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Lecture Notes for MATH 592A

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Lecture Notes for MATH 592A Lecture Notes for MATH 592A Vladislav Panferov February 25, 2008 1. Introduction As a motivation for developing the theory let’s consider the following boundary-value problem, u′′ + u = f(x) x Ω=(0, 1) − ∈ u(0) = 0, u(1) = 0, where f is a given (smooth) function. We know that the solution is unique, sat- isfies the stability estimate following from the maximum principle, and it can be expressed explicitly through Green’s function. However, there is another way “to say something about the solution”, quite independent of what we’ve done before. Let’s multuply the differential equation by u and integrate by parts. We get x=1 1 1 u′ u + (u′ 2 + u2) dx = f u dx. − x=0 0 0 h i Z Z The boundary terms vanish because of the boundary conditions. We introduce the following notations 1 1 u 2 = u2 dx (thenorm), and (f,u)= f udx (the inner product). k k Z0 Z0 Then the integral identity above can be written in the short form as u′ 2 + u 2 =(f,u). k k k k Now we notice the following inequality (f,u) 6 f u . (Cauchy-Schwarz) | | k kk k This may be familiar from linear algebra. For a quick proof notice that f + λu 2 = f 2 +2λ(f,u)+ λ2 u 2 0 k k k k k k ≥ 1 for any λ. This expression is a quadratic function in λ which has a minimum for λ = (f,u)/ u 2. Using this value of λ and rearranging the terms we get − k k (f,u)2 6 f 2 u 2. k k k k Taking the square roots we obtain the Cauchy-Schwarz. We use Cauchy-Schwarz to obtain u′ 2 + u 2 =(f,u) 6 f u . k k k k k kk k Feeling generous we can increase the right-hand side to get u′ 2 + u 2 6 f ( u′ 2 + u 2)1/2, k k k k k k k k k k so u′ 2 + u 2 6 f 2. (1) k k k k k k This is an example of an apriori estimate: by some formal manipulations with the equation we obtained quantitative information about the solution in terms of the “data” (the function f). Estimate (1) is also called an energy estimate as the expression u′ 2 + u 2 in some k k k k physical context can be viewed as the energy. Using the energy estimate we can give another proof of uniqueness. Theorem. (Uniqueness) Assume that u , u C2[0, 1] are two solutions of the boundary-value problem. 1 2 ∈ Then u (x)= u (x), x [0, 1]. 1 2 ∈ Proof. Take w = u u . By the apriori estimate w′ 2 + w 2 6 0, so w(x)=0, 1 − 2 k k k k x [0, 1] since w is smooth. ∈ The formal calculation that led to estimate (1) can be extended (by adding a little sophistication) to the general case of a differential operator u = (au′)′ + bu′ + cu A − if a, b, c are smooth functions, a(x) a > 0 and c(x) b′(x)/2 0. ≥ 0 − ≥ In fact, this approach has many other useful consequences, but before we get there we want to learn some facts about functions for which all terms in the apriori estimate (1) are finite. 2 2. The function space L2(0, 1) Definition. We say that a function f : (0, 1) R belongs to L2(0, 1), or that it is → square integrable, if the integral 1 f 2 = f(x)2 dx k k Z0 is finite. More generally, if Ω Rn we may define ⊆ f 2 = f(x)2 dx k k ZΩ (the n-dimensional integral) and use the notation L2(Ω) for the class of all functions for which f is finite. k k Examples: If α > 1 then f(x) = x α, or more generally f(x) = x a α are − 2 | | | − | in L2(0, 1). The limiting case f(x) = 1 is not in L2(0, 1). However, if x = √|x| 2 1 (x1, x2) R and x denotes the Euclidean length of the vector, then f(x) = ∈ | | √|x| is in L2(( 1, 1)2) − Exercise: For which α does the functionf(x) = x α, where Rn x = (x ,...,x ), | | ∋ 1 n satisfy f L2(( 1, 1)n)? ∈ − More examples: If f(x)=0, x< 1 , f(x)=1, x 1 then f L2(0, 1). Notice that 2 ≥ 2 ∈ f(x) is discontinuous. n 1 α If α = 3 (say) then f(x) = ci x ai , where ai (0, 1), and ci any, satisfies − i=1 | − | ∈ f L2(0, 1). Again, f is discontinuous.P For a more extreme example take f(x) = ∞∈ −i α 2 x ai , where ai is a countable set in (0, 1), for example, the set of all i=1 | − | { } rationalP numbers between 0 and 1. Then again, f L2(0, 1). ∈ Bottom line, L2(Ω) is a good place for discontinuous functions. In the above we assume that we know how to define integrals. In general this is done using Lebesgue’s theory of integration, of which we will have to know very little (reading, say, the article on wikipedia.org would suffice). The most frequently used concept from there would be the one of a set of measure zero which comes into play as follows. Definition. We say that two functions f and f L2(Ω) are identical, or equal in 1 2 ∈ the sense of L2(Ω), if f f = 0. k 1 − 2k 3 Now, we saw that functions in L2 do not have to be continuous. From the condition 1 f f 2 = (f (x) f (x))2 dx =0 k 1 − 2k 1 − 2 Z0 we cannot conclude that f (x)= f (x) for all x (0, 1). For example, if 1 2 ∈ 1 6 1 0, x< 2 0, x 2 f1(x)= and f2(x)= 1, x 1 1, x> 1 ( ≥ 2 ( 2 2 then f1 = f2 in L (0, 1). More generally, the value of the integral is not changed if a function is modified for x on a set of measure zero. Examples of such sets in R are isolated points, finite and countable sets, and even some uncountable ones (a famous example is the Cantor set). In R2 we could add lines, or more general curves to that list, and in R3 (smooth) two-dimensional surfaces have measure zero. Example: If f(x)=0, x (0, 1) and g(x) is the Dirichlet function ∈ 1, x Q g(x)= ∈ ( 0, otherwise then f = g in L2(0, 1). Notice that g is not Riemann-integrable on (0, 1) (it is 1 Lebesgue-integrable) but we will still say 0 g(x) dx = 0. If a relation (such as equality) holds forR all x except on a set of measure zero we say it holds almost everywhere or for almost all x (sometimes abbreviated as a. e. or a. a.). Another name for the same thing (borrowed from Probability) is “almost surely”. It is an easy observation that L2(Ω) is a vector space. Indeed, if f L2 and λ R ∈ ∈ then λf L2. Also, if f L2 and g L2 then ∈ ∈ ∈ 2 (f + g)2 dx = (f 2 +2fg + g2) dx 6 f 2 dx + g2 dx , Z Z qR qR so f + g L2 (we used the Cauchy-Schwarz). ∈ We define the inner product (also called scalar product) on L2(0, 1) as 1 (f,g)= f(x) g(x) dx. Z0 4 More generally, an inner product on a real vector space is defined by the properties (f + λg,h)=(f, h)+ λ(g, h), (f,g + λh)=(f, h)+ λ(f, h) (bilinear) (f,g)=(g, f) (symmetric) (f, f) 0, and (f, f)=0 f =0. (positive definite) ≥ ⇔ (For short, an inner product is a bilinear form that is symmetric and positive defi- nite). We also see that the norm f satisfies f = (f, f) and k k k k p f 0, and f =0 f =0, (positive definite) k k ≥ k k ⇔ λf = λ f , (positive homogeneous) k k | |k k and f + g 6 f + g (triangle inequality) k k k k k k These must be familiar from linear algebra. The proof of the trinagle inequality follows from the Cauchy-Schwarz as above. Definition. We say that a sequence f L2(Ω) converges (as n ) to n ∈ → ∞ f L2(Ω), denoted f f in L2, if f f 0. ∈ n → k n − k→ Other terms used frequently are “convergence in the mean square” or “convergence in the L2-norm”, or “convergence in L2”. Examples. Set 1 0, x< n 1+nx 1 1 0, x< 0 fn(x)= , 6 x 6 and f(x)= 2 − n n 1, x 0 1 ( 1, x> n ≥ Then f f in L2( 1, 1). Indeed, n → − 1 1 0 2 n 2 2 (1 + nx) (1 nx) 1 (fn(x) f(x)) dx = dx + − dx = 0. − 1 4 4 6n → Z−1 Z− n Z0 In this case f(x) is discontinuous and fn do not converge to f uniformly. Geometrically, it is pehaps easier to see what happens when the square is replaced 5 by the absolute value, 1 f f dx 0 | n − | → Z0 (in this case we say f f in L1; this is a different, but related type of convergence). n → The above simply means that the area between the graphs of fn and f goes to zero.
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