Preface

This section is devoted to boundary value problems.

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Boundary Value Problems

Consider a second order differential equation

$x'' (t) = f(t,x,x' ) \qquad \mbox{for} \quad a \le t \le b,$
subject to the boundary conditions of the first kind (also called the Dirichlet boundary conditions)
$x (a) = \alpha \qquad \mbox{and} \qquad x(b) = \beta .$
Generally speaking, a boundry value problem may have a unique solutions, may have many solutions, or may have no solution. The conditions that guarantee that a solution to the formulated above Dirichlet boundary value problem exists should be checked before any numerical scheme is applied; otherwise,a list of meaningless output may be generated. The general conditions are stated in the following theorem. div id="theorem1" class="theorem">

Theorem: Suppose that f(t,x,y) is continuous on the region $$R = \left\{ (t,x,y)\, : \, a \le t \le b, \ -\infty < x < \infty , \ -\infty < y < \infty \right\}$$ and that $$\partial f/\partial x = f_x \quad \partial f/\partial y = f_y$$ are continuous on R. If there exists a positive constant M for which fx and fy satisfy

$\begin{split} f_x (t,x,y) > 0 \qquad\mbox{for all} \quad (t,x,y) \in R , \\ \left\vert f_y (t,x,y) \right\vert \le M \qquad\mbox{for all} \quad (t,x,y) \in R , \end{split}$
then the boundary value problem
$x'' (t) = f(t,x,x' ) \qquad \mbox{subject} \quad u(a) = \alpha, \quad u(b) = \beta$
has a unique solution x = x(t) for $$a \le t \le b .$$

The notation $$y = x' (t)$$ has been used to distinguish the third variable of the function $$f(t,x,x' ) .$$ Recall that we also use the dot notation for derivatives with respect to time variable t: $$\dot{x} = x' (t) .$$ Finally, the special case of linear differential equations is worthy of mention.

Theorem: Assume that f in the previous theorem has the linear form $$f(t,x,x' ) = p(t)\,x' + q(t)\,x + r(t) ,$$ and that f and its partial derivatives $$q(t) = \partial f/\partial x$$ and p(t) are continuous on R. If there exists a positive constant M for which p(t) and q(t) satisfy

$\begin{split} q (t) > 0 \qquad\mbox{for all} \quad (t,x,y) \in R , \\ \left\vert p (t) \right\vert \le M = \max_{a\le t \le b} \, |p(t)| \qquad\mbox{for all} \quad t \in [a,b] , \end{split}$
then the linear boundary value problem
$x'' (t) = p(t)\,x' + q(t)\,x + r(t) \qquad \mbox{subject} \quad u(a) = \alpha, \quad u(b) = \beta$
has a unqiue solution x = x(t) for $$a \le t \le b .$$

Conditions of the above theorem are fulfilled for constant coefficient equations when p(t) = p and q(t) = q. In this case, a closed form formula for solution is possible to obtain.

L[x_, y_] = y'''[x] - 2 y''[x] - 5 y'[x] + 6 y[x]
yb[x_] = y[x] /. DSolve[L[x, y] == 36 x, y[x], x][[1]]
const = Solve[{yb[0] == 2, yb[1] == 1, yb[2] == -1}, {C[1], C[2], C[3]}]
z[x_] = Simplify[yb[x] /. const[[1]]]
Out[5]= 5 + (E^(-1 + 3 x) (10 - 3 E - 18 E^2 + 10 E^3))/((-1 + E)^2 (1 + E) (1 + E +
E^2 + E^3 + E^4)) + ( E^(-1 + x) (-10 + 18 E^2 + 3 E^3 - 10 E^5))/((-1 + E)^2 (1 + E) (1 +
E + E^2)) + (E^(4 - 2 x) (-18 + 10 E + 10 E^3 - 3 E^4))/(
1 - E^3 - E^5 + E^8) + 6 x
{p1, p2, p3} = N[{C[1], C[2], C[3]} /. const[[1]]]
zb[x_] = yb[x] /. {C[1] -> p1, C[2] -> p2, C[3] -> p3}
Plot[zb[x], {x, -1, 2.5}]

1. Agarwal, R.P., Sheng, Q., Wong, P.J.Y., Abel-Gontscharoff boundary value problems, Math Comput Modeling, 1993, Vol. 17, No. 7, pp. 37--55.