Nanoparticle-Mediated Heating: A Theoretical Study for Photothermal Treatment and Photo Immunotherapy

Abstract

Plasmonic metal nanoparticles are efficient absorbers of optical energy and thus can serve as a heat source for photothermal therapy. In this study, time-dependent solutions to the heat-flow equation are derived for the temperature elevation of tissue arising from the absorption of light by a suspension of plasmonic nanoparticles. Analytical solutions for the temperature are obtained assuming the diffusion approximation for light transport. Two types of heat sources are considered: heat production arising from a point source of light and from planewave illumination. The results will provide a theoretical basis for the design and optimization of experimental systems for photothermal treatment and photo immunotherapy.

Appendices

Appendix 1: Derivation of the Green’s Functions

For a spherically-symmetric source, the Green’s function, g(r, t|r′, t′), for the diffusion equation is defined as the solution to

$$\displaystyle \begin{aligned} {1\over r^2} {\partial~\over \partial r} \left(r^2 {\partial g \over\partial r} \right) - {1\over\kappa}{\partial g\over \partial t} - {g\over\kappa\tau} \,=\, -{1\over r^2}\delta(r\!-\!r') \delta(t\!-\!t'). {} \end{aligned} $$

(37)

To find g(r, t|r′, t′), we employ the completeness relation for the spherical Bessel function j ₀(rρ):

$$\displaystyle \begin{aligned} \int_0^\infty j_0(r\rho) j_0(r'\rho) \rho^2 \, d\rho \,=\, {\pi\over 2r^2} \delta(r \!-\! r'). {} \end{aligned} $$

(38)

We now let [52]

$$\displaystyle \begin{aligned} g(r,t|r',t') \,=\, \int_0^\infty C_\rho(t,t')\, j_0(r\rho) j_0(r'\rho)\, \rho^2\, d\rho, {} \end{aligned} $$

(39)

and substitute into (37). Noting the relation,

$$\displaystyle \begin{aligned} {1\over r^2} {\partial~\over \partial r} \left(r^2 {\partial j_0(\rho r) \over\partial r} \right) + \rho^2 j_0(\rho r) \,=\, 0, {} \end{aligned} $$

(40)

and using (38) for $\delta (r\!-\!r')/r^2$ on the right of (37), we obtain the following differential equation for C _ρ(t, t′):

$$\displaystyle \begin{aligned} {1\over \kappa} {dC_\rho\over dt} + (\rho^2 \!+\! 1/\kappa\tau)C_\rho \,=\, {2\over \pi}\delta(t \!-\! t'). \end{aligned}$$

The solution to this equation is

$$\displaystyle \begin{aligned} C_\rho(t,t') \,=\, {2\kappa\over\pi} u(t\!-\!t')\, e^{-\kappa(\rho^2 + 1/\kappa\tau)(t-t')}, {} \end{aligned}$$

where u(t) is the unit step function. Substituting into (39), we obtain (10).

For a planewave source, the Green’s function, g(z, t|z′, t′), for the diffusion equation is defined as the solution to

$$\displaystyle \begin{aligned} {\partial^2 g \over \partial z^2} - {1\over\kappa}{\partial g\over \partial t} - {g\over\kappa\tau} \,=\, -\delta(z\!-\!z') \delta(t\!-\!t'). \end{aligned}$$

Using the same method as above, the free-space green’s function is found to be

$$\displaystyle \begin{aligned} g(z,t|z',t') \,=\, {\kappa\over\pi}u(t\!-\!t')\int_0^\infty \cos[s(z\!-\!z')]\, e^{-\kappa(s^2 + 1/\kappa\tau)(t-t')}\, ds. \end{aligned}$$

To enforce boundary conditions 1 and 2 we employ linear combinations of the free-space Green’s functions as follows. The Green’s functions for boundary conditions 1 abd 2 are, respectively, $g_1(z,t|z',t') = g(z,t|z',t') - g(z,t|-\!z',t')$ and $g_2(z,t|z',t') = g(z,t|z',t') + g(z,t|-\!z',t')$, which result in (22) and (23).

Appendix 2: Temperature Elevation for Boundary Condition 3

We can derive the temperature elevation for boundary condition 3 using the same approach as above once we have the Green’s function g ₃(z, t|z′, t′) that enforces this boundary condition. One can show that

$$\displaystyle \begin{aligned} g_3(z,t|z',t') \,=\, g_2(z,t|z',t') + g_h(z,t|z',t'), \end{aligned}$$

where g ₂(z, t|z′, t′) is given by (23) and

$$\displaystyle \begin{aligned} \begin{array}{rcl} & &\displaystyle g_h(z,t|z',t')\\ & &\displaystyle \equiv {2h\kappa\over\pi}u(t\!-\!t')\int_0^\infty e^{-\kappa(t-t')(s^2 + 1/\kappa\tau)}\left\{{s\sin[(z\!+\!z')s] - h\cos[(z\!+\!z')s] \over h^2 \!+\! s^2} \right\} ds. \end{array} \end{aligned} $$

The most straightforward method of deriving this result is to use Laplace transforms; the general procedure is outlined in [53]. The Green’s function g ₃ reduces to g ₂ when h = 0. One can also show that g ₃ → g ₁ in the limit h →∞.

The temperature elevation T ₃(z, t) is now obtained by substituting g ₃(z, t|z′, t′) into (24) and integrating with respect to z′ and t′. This results in

$$\displaystyle \begin{aligned} T_3(z,t) \,=\, T_2(z,t) + T_h(z,t). \end{aligned}$$

Here T ₂(z, t) is given by (28) and

$$\displaystyle \begin{aligned} T_h(z,t) \,\equiv\, {2Kh\over\pi k} \int_0^\infty {F_s(z) \over (s^2 \!+\! h^2) (s^2 \!+\! \mu_e^2) (s^2 \!+\! 1/\kappa\tau)} \left[1 - e^{-\kappa(s^2 + 1/\kappa\tau)t} \right] ds. \end{aligned}$$

where

$$\displaystyle \begin{aligned} F_s(z) \,\equiv\, (\mu_e \!+\! h)s\sin{}(zs) + (s^2 \!-\! h\mu_e)\cos{}(zs). \end{aligned}$$

When h = 0, we have T ₃(z, t) = T ₂(z, t) and when h →∞, T ₃(z, t) → T ₁(z, t).

The steady-state temperature (t →∞) may be written

$$\displaystyle \begin{aligned} T_3^{(ss)}(z) \,=\, T_2^{(ss)}(z) + T_h^{(ss)}(z), \end{aligned}$$

where $T_2^{(ss)}(z)$ is given by (34) and

$$\displaystyle \begin{aligned} T_h^{(ss)}(z) \,\equiv\, {2Kh\over\pi k} \int_0^\infty {F_s(z)\, ds \over (s^2 \!+\! h^2) (s^2 \!+\! \mu_e^2) (s^2 \!+\! 1/\kappa\tau)}. \end{aligned}$$

This can be readily integrated after expanding the integrand in partial fractions. After some algebra, we obtain

$$\displaystyle \begin{aligned} T_3^{(ss)}(z) \,=\, {K\over k(1/\kappa\tau - \mu_e^2)} \left[ e^{-\mu_e z} - \left({\mu_e + h \over 1/\sqrt{\kappa\tau} + h} \right) e^{-z/\sqrt{\kappa\tau}} \right]. \end{aligned}$$

$T_3^{(ss)}(z)$ will have a peak temperature at some distance z _max > 0 that can be found by differentiating $T_3^{(ss)}(z)$ with respect to z and setting the derivative to zero. This results in

$$\displaystyle \begin{aligned} z_{max} \,=\, {1 \over (\mu_e - 1/\sqrt{\kappa\tau})} \ln\left[{1 + h\sqrt{\kappa\tau} \over 1 + h/\mu_e} \right]. \end{aligned}$$

This reduces to (35) when h →∞ and z _max = 0 when h = 0. Note that no finite z _max exists in the absence of perfusion.