An Introduction to Mathematical Physics

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If we start with the same solution as before, v = e^+Pv, we find that, in order to v=0 satisfy the first and second conditions, it must take the form v-ffx) A u < -'--U v=o sin- m-n-x ir = o Flo. 48. where m, as before, is any integer. In satisfying the third condition, we are apparently at a stop because neither e ' nor e l vanishes when y = 0. Their difference, however, vanishes. We have therefore . . miry . rmrx v = smh — t 2 - sm — =— • Taking every possible value of m and multiplying each term by a constant b m , we find for y = h, „, . . rmrh . rmrx v = zb m sum —j— sm — j— ■ This must be the half range sine expansion for f(x). Hence , . . rmrh 2 C l ,, . . rmrx , b m smh — = | f(x) sm -j- dx. If we include the sinh — j— in the b m , the solution can be written sinh rmry * = 2J. sinh Vttrh sin where 2 (*' .= 7 J /<*)» wwra; . sin —f— rfa;. 86 CONDUCTION OF HEAT (3) Let us now consider the case of the same rectangular plate with different boundary conditions, namely i> = for x = 0, v = for x = l, v = 4>(x) for y = 0, v=f(x) for y — h.

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