Week 9 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)

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

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

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...

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.

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.

Example 5: Find where x > 0 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: (8) where γ > 0. Close the contour in integral (8) as follows...

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

Thus, the limiting version of (9) is hence, Introduce p > 0, such that Recalling how the branch of s1/2 was fixed, we have

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

Finally, using the results of Example 4, obtain Comment: Consider the horizontal segments of the curves C3 and C5 (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.