Chem Differential Eq HW Solutions Fall 2011 35

Chem Differential Eq HW Solutions Fall 2011 35 - The...

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Section 3.4 D’Alembert’s Method 35 Solutions to Exercises 3.4 1. We will use (5), since g * = 0. The odd extension of period 2 of f ( x )= sin πx is f * ( x ) = sin πx .So u ( x, t )= 1 2 ± sin( π ( x + t π )) + sin( π ( x - t π )) ² = 1 2 ± sin( πx + t ) + sin( πx - t ) ² . 5. The solution is of the form u ( x, t )= 1 2 ± f * ( x - t )+ f ( x + t ) ² + 1 2 ± G ( x + t ) - G ( x - t ) ² = 1 2 ±( f * ( x - t ) - G ( x - t ) ) + ( f * ( x + t )+ G ( x + t ) , where f * is the odd extension of f and G is as in Example 3. In the second equality, we expressed u as the average of two traveling waves: one wave traveling to the right and one to the left. Note that the waves are not the same, because of the G term. We enter the formulas in Mathematica and illustrate the motion of the string.
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Unformatted text preview: The difficult part in illustrating this example is to define periodic functions with Mathematica. This can be done by appealing to results from Section 2.1. We start by defining the odd extensions of f and G (called big g) on the interval [-1, 1]. Clear f, bigg f x_ : 2 x ; 1 2 x 1 2 f x_ : 2 1 x ; 1 2 x 1 f x_ : 2 1 x ; 1 x 1 2 bigg x_ 1 2 x^2 1 2 Plot f x , bigg x , x, 1, 1-1-0.5 0.5 1-1-0.5 0.5 1 Here is a tricky Mathematica construction. (Review Section 2.1.)...
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