Unformatted text preview: 1 ) and has the correct initial conditions because, in general, the solution to ˙ u = Au with initial conditions u is u ( t ) = e tA u . To prove the claim and hence the result, it remains to show that z ( t ) also satisﬁes equation ( 1 ). It clearly satisﬁes the correct initial conditions. To do this, we diﬀerentiate using the product rule: d dt z ( t ) = d dt ( e tS e tN y ) = Se tS e tN y + e tS Ne tN y (2) Now since S and N commute, we have e tS N = Ne tS . This follows easily using the power series expression for e tS . Thus, equation ( 2 ) becomes d dt z ( t ) = Se tS e tN y + Ne tS e tN y = ( S + N ) e tS e tN y = ( S + N ) z This proves the claim. 1...
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 Fall '09
 Marsden
 Product Rule, Trigraph, Commuting, initial conditions

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