Chem Differential Eq HW Solutions Fall 2011 133

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online