (see Problem 39 of Section 10.4).
(b) Find the solution if f (y) = y(2b — y).
(c) Let a = 3 and b = 2; plot the solution in several ways.
> 14. (a) Find the solution u(x, y) ofLaplace’s equation in the rectangle 0 < x < a,0 < y < b,
also satisfying the boundary conditions
ux (0, y) = 0, ux (a, y) = 0, 0 < y < b,
u(x, 0) = 0, u(x, b) = g(x), 0 < x < a.
(b) Find the solution if g(x) = 1 + x2(x — a)2.
(c) Let a = 3 and b = 2; plot the solution in several ways.
15. By writing Laplace’s equation in cylindrical coordinates r, 6, and z and then assuming that
the solution is axially symmetric (no dependence on 6), we obtain the equation
urr + (1/r )ur + uzz = 0.
Assuming that u(r, z) = R(r)Z(z), showthat R and Z satisfy the equations
rR" + R' + X2rR = 0, Z" — ę2 Z = 0.
The equation for R is Bessel’s equation of order zero with independent variable ę.
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Chapter 10. Partial Differential Equations and Fourier Series
Derivation of the Heat Conduction Equation. In this section we derive the differ-APPENDIX ential equation that, to a first approximation at least, governs the conduction of heat
A in solids. It is important to understand that the mathematical analysis of a physical
situation or process such as this ultimately rests on a foundation of empirical knowledge of the phenomenon involved. The mathematician must have a place to start, so
to speak, and this place is furnished by experience. Consider a uniform rod insulated on the lateral surfaces so that heat can flow only in the axial direction. It has been demonstrated many times that if two parallel cross sections of the same area A and different temperatures Tl and T2, respectively, are separated by a small distance d, an amount of heat per unit time will pass from the warmer section to the cooler one. Moreover, this amount of heat is proportional to the area A, the temperature difference | T2 — T |, and inversely proportional to the separation distance d. Thus
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