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Chem Differential Eq HW Solutions Fall 2011 39

# Chem Differential Eq HW Solutions Fall 2011 39 - u x t =-4...

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Section 3.4 D’Alembert’s Method 39 (c) Continuing from (b) and using the notation in the text, we obtain 1 2 c x + ct x - ct g * ( s ) ds = 1 2 c [ G ( x + ct ) - G ( x - ct )] = n =1 - b * n 1 2 cos( L ( x + ct )) - cos( L ( x - ct )) = n =1 b * n sin( L x ) sin( L ct ) = n =1 b * n sin( L x ) sin( λ n t ) . (d) To derive d’Alembert’s solution from (8), Section 3.3, proceed as follows: u ( x, t ) = n =1 b n sin( L x ) cos( λ n t ) + n =1 b * n sin( L x ) sin( λ n t ) = 1 2 ( f * ( x - ct ) + f * ( x + ct ) ) + 1 2 c [ G ( x + ct ) - G ( x - ct )] , where in the last equality we used Exercise 16 and part (c). 21. Follow the labeling of Figure 8 in Section 3.4. Let P 1 = ( x 0 , t 0 ) be an arbitrary point in the region II . Form a characteristic parallelogram with vertices P 1 , P 2 , Q 1 , Q 2 , as shown in Figure 8 in Section 3.4. The vertices P 2 and Q 1 are on the characteristic line x + 2 t = 1 and the vertex Q 2 is on the boundary line x = 1. From Proposition 1, we have u ( P 1 ) = u ( Q 1 ) + u ( Q 2 ) - u ( P 2 ) = u ( Q 1 ) - u ( P 2 ) , because u ( Q 2 ) = 0. We will find u ( P 2 ) and u ( Q 1 ) by using the formula
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Unformatted text preview: u ( x, t ) =-4 t 2 + x-x 2 + 8 tx from Example 4, because P 2 and Q 1 are in the region I . The point Q 1 is the intersection point of the characteristic lines x-2 t = x-2 t and x + 2 t = 1. Adding the equations and then solving for x , we get x = x + 1-2 t 2 . The second coordinate of Q 1 is then t = 1-x + 2 t 4 . The point Q 2 is the intersection point of the characteristic line x + 2 t = x + 2 t and x = 1. Thus t = x + 2 t-1 2 . The point P 2 is the intersection point of the characteristic lines x +2 t = 1 and x-2 t = 1-( x + 2 t-1). Solving for x and t , we ±nd the coordinates of P 2 to be x = 3-x-2 t 2 and t =-1 + x + 2 t 4 ....
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