Chem Differential Eq HW Solutions Fall 2011 1

# Chem Differential Eq HW Solutions Fall 2011 1 - Section 1.1...

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Section 1.1 What Is a Partial Diﬀerential Equation? 1 Solutions to Exercises 1.1 1. If u 1 and u 2 are solutions of (1), then ∂u 1 ∂t + 1 ∂x = 0 and 2 + 2 =0 . Since taking derivatives is a linear operation, we have ( c 1 u 1 + c 2 u 2 )+ ( c 1 u 1 + c 2 u 2 )= c 1 1 + c 2 2 + c 1 1 + c 2 2 = c 1 =0 ± ²³ ´ µ 1 + 1 + c 2 =0 ± ²³ ´ µ 2 ++ 2 , showing that c 1 u 1 + c 2 u 2 is a solution of (1). 5. Let α = ax + bt , β = cx + dt , then = ∂α + ∂β = a + c = + = b + d . Recalling the equation, we obtain - ( b - a ) +( d - c ) . Let a =1 ,b =2 ,c ,d = 1. Then u = f ( β ) u ( x, t f ( x + t ) , where f is an arbitrary diﬀerentiable function (of one variable). 9. (a) The general solution in Exercise 5 is u ( x, t f ( x + t ). When t = 0, we get u ( x, 0) = f ( x )=1 / ( x 2 + 1). Thus u ( x, t f ( x + t 1 ( x + t ) 2 +1 . (c) As t increases, the wave f ( x 1 1+ x 2 moves to the left. -2 -1 0 1 2 0 1 2 3 0 1 Figure for Exercise 9(b). 13. To Fnd the characteristic curves, solve dy dx = sin x. Hence y = - cos x + C or y + cos x = C . Thus the solution of the partial diﬀerential equation is
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