A numerical method for solving the inverse heat conduction problem without initial value

Figures & data

Figure 1. Schematic representation of K = 0.5, without noise: (a) u(0, t) and (b) .

Figure 2. Schematic representation of K = 0.5, with 0.5% random noise: (a) u(0, t) and (b) .

Figure 3. Schematic representation of K = 0.5, with 1% random noise: (a) u(0, t) and (b) .

Figure 4. Schematic representation of K = 0.5, with 3% random noise: (a) u(0, t) and (b) .

Figure 5. Graph of K = 2, without noise: (a) u(0, t) and (b) .

Figure 6. Graph of K = 2, with 0.5% random noise: (a) u(0, t) and (b) .

Figure 7. Graph of K = 2, with 1% random noise: (a) u(0, t) and (b) .

Figure 8. Graph of K = 2, with 3% random noise: (a) u(0, t) and (b) .

Figure 9. Graph of K = 3, without noise: (a) u(0, t) and (b) .

Figure 10. Graph of K = 3, with 0.5% random noise: (a) u(0, t) and (b) .

Figure 11. Graph of K = 3, with 1% random noise: (a) u(0, t) and (b) .

Figure 12. Graph of K = 3, with 3% random noise: (a) u(0, t) and (b) .

Table 1. Absolute errors with different noise level and K.

K	Noise level	Errors for u(0, t)	Errors for
0.5	0	0.00095	0.0090
0.5	0.5%	0.0058	0.0270
0.5	1%	0.0080	0.0282
0.5	3%	0.0127	0.0296
2	0	0.0035	0.0079
2	0.5%	0.0070	0.0158
2	1%	0.0086	0.0316
2	3%	0.0348	0.0595
3	0	0.0011	0.0091
3	0.5%	0.0150	0.1132
3	1%	0.0157	0.1137
3	3%	0.1342	0.2568

Figure 13. Graph of (a) u(0, t) and (b) , without random noise.

Figure 14. Graph of (a) u(0, t) and (b) , with 0.5% random noise.

Figure 15. Graph of (a) u(0, t) and (b) , with 1% random noise.

Figure 16. Graph of (a) u(0, t) and (b) , with 3% random noise.

Table 2. Absolute errors with different noise level.

Noise level	Errors for u(0, t)	Errors for
0	0.0426	0.4233
0.5%	0.1604	0.8297
1%	0.2675	1.0645
3%	0.9866	2.4189