一类三维逆时热传导问题的数值求解

Abstract

Abstract:

In this paper, we consider the numerical solution of a three-dimensional inverse heat conduction problem. Based on the finite difference and the finite element method, the stiffness matrix and load vector are derived to solve the heat conduction problem. We use the variable separation method to establish the corresponding relationship between the temperature field at time T and the initial temperature field for the inverse problem. The inversion formulation is obtained. The local stability for the inverse problems is proved under certain priori assumptions. To overcome the ill-posedness for solving the inverse problem, we used the Tikhonov regularization and perturbation regularization method to reconstruct the initial temperature field. We verified the effectiveness of the algorithm through several numerical experiments.

Key words: 3-D inverse heat conduction problem, Finite element, Ill-posedness, Regularization method

CLC Number:

O241.8

Qingchun Meng,Lei Zhang. Numerical Solution of the Three-Dimensional Inverse Heat Conduction Problems[J].Acta mathematica scientia,Series A, 2022, 42(1): 187-200.

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Figures/Tables 8

References 19

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T	J=10			J=20
T	$\left\\|e_{1}\right\\|_{L^{2}}$	$\left\\|e_{1}\right\\|_{L ^{\infty}}$	$\left\\|e_{2}\right\\|$	$\left\\|e_{1}\right\\|_{L^{2}}$	$\left\\|e_{1}\right\\|_{L ^{\infty}}$	$\left\\|e_{2}\right\\|$
2	0.0972	0.0087	0.0111	0.0961	0.0030	0.0039
4	0.1537	0.0137	0.0222	0.1514	0.0048	0.0077
6	0.1824	0.0163	0.0335	0.1790	0.0057	0.0116
8	0.1923	0.0172	0.0449	0.1881	0.0059	0.0155

T	J=30			J=40
T	$\left\\|e_{1}\right\\|_{L^{2}}$	$\left\\|e_{1}\right\\|_{L ^{\infty}}$	$\left\\|e_{2}\right\\|$	$\left\\|e_{1}\right\\|_{L^{2}}$	$\left\\|e_{1}\right\\|_{L ^{\infty}}$	$\left\\|e_{2}\right\\|$
2	0.0961	0.0030	0.0039	0.1435	0.0016	0.0020
4	0.1514	0.0048	0.0077	0.2260	0.0025	0.0041
6	0.1790	0.0057	0.0116	0.2669	0.0030	0.0061
8	0.1881	0.0059	0.0155	0.2803	0.0031	0.0082

δ	α = 0.0001			α = 0.0005
δ	$\left\\|e_{3}\right\\|_{L^{2}}$	$ \left\\|e_{3}\right\\|_{L ^{\infty}}$	$\left\\|e_{4}\right\\|$	$\left\\|e_{3}\right\\|_{L^{2}}$	$ \left\\|e_{3}\right\\|_{L ^{\infty}}$	$\left\\|e_{4}\right\\|$
0.001	0.3704	0.0119	0.0117	0.3165	0.0100	0.0100
0.005	0.4378	0.0134	0.0138	0.3815	0.0126	0.0121
0.01	0.5333	0.0198	0.0169	0.4628	0.0156	0.0146
0.05	1.3499	0.0734	0.0427	1.1337	0.0432	0.0359

δ	α = 0.001			α = 0.005
δ	$\left\\|e_{3}\right\\|_{L^{2}}$	$ \left\\|e_{3}\right\\|_{L ^{\infty}}$	$\left\\|e_{4}\right\\|$	$\left\\|e_{3}\right\\|_{L^{2}}$	$ \left\\|e_{3}\right\\|_{L ^{\infty}}$	$\left\\|e_{4}\right\\|$
0.001	0.2490	0.0080	0.0079	0.2778	0.0088	0.0088
0.005	0.3122	0.0100	0.0099	0.2147	0.0068	0.0068
0.01	0.3950	0.0122	0.0125	0.1381	0.0048	0.0044
0.05	1.0394	0.0332	0.0329	0.5074	0.0152	0.0160

α	T=4			T=6
α	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$
0.0001	0.2490	0.0080	0.0079	0.2778	0.0088	0.0088
0.0005	0.3122	0.0100	0.0099	0.2147	0.0068	0.0068
0.001	0.3950	0.0122	0.0125	0.1381	0.0048	0.0044
0.005	1.0394	0.0332	0.0329	0.0377	0.0011	0.0013

α	T=8			T=10
α	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$
0.0001	0.4306	0.0141	0.0154	0.5406	0.0201	0.0193
0.0005	0.3987	0.0130	0.0142	0.4987	0.0159	0.0178
0.001	0.3616	0.0117	0.0129	0.4513	0.0142	0.0161
0.005	0.0686	0.0023	0.0024	0.0789	0.0024	0.0028

α	T=20			T=30
α	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$
0.0001	1.0701	0.0332	0.0382	1.5610	0.0492	0.0557
0.0005	0.9424	0.0297	0.0336	1.1354	0.0359	0.0405
0.001	0.7844	0.0247	0.0280	0.6202	0.0196	0.0221
0.005	0.4205	0.0133	0.0150	2.9273	0.0926	0.1044

α	T=40			T=50
α	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$
0.0001	1.8839	0.0595	0.0672	1.6395	0.0519	0.0585
0.0005	0.5125	0.0162	0.0183	2.3469	0.0742	0.0837
0.001	1.0346	0.0327	0.0369	6.0408	0.1910	0.2154
0.005	9.2018	0.2910	0.3281	17.8107	0.5632	0.6350

α	T=4			T=6
α	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$
0.0001	0.6014	0.0454	0.0214	0.9492	0.0748	0.0338
0.0005	0.2497	0.0135	0.0089	0.3730	0.0185	0.0133
0.001	0.2042	0.0101	0.0073	0.3015	0.0144	0.0108
0.005	0.0210	0.0015	0.0007	0.0585	0.0031	0.0021

α	T=8			T=10
α	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$
0.0001	1.0726	0.0650	0.0382	1.1132	0.0726	0.0397
0.0005	0.4502	0.0205	0.0161	0.5333	0.0230	0.0190
0.001	0.3932	0.0199	0.0140	0.4762	0.0163	0.0170
0.005	0.0865	0.0044	0.0031	0.0953	0.0037	0.0034

α	T=20			T=30
α	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$
0.0001	1.1529	0.0486	0.0411	1.5826	0.0521	0.0564
0.0005	0.9618	0.0333	0.0343	1.1500	0.0356	0.0401
0.001	0.7997	0.0240	0.0285	0.6346	0.0203	0.0226
0.005	0.4066	0.0131	0.0145	2.9149	0.0922	0.1039

α	T=40			T=50
α	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$	$\left\\|e_{5}\right\\|_{L^{2}}$	$\left\\|e_{5}\right\\|_{L ^{\infty}}$	$\left\\|e_{6}\right\\|$
0.0001	1.8994	0.0617	0.0677	1.6546	0.0525	0.0590
0.0005	0.5265	0.0166	0.0188	2.3341	0.0738	0.0832
0.001	1.0211	0.0325	0.0364	6.0298	0.1907	0.2150
0.005	9.1925	0.2907	0.3278	17.8057	0.5631	0.6349

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Numerical Solution of the Three-Dimensional Inverse Heat Conduction Problems

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