1 Week 7 where R(x) is a given complex function. It turns out that (1) doesn’t necessarily have a solution. Consider a problem of the form 4. Non-homogeneous boundary-value problems (1)

2 Theorem 1: Let the operator Ĥ + have a zero EValue in S +, i.e. there exists a function f + (x) such that (2) Proof: Consider, which yields... Then, problem (1) may have a solution only if f + (x) is orthogonal to R(x), i.e.

3 By definition of Ĥ +, Let g = f + and apply (4) to (3), (4) (3) (5) hence, = 0 due to (2) as required. █

4 Theorem 2: Consider a problem of the form (7) Problem (6)-(7) has a solution if and only if one of the following conditions holds... (6) where c 0,1 (x) and R(x) are complex functions.

5 (a) The adjoint problem, doesn’t have a solution. (b) The adjoint problem does have a solution, f + (x), and (8)

6 The proof of Theorem 2 consists of two parts: 1.If neither of conditions (a)-(b) holds, problem (6)-(7) has no solution. 2.If either (a) or (b) holds, problem (6)-(7) does have a solution. It’s not easy to prove part 2 – so we won’t, whereas part 1 doesn’t need to be proved as it’s a particular case of Theorem 1.

7 Theorem 3: Let BC (7) be replaced with its non-homogeneous analogue, Then, Theorem 2 remains intact, but only if (8) is replaced with We shall prove only “part 1” of this theorem, i.e., if the adjoint problem has a zero eigenvalue, the solution of the original problem exists only if condition (9) holds. (9)

8 Proof of part 1 (by contradiction): Let the adjoint problem have a solution, f +, and consider which yields I1I1 I2I2 Simplify I 2 and I 1 by integrating by parts:

9 hence, Taking into account the equation and BCs for f + (x), and the BCs for f(x), this equality yields the desired result [condition (9)]. █ ρa,bρa,b

10 Laplace transformation 1.The basics 2.The Shifting Theorems 3.Applications to ODEs 4.Inversion of Laplace transformation using complex integrals 5.Applications to PDEs 1. The basics Example 1: Why do we need the LT? Consider the following boundary-value problem: (10)

11 The BC suggests that the solution can be represented by This substitution, however, isn’t compatible with Eq. (10) (the cosines can’t be eliminated). The reason for the difficulty is that Eq. (10) is simply not separable... so what do we do with linear non-separable equations? An initial-boundary-value problem for an (n+1) -dimensional linear equation with time-independent coefficients can be reduced to an n - dimensional boundary-value problem via the Laplace transform.

12 ۞ Consider f(t) defined for t ≥ 0. Then, if the integral exists, it is called the Laplace transform of f(t). Alternative notation: Example 2:

13 (integrate by parts n times) Laplace transformation is a linear – hence,

14 Example 3: Theorem 4: Existence of Laplace transforms Let f(t) be a piece-wise continuous function for t ≥ 0 and where M and a are positve constants. Then L [f(t)] exists for s ≥ a.

15 If we are given a Laplace transform F(s), we can find the original function f(t) through the inverse LT, Since the forward LT doesn’t ‘know’ how f(t) behaves at t < 0, the inverse LT can’t restore this part of f(t). There are two ways to find inverse transforms: by using the table of transforms and by evaluating a certain complex integral. Comment: Find Example 4:

16 The inverse LT is linear, i.e. Find Example 5: ۞ The Heaviside step function, u(t – a), is hence...

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18 Oliver Heaviside (1850 –1925) was a self-taught English electrical engineer, mathematician, and physicist who adapted complex numbers to the study of electrical circuits, invented mathematical techniques for solving differential equations (later found to be equivalent to the Laplace transformation), and predicted the existence of the “Heaviside layer” in the ionosphere. He didn’t go to university and his only paid employment was a telegraph operator job (while he was years old). In later years his behaviour became quite eccentric: he would sign letters with the initials “W.O.R.M.” after his name, used granite blocks for furniture, and reportedly started painting his fingernails pink. For most of his life, Heaviside was at odds with the scientific establishment. Most of his recognition was gained posthumously.

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20 Comment: Observe that Theorem 5: 1 st Shifting Theorem If L -1 [F(s)] = f(t), then i.e. the inverse transforms of two different functions coincide... how come? 2. The Shifting Theorems Proof: By definition of the LT...

21 Apply L -1 to this equality to obtain the desired result. █ hence, Find Example 6:

22 Solution: F(s) = (s – 2) –4 is not in the table, but, using the 1 st Shifting Theorem with a = –2, we can reduce the problem to F(s) = s –4 : Find Example 7: The answer:

23 Theorem 6: 2 nd Shifting Theorem If L -1 [F(s)] = f(t), then Alternative formulation: Find Example 8: Solution: Use the 2 nd Shifting Theorem with a = 6 and F(s) = s –4...

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25 is called the convolution integral (or simply convolution) of f(t) and g(t). (11) Comment: In other contexts, the word “convolution” and the same notation are used for slightly different integrals, e.g. ۞ Given functions f(t) and g(t), an expression of the form

26 Theorem 7: Convolution is commutative, i.e. Proof: Change in (11) τ → τ 1, where τ 1 = t – τ. Then (11) becomes Interchanging the limits (and changing the sign accordingly), we omit 1 and obtain as required. █

27 Theorem 8: Proof: hence, Change (t 1, t 2 ) → (t, τ), where t = t 1 + t 2 and τ = t 2,

28 Change the order of integration, hence, as required. █