Analytical solution of the Pennes equation in time-domain for two-dimensions
Jan 30, 2017 by Hugo Milan
You can download the algorithm D2_BHE_f.m for Octave/Matlab that can solve this problem here and you can see instructions in how to use it here. If you want to see the final solution, go to Solution.
In this page, we will solve the dynamic Pennes equation in two-dimensions using the principles of superposition and separation of variables. We build on the previous solution of the Pennes equation in one-dimension described here to solve this two-dimensional problem. The problem we will solve is restricted to the following initial and boundary conditions:
The problem geometry for the pennes equation can be represented as shown below.
This geometry is a plate in a three-dimensional space, as you can see in the three-dimensional model below.
Governing equations
The Pennes equation is defined as:
\begin{equation} \rho c_{p}\frac{\partial T}{\partial t} = k\left( \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} \right) + S + Q_m + \omega_b\rho_b c_{b} (T_b - T) \label{eq:Pennes} \end{equation}
where \(T\) is temperature, \(t\) is time, \(\rho\) is density, \(c_{p}\) is specific heat, \(k\) is heat conductivity, and \(S\) is source, \(Q_m\) is metabolic heat generation, \(\omega_b\) is blood perfusion, \(\rho_b\) is blodd density, \(c_{b}\) is blood specific heat, and \(T_b\) is blood temperature.
Solving
I will use the principle of suporposition so that:
\begin{equation} T(y,t) = T_c + \phi_a(y) + \phi_b(y)\tau_b(t) + \phi_c(x, y) + \phi_d(x,y)\tau_d(t) \label{eq:Sup} \end{equation}
and the initial and boundary conditions for these problems are:
Defining \(W_b = \omega_b\rho_b c_{b} \) and applying Eq. \ref{eq:Sup} in \ref{eq:Pennes} yields
\begin{equation} \left( k\frac{\partial^2 \phi_a}{\partial y^2} - W_b\phi_a + S + Q_m + W_b(T_b - T_c) \right) \ldots \\ \ldots - \left( \rho c_{p}\phi_b\frac{\partial \tau_b}{\partial t} - k\tau_b\frac{\partial^2 \phi_b}{\partial y^2} + W_b\phi_b\tau_b \right) = \ldots \\ \ldots \left( - k \frac{\partial^2 \phi_c}{\partial x^2} - k \frac{\partial^2 \phi_c}{\partial y^2} + W_b\phi_c \right) \ldots \\ \ldots + \left( \rho c_{p}\phi_d\frac{\partial \tau_d}{\partial t} - k \tau_d\frac{\partial^2 \phi_d}{\partial x^2} - k \tau_d\frac{\partial^2 \phi_d}{\partial y^2} + W_b\phi_d\tau_d\right) \end{equation}
which implies that we can solve Eqs. \ref{eq:a}-\ref{eq:d} \begin{equation} k\frac{\partial^2 \phi_a}{\partial y^2} - W_b\phi_a + S + Q_m + W_b(T_b - T_c) = 0 \label{eq:a} \end{equation}
\begin{equation} \rho c_{p}\phi_b\frac{\partial \tau_b}{\partial t} - k\tau_b\frac{\partial^2 \phi_b}{\partial y^2} + W_b\phi_b\tau_b = 0 \label{eq:b} \end{equation}
\begin{equation} k \frac{\partial^2 \phi_c}{\partial x^2} + k \frac{\partial^2 \phi_c}{\partial y^2} - W_b\phi_c = 0 \label{eq:c} \end{equation}
\begin{equation} \rho c_{p}\phi_d\frac{\partial \tau_d}{\partial t} - k \tau_d\frac{\partial^2 \phi_d}{\partial x^2} - k \tau_d\frac{\partial^2 \phi_d}{\partial y^2} + W_b\phi_d\tau_d = 0 \label{eq:d} \end{equation}
Solving Eq. \ref{eq:a} and \ref{eq:b}
The solution of Eq. \ref{eq:a} and \ref{eq:b} was previously covered when we solved the pennes equation in one-dimension.
Solving Eq. \ref{eq:c}
We will use separation of variables to solve Eq. \ref{eq:c}. In separation of variables, we assume that the solution is the multiplication of a function of \(x\) ( \(\phi_{cx}(x)\) ) and \(y\) ( \(\phi_{cy}(y)\) ). Hence,
\begin{equation} \frac{1}{\phi_{cx}}\frac{\partial^2 \phi_{cx}}{\partial x^2} - \frac{W_b}{k}= -\frac{1}{\phi_{cy}}\frac{\partial^2 \phi_{cy}}{\partial y^2} = \gamma_n^2 \end{equation}
The solution of \(\phi_{cy}\) can be expressed as:
\begin{equation} \phi_{cy} = c_6 \sin{\left(\gamma_n y\right)} + c_7 \cos{\left(\gamma_n y\right)} \end{equation}
Applying the boundary condition at \(y = 0\) we get \(c_7 = 0\). Applying the boundary condition at \(y = H\) we get:
\begin{equation} \sin{\left(\gamma_n H\right)} = 0 \label{eq:gamman} \end{equation}
which implies that
\begin{equation} \gamma_n = \frac{n\pi}{H} \end{equation}
with \(n = 1, 2, 3, \dotsc\) We start the counting from \(1\) because \(n = 0\) yields \(\phi_{cy} = 0\). The solution of \(\phi_{cx}\) can be expressed as:
\begin{equation} \phi_{cx} = c_8 \sinh{\left(x \sqrt{\gamma_n^2 +\dfrac{W_b}{k}}\right)} + c_9 \cosh{\left(x\sqrt{\gamma_n^2 +\dfrac{W_b}{k}}\right)} \end{equation}
Applying the boundary condition at \(x = 0\) we get \(c_8 = 0\). Then, the solution that we are looking for is:
\begin{equation} \phi_c = \sum_{n=1}^{\infty} c_n \cosh{\left(x\sqrt{\gamma_n^2 +\dfrac{W_b}{k}}\right)}\sin{\left(\gamma_n y\right)} \label{eq:c:almost} \end{equation}
Now, we apply the non-homogeneous boundary condition at \(x = L\):
\begin{equation} \frac{\partial\phi_c}{\partial x}(x = L) = \sum_{n=1}^{\infty} c_n \sqrt{\gamma_n^2 +\dfrac{W_b}{k}}\sinh{\left( L\sqrt{\gamma_n^2 +\dfrac{W_b}{k}}\right)}\sin{\left(\gamma_n y\right)} = \dfrac{q_x}{k} \label{eq:c:cn} \end{equation}
To solve Eq. \ref{eq:c:cn}, we multiply both sides by an orthogonal function of the function that multiplies \(c_n\) and integrate it from \(y = 0\) to \(y = H\) to elimate the \(y\) dependence. That is, we need a function that is orthogonal to \(\sin{\left(\gamma_n y\right)}\). The function that we are looking for is \(\sin{\left(\gamma_m y\right)}\). Multiplying both sides by this function yields
\begin{equation} \int_{y=0}^{y=H}\sum_{n=1}^{\infty}c_n \sqrt{\gamma_n^2 +\dfrac{W_b}{k}}\sinh{\left(L\sqrt{\gamma_n^2 +\dfrac{W_b}{k}}\right)}\sin{\left(\gamma_n y\right)}\sin{\left(\gamma_m y\right)} dy = \ldots\\ \ldots \int_{y=0}^{y=H}\dfrac{q_x}{k}\sin{\left(\gamma_m y\right)} dy \label{eq:b:cn2} \end{equation}
Since we have
\begin{equation} \int_{y=0}^{y=H} \sin{\left(\gamma_n y\right)}\sin{\left(\gamma_m y\right)} dy = 0 \label{eq:mnn} \end{equation}
for \(n \neq m\), and
\begin{equation} \int_{y=0}^{y=H} \sin{\left(\gamma_n y\right)}\sin{\left(\gamma_m y\right)} dy = \frac{H}{2} \label{eq:mn} \end{equation}
for \(n = m\), we can drop the summation and solve Eq. \ref{eq:b:cn2} for each \(c_n\):
\begin{equation} c_n = \dfrac{\displaystyle \dfrac{q_x}{k}\int_{y=0}^{y=H}\sin{\left(\lambda_n y\right)} dy}{ \displaystyle \sqrt{\gamma_n^2 +\dfrac{W_b}{k}}\sinh{\left(L\sqrt{\gamma_n^2 +\dfrac{W_b}{k}}\right)}\int_{y=0}^{y=H}\sin^2{\left(\lambda_m y\right)} dy } \label{eq:b:cn3} \end{equation}
The solutions of the integral on the numerator is:
\begin{equation} \int_{y=0}^{y=H} \sin{\left(\gamma_n y\right)} dy = \frac{1}{\gamma_n}\left[1 - \cos{ \left( \gamma_n H\right) } \right] \end{equation}
From the definition of \(\gamma_n\) in Eq. \ref{eq:gamman},
\begin{equation} \int_{y=0}^{y=H} \sin{\left(\gamma_n y\right)} dy = \frac{1}{\gamma_n}\left[1 - \left( -1 \right)^n \right] \label{eq:c:int} \end{equation}
From Eq. \ref{eq:c:int} it is obvious that the summation in Eq. \ref{eq:c:almost} will be non-zero only for odd values of \(n\). Hence, we define
\begin{eqnarray} n &=& 2o - 1\\ \gamma_o &=& \frac{ \left( 2o - 1 \right) \pi}{H} \end{eqnarray}
Therefore, the final solution of Eq. \ref{eq:c} is:
\begin{equation} \phi_c = \sum_{o=1}^{\infty} \frac{4q_x\cosh{\left(x\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)}\sin{\left(\gamma_o y\right)}}{Hk\gamma_o\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\sinh{\left(L\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)} } \label{eq:c:final} \end{equation}
Although Eq. \ref{eq:c:final} can be used when \(W_b=0\), I recommend the use of the solutions for the heat equation shown here for these cases.
Solving Eq. \ref{eq:d}
We will use separation of variables to solve Eq. \ref{eq:d}. In separation of variables, we assume that the solution is the multiplication of a function of \(x\) ( \(\phi_{dx}(x)\) ) and \(y\) ( \(\phi_{dy}(y)\) ) and \(t\) \( ( \tau_d(t) ) \). Hence,
\begin{equation} \frac{\rho c_p}{k \tau_d}\frac{\partial \tau_d}{\partial t} + \frac{W_b}{k} = \frac{1}{\phi_{dx}}\frac{\partial^2 \phi_{dx}}{\partial x^2} + \frac{1}{\phi_{dy}}\frac{\partial^2 \phi_{dy}}{\partial y^2} = - \eta_p^2 \end{equation}
Defining \(\alpha = k/( \rho c_p ) \), the solution of \( \tau_d \) is expressed as:
\begin{equation} \tau_d = c_{10}\exp{\left[ -\alpha t\left(\eta_p^2 + \frac{W_b}{k} \right) \right]} \end{equation}
Now, to solve for \( \phi_d \), we define:
\begin{equation} \frac{1}{\phi_{dx}}\frac{\partial^2 \phi_{dx}}{\partial x^2} = - \frac{1}{\phi_{dy}}\frac{\partial^2 \phi_{dy}}{\partial y^2} - \eta_p^2 = - \beta_r^2 \end{equation}
Hence, the solution of \( \phi_{dx} \) is:
\begin{equation} \phi_{dx} = c_{11}\sin{ \left( \beta_r x \right) } + c_{12}\cos{ \left( \beta_r x \right) } \end{equation}
Applying the boundary condition at \(x=0\) yields \(c_{11}=0\). Applying the boundary condition at \(x=L\):
\begin{equation} \frac{\partial\phi_{dx}}{\partial x}(x=L) = 0 = \sin{ \left( \beta_r L \right) } \end{equation}
which implies that
\begin{equation} \beta_r = \frac{r\pi}{L} \label{eq:d:beta} \end{equation}
with \(r=0,1,2,\ldots\) We start counting in 0 because \(r=0\) does not necessarily imply \(\phi_{dx}=0\). Now, the solution of \(\phi_{dy}\) is:
\begin{equation} \phi_{dy} = c_{13}\sin{ \left( y\sqrt{\eta_p^2 - \beta_r^2} \right) } + c_{14}\cos{ \left( y\sqrt{\eta_p^2 - \beta_r^2} \right) } \end{equation}
Applying the boundary condition at \(y=0\) yields \(c_{13}=0\). Defining \(\delta_s^2 = \eta_p^2 - \beta_r^2\) and applying the boundary condition at \(y=H\) yields:
\begin{equation} \sin{ \left( H\delta_s \right) } = 0 \end{equation}
which implies that
\begin{equation} \beta_s = \frac{s\pi}{H} \end{equation}
with \(s=1,2,3,\ldots\) Therefore, the solution that we are looking for is:
\begin{equation} \phi_d\tau_d = \sum_{r=0}^{\infty}\sum_{s=1}^{\infty}c_{rs}\cos{ \left( \beta_r x \right) }\sin{ \left( \delta_s y\right) }\exp{\left[ -\alpha t \left( \beta_r^2 + \delta_s^2 + \frac{W_b}{k} \right) \right]} \end{equation}
Now, we apply the initial condition
\begin{equation} \sum_{r=0}^{\infty}\sum_{s=1}^{\infty}c_{rs}\cos{ \left( \beta_r x \right) }\sin{ \left( \delta_s y\right) } = \ldots \\ \ldots -\sum_{o=1}^{\infty} \frac{4q_x\cosh{\left(x\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)}\sin{\left(\gamma_o y\right)}}{Hk\gamma_o\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\sinh{\left(L\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)} } \label{eq:d:crs} \end{equation}
Once more, to solve Eq. \ref{eq:d:crs}, we are looking for orthogonal function of the functions that multiplies \(c_{rs}\). We want to integrate these functions from \(x = 0\) to \(x = L\) to elimate the \(x\) dependence and from \(y = 0\) to \(y = H\) to elimate the \(y\) dependence. That is, we need a function that is orthogonal to \(\cos{\left(\beta_r x\right)}\sin{\left(\delta_s y\right)}\). The function that we are looking for is \(\cos{\left(\beta_u x\right)}\sin{\left(\delta_v y\right)}\). Multiplying both sides by this function yields
\begin{equation} \int_{x=0}^{x=L}\int_{y=0}^{y=H}\sum_{r=0}^{\infty}\sum_{s=1}^{\infty}c_{rs}\cos{ \left( \beta_r x \right) }\sin{ \left( \delta_s y\right) }\cos{\left(\beta_u x\right)}\sin{\left(\delta_v y\right)} dydx= \ldots \\ \ldots -\int_{x=0}^{x=L}\int_{y=0}^{y=H}\sum_{o=1}^{\infty} \frac{4q_x\cosh{\left(x\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)}\sin{\left(\gamma_o y\right)}}{Hk\gamma_o\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\sinh{\left(L\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)} }\cos{\left(\beta_u x\right)}\sin{\left(\delta_v y\right)}dydx \label{eq:int1} \end{equation}
From Eq. \ref{eq:mnn} and \ref{eq:mn}, the left-hand side of Eq. \ref{eq:int1} will be non-zero when \(\delta_s = \delta_v\). Moreover, the right-hand side will be non-zero when \(\gamma_o = \delta_v\), so, we have \(\delta_s = \gamma_o\), which implies that Eq. \ref{eq:int1} will be non-zero for \(s = o\) and simplified to
\begin{equation} \int_{x=0}^{x=L}\sum_{r=0}^{\infty}c_{rs}\cos{ \left( \beta_r x \right) }\cos{\left(\beta_u x\right)} dx = \ldots \\ \ldots -\int_{x=0}^{x=L} \frac{4q_x\cosh{\left(x\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)}}{Hk\gamma_o\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\sinh{\left(L\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)} } \cos{\left(\beta_u x\right)}dx \label{eq:int2} \end{equation}
Also note that for \(r=0\) we do not need orthogonal functions and we can integrate Eq. \ref{eq:int2} directly without using cossines. This is,
\begin{equation} c_{0s} = -\dfrac{\displaystyle \int_{x=0}^{x=L} \frac{4q_x\cosh{\left(x\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)}}{Hk\gamma_o\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\sinh{\left(L\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)} } dx }{ \displaystyle \int_{x=0}^{x=L}dx} \end{equation}
which yields
\begin{equation} c_{0s} = -\frac{4q_x}{HLk\gamma_o \left( \gamma_o^2 + \dfrac{W_b}{k} \right)} \end{equation}
Back to the case \(r\neq 0\), we have
\begin{equation} \int_{x=0}^{x=L} \cos{\left(\beta_r x\right)}\cos{\left(\beta_u x\right)} dx = 0 \label{eq:ruu} \end{equation}
for \(r \neq u\), and
\begin{equation} \int_{x=0}^{x=L} \cos{\left(\beta_r x\right)}\cos{\left(\beta_u x\right)} dx = \frac{L}{2} \label{eq:ru} \end{equation}
for \(r = u\). Then we can drop the summation over \(r\) and solve Eq. \ref{eq:int2} for each \(c_{rs}\):
\begin{equation} c_{rs} = -\dfrac{\displaystyle \int_{x=0}^{x=L} \frac{4q_x\cosh{\left(x\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)}}{Hk\gamma_o\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\sinh{\left(L\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)} } \cos{\left(\beta_u x\right)}dx }{ \displaystyle \int_{x=0}^{x=L}\cos^2{ \left( \beta_r x \right) } dx } \label{eq:int3} \end{equation}
The solution of the integral in the numerator is \begin{equation} \int_{x=0}^{x=L} \cosh{\left(x\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)}\cos{\left(\beta_r x\right)}dx = \ldots\\ \ldots \frac{\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\sinh{\left( L\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)}\cos{\left(\beta_r L\right)} + \beta_r\cosh{\left(L\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)}\sin{\left(\beta_r L\right)}}{\gamma_o^2 + \dfrac{W_b}{k} + \beta_r^2} \end{equation}
which, from the definition of \(\beta_r\) (Eq. \ref{eq:d:beta}), becomes: \begin{equation} \int_{x=0}^{x=L} \cosh{\left(x\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)}\cos{\left(\beta_r x\right)}dx = \ldots\\ \ldots\left( -1 \right)^r\frac{\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\sinh{\left(L\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)}}{\gamma_o^2 + \dfrac{W_b}{k} + \beta_r^2} \end{equation}
Therefore, the final solution of Eq. \ref{eq:d} is:
\begin{equation} \phi_d\tau_d = \sum_{o=1}^{\infty} -\frac{4q_x\sin{ \left( \gamma_o y\right) }}{HLk\gamma_o} \left\{ \frac{\exp{\left[ -\alpha t \left( \gamma_o^2 + \dfrac{W_b}{k}\right) \right]}}{ \gamma_o^2 + \dfrac{W_b}{k} } \ldots\\ \ldots + \sum_{r=1}^{\infty} \left( -1 \right)^r\frac{2\cos{ \left( \beta_r x \right) }}{\beta_r^2 + \gamma_o^2 + \dfrac{W_b}{k}} \exp{\left[ -\alpha t \left( \beta_r^2 + \gamma_o^2 + \dfrac{W_b}{k}\right) \right]}\right\} \label{eq:d:final} \end{equation}
Although Eq. \ref{eq:d:final} can be used when \(W_b=0\), I recommend the use of the solutions for the heat equation shown here for these cases.
Solution
The final solution is:
with \(\alpha = k/( \rho c_p ) \), and \(\phi_a(y)\) and \(\phi_b(y)\tau_b(t)\) expressed as:
\begin{equation} \phi_a = \frac{Q_t}{W_b} + \left\{ \left( T_s - T_c \right) + \dfrac{Q_t}{W_b}\left[ \cosh{ \left( H\sqrt{\dfrac{W_b}{k}} \right) } - 1\right] \right\}\frac{\sinh{ \left( y\sqrt{\dfrac{W_b}{k}} \right)}}{\sinh{ \left( H\sqrt{\dfrac{W_b}{k}} \right) }} \ldots\\ \ldots - \frac{Q_t}{W_b} \cosh{ \left( y\sqrt{\dfrac{W_b}{k}} \right) } \end{equation}
\begin{equation} \lambda_m = \frac{m\pi}{H} \end{equation}
\begin{eqnarray} c_m &=& c_{m1} + c_{m2} + c_{m3}\\ c_{m1} &=& \frac{2Q_t}{HW_b\lambda_m}\left[\left( -1 \right)^m - 1 \right]\\ c_{m2} &=& \left( -1 \right)^m \frac{2\lambda_m c_1}{H} \frac{\sinh{ \left( H\sqrt{\dfrac{W_b}{k}} \right)} }{ \dfrac{W_b}{k} + \lambda_m^2}\\ c_{m3} &=& \frac{2Q_t\lambda_m}{HW_b} \frac{ 1 - \left( -1 \right)^m\cosh{ \left( H\sqrt{\dfrac{W_b}{k}} \right)} }{ \dfrac{W_b}{k} + \lambda_m^2} \end{eqnarray}
\begin{equation} \phi_b\tau_b = \sum_{m=1}^{\infty}c_m \sin{\left(\lambda_m y\right)} \exp{\left[-\alpha t \left(\lambda_m^2 + \dfrac{W_b}{k} \right)\right]} \end{equation}
\begin{equation} \gamma_o = \frac{ \left( 2o - 1 \right) \pi}{H} \end{equation}
\begin{equation} \phi_c = \sum_{o=1}^{\infty} \frac{4q_x\cosh{\left(x\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)}\sin{\left(\gamma_o y\right)}}{Hk\gamma_o\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\sinh{\left(L\sqrt{\gamma_o^2 +\dfrac{W_b}{k}}\right)} } \label{eq:c:solution} \end{equation}
\begin{equation} \beta_r = \frac{r\pi}{L} \end{equation}
\begin{equation} \phi_d\tau_d = \sum_{o=1}^{\infty} -\frac{4q_x\sin{ \left( \gamma_o y\right) }}{HLk\gamma_o} \left\{ \frac{\exp{\left[ -\alpha t \left( \gamma_o^2 + \dfrac{W_b}{k}\right) \right]}}{ \gamma_o^2 + \dfrac{W_b}{k} } \ldots\\ \ldots + \sum_{r=1}^{\infty} \left( -1 \right)^r\frac{2\cos{ \left( \beta_r x \right) }}{\beta_r^2 + \gamma_o^2 + \dfrac{W_b}{k}} \exp{\left[ -\alpha t \left( \beta_r^2 + \gamma_o^2 + \dfrac{W_b}{k}\right) \right]}\right\} \label{eq:d:solution} \end{equation}
The flux/heat flux can be obtained by derivating these equations with respect to \(x\) and \(y\).
Note that these solutions are not suitable for \(W_b = 0\). In this case, the correct solutions are for the heat equation shown here.
Using the Algorithm
The algorithm D2_BHE_f.m was written as a function in Octave/Matlab and you can obtain it here. To use it, you will need to define the material properties and the maximum values for \(m\), \(o\), and \(r\).
As an example, we can use the following script:
% Thermal properties
Tc = 37; % core and initial temperature (oC)
Ts = 150; % superficial temperature (oC)
qx = 1e5; % heat flux at x = L (W/m2)
rho = 1200; % tissue density (kg/m3)
cp = 3200; % specific heat (J/(K-kg))
k = 0.3; % thermal conductivity (W/(K-m))
wb = 1e-4; % blood perfusion (s-1)
rhob = 1052; % blood density (kg/m3)
cb = 3600; % blood specific heat (J/(K-kg))
Tb = 37; % temperature of the blood (oC)
Qmet = 500; % internal heat generation (W/m3)
% If there is another heat source S besides Qmet, sum it to Qmet.
% For example, if Qmet = 250 and S = 250, you would input Qmet = 500.
% Material properties
H = 1e-3; % m
L = 0.75e-3; % m
yprime = 0:(H/100):H;
xprime = 0:(L/100):L;
y = repmat( yprime, 1, size(xprime,2)); % creating a subspace for y
x = repelems( xprime, [1:size(xprime,2); ...
size(yprime,2)*ones(1,size(xprime,2))]); % creating a subspace for x
% number of iterations
minf = 50; % for m
oinf = 100; % for o
rinf = 50; % for r
% time of the solution
t = 0.5;
[Tprime, qxprime, qyprime] = D2_BHE_f(x, y, L, H, t, Ts, Tc, qx, ...
k, rho, cp, wb, rhob, cb, Tb, Qmet, minf, oinf, rinf);
T = reshape(Tprime, size(yprime,2), size(xprime,2));
qxout = reshape(qxprime, size(yprime,2), size(xprime,2));
qyout = reshape(qyprime, size(yprime,2), size(xprime,2));
surf(xprime, yprime, T)
figure;
surf(xprime, yprime, qxout)
figure;
surf(xprime, yprime, qyout)
figure;
surf(xprime, yprime, sqrt(qxout.^2 + qyout.^2) )which generates
Limitations
Every solution method has limitations. The most noticeable limitation is that we have to perform infinity sums to get predictions using solutions from separation of variables. Infinity sums, however, cannot be performed in numerical computations, which leads us to truncate the calculations for large values of \(m\), \(o\), and \(r\). How large these numbers have to be will depend on the problem geometry and on the material properties but, if theses number are not large enough, the calculations will manifest oscillations that will be most evident for predicted fluxes positioned near the flux boundary conditions (that is, near \(x = L\) in this solution). When observed that \(m\), \(o\), and/or \(r\) have to be increased, a rule-of-thumb is to double their values, re-calculate the predicionts, and observe if the oscillations have decreased to acceptable intensities.
Another surprising limitation is that the flux boundary condition at \(x = L\) is not satisfied when \(y = 0\) or \(y = H\). You can derivate Eqs. \ref{eq:c:solution} and \ref{eq:d:solution} with respect to \(x\) and observe that the term \(\sin{\left(\gamma_o y\right)}\) vanishes when \(y = 0\) or \(y = H\) no matter what is the value of the boundary condition. The solution, however, is accurate for \(y \rightarrow 0\) and \(y \rightarrow H\) and the temperature is not affected by this peculiarity.
Done
Now, you can go to: