Chem Differential Eq HW Solutions Fall 2011 37

# Chem Differential Eq HW Solutions Fall 2011 37 - 1 2 c ³ x...

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Section 3.4 D’Alembert’s Method 37 u x_, t_ : 1 2 periodicf x t periodicf x t 1 2 periodicbigg x t periodicbigg x t tt Table Plot Evaluate u x, t , x, 0, 1 , PlotRange 0, 1 , 1, 1 , Ticks .5 , 1, .5, .5, 1 , DisplayFunction Identity , t, 0, 2.3, 1 5 ; Show GraphicsArray Partition tt, 4 9. You can use Exercise 11, Section 3.3, which tells us that the time period of motion is T = 2 L c . So, in the case of Exercise 1, T =2 π , and in the case of Exercise 5, T = 2. You can also obtain these results directly by considering the formula for u ( x, t ). In the case of Exercise 1, u ( x, t )= 1 2 ± sin( πx + t )+ sin( πx - t ) ² so u ( x, t +2 π )= 1 2 ± sin( πx + t 2 π ) + sin( πx - t 2 π ) ² = u ( x, t ). In the case of Exercise 5, use the fact that f * and G are both 2-periodic. 13. We have u ( x, t )= 1 2 [ f * ( x + ct )+ f * ( x - ct )] +
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Unformatted text preview: 1 2 c ³ x + ct x-ct g * ( s ) ds, where f * and g * are odd and 2 L-periodic. So u ( x, t + L c ) = 1 2 [ f * ( x + ct + L ) + f * ( x-ct-L )] + 1 2 c ³ x + ct + L x-ct-L g * ( s ) ds. Using the fact that f * is odd, 2 L-period, and satisFes f * ( L-x ) = f * ( x ) (this property is given for f but it extends to f * ), we obtain f * ( x + ct + L ) = f * ( x + ct + L-2 L ) = f * ( x + ct-L ) =-f * ( L-x-ct ) =-f * ( L-( x + ct )) =-f * ( x + ct ) . Similarly f * ( x-ct-L ) =-f * ( L-x + ct ) =-f * ( L-x + ct ) =-f * ( L-( x-ct )) =-f * ( x-ct ) ....
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