Chem Differential Eq HW Solutions Fall 2011 133

# Chem Differential Eq HW Solutions Fall 2011 133 - Section...

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Section 7.9 The Nonhomogeneous Heat Equation 133 Solutions to Exercises 7.9 1. Proceed as in Example 1 with c =1 / 2. Equation (3) becomes in this case u ( x, t )= 2 2 t e - x 2 /t * δ 1 ( x ) = 1 πt e - ( x - 1) 2 /t , since the eﬀect of convolution by δ 1 is to shift the function by 1 unit to the right and multiply by 1 2 π . 5. We use the superposition principle (see the discussion preceeding Example 4). If φ is the solution of u t = 1 4 u xx + δ 0 , u ( x, 0) = 0 and ψ is the solution of u t = 1 4 u xx , u ( x, 0) = U 0 ( x ), then you can check that φ + ψ is the solution of u t = 1 4 u xx + δ 0 , u ( x, 0) = U 0 ( x ). By Examples 1, φ ( x, t 2 t π e - x 2 /t - 2 | x | π Γ ± 1 2 , x 2 t ² and by Exercise 20, Section 7.4, ψ ( x, t 1 2 erf ± x t ² . 9. Apply Theorem 2 with
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