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HARVARD MATH 19 - ps11

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HW #11Ex. 1, 3 (a,c), 4(a,c,e)Ex 1.u(t, x) = R1√te−x24µt.We have the graph of u(1, x) and of u(2, x). We can find the value of u(1, x)and u(2, x) for any x.Determine R and µ in ea ch case:u(1, x) = Re−x2/4µIf we take x = 0 then u(1 , 0) = e−0/4µ= R. Therefore, R = u(1, 0).Now, take another value for x.u(1, 2) = u (1, 0)e−44µ⇒ u(1, 2) = u(1, 0)e−1µ.−1µ= lnu(1,2)u(1,0)⇒ µ = −1lnu(1,2)u(1,0).For u(2, x), we do the same:u(2, 0) =R√2e02/8µ⇒ R =√2u(2, 0).u(2,√8) =√2u(2,0)√2e−8/8µ= u(2, 0 )e−1µ.Thus, µ = −1lnu(2,√8)u(2,0)Ex. 3a) See the graphs on the next page.1Figure 1: sin(πx) – positive on 0 ≤ x ≤ 1 since the sinus function is positive inthe interval [0, π]Figure 2: sin(3πx) – positive only in the intervals 0 ≤ x ≤ 1/3 and 2/3 ≤ x ≤ 1,but negative in between 1/3 and 2/3Figure 3: cos(πx/2) – positive in the interval 0 ≤ x ≤ 1c)2Figure 4: sin(πx/4) – positive in the interval 0 ≤ x ≤32Figure 5: sin(πx/2) – pos itive in the interval 0 ≤ x ≤32Figure 6: sin(3πx/4) – positive in the interval 0 ≤ x ≤ 4/3, but negative in4/3 < x ≤ 3/23Figure 7: sin(πx) – positive in the interval 0 ≤ x ≤ 1, but negative in theinterval 1 < x ≤ 3/24Ex. 4a) B(x) = αe5x+ βe−5x.x = 0x = 1α + β = 0αe5+ βe−5= 0α = −βα = −βe−10The solution is α = β = 0.c) B(x) = α cos(πx) + β sin(πx).x = 0x = 1α cos 0 + β sin 0 = 0α cos π + β sin π = 0α = 0−α = 0Thus, α = 0, β ∈Re) B(x) = αe3π x+ βe−3πx.x = 0x = 1α + β = 0αe3π+ βe−3π= 0α = −βα = −βe−6πThe solution is α = β =


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HARVARD MATH 19 - ps11

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