Week 8

1. Week 8 5. Applications of the LT to PDEs PDEs can be solved via the LT using (more or less) the same approach as that for ODEs. Example 1: Solve the following initial-boundary-value problem: (1) (2) (3)

2. Solution: Step 1: Take the LT of Eq. (1): hence, using (2), Step 2: Solve this ODE, (4)

3. Step 3: Take the LT of BC (3): hence, (4) yields Step 4: Take the inverse LT: hence, (5)

4. If t≤ ½ x2, the PoI in (5) can be closed to the right – hence, (6) If t≥ ½ x2, the PoI in (5) has to be closed to the left – hence, (7) Comments: • For t= ½ x2, both formulae (6)-(7) yield the same result, t= 0, as the should. • Formulae (6)-(7) can be written as a single expression, • The graphs of the solution for increasing values of t look like...

5. Some important integrals and inverse Laplace transforms Example 2: Show that where t > 0 is a real parameter. Hint: consider and change (p, q) to polar coords. (r, φ): p = r cos φ, q = r sin φ.

6. Example 3: Show that where t > 0 and x (of arbitrary sign) are real parameters. Example 4: Find where t > 0 and x (of arbitrary sign) are real parameters.

7. Example 5: Find where x is a real parameter, and the branch of s1/2 on the plane of complex s is fixed by the condition 11/2 = +1 and a cut along the negative part of the real axis. Solution: where γ > 0.

8. The easiest way to make I real (which should be one’s general strategy when evaluating complex integrals) is to ‘wrap’ the PoI around the cut, which yields Introduce p > 0, such that Recalling how the branch of s1/2 was fixed, we have

9. Then, hence, In the 1st integral, change p to pnew= –p, then omit the new and ‘join’ the two integrals together, which yields

10. Finally, using the results of Example 4, obtain