**3. Two-tissue reversible compartment model of [11***C***]***PIB*

The mathematical model for the representation of the dynamics of [11*C*]*PIB* radiotracer, [11, 15, 20], is expressed by the system of two differential equations:

$$\begin{cases} \frac{d\mathbf{C}\_1}{dt} = \mathbf{K}\_1 \mathbf{C}\_a(t) - (k\_2 + k\_3)\mathbf{C}\_1(t) + k\_4 \mathbf{C}\_2(t) \\\frac{d\mathbf{C}\_2}{dt} = k\_3 \mathbf{C}\_1(t) - k\_4 \mathbf{C}\_2(t) \\\ \mathbf{C}\_1(\mathbf{0}) = \mathbf{0}, \mathbf{C}\_2(\mathbf{0}) = \mathbf{0}. \end{cases} \tag{4}$$

where *Ca*ð Þ*t* is the arterial input function (AIF) considered to be known, *C*1ð Þ*t* and *C*2ð Þ*t* are, respectively, the concentration within the non-displaceable and displaceable compartments and *K*<sup>1</sup> and *k*2, *k*3, *k*<sup>4</sup> are kinetic rate constants which have to be determined.

The Laplace transform with respect to *t* is applied to solve the system of differential equations Eq. (4), with the notation

*Biomechanical Model Improving Alzheimer's Disease DOI: http://dx.doi.org/10.5772/intechopen.92047*

$$\mathfrak{s}\{\mathbf{C}\_{i}(t)\} = \overline{\mathbf{C}}\_{i}(s) = \int\_{0}^{\infty} e^{-st} \mathbf{C}\_{i}(t)dt$$

and

**2.1 Estimation of rate constants**

*and and two tissues (C1 and C2).*

*Recent Advances in Biomechanics*

*∂ ∂Kij*

8 >>>>>><

>>>>>>:

to time.

**Figure 3.**

*∂ ∂Kji*

*dC*<sup>1</sup>

8 >>>><

>>>>:

have to be determined.

**80**

*dC*<sup>2</sup>

differential equations Eq. (4), with the notation

*d*

*d*

0 @

0 @ *dtCi*ðÞ¼ *<sup>t</sup>* <sup>X</sup>

*dtCi*ðÞ¼ *<sup>t</sup>* <sup>X</sup>

**3. Two-tissue reversible compartment model of [11***C***]***PIB*

*dt* <sup>¼</sup> *<sup>k</sup>*3*C*1ðÞ�*<sup>t</sup> <sup>k</sup>*4*C*2ð Þ*<sup>t</sup> C*1ð Þ¼ 0 0,*C*2ð Þ¼ 0 0*:*

In order to estimate the parameters *kij*, *kji*, a nonlinear regression problem is solved using the Levenberg-Marquardt method [18, 19]. The sensitivity equations are generated partially deriving Eq. (2) with respect to the parameters *kij*, *kji*

*A schematic diagram of a reversible two compartments model to illustrate the flux of tracer between blood (Ca)*

*N*

*j*¼1, *j*6¼*i*

*N*

*j*¼1, *j*6¼*i*

Over which region of interest (ROI) is defined discrete TAC using the image processing. The Jacobian matrix it consists of the column vectors whose values resulting from the numerical integration of the sensitivity equations with respect

The mathematical model for the representation of the dynamics of [11*C*]*PIB* radiotracer, [11, 15, 20], is expressed by the system of two differential equations:

*dt* <sup>¼</sup> *<sup>K</sup>*1*Ca*ðÞ�*<sup>t</sup>* ð Þ *<sup>k</sup>*<sup>2</sup> <sup>þ</sup> *<sup>k</sup>*<sup>3</sup> *<sup>C</sup>*1ð Þþ*<sup>t</sup> <sup>k</sup>*4*C*2ð Þ*<sup>t</sup>*

where *Ca*ð Þ*t* is the arterial input function (AIF) considered to be known, *C*1ð Þ*t* and *C*2ð Þ*t* are, respectively, the concentration within the non-displaceable and displaceable compartments and *K*<sup>1</sup> and *k*2, *k*3, *k*<sup>4</sup> are kinetic rate constants which

The Laplace transform with respect to *t* is applied to solve the system of

*KijCj*ðÞ�*<sup>t</sup> KjiCi*ð Þ*<sup>t</sup>* � �

*KijCj*ðÞ�*<sup>t</sup> KjiCi*ð Þ*<sup>t</sup>* � �

1 A

1 A (3)

(4)

$$\kappa \left\{ \frac{d\mathbf{C}\_k(t)}{dt} \right\} = s\overline{\mathbf{C}}\_i(s) - \mathbf{C}\_i(\mathbf{0}).$$

An algebraic system is obtained

$$\begin{cases} \left(s + k\_2 + k\_3\right) \overline{C}\_1(s) - k\_4 \overline{C}\_2(s) = K\_1 \overline{C}\_a(s) \\ -k\_3 \overline{C}\_1(s) + \left(s + k\_4\right) \overline{C}\_2(s) = \mathbf{0} \end{cases},\tag{5}$$

that can be written in matrix form as

$$
\begin{bmatrix} s + k\_2 + k\_3 & -k\_4 \\ -k\_3 & s + k\_4 \end{bmatrix} \begin{bmatrix} \overline{C}\_1(s) \\ \overline{C}\_2(s) \end{bmatrix} = \begin{bmatrix} K\_1 \overline{C}\_a(s) \\ \mathbf{0} \end{bmatrix}. \tag{6}
$$

The solution of the algebraic system (6) is

$$
\begin{bmatrix}
\overline{\mathbf{C}}\_1(\boldsymbol{\varepsilon}) \\
\overline{\mathbf{C}}\_2(\boldsymbol{\varepsilon})
\end{bmatrix} = \begin{bmatrix}
\boldsymbol{s} + \boldsymbol{k}\_2 + \boldsymbol{k}\_3 & -\boldsymbol{k}\_4 \\
\end{bmatrix}^{-1} \begin{bmatrix}
\boldsymbol{K}\_1 \overline{\mathbf{C}}\_a(\boldsymbol{\varepsilon}) \\
\mathbf{0}
\end{bmatrix}.\tag{7}
$$

The inverse matrix is

$$
\begin{bmatrix} s + k\_2 + k\_3 & -k\_4 \\ -k\_3 & s + k\_4 \end{bmatrix}^{-1} = \frac{\mathbf{1}}{(s + k\_2 + k\_3)(s + k\_4) - k\_3k\_4} \begin{bmatrix} s + k\_4 & k\_4 \\ k\_3 & s + k\_2 + k\_3 \end{bmatrix}. \tag{8}
$$

Therefore,

$$\begin{aligned} \overline{\mathbf{C}}\_{1}(\mathbf{s}) &= \frac{(\mathbf{s} + \mathbf{k}\_{4})K\_{1}\mathbf{C}\_{4}(\mathbf{s})}{s^{2} + (k\_{2} + k\_{3} + k\_{4})\mathbf{s} + k\_{2}k\_{4}} \\ \overline{\mathbf{C}}\_{2}(\mathbf{s}) &= \frac{k\_{3}K\_{1}\overline{\mathbf{C}}\_{4}(\mathbf{s})}{s^{2} + (k\_{2} + k\_{3} + k\_{4})\mathbf{s} + k\_{2}k\_{4}} \end{aligned} \tag{9}$$

Using the inverse Laplace in Eq. (9), results

$$\begin{split} \mathbf{C}\_{1}(t) &= \boldsymbol{\varepsilon}^{-1} \left\{ \frac{(s+k\_{4})K\_{1}\overline{C}\_{a}(s)}{s^{2}+(k\_{2}+k\_{3}+k\_{4})s+k\_{2}k\_{4}} \right\} \\ \mathbf{C}\_{2}(t) &= \boldsymbol{\varepsilon}^{-1} \left\{ \frac{k\_{3}K\_{1}\overline{C}\_{a}(s)}{s^{2}+(k\_{2}+k\_{3}+k\_{4})s+k\_{2}k\_{4}} \right\}. \end{split} \tag{10}$$

Now, the proprieties inverse Laplace transform are used, considering \* to denote the convolution<sup>1</sup> .

<sup>1</sup> The property of commutativity is valid in convolution operation for Laplace transform of *f t*ð Þ and *g t*ð Þ functions, defined by *f t*ð Þ <sup>∗</sup> *g t*ðÞ¼ <sup>Ð</sup>*<sup>t</sup>* <sup>0</sup> *f u*ð Þ*g t*ð Þ � *<sup>u</sup> du* <sup>¼</sup> <sup>Ð</sup>*<sup>t</sup>* <sup>0</sup> *f t*ð Þ � *u g u*ð Þ*du*.

$$\begin{aligned} \mathbf{C}\_{1}(t) &= \mathfrak{s}^{-1} \left\{ \frac{(s + k\_{4})}{s^{2} + (k\_{2} + k\_{3} + k\_{4})s + k\_{2}k\_{4}} \right\} \ast K\_{1} \mathfrak{s}^{-1} \left\{ \overline{\mathbf{C}}\_{a}(s) \right\} \\\\ \mathbf{C}\_{2}(t) &= \mathfrak{s}^{-1} \left\{ \frac{k\_{3}}{s^{2} + (k\_{2} + k\_{3} + k\_{4})s + k\_{2}k\_{4}} \right\} \ast K\_{1} \mathfrak{s}^{-1} \left\{ \overline{\mathbf{C}}\_{a}(s) \right\}. \end{aligned} \tag{11}$$

*d*

8 >>>>><

*Biomechanical Model Improving Alzheimer's Disease DOI: http://dx.doi.org/10.5772/intechopen.92047*

>>>>>:

*d*

outflow from the compartment (transport constants).

(

Eq. (18) is represented matrically

*C*1ð Þ*s C*2ð Þ*s*

Then,

**83**

" #

*C*1ð Þ*s C*2ð Þ*s*

<sup>¼</sup> <sup>1</sup>

*<sup>C</sup>*1ðÞ¼ *<sup>t</sup> £*�<sup>1</sup> *<sup>K</sup>*1*Ca*ð Þ*<sup>s</sup>*

*<sup>C</sup>*2ðÞ¼ *<sup>t</sup> £*�<sup>1</sup> *<sup>k</sup>*3*C*1ð Þ*<sup>s</sup>*

*s s*ð Þ þ *k*<sup>2</sup> þ *k*<sup>3</sup>

*C*1ðÞ¼ *s*

*C*2ðÞ¼ *s*

ð Þ *s* þ *k*<sup>2</sup> þ *k*<sup>3</sup> � �

*s* � �

The representation Eq. (22) implies that

*C*2ðÞ¼ *t k*<sup>3</sup> ∗*C*1ðÞ¼ *t k*<sup>3</sup>

*C*1ðÞ¼ *t K*<sup>1</sup> *e*

*C*2ðÞ¼ *t k*<sup>3</sup>

*C*1ðÞ¼ *t K*<sup>1</sup> *e*

ment model for [18*F*]*FDG* Eq. (17) is

" #

*dtC*1ðÞ¼ *<sup>t</sup> <sup>K</sup>*1*Ca*ðÞ�*<sup>t</sup>* ð Þ *<sup>k</sup>*<sup>2</sup> <sup>þ</sup> *<sup>k</sup>*<sup>3</sup> *<sup>C</sup>*1ð Þ*<sup>t</sup>*

*dtC*2ðÞ¼ *<sup>t</sup> <sup>k</sup>*3*C*1ð Þ*<sup>t</sup>* ,

where *Ca*ð Þ*t* is the input function and is considered to be known, *C*1ð Þ*t* and *C*2ð Þ*t* are the concentration in C1 and C2 compartments, respectively, and *K*1, *k*2, *k*<sup>3</sup> are positives proportionality rates describing, the tracer influx into and the tracer

Similarly to that developed in the previous section, considering *k*<sup>4</sup> ¼ 0, applying the Laplace transform with respect to *t* in Eq. (17), appear the algebraic system

ð Þ *s* þ *k*<sup>2</sup> þ *k*<sup>3</sup> *C*1ðÞ¼ *s K*1*Ca*ð Þ*s*

*s* 0 *k*<sup>3</sup> *s* þ *k*<sup>2</sup> þ *k*<sup>3</sup>

� ��<sup>1</sup> *<sup>K</sup>*1*Ca*ð Þ*<sup>s</sup>*

� � *<sup>K</sup>*1*Ca*ð Þ*<sup>s</sup>*

<sup>¼</sup> *<sup>k</sup>*3*C*1ð Þ*<sup>s</sup> s*

� � <sup>∗</sup> *£*�<sup>1</sup> *Ca*ð Þ*<sup>s</sup>* � �

�ð Þ *<sup>k</sup>*2þ*k*<sup>3</sup> ð Þ *<sup>t</sup>*�*<sup>u</sup> Ca*ð Þ *<sup>u</sup> du*

ð Þ *s* þ *k*<sup>2</sup> þ *k*<sup>3</sup>

ð*t* 0 *e*

*C*1ð Þ *u du:*

Then, with *k*<sup>2</sup> þ *k*<sup>3</sup> >0, the analytical solution of the irreversible two compart-

ð*t* 0 *e*

ð Þ *<sup>k</sup>*2þ*k*<sup>3</sup> *uCa*ð Þ *<sup>u</sup> du*

�*k*3*C*1ð Þþ*s sC*2ðÞ¼ *s* 0

<sup>¼</sup> *<sup>s</sup>* <sup>þ</sup> *<sup>k</sup>*<sup>2</sup> <sup>þ</sup> *<sup>k</sup>*<sup>3</sup> <sup>0</sup> �*k*<sup>3</sup> *s*

> *K*1*Ca*ð Þ*s s* þ *k*<sup>2</sup> þ *k*<sup>3</sup>

*k*3*K*1*Ca*ð Þ*s* ð Þ *s* þ *k*<sup>2</sup> þ *k*<sup>3</sup> *s*

<sup>¼</sup> *<sup>k</sup>*<sup>3</sup> <sup>∗</sup> *£*�<sup>1</sup> *<sup>C</sup>*1ð Þ*<sup>s</sup>* � �*:*

�ð Þ *<sup>k</sup>*2þ*k*<sup>3</sup> *<sup>t</sup>* <sup>∗</sup>*Ca*ðÞ¼ *<sup>t</sup> <sup>K</sup>*<sup>1</sup>

ð*t* 0

�ð Þ *k*2þ*k*<sup>3</sup> *t*

*C*1ð Þ *u du:*

ð*t* 0

<sup>¼</sup> *<sup>K</sup>*1*£*�<sup>1</sup> <sup>1</sup>

(17)

*:* (18)

*:* (19)

*:* (20)

(21)

(22)

(23)

(24)

0 " #

> 0 " #

*C*1ð Þ¼ 0 0,*C*2ð Þ¼ 0 0

Then,

$$\begin{aligned} \mathbf{C}\_{1}(t) &= K\_{1}\varepsilon^{-1} \left\{ \frac{s+k\_{4}}{s^{2}+(k\_{2}+k\_{3}+k\_{4})s+k\_{2}k\_{4}} \right\} \ast \mathbf{C}\_{a}(t) \\\\ \mathbf{C}\_{2}(t) &= K\_{1}\varepsilon^{-1} \left\{ \frac{k\_{3}}{s^{2}+(k\_{2}+k\_{3}+k\_{4})s+k\_{2}k\_{4}} \right\} \ast \mathbf{C}\_{a}(t) . \end{aligned} \tag{12}$$
 
$$\begin{aligned} \frac{s+k\_{4}}{s^{2}+(k\_{2}+k\_{3}+k\_{4})s+k\_{2}k\_{4}} &= \frac{A}{s-s\_{1}} + \frac{B}{s-s\_{2}} \\\\ \frac{k\_{3}}{s^{2}+(k\_{2}+k\_{3}+k\_{4})s+k\_{2}k\_{4}} &= \frac{C}{s-s\_{1}} + \frac{D}{s-s\_{2}} \end{aligned} \tag{13}$$

In Eq. (13) *s*<sup>1</sup> and *s*<sup>2</sup> are the roots of *s* <sup>2</sup> <sup>þ</sup> ð Þ *<sup>k</sup>*<sup>2</sup> <sup>þ</sup> *<sup>k</sup>*<sup>3</sup> <sup>þ</sup> *<sup>k</sup>*<sup>4</sup> *<sup>s</sup>* <sup>þ</sup> *<sup>k</sup>*2*k*<sup>4</sup> <sup>¼</sup> 0, dependent on transport constants *k*2, *k*3, and *k*4. The parameters *A*, *B*, *C*, and *D* are obtained by partial fraction decomposition technique. Then, because that the inverse Laplace transforms are simply linear combinations of exponential functions with the exponents *s*<sup>1</sup> and *s*<sup>2</sup> depending on *k*2, *k*3, and *k*4. Applying the linearity property of the inverse Laplace transform is obtained

$$\begin{aligned} \mathbf{C}\_{1}(t) &= K\_{1} \left\{ A e^{\ell\_{1}t} + B e^{\ell\_{2}t} \right\} \ast \mathbf{C}\_{a}(t) \\ \mathbf{C}\_{2}(t) &= K\_{1} \left\{ C e^{\ell\_{1}t} + D e^{\ell\_{2}t} \right\} \ast \mathbf{C}\_{a}(t) \end{aligned} \tag{14}$$

The analytical solution of the reversible two-compartment model for [11*C*]*PIB* (4) is

$$\begin{aligned} \mathbf{C}\_{1}(t) &= K\_{1} \left\{ A e^{\gamma\_{1}t} \Big|\_{0}^{t} e^{-\varepsilon\_{1}u} \mathbf{C}\_{a}(u) du + B e^{\varepsilon\_{2}t} \Big|\_{0}^{t} e^{-\varepsilon\_{2}u} \mathbf{C}\_{a}(u) du \right\} \\ \mathbf{C}\_{1}(t) &= K\_{1} \left\{ C e^{\gamma\_{1}t} \Big|\_{0}^{t} e^{-\varepsilon\_{1}u} \mathbf{C}\_{a}(u) du + D e^{\varepsilon\_{2}t} \Big|\_{0}^{t} e^{-\varepsilon\_{2}u} \mathbf{C}\_{a}(u) du \right\} \end{aligned} \tag{15}$$

In Eq. (15), it is visible the importance of construction of input function *Ca*ð Þ*t* in order to make it possible to calculate the integral

$$I = \int\_0^t \varepsilon^{\epsilon\_i u} \mathcal{C}\_a(u) \, du. \tag{16}$$
