Contemporary Meshfree Methods for Three Dimensional Heat Conduction Problems

Abstract

This work aims to be a fairly comprehensive study on the respective performance of several meshfree schemes selected for 3D heat conduction problems. A wide array of such methods is implemented in this paper, two of which are employed in 3D for the first time. These methods are compared in a systematic fashion: First, their ability to approximate the Laplacian operator, the key ingredient of the heat equation, is examined. Synthetic benchmarks as well as a real-world engineering problem where experimental data is available follow. In the interest of reproducibility and knowledge dissemination, the complete source code is made public and can be downloaded from: https://github.com/mroethli/thermal_iwf.

Appendices

Appendices

Note: throughout this section it is assumed that $1 = \int _{-\infty }^{\infty } W(x-\tilde{x})d\tilde{x}$ holds in all cases. This can be trivially ensured by choosing a normalized kernel, which is one of the “golden rules of SPH” [59]. All constants in front of monomials were dropped in the following sections, i.e. instead of the normal definition of a polynomial:

$$\begin{aligned} P(n)&= \sum _i^N = a_ix^i \end{aligned}$$

(118)

This one is used:

$$\begin{aligned} P(n)&= \sum _i^N = x^i \end{aligned}$$

(119)

This is merely due to convenience and brevity. All proofs could be repeated with the first definition.

Furthermore, all proofs in this section were carried out for one dimension only. It is believed that the proofs would extend to three dimensions. One obvious complication one would face if attempting to construct analogous proofs in 3 dimensions is that the n-th order moment about some kernel $W(\cdot )$ becomes a tensor of order n instead of a scalar. However, all entries of said tensors are monomials of degree n, such that central argument of the one dimensional proofs still hold.

1.1 Appendix 1: Proof That Reproducing the First $n+1$ Monomials Exactly Entails Completeness of Order n

Assuming:

$$\begin{aligned} x^i&= \int _{-\infty }^{\infty } \tilde{x}^i W(x-\tilde{x})d\tilde{x} \end{aligned}$$

(120)

for $i=0 \cdots N$. Then

$$\begin{aligned} \sum _{i=0}^N x^i&= \int _{-\infty }^{\infty } \sum _{i=0}^N \tilde{x}^i W(x-\tilde{x})d\tilde{x} \end{aligned}$$

(121)

$$\begin{aligned}&= \sum _{i=0}^N \int _{-\infty }^{\infty } \tilde{x}^i W(x-\tilde{x})d\tilde{x} \end{aligned}$$

(122)

$$\begin{aligned}&= \sum _{i=0}^N x^i \end{aligned}$$

(123)

1.2 Appendix 2: Proof That Satisfying the First n Moment Conditions entails Reproducing the First n Monomials Exactly

Some auxiliary properties need to be proven beforehand. The first being:

$$\begin{aligned} \begin{pmatrix}R \\ k \end{pmatrix}&= (-1)^k \begin{pmatrix} k-r-1 \\ k\end{pmatrix} \end{aligned}$$

(124)

The proof follows more or less directly from the definition for the binomial coefficient:

$$\begin{aligned} \begin{pmatrix} R \\ k \end{pmatrix}&= \frac{R^{\underline{k}}}{k!} \end{aligned}$$

(125)

$$\begin{aligned}&= \frac{R(R-1)(R-2)\cdots (R-k+1)}{k!} \end{aligned}$$

(126)

Factoring out $(-1)$ and subsequently reversing the order:

$$\begin{aligned} \begin{pmatrix} R \\ k \end{pmatrix}&= (-1)^k\frac{-R(-(R-1))(-(R-2))\cdots (-(R-k+1))}{k!} \end{aligned}$$

(127)

$$\begin{aligned}&= (-1)^k\frac{(k-R-1)(k-R-2) \cdots (k - R - k)}{k!} \end{aligned}$$

(128)

Expanding the last term with a null term ($+1-1$):

$$\begin{aligned} \begin{pmatrix} R \\ k \end{pmatrix}&= (-1)^k\frac{(k-R-1)(k-r-2) \cdots (k - R -1 - k+1)}{k!} \end{aligned}$$

(129)

$$\begin{aligned}&= \frac{(k-R-1)^{\underline{k}}}{k!} \end{aligned}$$

(130)

$$\begin{aligned}&= (-1)^k \begin{pmatrix} k-R-1 \\ k \end{pmatrix} \end{aligned}$$

(131)

The second property needed reads:

$$\begin{aligned} \sum _{k=0}^n \begin{pmatrix} R + k \\ k \end{pmatrix}&= \begin{pmatrix} R + n + 1 \\ n \end{pmatrix} \end{aligned}$$

(132)

A proof by induction follows:

• The base case ($n=0$) is trivial:

  $$\begin{aligned} \begin{pmatrix}R \\ 0\end{pmatrix} = \begin{pmatrix}R+1 \\ 0\end{pmatrix} = 1 \end{aligned}$$

  (133)

• The induction hypothesis reads: if

  $$\begin{aligned} \sum _{k=0}^n \begin{pmatrix} R + k \\ k \end{pmatrix}&= \begin{pmatrix} R + n + 1 \\ n \end{pmatrix} \end{aligned}$$

  (134)

  holds, then

  $$\begin{aligned} \sum _{k=0}^{n+1} \begin{pmatrix} R + k \\ k \end{pmatrix}&= \begin{pmatrix} R + n + 2 \\ n+1 \end{pmatrix} \end{aligned}$$

  (135)

  hold as well.

• Beginning the induction step by consuming the uppermost index of the sum:

  $$\begin{aligned} \sum _{k=0}^{n+1} \begin{pmatrix} R + k \\ k \end{pmatrix}&= \sum _{k=0}^n \begin{pmatrix} R + k \\ k \end{pmatrix} + \begin{pmatrix} R+n+1 \\ n + 1\end{pmatrix} \end{aligned}$$

  (136)

  Using the induction hypothesis

  $$\begin{aligned} \sum _{k=0}^{n+1} \begin{pmatrix} R + k \\ k \end{pmatrix}&= \begin{pmatrix} R + n + 1 \\ n \end{pmatrix} + \begin{pmatrix} R + n + 1 \\ n + 1\end{pmatrix} \end{aligned}$$

  (137)

  And finally applying Pascal’s rule:

  $$\begin{aligned} \sum _{k=0}^{n+1} \begin{pmatrix} R + k \\ k \end{pmatrix}&= \begin{pmatrix} R + n + 2 \\ n + 1\end{pmatrix} \end{aligned}$$

  (138)

The first and second properties can now be used to prove a third one:

$$\begin{aligned} \sum _{k=0}^{n+1} (-1)^k \begin{pmatrix} R \\ k \end{pmatrix}&= (-1)^n \begin{pmatrix} r-1 \\ n \end{pmatrix} \end{aligned}$$

(139)

Applying the property number one, followed by property number two, then one again:

$$\begin{aligned} \sum _{k=0}^{n+1} (-1)^k \begin{pmatrix} R \\ k \end{pmatrix}&= \sum _{k=0}^{n+1} \begin{pmatrix} k - R - 1 \\ k \end{pmatrix} \end{aligned}$$

(140)

$$\begin{aligned}&= \begin{pmatrix} -R + n \\ n \end{pmatrix} \end{aligned}$$

(141)

$$\begin{aligned}&= (-1)^n \begin{pmatrix}R-1 \\ n\end{pmatrix} \end{aligned}$$

(142)

The property is not quite in the form needed later later yet. In fact, the special case of $R=n$ with shifted boundaries is required. Shifting the top boundary:

$$\begin{aligned} \sum _{k=0}^n (-1)^k \begin{pmatrix} R \\ k \end{pmatrix}&= (-1)^n \begin{pmatrix} R-1 \\ n \end{pmatrix} \end{aligned}$$

(143)

$$\begin{aligned} \sum _{k=0}^{n-1} (-1)^k \begin{pmatrix} R \\ k \end{pmatrix}&= (-1)^{n-1} \begin{pmatrix} R-1 \\ n-1 \end{pmatrix} \end{aligned}$$

(144)

setting $R = n$:

$$\begin{aligned} \sum _{k=0}^{n-1} (-1)^k \begin{pmatrix} n \\ k \end{pmatrix}&= (-1)^{n-1} \begin{pmatrix} n-1 \\ n-1 \end{pmatrix} \end{aligned}$$

(145)

$$\begin{aligned} \sum _{k=0}^{n-1} (-1)^k \begin{pmatrix} n \\ k \end{pmatrix}&= -(-1)^{n} \end{aligned}$$

(146)

and finally shifting the lower boundary (and moving the one term to the right-hand side):

$$\begin{aligned} \sum _{k=1}^{n-1} (-1)^k \begin{pmatrix} n \\ k \end{pmatrix}&= -(-1)^{n} -1 \end{aligned}$$

(147)

On to the statement actually to be proven: $0 = \int \! (x-\tilde{x})^n W(x-\tilde{x}){{\mathrm {d}}}\tilde{x}$ implies that $x^n = \int \! \tilde{x}^n W(x-\tilde{x}){{\mathrm {d}}}\tilde{x}$ for $n>1$. The proof is again to be by induction.

• The base case ($n=1$) is quite straightforward:

  $$\begin{aligned} 0&= \int \! (x-\tilde{x}) W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (148)
  $$\begin{aligned}&= \int \! x W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} - \int \! \tilde{x} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (149)
  $$\begin{aligned}&= x \underbrace{\int \! W(x-\tilde{x}){{\mathrm {d}}}\tilde{x}}_{1} - \int \! \tilde{x} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (150)

  Thus (underbraced term by moment condition number 1, which can be trivially ensured using a normalized W):

  $$\begin{aligned} x = \int \! \tilde{x} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (151)

• The induction hypothesis reads: if $0 = \int \! (x-\tilde{x})^{q-1} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x}$ implies $x^{q-1} = \int \! \tilde{x}^{q-1} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x}$ for $q=1\cdots n-1$, then $0 = \int \! (x-\tilde{x})^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x}$ implies $x^{n} = \int \! \tilde{x}^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x}$.

• The induction step starts by expanding the binomial term in the moment condition into Pascal’s triangle:

  $$\begin{aligned} 0&= \int \! (x-\tilde{x})^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (152)
  $$\begin{aligned}&= \int \! \sum _{k=0}^n \begin{pmatrix}n \\ k \end{pmatrix} x^{n-k}(-\tilde{x})^k W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (153)

  The first and last term of the sum is now consumed:

  $$\begin{aligned} 0 =&\int \! x^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (154)
  $$\begin{aligned}&+ \int \! \sum _{k=1}^{n-1} \begin{pmatrix}n \\ k \end{pmatrix} x^{n-k}(-\tilde{x})^k W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (155)
  $$\begin{aligned}&+ \int \! (-\tilde{x})^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (156)

  Moving the integral into the the summation as far as possible:

  $$\begin{aligned} 0 =&\int \! x^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (157)
  $$\begin{aligned}&+ \sum _{k=1}^{n-1} \begin{pmatrix}n \\ k \end{pmatrix} x^{n-k}\int \! (-\tilde{x})^k W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (158)
  $$\begin{aligned}&+ \int \! (-\tilde{x})^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (159)

  Since $k\le n-1$ the induction hypothesis $\int (-\tilde{x})^k W(x-\tilde{x}) {{\mathrm {d}}}\tilde{x} = - x^k$ can be applied:

  $$\begin{aligned} 0 =&\int \! x^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (160)
  $$\begin{aligned}&+ \sum _{k=1}^{n-1} \begin{pmatrix}n \\ k \end{pmatrix} x^{n-k}(-x)^k \end{aligned}$$

  (161)
  $$\begin{aligned}&+ \int \! (-\tilde{x})^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (162)

  Using that $x^{n-k}(-x)^k = (-1)^k x^n$:

  $$\begin{aligned} 0 =&\int \! x^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (163)
  $$\begin{aligned}&+ x^n\sum _{k=1}^{n-1} \begin{pmatrix}n \\ k \end{pmatrix} (-1)^k \end{aligned}$$

  (164)
  $$\begin{aligned}&+ \int \! (-\tilde{x})^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (165)

  Using the (manipulated) third property from above:

  $$\begin{aligned} 0 =&\int \! x^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (166)
  $$\begin{aligned}&+ x^n (-(-1)^n-1) \end{aligned}$$

  (167)
  $$\begin{aligned}&+ \int \! (-\tilde{x})^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (168)

  The proof is readily completed:

  $$\begin{aligned} 0 =&x^{n} \underbrace{\int \! W(x-\tilde{x}){{\mathrm {d}}}\tilde{x}}_{1} + x^n (-(-1)^n-1)\nonumber \\&+ \int \! (-\tilde{x})^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (169)
  $$\begin{aligned} 0 =&x^n + x^n (-(-1)^n) - x^n + \int \! (-\tilde{x})^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (170)
  $$\begin{aligned} x^n =&\int \! (\tilde{x})^{n} W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} \end{aligned}$$

  (171)

1.3 Appendix 3: Proof That Completeness of Order n Ensures a Convergence Rate of Order $n+1$

To proof is started with a Taylor approximation of some function f(.) about a point x:

$$\begin{aligned} f(\tilde{x}) = f(x) + (\tilde{x}-x)\cdot f'(x) + \frac{1}{2}(\tilde{x}-x)^2\cdot f''(x) + \cdots \end{aligned}$$

(172)

Multiplying this by some kernel function $W(x-\tilde{x})$ and integrate over the whole domain:

$$\begin{aligned} \int \!f(\tilde{x}) W(x-\tilde{x}) {{\mathrm {d}}}\tilde{x} =&\int \! W(x-\tilde{x})f(x) {{\mathrm {d}}}\tilde{x} \end{aligned}$$

(173)

$$\begin{aligned}&+ \int \! (\tilde{x}-x) \cdot W(x-\tilde{x})\cdot f'(x) {{\mathrm {d}}}\tilde{x} \end{aligned}$$

(174)

$$\begin{aligned}&+ \frac{1}{2!} \int \! (\tilde{x}-x)^2 \cdot W(x-\tilde{x})\cdot f''(x){{\mathrm {d}}}\tilde{x} + \cdots \end{aligned}$$

(175)

Realizing that the derivative terms are not bound by the differential ${{\mathrm {d}}}\tilde{x}$:

$$\begin{aligned} \int \!f(\tilde{x}) W(x-\tilde{x}) {{\mathrm {d}}}\tilde{x} =&\int \! W(x-\tilde{x})f(x) {{\mathrm {d}}}\tilde{x} \end{aligned}$$

(176)

$$\begin{aligned}&+ f'(x)\cdot \int \! (\tilde{x}-x) \cdot W(x-\tilde{x}) {{\mathrm {d}}}\tilde{x} \end{aligned}$$

(177)

$$\begin{aligned}&+ \frac{1}{2!} f''(x)\cdot \int \! (\tilde{x}-x)^2 \cdot W(x-\tilde{x}){{\mathrm {d}}}\tilde{x} + \cdots \end{aligned}$$

(178)

The familiar moment conditions are found again on the right-hand side (albeit in negative form). It is apparent that if the first n moment conditions hold, the above expression (which is equal to the function approximation by a kernel function) is exact of to a term in the order of $\mathcal {O}((x-\tilde{x})^{n+1})$.