Comparative Study on Sixth Order Boundary Value Problems with Application to Linear Hydrodynamic Stability Problem and Benard Layer Eigenvalue Problem

Abstract

Numerical estimation for higher order eigenvalue problems are promising and has accomplished significant importance, mainly due to existence of higher order derivatives and boundary conditions relating to higher order derivatives of the unknown functions. In this article, we perform a numerical study of linear hydrodynamic stability of a fluid motion caused by an erratic gravity field. We employ two methods, collocation and spectral collocation based on Bernstein and Legendre polynomials to solve the linear hydrodynamic stability problems and Benard type convection problems. In order to handle boundary conditions, our techniques state all the unknown coefficients of boundary conditions derivatives in terms of known co-efficient. The schemes have been carried out to several test problems to establish the efficiency of the two methods.

Appendix

Lemma

Lemma [23] Suppose that the function$u:[0,1] \to R$is n + 1 times continuously differentiable i.e., $u \in C^{n + 1} \left. {\left( {[0,1]} \right)} \right)$and$\,\,V_{n} = span\,\,\,\left\{ {B_{0,n} (x) \, B_{1,n} (x) B_{2,n} (x)\ldots \ldots B_{n,n} (x)} \right\}$. If$C^{T} B$be the best approximation$u$out of${\text{V}}_{\text{n}}$, then

$$ \left\| {u - C^{T} B} \right\|_{{L^{2} \left[ {0,1} \right]}} \le \frac{\gamma }{{\left( {n + 1} \right)!\,\sqrt {2n + 3} }}[25]; $$

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where, $\gamma = \mathop {\hbox{max} }\limits_{x \in [0,1]} \,\left| {u^{n + 1} (x)} \right|\,$, $C = \left[ {c_{.0} ,c_{1.} ,c_{2.} , \ldots ,c_{n} } \right]^{T}$

Proof

Since, the set $p_{n} = \left\{ {x_{1} ,x_{2} ,x_{3} , \ldots ,x_{n} } \right\}$ is a basis polynomials space of degree n.

Therefore, we define the function

$$ {\text{u}}_{1} ({\text{x}}) = {\text{u}}\,(0) + {\text{x}}\,{{u^{\prime}}}\,(0) + \frac{{{\text{x}}^{ 2} }}{2\,!}{{u^{\prime\prime}}}\left( 0 \right) + \frac{\text{x}}{3\,!}^{ 3} {{u^{\prime\prime\prime}}}\left( 0 \right) + \cdots + \frac{{{\text{x}}^{\text{n}} }}{{{\text{n}}\,\,!}}{{u^{\prime\prime}}}\left( 0 \right) $$

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From the Taylor expansion, we have

$$ \left| {u(x) - u_{1} (x)} \right| = \int\limits_{0}^{1} {\left| {u^{n + 1} \left( {\alpha_{x} } \right)\,\frac{{x^{n + 1} }}{(n + 1)\,\,!}} \right|} $$

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where, ${{\alpha }}_{\text{x}} \in \left( {0,\,\,1} \right)$. Since $C^{T} B$ be the finest estimation of ${\text{u}}$ out of ${\text{V}}_{\text{n}}$, $u_{1} \in V_{n}$.

$$ \begin{aligned} \left\| {\,u - C^{T} } \right\|_{{L^{2} (0,1)}}^{2} \le\, & \left\| {u - u_{1} } \right\|_{{L^{2} (0,1)}}^{2} = \int\limits_{0}^{1} {\left| {u(x) - u_{1} (x)} \right|^{2} } dx \\ = & \int\limits_{0}^{1} {\left| {u^{n + 1} \left( {\alpha_{x} } \right)} \right|^{2} } \left( {\frac{{x^{n + 1} }}{(n + 1)\,\,!}} \right)^{2} dx \\ \le & \frac{{x^{n + 1} }}{{(n + 1)\,\,!^{2} }}\int\limits_{0}^{1} {x^{2n + 2} } dx = \frac{\gamma }{{(n + 1)!^{2} (2n + 3)}} \\ \end{aligned} $$

Taking the square root on both sides,

$$ \left\| {\,u - C^{T} } \right\|_{{L^{2} (0,1)}}^{2} \le \frac{\gamma }{{(n + 1)!^{2} (2n + 3)}} $$

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We thus proved.