Inverse determination of blood perfusion coefficient by using different deterministic and heuristic techniques

Abstract

This work aims to establish an estimation of the blood perfusion coefficient in cancerous tissues, using inverse problem techniques. The blood perfusion coefficient is modeled as a constant parameter or as a function of the position. Several optimization methods are studied for the constant parameter model. The function estimation approach is performed by the conjugate gradient method with adjoint problem to evaluate the model with spatial variation of the blood perfusion coefficient. The heat transfer problem is represented by the Pennes’ equation. It is a standard heat diffusion equation with a sink term and a source term. The sink term accounts for the blood flow within the biological tissue and the source term accounts for the combined effect of the internal metabolic heat generation together with an external heat flux associated to the cancer treatment. The results are obtained for different functions of blood perfusion coefficients, simulating measurements with and without errors. The performances of the studied inverse problem techniques are analyzed.

Appendix

In this section, it is reported the use of the integral transform technique in the solution of the proposed problem.

$$ \frac{{\partial^{2} \theta }}{{\partial X^{2} }} + \frac{{\partial^{2} \theta }}{{\partial Y^{2} }} - P_{\text{f}} \theta + G = \frac{\partial \theta }{\partial \tau },\quad 0 < Y < {\text{H}},\;0 < X < 1,\;\tau > 0 $$

(48)

$$ \frac{\partial \theta }{\partial X} = 0,\quad X = 0,\;\tau > 0 $$

(49)

$$ \frac{\partial \theta }{\partial X} = 0,\quad X = 1,\;\tau > 0 $$

(50)

$$ - \frac{\partial \theta }{\partial Y} + {\text{Bi}}\theta = {\text{Bi}}\theta_{\infty },\quad Y = 0,\;\tau > 0 $$

(51)

$$ \theta = 0,\quad Y = H,\;\tau > 0 $$

(52)

$$ \theta = \theta_{0},\quad 0 < Y < 1/L,\;0 < X < 1,\;\tau = 0 $$

(53)

where H = 1/L.

The integral transform pair for the function $\theta \left( {X ,Y,\tau } \right)$ with respect to the X variable is readily obtained by splitting up the representation into two parts as

Integral transform:

$$ \overline{\theta } \left( {\mu_{i} ,Y,\tau } \right) = \int\limits_{0}^{1} {\Uppsi \left( {\mu_{i} ,X} \right)} \theta \left( {X,Y,\tau } \right){\text{d}}X $$

(54)

Inversion formula:

$$ \theta \left( {X,Y,\tau } \right) = \sum\limits_{i = 0}^{\infty } {\frac{{\Uppsi \left( {\mu_{i} ,X} \right)}}{{N\left( {\mu_{i} } \right)}}\overline{\theta } \left( {\mu_{i} ,Y,\tau } \right)} $$

(55)

The eigenvalue problem associated is given by the equations above,

$$ \frac{{{\text{d}}^{2} \Uppsi }}{{{\text{d}}X^{2} }} + \mu_{i}^{2} \Uppsi = 0,\quad 0 < X < 1 $$

(56)

$$ \frac{{{\text{d}}\Uppsi }}{{{\text{d}}X}} = 0,\quad X = 0 $$

(57)

$$ \frac{{{\text{d}}\Uppsi }}{{{\text{d}}X}} = 0,\quad X = 1 $$

(58)

The functions $\Uppsi \left( {\mu_{i} ,X} \right)$, $N\left( {\mu_{i} } \right)$ and the eigenvalues $\mu_{\text{i}}$ from table 2-2 case 5 Özisik [17].

The associated eigenfunction is

$$ \Uppsi \left( {\mu_{i} ,X} \right) = \cos \left( {\mu_{i} X} \right),\quad i = 0,1,2, \ldots $$

(59)

and the norm is

$$ \frac{1}{{N\left( {\mu_{i} } \right)}} = \left\{ {\begin{array}{*{20}c} 2,& \quad {\mu_{i} \ne 0} \\ 1, & \quad {\mu_{i} = 0} \\ \end{array} } \right. $$

(60)

The eigenvalues are the positive roots of the following equation

$$ {\text{sen}}\left( {\mu_{i} X} \right) = 0,i = 0,1,2, \ldots $$

(61)

$$ \therefore \mu_{i} = i\pi ,i = 0,1,2, \ldots $$

(62)

Applying $\int\nolimits_{0}^{1} {\Uppsi \left( {\mu_{i} ,X} \right)} \left( \cdot \right){\text{d}}X$ into Eq. (48).

$$ \begin{gathered} \int\limits_{0}^{1} {\Uppsi \left( {\mu_{\text{i}} ,X} \right)} \frac{{\partial^{2} \theta }}{{\partial X^{{^{2} }} }}{\text{d}}X + \int\limits_{0}^{1} {\Uppsi \left( {\mu_{i} ,X} \right)} \frac{{\partial^{2} \theta }}{{\partial {\text{Y}}^{{^{2} }} }}{\text{d}}X - \int\limits_{0}^{1} {\Uppsi \left( {\mu_{\text{i}} ,{\text{X}}} \right)} P_{\text{f}} \theta {\text{d}}X + \underbrace {{\int\limits_{0}^{1} {\Uppsi \left( {\mu_{\text{i}} ,X} \right)} {\text{Gd}}X}}_{{\overline{\text{G}}_{i} }} \hfill \\ = \int\limits_{0}^{1} {\Uppsi \left( {\mu_{i} ,X} \right)} \frac{\partial \theta }{\partial \tau }{\text{d}}X \hfill \\ \end{gathered} $$

(63)

$$ \underbrace {{\int\limits_{0}^{1} {\Uppsi \left( {\mu_{i} ,X} \right)} \frac{{\partial^{2} \theta }}{{\partial X^{2} }}{\text{d}}X}}_{{({\text{I}})}} + \frac{{\partial^{2} \overline{\theta } }}{{\partial Y^{{^{2} }} }} - P_{\text{f}} \overline{\theta } + \overline{G}_{i} = \frac{{\partial \overline{\theta } }}{\partial \tau } $$

(64)

Solving the integral (I) and using the Eqs. (49), (50) and Eqs. (57), (58).

$$ ({\text{I}}) = - \mu_{i}^{2} \overline{\theta } \left( {\mu_{i} ,Y,\tau } \right) $$

(65)

Replacing the Eq. (65) into Eq. (64) it is obtained the following partial differential equation for the transform $\overline{\theta } \left( {\mu_{i} ,Y,\tau } \right),$

$$ - \mu_{i}^{2} \overline{\theta } + \frac{{\partial^{2} \overline{\theta } }}{{\partial Y^{{^{2} }} }} - P_{\text{f}} \overline{\theta } + \overline{G}_{i} = \frac{{\partial \overline{\theta } }}{\partial \tau } $$

(66)

Transforming the Eqs. (51)–(53) in “X”:

$$ - \frac{{\partial \overline{\theta } \left( {\mu_{i} ,Y,\tau } \right)}}{\partial Y} + {\text{Bi}}\overline{\theta } \left( {\mu_{i} ,Y,\tau } \right) = {\text{Bi}}\int\limits_{0}^{1} {\Uppsi \left( {\mu_{\text{i}} ,{\text{X}}} \right)} \theta_{\infty } {\text{d}}X\quad Y = 0,\;\tau > 0 $$

(67)

$$ \overline{\theta } \left( {\mu_{i} ,Y,\tau } \right) = 0,\quad Y = H,\;\tau > 0 $$

(68)

$$ \overline{\theta } \left( {\mu_{i} ,Y,\tau } \right) = \int\limits_{0}^{1} {\Uppsi \left( {\mu_{i} ,X} \right)} \, \theta_{0} {\text{d}}X,\quad 0 < Y < H,\;0 < X < 1,\;\tau = 0 $$

(69)

The integral transform pair for the function $\overline{\theta } \left( {\mu_{i} ,Y,\tau } \right)$ with respect to the Y variable is readily obtained by splitting up the representation into two parts as

Integral transform:

$$ \overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,\tau } \right) = \int\limits_{0}^{H} {\lambda \left( {\gamma_{j} ,Y} \right)} \overline{\theta } \left( {\mu_{i} ,Y,\tau } \right){\text{d}}Y $$

(70)

Inversion formula:

$$ \overline{\theta } \left( {\mu_{i} ,Y,\tau } \right) = \sum\limits_{j = 1}^{\infty } {\frac{{\lambda \left( {\gamma_{j} ,Y} \right)}}{{{\text{M}}\left( {\gamma_{j} } \right)}}\overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,\tau } \right)} $$

(71)

The eigenvalue problem associated is given by the equations above,

$$ \frac{{{\text{d}}^{2} \lambda }}{{{\text{d}}Y^{2} }} + \left( {\gamma_{j}^{2} - P_{\text{f}} } \right) \, \lambda = 0,\,0 < Y < H $$

(72)

$$ - \frac{{{\text{d}}\lambda }}{{{\text{d}}Y}} + {\text{Bi}}\lambda = 0,Y = 0 $$

(73)

$$ \lambda = 0,Y = H $$

(74)

where,

$$ \gamma_{j}^{2} - P_{\text{f}} = \beta_{j}^{2} $$

(75)

The functions $\lambda \left( {\gamma_{j} ,Y} \right)$, $M\left( {\gamma_{j} } \right)$ and the eigenvalues $\gamma_{\text{j}}$ from table 2-2 case 3 Özisik [17]. The associated eigenfunction is

$$ \lambda \left( {\gamma_{j} ,Y} \right) = {\text{sen}}\left( {\gamma_{j} Y} \right),\quad j = 0,1,2, \ldots $$

(76)

and the norm is

$$ \frac{1}{{M\left( {\gamma_{j} } \right)}} = \frac{{H\left( {\beta_{j}^{2} + {\text{Bi}}^{2} } \right) + {\text{Bi}}}}{{2\left( {\beta_{\text{j}}^{2} + {\text{Bi}}^{2} } \right)}} $$

(77)

The eigenvalues are the positive roots of the following equation

$$ \beta_{j} \cot \left( {\beta_{j} H} \right) = - {\text{Bi}},\quad i = 0,1,2, \ldots $$

(78)

Applying $\int\nolimits_{0}^{H} {\lambda \left( {\gamma_{j} ,Y} \right)} \left( \cdot \right){\text{d}}Y$ into Eq. (66)

$$ - \mu_{i}^{2} \int\limits_{0}^{H} {\lambda \left( {\gamma_{j} ,Y} \right)} \overline{\theta } {\text{d}}Y + \int\limits_{0}^{H} {\lambda \left( {\gamma_{j} ,Y} \right)} \frac{{\partial^{2} \overline{\theta } }}{{\partial Y^{{^{2} }} }}{\text{d}}Y - P_{\text{f}} \int\limits_{0}^{H} {\lambda \left( {\gamma_{j} ,Y} \right)} \overline{\theta } {\text{d}}Y + \underbrace {{\int\limits_{0}^{H} {\lambda \left( {\gamma_{j} ,Y} \right)} \overline{\text{G}}_{i} {\text{d}}Y}}_{{\overline{\overline{\text{G}}}_{ij} }} = \int\limits_{0}^{\text{H}} {\lambda \left( {\gamma_{j} ,Y} \right)} \frac{{\partial \overline{\theta } }}{\partial \tau }_{i} {\text{d}}Y $$

(79)

Then,

$$ - \mu_{i}^{2} \overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,\tau } \right) + \underbrace {{\int\limits_{0}^{H} {\lambda \left( {\gamma_{j} ,Y} \right)} \frac{{\partial^{2} \overline{\theta } }}{{\partial Y^{{^{2} }} }}{\text{d}}Y}}_{{({\text{II}})}} - P_{\text{f}} \overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,\tau } \right) + \overline{\overline{\text{G}}}_{ij} = \frac{{{\text{d}}\overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,\tau } \right)}}{{{\text{d}}\tau }} $$

(80)

Solving the integral (II) and using the Eqs. (67), (68) and Eqs. (72), (73).

$$ \begin{gathered} - \mu_{i}^{2} \overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,\tau } \right) - \beta_{j}^{2} \overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,\tau } \right) + {\text{Bi}}\theta_{\infty } \lambda \left( {\gamma_{j} ,0} \right)\int\limits_{0}^{1} {\Uppsi \left( {\mu_{i} ,X} \right)} {\text{d}}X \hfill \\ - \gamma_{j}^{2} \overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,\tau } \right) + \beta_{j}^{2} \overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,\tau } \right) + \overline{\overline{\text{G}}}_{ij} = \frac{{{\text{d}}\overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,\tau } \right)}}{{{\text{d}}\tau }} \hfill \\ \end{gathered} $$

(81)

Considering

$$ P_{ij} = {\text{Bi}}\theta_{\infty } \lambda \left( {\gamma_{j} ,0} \right)\int\limits_{0}^{1} {\Uppsi \left( {\mu_{i} ,X} \right)} {\text{d}}X + \overline{\overline{\text{G}}}_{ij} $$

(82)

The Eq. (81) is rewritten as

$$ \frac{{{\text{d}}\overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,\tau } \right)}}{{{\text{d}}\tau }} + \left( {\mu_{i}^{2} + \gamma_{j}^{2} } \right)\overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,\tau } \right) = P_{ij} $$

(83)

Transforming the Eq. (69) in “Y”,

$$ \overline{\overline{{\overline{\theta } }}} \left( {\mu_{i} ,\gamma_{j} ,0} \right) = \int\limits_{0}^{\text{H}} {\int\limits_{0}^{1} {\lambda \left( {\gamma_{j} ,Y} \right) \, \Uppsi \left( {\mu_{i} ,X} \right)} \, \theta_{0} {\text{d}}X\;{\text{d}}Y} $$

(84)

Solving the Eqs. (83), (84) with the integrating factor, e.g., multiplying the Eq. (83) by ${\text{e}}^{{\left( {\mu_{i}^{2} + \gamma_{j}^{2} } \right)}}$ and integrating from 0 to τ,

$$ \overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,\tau } \right) = \frac{{P_{ij} }}{{\mu_{i}^{2} + \gamma_{j}^{2} }} - \frac{{{\text{e}}^{{ - \left( {\mu_{i}^{2} + \gamma_{j}^{2} } \right)\tau }} }}{{\mu_{i}^{2} + \gamma_{j}^{2} }}P_{ij} + {\text{e}}^{{ - \left( {\mu_{i}^{2} + \gamma_{j}^{2} } \right)\tau }} \overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,0} \right) $$

(85)

Replacing the Eq. (85) into Eq. (71) and using the Eq. (55) is obtained the analytical solution

$$ \theta \left( {X ,Y,\tau } \right) = \sum\limits_{i = 0}^{\infty } {\sum\limits_{j = 1}^{\infty } {\frac{{\Uppsi \left( {\mu_{\text{i}} ,X} \right)}}{{N\left( {\mu_{i} } \right)}}\frac{{\lambda \left( {\gamma_{j} ,Y} \right)}}{{M\left( {\gamma_{j} } \right)}} \cdot \left[ {\frac{{P_{ij} }}{{\mu_{i}^{2} + \gamma_{j}^{2} }} - \frac{{{\text{e}}^{{ - \left( {\mu_{i}^{2} + \gamma_{j}^{2} } \right)\tau }} }}{{\mu_{i}^{2} + \gamma_{j}^{2} }}P_{ij} + {\text{e}}^{{ - \left( {\mu_{i}^{2} + \gamma_{j}^{2} } \right)\tau }} \overline{{\overline{\theta } }} \left( {\mu_{i} ,\gamma_{j} ,0} \right)} \right]} } $$

(86)