ps08sol

# ps08sol - 18.03 Problem Set 8 Fall 2009 Solutions All the...

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All the functions of t appearing here are assumed to be zero for t < 0 . Thus for example e ta means u 0 ( t ) · e ta = ( 0 ( t < 0) , e ta ( t > 0) . 1. (4 points) Suppose that f and g are piecewise continuous and of exponential types A and B : | f ( t ) | ≤ Ce At , | g ( t ) | ≤ De Bt ( t 0) . Show that f * g is also of exponential type: that is, that there are constants M and N so that ( f * g )( t ) Me Nt . ( Exponential type of f * g ) Give some explicit formulas for M and N in terms of A , B , C , and D . (There are lots of values of M and N that work. A good answer is one with N small and M small; that means the estimate (Exponential type of f * g ) is giving you as much information as possible.) We have ( f * g )( t ) = Z t 0 e - st f ( τ ) g ( t - τ ) dτ, and so, assuming that A 6 = B , | ( f * g )( t ) | ≤ Z t 0 | Ce De B ( t - τ ) | = CDe Bt Z t 0 | e ( A - B ) τ | CDe Bt | A - B | | e ( A - B ) t - 1 | = CD | A - B | | e At - e Bt | ≤ 2 CD | A - B | e max( A,B ) t . Hence, when A 6 = B , we may take M to be 2 CD | A - B | , and N to be the maximum of A and B . When A = B , we have | ( f * g )( t ) | ≤ Z t 0 | Ce De B ( t - τ ) | = CDe At Z t 0 1 = CDte At CD e ( A + ± ) t for any ± > 0, for suﬃciently large t . Hence we may take M to be CD , and N to be anything bigger than A . This makes sense, since if g is exponential of type A , then g is also exponential of type B := A + ± for any ± > 0. 2. (5 points) a) From the deﬁnition of convolution, calculate e at * e bt whenever a 6 = b . We have e at * e bt = Z t 0 e e b ( t - τ ) = e bt Z t 0 e ( a - b ) τ = e bt a - b ( e ( a - b ) t - 1) = e at - e bt a - b . b) From the deﬁnition of convolution, calculate e at * e at . Now we have

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ps08sol - 18.03 Problem Set 8 Fall 2009 Solutions All the...

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