Effectiveness of Cold Perfusion vs Surface CoolingMethods for Myocardial Protection during Open-Heart

Surgery

Bryan Felix, Ewnet Gebrehiwot and Scott Hansen

August 21, 2019

Abstract

Methods of cold blood perfusion and surface cooling saline solution are used forthe purpose of lowering oxygen and metabolism need of the heart muscle during open-heart surgery. These two approaches were analyzed as a 1-D transient plane wall forthe cold blood perfusion and 1-D transient hollow cylindrical for the surface cooling.Both Metabolism and Perfusion rates used/calculated for the plane wall 1-D model ofcold blood perfusion were 170 W/m3 and 0.0188 m3/m3−s, respectively. In the case ofthe surface cooling, perfusion rate was not applicable and the metabolism was assumedto decrease linearly as the temperature decreases. For the first method of cold bloodperfusion, the equilibration length was calculated to be 64mm and compared to 90mmlength of the coronary artery length from aorta to left Ventricle apex. An increase eitherin density or flow rate is suggested to achieve higher equilibration length. Finally, thebio-heat equation analysis resulted in 1-2C cooling of the whole heart within 5 minutesusing cold blood perfusion method and 3C cooling of most areas of the heart usingthe surface cooling method. The 3-D model also aligned with the 1-D analysis byconfirming the type of temperature distributions both in steady and time dependantstates of the heart tissue cooling. In conclusion, the cold blood perfusion method seemsto be more effective in cooling the heart tissue evenly compared to the surface coolingmethod.

1

1 Introduction

The heart muscle must be protected from tissue degradation and edema during open-heartsurgery procedure. During open-heart surgery, the blood circulation is maintained by aheart-lung machine, which stops blood flow through the heart. The left atrium and aortaare cannulated such that the de-oxygenated blood coming through superior and inferior venacava exits through the left atrium to the heart-lung machine and the oxygenated blood fromthe heart-lung machine enters the circulatory system through the ascending aorta. Justbelow the cannulated aorta, a clamp is placed to make sure the oxygenated blood enteringthe ascending aorta do not flow back into the heart. Once global ischemia (the clamping ofthe aorta) is achieved, the heart muscle is ready for a cardioplegia procedure.

Cardioplegia is a collective name for a cold infusion of any solution into the heart separatefrom the total body perfusion. The main purpose of cardioplegia is to conserve energy, slowdown the metabolic demand and degenerative processes and preventing myocardial edema,through adequate venting and drainage of the heart tissue. In early 1960s and 70s, lack ofsuch proper myocardial protection during open-heart surgery caused post-operative cardiacdysfunction and high mortality rate [3].

There are several methods of cardioplegia such as Crystalloid solution, St. Thomas IIsolution, Bretschiender (Custodiol) solution, Cold Blood and Topical Hypothermia as it isdiscussed in depth in [3]. This paper will focus on two methods of cooling down the hearttemperature that are the Cold Blood and Topical Hypothermia. For these two methods, thebioheat transfer governing equations, equilibration length and the temperature distributionwill be analyzed.

2 Method 1: Cold Blood Cardioplegia

A Cold Blood Cardioplegia is a mixture of blood with Crystalloid Solution administeredvia cannulas placed into the aortic root below the aortic cross-clamp as shown on Figure 1below. The advantage of Cold Blood cardioplegia is that it provides oxygen and nutrientsto the ischemic myocardium and buffering capacity.

2.1 Governing Equation

Harry Pennes developed the original bioheat equation in 1948 under the assumption thatheat transfer occurs primarily in the capillaries. It is a widely used model even though ithas limitations that have been critiqued and modified over the years. Primarily, it has beenshown that very little heat transfer occurs in the capillaries. Rather, it is in the arteriolesthat significant heat transfer takes place. Also, it is typically not valid to assume that thearterial temperature is constant, but is equilibrating with the surrounding tissue over someequilibration length of the vessel. Further, the blood perfusion rate is generally unknown.For the purpose of this paper, it provides a suitable model for blood perfusion in the tissue ifthese limitations are kept in mind. Below is the general form of the Pennes heat sink model:

ρc∂T

∂t= ∇ · (k∇T ) + wbcb(Ta − T ) + qm (1)

2

where wb is the cold blood perfusion rate in volume of tissue per units time, cb is the spe-cific heat of the tissue, Ta is a cold perfusate supply temperature and qm is the metabolismrate of the heart. For this perfusion approach we will assume that the heart wall can bemodeled as a plane wall with the exterior of the heart and the half width of the heart servingas the geometric boundaries. The control volume used in this analysis is given in Figure 2.

Figure 1: Cold Blood Infusion Through Coronary Artery (Source: Iaizzo 3rd Edition, 2015)

Figure 2: Control volume for perfused tissue case.

2.1.1 Constraints and Conditions

The constants in Equation (1) are tissue properties for the myocardium and the perfusate.From [1], the thermal conductivity of the myocardium is given as 0.5363 W

mK. From [7], we

obtained values of 1081 kgm3 and 3686 J

kgCfor the density and specific heat of the myocardium

3

respectively. For the perfusate, we assumed that its thermal and physical properties weresimilar to those of water. Therefore, we applied a density of 1000 kg

m3 and a specific heat of4200 J

kgCto the perfusate for this analysis.

Finally, the bioheat equation we propose makes use of a both a blood perfusion term anda metabolic heat generation term. Blood perfusion is difficult to measure, but [1] proposes0.0188 m3

m3−s as a realistic perfusion rate. In [1] also proposes a temperature dependent heat

generation term q. This value is minimized at 170 Wm3 . Here, we will assume that this is the

value for a constant heat generation term.

First, we solved the Pennes bioheat equation using the simplified control volume describedearlier. The solution here is based on the assumptions that steady state has been reached,so there is no time dependence. We divided the control volume in half so as to be onlyfocusing on the half width of the heart wall. We did this because we are also assuming thatthere is symmetry about this plane with both the internal and external surfaces of the heartremaining at 37C. The weakness with this assumption is that the interior of the heart willnot remain at 37C if the heart is on bypass and warm blood is not filling the chambers.However, this allows us to make an assumption about symmetry across the center of theheart, which leads to the following conditions:

x = 0, Tt = 37C, and (2)

x = 0.75cm,dT

dx= 0 (3)

Next, in order to observe the rate of cooling, we took a lumped approach to the problem.Under the lumped method, the thermal conductivity is assumed to be so large so that theentire control volume is at the same temperature at each point in time-there is no thermalgradient. The general bioheat equation then simplifies to Equation 7 below. The differentialequation can be solved readily by substituting changes in temperature with the vriable θ. If,after the heat transfer to the perfusate through the catheter and the coronary arteries (moreon this later), the perfusate reaches the heart wall at a temperature of 34C, the initialboundary condition at t = 0 will be the following.

t = 0, θ = T − Ta = 3C (4)

2.2 Model Scaling and Analysis

Now that the tissue properties have been established and the boundary conditions defined inthe previous section, both the 1-D analytical approach and the lumped approach are readilysolvable. In this case, we used MATLAB to solve the differential equations in each case. Thecode can be found at the end of this document. Equation (5) gives the solution to the 1-Dapproach, with the constants provided below. This is the steady state case with constant

4

Figure 3: Steady-state, 1-D solution with perfusion

metabolic heat generation and perfusion.

T (x) =c1e

c1/22 ( x

c3− 3

c4)

c5e3(c

1/26 )

c7+1

+c8e

c1/29 x

c10

c11e3(c

1/212 )

c13+1

)

+c14

c15

(5)

where the constants have values of:

c1 = 2.36× 1018

c2 = −9.23× 1015

c3 = 2.62× 105

c4 = 5.24× 107

c5 = 7.88× 1017

c6 = 9.23× 1015

c7 = 5.24× 107

c8 = 2.36× 1018

c9 = 9.23× 1015

c10 = 2.62× 105

c11 = 7.88× 1017

c12 = 9.23× 1015

c13 = 5.24× 107

c14 = 1.86× 1015

c15 = 7.88× 1017

(6)

Figure 3 provides a plot of the steady state solution to the 1-D approach. Heat generationin the tissue prevents the tissue temperature from reaching the temperature of the perfusate.Of not, however, is that there is fairly uniform cooling across the thickness of the tissue.

We also employed the lumped approach to solve for the transient solution to Equation 1. Inthis case, the temperature gradient across the tissue goes to zero as the thermal conductivitygoes to infinity and the general form of the equation simplifies to:

ρc∂T

∂t= −wbcb(Ta − T ) + qm (7)

5

Figure 4: Lumped solution with perfusion

For this, the transient case, the solution to the differential equation is:

T (t) =8.00× 1018e−

2.61 × 1015t1.44 × 1017

2.67× 1018 +6.30× 1015

2.67× 1018 (8)

This solution is plotted in Figure 4, showing that steady state conditions are reachedafter about 5 minutes of tissue perfusion.

2.3 Equilibration Length

The characteristic length where the cold blood changes from the cold temperature to tissuetemperature is calculated by the following equation.

Le =π(ra)

2ρbcbu

(ktσ)(9)

where ra is the radius of the coronary artery, ρ and cb are density and heat capacity of thecold blood, u is the blood velocity and kt is the thermal conductivity of the tissue and σ isthe shape factor of the artery. Based on Baish [1] shape factor formula and artery radius of0.5 mm, the equilibration length comes up to be 64.9mm as shown below.

Le =3.14X(0.5)2X1000X4200X130

0.5363X12.3= 64.9mm (10)

According to Handbook of Advanced Cardiac [3], the average length measurement forcoronary artery left ventricle (LV) outflow from aortic valve to LV apex is approximately90mm. In this case, the cold blood perfusion will reach equilibration before it reaches theapex of the heart. In order to mitigate this deficiency, either the density of the cold bloodor the flow rate would have to be increased.

6

3 Method 2: Topical Hypothermia

Topical Hypothermia is the second method for protecting the heart during open-heartsurgery. This methods is a continual flowing of cold Saline on top surface of the heartand eliminating the overflow solution using wall suction.

3.1 Governing Equation

To place our approach in a 2 dimensional geometry we take a cross-section of the heart asshown in Figure 5. Then, we simplify our analysis by assuming a lumped contribution ofheat coming from the right and left ventricle, as shown in Figure 6. Boundary conditions aredependant on the angle, measured counter-clock-wise from the horizontal. Thus we come upwith the following governing equation,

ρc∂

∂tT (r, θ, t) = k∆T (r, θ, t) + µ(r) (11)

where ρ is the density the heart muscle tissue, c is its specific heat constant, k is its tem-perature diffusivity, and µ(r) the metabolic contribution. Temperature (T ) is a functionthat depends on the distance from the center of the lumped ventricles (r), and the angle isdenoted by θ. We impose the boundary conditions at the interface between the interior ofthe ventricles with the muscle tissue (denoted by R1), and at the interface between the heartand the exterior (denoted by R2). Explicitly

T (R1, θ) = f1(θ)T (R2, θ) = f2(θ)

(12)

Figure 5: Cross section of the heart. Taken from [5].

3.1.1 Constraints and Conditions

The paper by Feigenbaum [6] studies the wall thickness of the left ventricle in the humanheart. Typical samples range in the order of 15mm. Moreover we assume tissue diffusivityto be approximately 0.5367 W/mK, in agreement with [7]. The heart muscle density (1081

7

f1(θ)

f2(θ)

R2

R1

@tT = ∆T + µ(r)

Figure 6: Geometry imposed

Core Temperature

Tempreature Drop

1 2 3 4 5 6

30

32

34

36

38

Profile Temperature of External Agent (T_a)

Figure 7: Temperature of the Saline Solution

kg/m3) and heat capacity (3686 J/kg/C) are taken from [7]. To establish the boundaryconditions, first, we take the interior to have a radius in the order of 5mm as suggested bymeasurements from [8]. We will assume that the exogenous agent solution is applied onlyon the top portion of the heart; that is around π < θ < 0. The temperature of the agent ismodeled by a step function with arbitrary amplitude as shown in Figure 7. We deliberatelyuse a linear combination of complex exponential functions to make the analysis exact. Forthe metabolic function, we will assume that the metabolic rate is a function that dependslinearly of the distance from the core. The amplitude of the metabolic rate is taken from [1].Assuming the core remains at steady state temperature 37C we have that the governingequation (11) is of the form

(3686)(1081)∂

∂tT (r, θ, t) = 0.55∆T (r, θ, t) + 170

r − 0.005

0.015, (13)

together with boundary conditions,

T (0.02, θ, t) = 18.5 + 0.5Ta + sin(θ)(0.63662Ta − 23.5549)

andT (0.005, θ, t) = 37.

8

3.2 Model Scaling and Analysis

We inspect the model at the steady state time-scale. For ε 1 take the long-time scale tobe given by the substitution t 7→ 1/εt for which equation (13) becomes:

ε(3686)(1081)∂

∂tT (r, θ, t) = 0.55∆T (r, θ, t) + 170

r − 0.005

0.015.

To leading order O(1) we have the steady state equation

0 = 0.55∆T (r, θ) + 170r − 0.005

0.015. (14)

This is an instance of Poisson’s equation or an inhomogenous Laplace problem in polarcoordinates. The solution to equation (14) is deduced in Appendix A. Note that, for ourchoice of boundary conditions, the solution will be exact. It follows that the steady stateprofile is given as

T (r, θ) = sin(θ)

(0.03 − Ta0.0008

r+ (Ta33.95− 1256.26)r

)+ (Ta0.36− 13.34) ln(r) + Ta1.91− 675.84r3 + 7.60r2 − 33.70

(15)

With the formula at hand we can start inspecting the behavior of different choices of agenttemperature. Figure 8 shows the effect of distinct temperatures applied to the upper hemi-sphere of the heart cross-section (blue indicates cooler temperatures).

-0.02 -0.01 0.00 0.01 0.02

-0.02

-0.01

0.00

0.01

0.02T_a=10 C

-0.02 -0.01 0.00 0.01 0.02

-0.02

-0.01

0.00

0.01

0.02T_a=20 C

-0.02 -0.01 0.00 0.01 0.02

-0.02

-0.01

0.00

0.01

0.02T_a=30 C

Figure 8: Different heat maps for several choices of Ta

We now give predictions to what the appropriate choice for Ta should be. We focus ourattention to the mid-range radius (in between the core and outer layer of tissue), that is atr = 12.5mm. Substitution into (15) gives

T (θ) = (0.356507Ta − 13.1908) sin(θ) + 0.330482Ta + 24.7769.

Figure 9 shows the contour plot for the temperature profile at r = 12.5mm plotted againstthe angle θ, and Ta. For the purposes of our project it is of interest to see the curve generated

9

Figure 9: Contour map for the temperature at r = 12mm as a function of Ta and θ

π

2π

3 π

22 π

θ (rad)

-20

-10

10

20

30

40T_a (C)

Figure 10: Profile of Ta with respect to the angle θ.

by a desired temperature drop from 37C to 34C. To that end, we take equation (15) andset r = 0.0125, and T (r, θ) = 34. The temperature profile of Ta satisfies

Ta =−13.19 sin(θ)− 9.22

−0.36 sin(θ)− 0.33

Figure 10 gives the plot of the relation of Ta at a given angle. It is worth pointing out thatfor most areas in the tissue, an agent with temperature between 25 and 31 degrees Celsiuswill be sufficient to reduce the steady state temperature by three degrees. On the otherhand, near the back of the heart, for θ close to 3π/2 our model predicts that the action ofthe solution is insignificant and that other measures are to be taken.

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3.3 Numerical analysis

3.3.1 Finite element background

We briefly describe the finite element method for a parabolic problem on a finite domain Ωof the form

ρc∂u

∂t= ∇(k∇u) + f (16)

where, in our context, ρ is the material density, c its specific heat constant, k denotes thethermal conductivity, and f is a source of temperature generation. We assume boundaryconditions are sufficient to admit a solution.

Weak formulation For brevity we consider the class of smooth functions vanishing at theboundary C∞0 (Ω). We remark that “smaller” function spaces are valid, what we require isthat for any v in the chosen space

v∣∣∂Ω

= 0,

and that the integrals to appear are bounded. Now, given a basis ϕ∞i=0 for C∞0 (Ω), we mayassume that a solution to (16) u, for fixed t is well approximated by the linear combination

u(x, t) =m∑i=0

v(t)ϕi(x). (17)

Before plugin (17) into (16) we take an arbitrary basis function ψ ∈ ϕ∞i=0 and consider theproduct (

ρc∂u

∂t= ∇(k∇u) + f

)· ψ. (18)

In particular, note that, by integrating by parts over Ω, and making use of the propertyψ |∂Ω= 0, we get the relation

ρc

∫Ω

∂u

∂tψ dx =

∫Ω

k∇u∇ψ dx+

∫Ω

fψ dx. (19)

To satisfy equation (19), for any choice of ψ is termed the weak formulation of the problem.Now we are ready to plugin the form of u(x, t) given by (17) into the weak form of theproblem. We have

ρc

∫Ω

ϕi(x)m∑i=0

dv(t)

dtψ dx =

∫Ω

k∇m∑i=0

v(t)ϕi(x)∇ψ dx+

∫Ω

fψ dx. (20)

One can prove that satisfying the weak formulation for a finite subset of functions ψ ∈ ϕni=0

converges to the true solution as n→∞ [9]. Thus, for fixed n we are able to rewrite (20) inmatrix notation as

Mdv

dt+Kv = F,

11

where

Ki,j =

∫Ω

(k∇ψj) · ∇ϕi dx, Mi,j =

∫Ω

aψjϕi dx, and Fi =

∫Ω

fϕi.

We have reduced the problem to a system of ordinary differential equations, for which manyalgorithms exist.

3.3.2 Transient solution to Topical Hypothermia

To elaborate on the approximate time to equilibrium we use finite element analysis to solveequation (13) with its corresponding boundary conditions. We discretize the geometry bya triangular mesh as depicted in Figure 11. For our basis function we use piece-wise linearfunctions defined over the nodes of the triangulation.

-0.03 -0.02 -0.01 0 0.01 0.02 0.03

-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02

Figure 11: Mesh definition

Figures 12, 13, 14 show the temporal convergence to steady state within 1 degree (blue).The figures are plotted for times t = 15, 150 and 500 seconds. Interestingly the figures revealthat a choice of Ta closer to core temperature converges faster. Moreover convergence is fastfor −π/4 < θ < −3π/4. On the other hand, convergence is slower for π/4 < θ < 3π/4.

-0.015 -0.01 -0.005 0 0.005 0.01 0.015-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02Time=15 Color: indicador(abs(u-sstate(x,y)))

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-0.015 -0.01 -0.005 0 0.005 0.01 0.015-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02Time=150 Color: indicador(abs(u-sstate(x,y)))

0

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0.6

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-0.015 -0.01 -0.005 0 0.005 0.01 0.015-0.02

-0.015

-0.01

-0.005

0

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0.015

0.02Time=500 Color: indicador(abs(u-sstate(x,y)))

0

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0.4

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0.6

0.7

0.8

0.9

1

Figure 12: Convergence to steady state for Ta = 10

12

-0.015 -0.01 -0.005 0 0.005 0.01 0.015-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02Time=15 Color: indicador(abs(u-sstate(x,y)))

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-0.015 -0.01 -0.005 0 0.005 0.01 0.015-0.02

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-0.005

0

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0.015

0.02Time=150 Color: indicador(abs(u-sstate(x,y)))

0

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-0.015 -0.01 -0.005 0 0.005 0.01 0.015-0.02

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0

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0.02Time=500 Color: indicador(abs(u-sstate(x,y)))

0

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1

Figure 13: Convergence to steady state for Ta = 20

-0.015 -0.01 -0.005 0 0.005 0.01 0.015-0.02

-0.015

-0.01

-0.005

0

0.005

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0.015

0.02Time=15 Color: indicador(abs(u-sstate(x,y)))

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0.02Time=150 Color: indicador(abs(u-sstate(x,y)))

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-0.015 -0.01 -0.005 0 0.005 0.01 0.015-0.02

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0.02Time=500 Color: indicador(abs(u-sstate(x,y)))

0

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Figure 14: Convergence to steady state for Ta = 30

4 3-D Modeling

We used a simplified geometry to model each cardiac cooling method. This allows for adirect comparison between the two. We used a 1.5cm cube as the control volume to repre-sent the heart wall of the left ventricle. We assumed that the chambers of the heart wereat a constant temperature of 37C with a constant metabolic heat generation in the tissueof 170 W

m3 . The model assumes that both the perfusate and the topical saline solution arereaching the tissue at a temperature of 34C. Both the tissue properties and the propertiesof the perfusate are the same as in the earlier analyses. The results in Figure 15 and Figure16 agree fairly well with the 1-D analytical solutions found earlier in this discussion.

The profile of the 3D model for the perfusion case in Figure 15 matches well with resultsof the analytical solution obtained earlier. Here, however, the tissue is cooled to a muchlarger degree. This can be attributed to the fact that the perfusion term in the model isnot temperature dependent, but rather relies on a constant temperature difference of 3C.Therefore, cooling will be constant in the tissue rather than diminishing as the temperatureof the tissue approaches the temperature of the artery. This underscores the importance ofnot relying solely on modeling. By intuition, one can readily observe that the temperaturein the tissue should never exceed that of the perfusate.

The model for the topical hypothermia case more closely resembles the results from our

13

analysis than for the perfused tissue case. As expected, the gradient is more or less linearsince the rate of metabolic generation is small. As observed in the 1-D approach, cooling ismost effective on the exterior wall of the heart with diminishing effect closer to the chamberwalls. It is also interesting to note the curved temperature gradient before reaching equilib-rium as a result of the heat generation term. The heat generation term could also be thecause of the small increase in temperature at the initial time point. Otherwise, it is likelyan error somewhere in the model that is resulting in this discontinuous behavior.

Figure 15: Perfused tissue steady state (left) and time dependent (right) temperature profiles.

Figure 16: Topically applied cooling steady state (left) and time dependent (right) temper-ature profiles.

5 Conclusions

In general, both methods of cooling the heart tissue during open-heart surgery are beneficial.Depending on the procedure type and time duration, one type of cooling can be preferred over

14

the other. For longer surgical procedures using the cold blood perfusion would provide anevenly distributed cooling of the heart tissue. For shorter and minor procedures, the topicalhypothermia can provide the surface cooling for the heart tissue. Furthermore, there areother types of perfusion solutions than cold blood. Different types of solutions can providevarious levels of cooling rate and equilibration length to get the desired clinical outcome.Considering all of the above, the perfusion method of cooling the heart tissue is more effectivethan the surface cooling method.

15

Appendices

A Solving the Laplace equation in an Annulus

Consider the inhomogeneous Laplace equation first, in a two dimensional Cartesian plane.Explicitly, the function u(x, y) satisfies

uxx + uyy = µ(|x, y|) for R1 <√x2 + y2 < R2 (21)

where µ is a function that depends on the magnitude of the vector (x, y). Additionally, weimpose the following boundary conditions

u(x, y) = f1(x, y) for√x2 + y2 = R1

u(x, y) = f2(x, y) for√x2 + y2 = R2

(22)

It is natural to translate the problem to polar coordinates by the change of variables

x = r cos(θ) and y = r sin(θ).

Then, equation (21) is equivalent to

r2urr + rur + uθθ = r2µ(r) for (r, θ) ∈ (R1, R2)× [0, 2π) (23)

subject to the boundary conditionsu(R1, θ) = f1(θ) for r = R1

u(R2, θ) = f2(θ) for r = R2(24)

We look for a solution of the form u = uh+up. Where uh solves the homogeneous Laplaceequation (that is, when µ = 0) and where up is a particular solution of the equations (23)and (24). We now provide constructive methods for uh and up.

Solution of the homogeneous problem Let u(r, θ) satisfy

r2urr + rur + uθθ = 0 for (r, θ) ∈ (R1, R2)× [0, 2π) (25)

with boundary conditions u(R1, θ) = f1(θ) for r = R1

u(R2, θ) = f2(θ) for r = R2(26)

We assume that separation of variables is possible and pressume that u has the form

u(r, θ) = P (r)Q(θ). (27)

Then, substitution into (25) leads to the equivalent form

r2P ′′ + rP ′

P= −Q

′′

Q. (28)

16

Note that the left hand side depends on r only, while the right hand side depends on θ. If(28) is to be true for all r and θ, it must follow that the quantity is a constant. In otherwords, we have that

λ =r2P ′′ + rP ′

P= −Q

′′

Q.

In particular Q(θ) satisfies Q′′+λQ = 0. The theory of ordinary differential equations gives,for a fixed value of λ, the solution

Q(θ) = A cos(√λθ) +B sin(

√λθ),

where A and B are constants to be determined. Since we wish Q(θ) to be 2π periodic inorder to have continuity at θ = 0, the period of Q must divide 2π evenly. That is

2π

Period of Q= n

with n ≥ 0 an integer. Then, it follows that λ = n2, and the true form of Q is

Qn(θ) = A cos(nθ) +B sin(nθ).

Back to (28), we now solve for P . With λ as computed above, it follows that P satisfies thedifferential equation

r2P ′′ + rP ′ − n2P = 0.

Via the theory of differential equations we have the following family of solutions dependentof n (we reuse the arbitrary constants A and B to avoid unnecessary notation),

Pn(r) =

A+B ln(r) for n = 0Arn +Br−n for n > 0

Having solved for Pn(r) and Qn(θ) explicitly, we make use of the principle of superpositionto presume u(r, θ) is of the form

u(r, θ) =∞∑n=0

Pn(r)Qn(θ).

Explicitly,

u(r, θ) = A0 +B0 ln(r) +∞∑n=1

(Anr

n +Bnr−n) cos(nθ) +

∞∑n=1

(Cnr

n +Dnr−n) sin(nθ), (29)

where Ai, Bi, Ci, Di are constants to be determined. In order to do so, recall that Fouriertheory asserts that the set

W = cos(nθ), sin(nθ)∞n=0

is an orthogonal basis for functions in the Hilbert Space L2[0, 2π). Thus we consider theinner product of u(r, θ) against these basis functions. Denote the L2[0, 2π) inner product

(∫ 2π

0fg dx)as 〈f, g〉 and note that

〈u(r, θ), cos(nx)〉 =(Anr

n +Bnr−n) 〈cos(nθ), cos(nθ)〉

17

and,

〈u(r, θ), sin(nx)〉 =(Cnr

n +Dnr−n) 〈sin(nθ), sin(nθ)〉 .

Now, to be more explicit, we look at the boundary conditions. Note that if r = R1 in theabove expressions, we have

〈f1(θ), cos(nx)〉 =(AnR

n1 +BnR

−n1

)〈cos(nθ), cos(nθ)〉 ,

〈f1(θ), sin(nθ)〉 =(CnR

n1 +DnR

−n1

)〈sin(nθ), sin(nθ)〉 .

Analogously, for r = R2 we have

〈f2(θ), cos(nx)〉 =(AnR

n2 +BnR

−n2

)〈cos(nθ), cos(nθ)〉 ,

〈f2(θ), sin(nθ)〉 =(CnR

n2 +DnR

−n2

)〈sin(nθ), sin(nθ)〉 .

Noting that 〈cos(nθ), cos(nθ)〉 = π and being conscious of the case n = 0 we may group theabove expressions into systems for which we can solve for the unknown coefficients. Namely,we get A0 and B0 by solving the system

12π〈f1, 1〉 = A0 +B0 ln(R1)

12π〈f2, 1〉 = A0 +B0 ln(R2)

An and Bn are given by solving1π〈f1, cos(nθ)〉 = AnR

n1 +BnR

−n1

1π〈f2, cos(nθ)〉 = AnR

n2 +BnR

−n2

and finally, Cn and Dn are given by the system1π〈f1, sin(nθ)〉 = CnR

n1 +DnR

−n1

1π〈f2, sin(nθ)〉 = CnR

n2 +DnR

−n2

Solution to the particular equation Let u(r, θ) satisfy

r2urr + rur + uθθ = r2µ(r) for (r, θ) ∈ (R1, R2)× [0, 2π) (30)

with boundary conditions u(R1, θ) = 0 for r = R1

u(R2, θ) = 0 for r = R2(31)

Note that these are distinct from (24), the reason being that the homogeneous solution wasalready chosen to satisfy the original boundary constraints. We proceed to assume that theparticular solution is a function of r exclusively. Thus, from (31) and (30) we get the ODE:

r2u′′ + ru′ = r2µ(r), (32)

with constraints, u(R1) = 0 and u(R2) = 0. Now, let v = u′ so that (32) yields

r2v′ + rv = r2µ(r),

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equivalentlyrv′ + v = rµ(r).

Note that the left hand side of the above expression is equivalent to (rv)′. Thus, via inte-gration on both sides, the following holds

rv =

∫ r

r0

sµ(s) ds+ c1 (33)

(34)

where r0 is chosen so that the definite integral evaluates to 0 at r0. Recalling the substitutionv′ = u we get

ru′ =

∫ r

r0

sµ(s) ds+ c1

⇒ u′ =1

r

∫ r

r0

sµ(s) ds+c1

r

⇒ u =

∫ r

r0

1

t

∫ t

t0

sµ(s) ds dt+ c1 ln(r) + c2.

As before, s0 and t0 are chosen so that the corresponding definite integrals evaluate to 0.Lastly, to find the values of c1 and c2, we impose the boundary conditions. Thus, c1 and c2

satisfy the system 0 =

∫ R1

r0

1t

∫ tt0sµ(s) ds dt+ c1 ln(R1) + c2

0 =∫ R2

r0

1t

∫ tt0sµ(s) ds dt+ c1 ln(R2) + c2

(35)

References

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[2] N. Cyr, M. Tetu, and M. Breton, IEEE Trans. Instrum. Meas. 42, 640 (1993).

[3] Paul A. Iaizzo Handbook of Cardiac Anatomy, Physiology and Devices. Third Edition.2015

[4] K. Yue, X. Zhang, F. Yu. An Analytic Solution of One-Dimensional Steady-state Pennes’Bioheat Transfer Equation in Cylindrical Coordinates. Journal of Thermal Science. Vol.13:3. 2004.

[5] Rene St.-Jacques. My Human Body. http://www.corpshumain.ca

[6] Feigenbaum H, Popp RL, Chip JN, Haine CL. Left Ventricular WallThickness Measured by Ultrasound. Arch Intern Med. 1968;121(5):391–395.doi:10.1001/archinte.1968.03640050001001

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[7] Hasgall PA, Di Gennaro F, Baumgartner C, Neufeld E, Lloyd B, Gosselin MC, PayneD, Klingenbock A, Kuster N, “IT’IS Database for thermal and electromagnetic param-eters of biological tissues,” Version 4.0, May 15, 2018, DOI: 10.13099/VIP21000-04-0.itis.swiss/database

[8] Yıldız, Banu Sahin, et al. ”The Prevalence of Atrial Fibrillation and Related Factors In-cluding Anthropometric, Hemodynamic and Echocardiographic Parameters in Patientswith Hemodialysis.” 229-232.

[9] Zienkiewicz, Olgierd Cecil, et al. The finite element method. Vol. 3. London: McGraw-hill, 1977.

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