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L ’ 2 Tn(t)=fn(t) (917) In addition, u(x,0) = (n≥1 Tn(0)sin nπ L x = (n≥1 φn sin &nπx L ’ = φ(x) so that Tn(0) = φn,n≥ 1 (918) Solving (917)–(918), we ﬁnd Tn(t)=φne −k(nπ L) 2t t 0 e−k(nπ L) 2(t−s)f n(s) ds and u(x,t)= (n≥1 φne −k(nπ L) 2t sin &nπx L ’ (n≥1 sin &nπx L ’ . . 9mm Z l iG j ̏ ̂ 6 ̂ I ׂ܂ B C N ł A ꂢ ɖ 3000 ł ܂ B g ň ʒ ł 郁 p b N B 蕨 ɂ œK ł B.

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