1. 1 Week 9 PDEs can be solved via the LT using (more or less) the same approach as that for ODEs. 5. Applications of the LT to PDEs Example 1: Solve the following initial-boundary-value problem: (1) (2) (3)

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

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

4. 4 If t ≤ ½ x 2, the PoI in (5) can be closed to the right – hence, If t ≥ ½ x 2, the PoI in (5) has to be closed to the left – hence, (6) (7) Comments:  For t = ½ x 2, 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. 5 Some important integrals and inverse Laplace transforms Example 2: Hint: consider and change (p, q) to polar coords. (r, φ) : p = r cos φ, q = r sin φ. Show that where t > 0 is a real parameter.

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

7. 7 Example 5: Find where x > 0 is a real parameter, and the branch of s 1/2 on the plane of complex s is fixed by the condition 1 1/2 = +1 and a cut along the negative part of the real axis. Solution: where γ > 0. Close the contour in integral (8) as follows... (8)

8. 8 The integrand is analytic inside the contour – hence, Next, let R → ∞, hence (9) and also (due to Jordan’s Lemma) Let ε → 0, hence (why?)

9. 9 Introduce p > 0, such that Recalling how the branch of s 1/2 was fixed, we have Thus, the limiting version of (9) is hence,

10. 10 Then, hence, In the 1 st integral, change p to p new = –p, then omit the new and ‘join’ it to the 2 nd integral, which yields

11. 11 Finally, using the results of Example 4, obtain Comment: Consider the horizontal segments of the curves C 3 and C 5 (let’s call these segments C' 3 and C' 5 ), and note that Jordan’s Lemma does not guarantee that their contributions vanish as R → ∞. Note, however, that, for z  C' 3,5, the integrand decays as R → ∞, whereas the lengths of C' 3 and C' 5 remain constant (both equal to γ ). Hence, the integrals over C' 3 and C' 5 do decay as R → ∞. Alternatively, we can move γ infinitesimally close to zero: this wouldn’t affect the original integral (why?), but would make the lengths (and, thus, contributions) of C' 3 and C' 5 equal to zero.