SI-method for solving stiff nonlinear boundary value problems

First of all, let us point out that

Otherwise, according to the the Pickard-Lindelof Theorem (see, for example, [1, p.350]), whose conditions are fulfilled, $u\left(x\right)$ must totally coincide with $0$ on $[a,b],$ which contradicts the condition $u\left(b\right)=u_b>0.$

Second, let us prove that $u\left(x\right)>0,$ $\forall x\in (a,b].$ Assume that the latter is not true and there exists at least one point $x_1\in (a,b)$ such that $u\left(x_1\right)\le 0.$ This immediately implies the existence of a point $b_1\in [x_1,b),$ such that

Obviously, function $u\left(x\right)\equiv 0$ also satisfies equation (1) which, in conjunction with condition (3), allows us to apply the result of Theorem 21 from [3, p. 48] (the maximum principle) and prove that neither $u\left(x\right)$ nor $-u\left(x\right)$ can achieve positive maximum on $[a,b_1]$ and, hence, $u\left(x\right)=0,$ $\forall x\in [a,b_1].$ The latter means that $u\left(0\right)=u^{\prime }\left(0\right)=0,$ which contradicts to (5)!?

The fact that $u\left(x\right)$ is positive on $(a,b]$ together with condition (4) means that

On the other hand, in the light of (5), the positiveness of $u\left(x\right)$ on $(a,b]$ immediately yields us

Combining (7) and (8) we get the statement of the Lemma about monotonicity of $u\left(x\right)$ on $[a,b].$ ∎

Indeed, from (16), (17) it follows that

The fact that $\tilde{u}\left(x\right)$ is continuously differentiable on $[a,c]$ and condition (20) yield us

which, in the light of formula (21), automatically guarantees that $\tilde{x}\left(u\right)$ is monotone on $\left[\tilde{u}\right(c),u_b].$ This completes the proof. ∎

Let ${\tilde{u}}_{\nu ,\mu ,i}\left(x\right)\in C^1\left(\right[a,c\left]\right)$ denote the solution of equation (13), (14) subjected to initial conditions

so that

Let us fix some arbitrary

and assume that

Under the conditions of Lemma and assumptions (29), (30) we are going to prove that

By definition, functions ${\tilde{u}}_{{\nu }_j,{\mu }_j,j}\left(x\right),$ ${\tilde{u}}_{{\overline{\nu }}_j,{\overline{\mu }}_j,j}\left(x\right)$ satisfy equations

respectively. It is easy to verify, that under conditions (3), (4), (23) the inequality

holds true.

Subtracting (34) from (33) and using inequalities (35), we get the estimate

From (29) and (30) it follows that

which, in conjunction with the maximum principle (which is applicable to $w\left(x\right)$ as well, see, for example, [3, Theorem 3, 4, p. 6–7])), yields us a fact that

The latter automatically implies inequalities (31) (32).

By now we proved that if conditions (29), (30) hold true for some $j\in \overline{0,N_1-1}$ then they are also fulfilled for $j+1$ with

Under the assumptions of the Lemma, inequality (27) implies conditions (29), (30) for $j=0$ with

and the rest obviously follows from what was proved above and the principle of mathematical induction. This completes the proof. ∎

The statement of the Lemma almost immediately follows from the corresponding theorem about continuity of solutions of IVPs with respect to initial conditions and parameters (see, for example, [1, Theorem 8.40, p 372]). ∎

From conditions (3), (4) and (23) and the maximum principle it follows that

The latter yields us the inequality

provided that

which, in conjunction with the obvious equality

Lemma 4 and the Bolzano’s theorem, provides us the existence of ${\nu }^{\ast }$ mentioned in the Lemma. The uniqueness follows from the monotonicity properties of ${\tilde{u}}_{\nu }\left(x\right)$ as a function of $\nu$ (Lemma 3). ∎

We start by proving that the function $\varphi \left(\nu \right)={\tilde{x}}_{\nu }\left(u_b\right)$ is continuous on $(0,{\nu }^{\ast }).$

It is easy to see that on each interval $[{\overline{u}}_i,{\overline{u}}_{i+1}],$ $i\in \overline{0,N_2-1}$ function ${\tilde{x}}_{\nu }\left(u\right)$ can be expressed in a recursive way

where

According to the definition of ${\overline{u}}_i$ given in (18), some intervals $[{\overline{u}}_i,{\overline{u}}_{i+1}]$ have zero measure, containing a single point ${\tilde{u}}_{\nu }\left(c\right).$ This, however, does not affect correctness of the reasoning below.