In this section, we present the proposed method for the solution of dynamics of the Heartbeat Model of the form

?x??(t) = -(x?³-Tx?+x?), T>0,

x??(t) = x?-x_{d},

x?(0) = c?,x?(0)=c?,

where x(t) represents the characteristics of heartbeat in terms of length of the muscle fiber x?(t) and electrochemical activity x?(t), the parameter ? is a small positive constant associated with the fast perturbation factor the system, x_{d} is a scalar quantity representing a typical length of muscle fiber in the diastolic state, T represents tension in the muscle fiber, c? and c? are constant representing the initial conditions. We can rewrite Eq. 2 as

x?(t) = ??^{t}(x?(s)-x_{d})ds

= ??^{t}x?(s)ds-x_{d}t+c.

Since x?(0)=c?, then

x?(t)=??^{t}x?(s)ds-x_{d}t+c?.

Thus, System (1)-(3) becomes

x??(t) = ((-1)/?)(x?³-Tx?+??^{t}x?(s)ds-x_{d}t+c?), T>0,

x?(0) = c?.

Let y(t)=x?(t)-c?. Then,

y?(t) = ((-1)/?)((y(t)+c?)³-Ty(t)-Tc?+??^{t}(y(s)+c?)ds-x_{d}t+c?),

y(0) = 0.

Let

?y = y?(t)+(1/?)((3c?²-T)y(t)+??^{t}y(s)ds)

= (1/?)(-c?³+Tc?-c?+(x_{d}-c?)t)=f(t),

y(0) = 0.

In order to solve Problem (9)-(10), we construct the kernel Hilbert spaces W?²0,T and W?¹0,T . Let

W?¹0,T={y(s): y is absolutely continuous real value function, y??L²0,T}.

The inner product in W?¹0,T is defined as

(u(t),v(t))_{W?¹0,T}=u(0)v(0) +??^{T}u?(t)v?(t)dy,

and the norm ?u?_{W?¹0,T} is given by

?u?_{W?¹0,T}=?((u(y),u(y))_{W?¹0,T})

where u,v?W?¹0,T.

Theorem 1. The space W?¹0,T is a reproducing kernel Hilbert space, i.e.; there exists R(s,t)?W?¹0,T and its second partial derivative with respect to t exists such that for any u?W?¹0,T and each fixed t,s?0,T, we have

(u(t),R(s,t))_{W?¹0,T}=u(s).

In this case, R(s,t) is given by

R(s,t)={}.

1+t, t?s

1+s , t>s

Proof: Using integration by parts and the fact that R(s,y) is a reproducing kernel of W?¹0,T, we get

R(s,t)={?

c?(s)+c?(s)t, t?s

d?(s)+d?(s)t, t>s

which implies that

c?(s)-c?(s) = 0,

d?(s) = 0,

c?(s)+c?(s) s = d?(s)+d?(s) s,

d?(s)- c?(s) = -1.

Thus,

c?(s)=1, c?(s)=1,d?(s)=1+s, d?(s)=0.

Next, we study the space W?²0,t. Let

W?²0,T = {y(s): y and y??? are absolutely continuous real value functions,

y?? ? L²0,T1,f(0)=0}.

The inner product in W?²0,T is defined as

(u(t),v(t))_{W?²0,T}=u(0)v(0)+u(T)v(T)+??^{T}u?²?(t)v?²?(t)dt,

and the norm ?u?_{W?²0,T} is given by

?u?_{W?²0,T}=?((u(t),u(t))_{W?³0,T})

where u,v?W?²0,T.

Theorem 2. The space W?²0,T is a reproducing kernel Hilbert space, i.e.; there exists K(s,t)?W?²0,T which has its four partial derivative with respect to t such that for any u?W?²0,T and each fixed t,s?0,T, we have

(u(y),K(s,y))_{W?²0,T}=u(s).

In this case, K(s,t) is given by

K(s,y)={}

?_{i=0}³c_{i}(s)t^{i}, t?s

?_{i=0}³d_{i}(s)t^{i}, t>s

where

c? = 0, c?=-((6s+s³T-3s²T²+2sT³)/(6T²)), c?=0,c?= -((s-T)/(6T)),

d? = ((s³)/6), d?=-((6s+s³T+2sT³)/(6T²)), d?=(s/2),d?= -(s/(6T)).

Proof: Using integration by parts and the fact that K(s,t) is a reproducing kernel of W?²0,T, we get

K(s,y)={}.

?_{i=0}³c_{i}(s)t^{i}, t?s

?_{i=0}³d_{i}(s)t^{i}, t>s

Thus,

c?(s) = 0, ?_{i=0}³d_{i}(s)T^{i}-6d?(s)=0,

c?(s) = 0, 6d?(s)T+2d?(s)=0,

?_{i=0}³c_{i}(s)s^{i}=?_{i=0}³d_{i}(s)s^{i},

?_{i=1}³ic_{i}(s)s^{i-1} = ?_{i=i}³id_{i}(s)s^{i-1},

?_{i=2}³i(i-1)c_{i}(s)s^{i-2} = ?_{i=1}³i(i-1)d_{i}(s)s^{i-2},

6d?(s)-6c?(s)=-1.

Therefore,

c? = 0, c?=-((6s+s³T-3s²T²+2sT³)/(6T²)), c?=0,c?= -((s-T)/(6T)),

d? = ((s³)/6), d?=-((6s+s³T+2sT³)/(6T²)), d?=(s/2),d?= -(s/(6T)).

Now, we present how to solve Problem (9)-(10) using the reproducing kernel method. Let

?_{i}(s)=R(s_{i},s)

for i=1,2,… where {s_{i}}_{i=1}^{?} is dense on 0,T. It is clear that L: W?²0,T?W?¹0,T is bounded linear operator. Let

?_{i}(s)=L^{?}?_{i}(s)

where L(?_{i}(s))=((?³?_{i}(s))/(?y³))+a?(s)((??_{i}(s))/(?y))+a?(s)?_{i}(s) and L^{?} is the adjoint operator of L. Using Gram-Schmidt orthonormalization to generate orthonormal set of functions {?_{i}(s)}_{i=1}^{?} where

?_{i}(s)=?_{j=1}^{i}?_{ij}?_{j}(s) 3.17

and ?_{ij} are coefficients of Gram-Schmidt orthonormalization. In the next theorem, we show the existence of the solution of Problem (9)-(10).

Theorem 3. If {s_{i}}_{i=1}^{?} is dense on 0,T, then

f(s)=?_{i=1}^{?}?_{j=1}^{i}?_{ij}h(s_{j})?_{i}(s). 2.nnn

Proof: One can see that

?_{i}(s) = L^{?}?_{i}(s)=(L^{?}?_{i}(s),K(s,y))_{W?²0,T}

= (?_{i}(s),L(K(s,y)))_{W?²0,T}=L(K(s,s_{i})).

For each fixed f(s)?W?²0,T, let

(f(s),?_{i}(s))_{W?²0,T}=0, i=1,2,….

Then,

(f(s),?_{i}(s))_{W?²0,T} = (f(s),L^{?}?_{i}(s))_{W?²0,T}

= (Lf(s),?_{i}(s))_{W?²0,T}

= Lf(s_{i})=0.

Since {s_{i}}_{i=1}^{?} is dense on 0,T, Lf(s)=0. Since L?¹ exists, u(s)=0. Thus, {?_{i}(s)}_{i=1}^{?} is the complete system of W?²0,T.

Moreover,