text stringlengths 270 6.81k |
|---|
Now suppose that f (m) is a contraction for some m. Hence by the first part, there is a unique x ∈ X such that f (m)(x) = x. But then f (m)(f (x)) = f (m+1)(x) = f (f (m)(x)) = f (x). So f (x) is also a fixed point of f (n)(x). By uniqueness of fixed points, we must have f (x) = x. Since any fixed point of f is clearl... |
(t)) such that fj(t0) = x(j) 0 for all j = 1,..., n and t ∈ (t0 − ε, t0 + ε). We can imagine this scenario as a particle moving in Rn, passing through x0 at time t0. We then ask if there is a trajectory f (t) such that the velocity of the particle at any time t is given by F(t, f (t)). This is a complicated system, sin... |
] → Rn that satisfies the differential equation and boundary conditions above. Even n = 1 is an important, special, non-trivial case. Even if we have only one dimension, explicit solutions may be very difficult to find, if not impossible. For example, df dt = f 2 + sin f + ef 53 5 Metric spaces IB Analysis II would be ... |
solutions. In fact, for any α ∈ [0, b], the function fα(tt − α)2 α ≤ t ≤ b is also a solution. So we have an infinite number of solutions. We are now going to use the contraction mapping theorem to prove this. In general, this is a very useful idea. It is in fact possible to use other fixed point theorems to show the ... |
um metric ∥g − h∥ = sup t∈[a,b] ∥g(t) − h(t)∥2. We see that X is a closed subset of the complete metric space C([a, b], Rn) (again taken with the supremum metric). So X is complete. For every g ∈ X, we define a function T g : [a, b] → Rn by (T g)(t) = x0 + t t0 F(s, g(s)) ds. Our differential equation is thus f = T f. ... |
contraction map. We will in fact show that this map satisfies the bound ∥T (m)g1(t) − T (m)g2(t)∥ ≤ sup t∈[a,b] (b − a)mκm m! sup t∈[a,b] ∥g1(t) − g2(t)∥. (‡) The key is the m!, since this grows much faster than any exponential. Given this bound, we know that for sufficiently large m, we have (b − a)mκm m! < 1, i.e. T... |
(m)g2(t)∥ ≤ κ t km−1(s − t0)m−1 (m − 1)! t0 κm sup (m − 1)! [t0,t] κm(t − t0)m m! ≤ = ∥g1 − g2∥2 t0 ∥g1 − g2∥2. sup [t0,t] sup [t0,s] t ∥g1 − g2∥2 ds (s − t0)m−1 ds So done. Note that to get the factor of m!, we had to actually perform the integral, instead of just bounding (s − t0)m−1 by (t − t0). In general, this is... |
∥ < ε. As in the case of R in IA Analysis I, we do not impose any requirements on F when x = a. In particular, we don’t assume that a is in the domain E. We would like a definition of differentiation for functions f : Rn → R (or more generally f : Rn → Rm) that directly extends the familiar definition on the real line.... |
= 0, and we call A the derivative. We are now in a good shape to generalize. Note that if f : Rn → R is a real-valued function, then f (a + h) − f (a) is a scalar, but h is a vector. So A is not just a number, but a (row) vector. In general, if our function f : Rn → Rm is vector-valued, then our A should be an m × n m... |
f (a + h) − f (a) − Bh∥. So ∥(B − A)h∥ ∥h∥ → 0 as h → 0. We set h = tu in this proof to get ∥(B − A)tu∥ ∥tu∥ → 0 59 6 Differentiation from Rm to Rn IB Analysis II as t → 0. Since (B − A) is linear, we know ∥(B − A)tu∥ ∥tu∥ = ∥(B − A)u∥ ∥u∥. So (B − A)u = 0 for all u ∈ Rn. So B = A. Notation. We write L(Rn; Rm) for the ... |
existential. This is unlike the definition of differentiability of real functions, where we are asked to compute an explicit limit — if the limit exists, that’s the derivative. If not, it is not differentiable. In the higher-dimensional world, this is not the case. We have completely no idea where to find the derivati... |
). Often, it is convenient to focus on the special cases where u = ej, a member of the standard basis for Rn. This is known as the partial derivative. By convention, this is defined for real-valued functions only, but the same definition works for any Rm-valued function. Definition (Partial derivative). The jth partial... |
standard basis for Rn and Rm, i.e. for any h ∈ Rn, Then A is given by Df (a)h = Ah. Aij = ⟨Df (a)ej, bi⟩ = Djfi(a). where {e1, · · ·, en} is the standard basis for Rn, and {b1, · · ·, bm} is the standard basis for Rm. The second property is useful, since instead of considering arbitrary Rm- valued functions, we can ju... |
of all partial derivatives. The converse implication does not hold in either of these. Example. Let f 2 : R2 → R be defined by f (x, y) = 0 xy = 0 1 xy ̸= 0 Then the partial derivatives are df dx (0, 0) = df dy (0, 0) = 0, In other directions, say u = (1, 1), we have f (0 + tu) − f (0) t = 1 t which diverges as t → 0.... |
uf (0) = Df (0)u, which should be a linear expression in u, but this is not. Alternatively, if f were differentiable, then we have Df (0)h = 1 0 h1 h2 = h1. However, we have f (0 + h) − f (0) − Df (0)h ∥h∥ = h3 1 h2 1+h2 2 h2 − h1 1 + h2 2 = − h1h2 2 1 + h2 2 h2 3, which does not tend to 0 as h → 0. For example, if h =... |
· · ·, hj, 0, · · ·, 0). Then we have f (a + h) − f (a) = = n j=1 n j=1 f (a + h(j)) − f (a + h(j−1)) f (a + h(j−1) + hjej) − f (a + h(j−1)). Note that in each term, we are just moving along the coordinate axes. Since the partial derivatives exist, the mean value theorem of single-variable calculus applied to g(t) = f... |
) → L(Rn; Rm) given by x → Df (x). To do so, we need a metric on the space L(Rn; Rm). In fact, we will use a norm. Let L = L(Rn; Rm). This is a vector space over R defined with addition and scalar multiplication defined pointwise. In fact, L is a subspace of C(Rn, Rm). To prove this, we have to prove that all linear ma... |
Then BA = B ◦ A ∈ L(Rn; Rp) and Proof. ∥BA∥ ≤ ∥B∥∥A∥. (i) This is since A is continuous and {x ∈ Rn : ∥x∥ = 1} is compact. (ii) The only non-trivial part is the triangle inequality. We have ∥A + B∥ = sup ∥x∥=1 ∥Ax + Bx∥ ≤ sup ∥x∥=1 (∥Ax∥ + ∥Bx∥) ≤ sup ∥x∥=1 ∥Ax∥ + sup ∥x∥=1 ∥Bx∥ = ∥A∥ + ∥B∥ (iii) This follows from lin... |
(U ) ⊆ V and g : V → Rp is differentiable at f (a). Then g ◦ f : U → Rp is differentiable at a, with derivative D(g ◦ f )(a) = Dg(f (a)) Df (a). 67 6 Differentiation from Rm to Rn IB Analysis II Proof. The proof is very easy if we use the little o notation. Let A = Df (a) and B = Dg(f (a)). By differentiability of f, ... |
− a) for some c ∈ (a, b). This is our favorite theorem, and we have used it many times in IA Analysis. Here we have an exact equality. However, in general, for vector-valued functions, i.e. if we are mapping to Rm, this is no longer true. Instead, we only have an inequality. We first prove it for the case when the dom... |
Theorem (Mean value inequality). Let a ∈ Rn and f : Br(a) → Rm be differentiable on Br(a) with ∥Df (x)∥ ≤ M for all x ∈ Br(a). Then ∥f (b1) − f (b2)∥ ≤ M ∥b1 − b2∥ for any b1, b2 ∈ Br(a). Proof. We will reduce this to the previous theorem. Fix b1, b2 ∈ Br(a). Note that tb1 + (1 − t)b2 ∈ Br(a) for all t ∈ [0, 1]. Now c... |
can have two disjoint intervals [1, 2] ∪ [3, 4], and define f (t) to be 1 on [1, 2] and 2 on [3, 4]. Then Df = 0 but f is not constant. f is just locally constant on each interval. The problem with this is that the sets are disconnected. We cannot connect points in [1, 2] and points in [3, 4] with a line. If we can do... |
iable at s with derivative 0. This is true for all s. So the map g : [0, 1] → R has zero derivative on (0, 1) and is continuous on (0, 1). So g is constant. So g(0) = g(1), i.e. f (a) = f (b). If γ were differentiable, then this is much easier, since we can show g′ = 0 by the chain rule: g′(t) = Df (γ(t))γ′(t). 6.4 Inv... |
unit vectors. Hence if Df is continuous, so is Djfi. (⇐) Since the partials exist and are continuous, by our previous theorem, we know that the derivative Df exists. To show Df : U → L(Rm; Rn) is continuous, note the following general fact: For any linear map A ∈ L(Rn; Rm) represented by (aij) so that Ah = aijhj, then... |
Rn with a ∈ V, f (a) ∈ W, V ⊆ U such that f |V : V → W is a bijection. Moreover, the inverse map f |−1 V : W → V is also C 1. We have a fancy name for these functions. Definition (Diffeomorphism). Let U, U ′ ⊆ Rn are open, then a map g : U → U ′ is a diffeomorphism if it is C 1 with a C 1 inverse. Note that different ... |
there is a unique x ∈ V such that f (x) = y. We are going to use the contraction mapping theorem to 72 6 Differentiation from Rm to Rn IB Analysis II prove this. This statement is equivalent to proving that for each y ∈ W, the map T (x) = x − f (x) + y has a unique fixed point x ∈ V. Let h(x) = x − f (x). Then note th... |
V. So we have shown that for each y ∈ W, there is a unique x ∈ V such that f (x) = y. So f |V : V → W is a bijection. We have done the hard work now. It remains to show that f |V is invertible with C 1 inverse. Claim. The inverse map g = f |−1 In fact, we have V : W → V is Lipschitz (and hence continuous). ∥g(y1) − g(... |
1 2. ∥v∥ = ∥Df (x)v − v∥ ≤ ∥Df (x) − I∥∥v∥ ≤ 1 2 ∥v∥. So we must have ∥v∥ = 0, i.e. v = 0. So ker Df (x) = {0}. So Df (g(y))−1 exists. Let x ∈ V be fixed, and y = f (x). Let k be small and In other words, h = g(y + k) − g(y). f (x + h) − f (x) = k. Since g is invertible, whenever k ̸= 0, h ̸= 0. Since g is continuous,... |
a, b) → f ((a, b)) is a bijection. In higher dimensions, this is not true. Even if we know that Df (x) is invertible for all x ∈ U, we cannot say f |U is a bijection. We still only know there is a local inverse. Example. Let U = R2, and f : R2 → R2 be given by f (x, y) = ex cos y ex sin y. Then we can directly compute ... |
) is just a finite-dimensional space, and is isomorphic to Rnm. So Df is 75 6 Differentiation from Rm to Rn IB Analysis II differentiable if and only if Df : U → Rnm is differentiable with A ∈ L(Rn; Rnm). This allows use to recycle our previous theorems about differentiability. In particular, we know Df is differentiab... |
= D(Df )(a)(u)(v). We know D2f (a) is a bilinear map. In coordinates, if u = n j=1 ujej, v = n j=1 vjej, where {e1, · · ·, en} are the standard basis for Rn, then using bilinearity, we have D2f (a)(u, v) = n n i=1 j=1 D2f (a)(ei, ej)uivj. This is very similar to the case of first derivatives, where the derivative can ... |
partials). Let U ⊆ Rn be open, f : U → Rm, a ∈ U, and ρ > 0 such that Bρ(a) ⊆ U. Let i, j ∈ {1, · · ·, n} be fixed and suppose that DiDjf (x) and DjDif (x) exist for all x ∈ Bρ(a) and are continuous at a. Then in fact DiDjf (a) = DjDif (a). The proof is quite short, when we know what to do. Proof. wlog, assume m = 1. ... |
. By continuity of the partial derivatives, we get DjDif (a) = DiDjf (a). This is nice. Whenever the second derivatives are continuous, the order does not matter. We can alternatively state this result as follows: Proposition. If f : U → Rm is differentiable in U such that DiDjf (x) exists in a neighbourhood of a ∈ U a... |
of the second derivative, as h → 0, we get ∥D2f (a + sh) − D2f (a)∥ → 0. So E(h) = o(∥h∥2). So done. 79 definition of a weak = U ηε(x − y)Dαu(y) dy = ηε ∗ Dαu. It is an exercise to verify that we can indeed move the derivative past the integral. Thus, if we fix V U. Then by the previous parts, we see that Dαuε → Dαu in ... |
= ∞ i=0 ui ∈ C∞(U ). Note that we do not know (yet) that v ∈ W k.p(U ). But it certainly is when we restrict to some V U. In any such subset, the sum is finite, and since u = ∞ i=0 ζiu, we have v − uW k,p(V ) ≤ ∞ i=0 ui − ζiuW k.p(V ) ≤ δ ∞ i=0 2−(i+1) = δ. Since the bound δ does not depend on V, by taking the supremum... |
fined on U, we can shift it downwards by some ε. It is a known result that translation is continuous, so this only changes u by a tiny bit. We can then mollify with a ¯ε < ε, which would then give a function defined on U (at least locally near x0). So fix some x0 ∈ ∂U. Since ∂U is C 0,1, there exists r > 0 such that γ ∈ C... |
x0 ∈ ∂U, we can find a neighbourhood V ⊆ U of x0 in U such that for any u ∈ W k,p(U ) and δ > 0, there exists v ∈ C∞( ¯V ) such that u − vW k,p(V ) ≤ δ. It remains to patch all of these together using a partition of unity. By the compactness of ∂U, we can cover ∂U by finitely many of these V, say V1,..., VN. We further ... |
almost everywhere in U (ii) Eu has support in V (iii) EuW 1,p(Rn) ≤ CuW 1,p(U ), where the constant C depends on U, V, p but not u. Proof. First note that C 1( ¯U ) is dense in W 1,p(U ). So it suffices to show that the above theorem holds with W 1,p(U ) replaced with C 1( ¯U ), and then extend by continuity. We first sh... |
C 1-diffeomorphism Φ : Rn → Rn given by Φ(x)i = xi Φ(x)n = xn − γ(x1,..., xn) i = 1,..., n − 1 Then since C 1 diffeomorphisms induce bounded isomorphisms between W 1,p, this gives a local extension. Since ∂U is compact, we can take a finite number of points x0 i ∈ ∂W, sets Wi and extensions ui ∈ C 1(Wi) extending u such ... |
C 1 boundary. Then there exists a bounded linear operator T : W 1,p(U ) → Lp(∂U ) for 1 ≤ p < ∞ such that T u = u|∂U if u ∈ W 1,p(U ) ∩ C( ¯U ). We say T u is the trace of u. Proof. It suffices to show that the restriction map defined on C∞ functions is a bounded linear operator, and then we have a unique extension to W ... |
). 0 0 3.5 Sobolev inequalities Before we can move on to PDE’s, we have to prove some Sobolev inequalities. These are inequalities that compare different norms, and allows us to “trade” different desirable properties. One particularly important thing we can do is to trade differentiability for continuity. So we will know... |
(x2)f2(x1). So |f (x1, x2)| dx = |f1(x2)| dx2 |f2(x1)| dx1 R2 = f1L1(R1)f2L1(R1). Suppose that the result is true for n ≥ 2, and consider the n + 1 case. Write f (x) = fn+1(˜xn+1)F (x), where F (x) = f1(˜x1) · · · fn(˜xn). Then by H¨older’s inequality, we have x1,...,xn |f ( ·, xn+1)| dx ≤ fn+1Ln(Rn)F ( ·, xn+1)Ln/(n−1... |
> p, and there exists c > 0 depending on n, p such that uLp∗ (Rn) ≤ cuW 1,p(Rn). In other words, W 1,p(Rn) is continuously embedded in Lp∗ (Rn). Proof. Assume u ∈ C∞ c (Rn), and consider p = 1. Since the support is compact, xi u(x) = uxi(x1,..., xi−1, yi, xi+1,..., xn) dyi. −∞ So we know that |u(x)| ≤ ∞ −∞ |Du(x1,...,... |
n − 1 = np n − p = p∗. So So Rn n−1 n |u|p∗ dx ≤ p(n − 1) n − p Rn p−1 p |u|p∗ dx DuLp(Rn). Rn 1/p∗ |u|p∗ dx ≤ p(n − 1) n − p DuLp(Rn). This argument is valid for u ∈ C∞ to W 1,p(Rn). c (Rn), and by approximation, we can extend We can deduce some corollaries of this result: Corollary. Suppose U ⊆ Rn is open and bounde... |
PDEs So we know that umLp∗ (U ) ≤ CDumLp(U ). Sending m → ∞, we obtain Since U is bounded, by H¨older, we have uLp∗ (U ) ≤ CDuLp(U ). |u|q dx 1/q ≤ 1/rq 1 dx 1/sq |u|qs ds U U U ≤ CuLp∗ (U ) provided q ≤ p∗, where we choose s such that qs = p∗, and r such that r + 1 1 s = 1. The previous results were about the case n ... |
tQ|1/p dt. 3 Function spaces III Analysis of PDEs where 1 p + 1 p = 1. Using that |Q| = rn, we obtain |¯u − u(0)| ≤ cr1−n+ n p DuLp(Rn) 1 0 t−n+ n p dt ≤ c 1 − n/p r1−n/pDuLp(Rn). Note that the right hand side is decreasing in r. So when we take r to be very small, we see that u(0) is close to the average value of u ar... |
s 4 Elliptic boundary value problems 4.1 Existence of weak solutions In this chapter, we are going to study second-order elliptic boundary value problems. The canonical example to keep in mind is the following: Example. Suppose U ⊆ Rn is a bounded open set with smooth boundary. Suppose ∂U is a perfect conductor and ρ :... |
An operator n Lu = − (aij(x)uj)xi + i,j=1 n i=1 bi(x)uxi + c(x)u 36 4 Elliptic boundary value problems III Analysis of PDEs is uniformly elliptic if n i,j=1 aij(x)ξiξj ≥ θ|ξ|2 for some θ > 0 and all x ∈ U, ξ ∈ Rn. We shall consider the boundary value problem Lu = f on U u = 0 on ∂U. This form of the equation is not ve... |
U ). Definition (Weak solution). We say u ∈ H 1 0 (U ) is a weak solution of Lu = f on U u = 0 on ∂U B[u, v] = (f, v)L2(U ) for f ∈ L2(U ) if for all v ∈ H 1 0 (U ). We’ll exploit the Hilbert space structure of H 1 0 (U ) to find weak solutions. Theorem (Lax–Milgram theorem). Let H be a real Hilbert space with inner prod... |
is clear, and if Aum → v for some v, then um − un ≤ 1 β Aum − Aun → 0 as m, n → ∞. So (un) is Cauchy, and hence has a limit u. Then by continuity, Au = v, and in particular, v ∈ im A) – Since im A is closed, we know H = im A ⊕ im A⊥. Now let w ∈ im A⊥. Then we can estimate βw2 ≤ B[w, w] = (Aw, w) = 0. So w = 0. Thus, ... |
DuL2(U )vL2(U ) + c3uL2(U )vL2(u) ≤ αuH 1(U )vH 1(U ) for some α. (ii) We start from uniform ellipticity. This implies θ U |Du|2 dx ≤ n U i,j=1 aij(x)uxiuxj dx = B[u, u] − n U i=1 biuxiu + cu2 dx ≤ B[u, u] + n biL∞(U ) |Du||u| dx i=1 + cL∞(U ) U |u|2 dx. 39 4 Elliptic boundary value problems III Analysis of PDEs Now by... |
, U ) ≥ 0. Again, if bi ≡ 0 and c ≥ 0, then we may take γ = 0. Proof. Take γ from the previous theorem when applied to L. Then if µ ≥ γ and we set Bµ[u, v] = B[u, v] + µ(u, v)L2(U ), This is the bilinear form corresponding to the operator Lµ = L + µ. Then by the previous theorem, Bµ satisfies boundedness and coercivity.... |
. We would like to think a bit more about it. 4.2 The Fredholm alternative To understand the second problem, we shall seek to prove the following theorem: Theorem (Fredholm alternative). Consider the problem Lu = f, u|∂U = 0. (∗) For L a uniformly elliptic operator on an open bounded set U with C 1 boundary, either (i)... |
(scalar multiple of) (L + γ)−1 (plus some bookkeeping). So let us show that (L+γ)−1 is compact. Note that this maps sends f ∈ L2(U ) 0 (U ). To make it an endomorphism, we have to compose this with the 0 (U ) → L2(U ). The proof that (L + γ)−1 is compact will not involve 0 (U ) → L2(U ) is to u ∈ H 1 inclusion H 1 (L ... |
is some ci such that (ei, v) → ci as → ∞. 42 4 Elliptic boundary value problems III Analysis of PDEs We would expect the weak limit to be ciei. To prove this, we need to first show it converges. We have p j=1 |cj|2 = lim k→∞ p |(ej, v)|2 j=1 p ≤ sup |(ej, v)|2 using Bessel’s inequality. So j=1 ≤ sup vk2 ≤ K 2, u = ∞ j=... |
] × · · · × [ξn, ξn + L] be a cube of length L. Then we have u2 L2(Q) ≤ 1 |Q| Q 2 u(x) dx + nL2 2 Du2 L2(Q). We can improve this to obtain better bounds by subdividing Q into smaller cubes, and then applying this to each of the cubes individually. By subdividing enough, this leads to a proof that um u in H 1 implies um... |
over all x, y ∈ Q, we get Q×Q dx dy I1 ≤ L2|Q|D1u2 L2(Q). Similarly estimating the terms on the right-hand side, we find that 2|Q|uL2(Q) − 2 2 u(x) dx ≤ n|Q| Q n i=1 It now follows that Diu2 L2(Q) = n|Q|L2Du2 L2(Q). Theorem (Rellich–Kondrachov). Let U ⊆ Rn be open, bounded with C 1 m=1 is a sequence in H 1(U ) with um ... |
enough, the first term is also < ε 2. The same result holds with H 1(U ) replaced by H 1 0 (U ). The proof is in fact simpler, and we wouldn’t need the assumption that the boundary is C 1. Corollary. Suppose K : L2(U ) → H 1(U ) is a bounded linear operator. Then the composition L2(U ) K H 1(U ) L2(U ) is compact. The ... |
�)L2(U ) = (φ, L†ψ)L2(U ) for all φ, ψ ∈ C∞ c (U ). By integration by parts, we know L† should be given by n L†v = − (aijvxj )xi − i,j=1 n i=1 bi(x)vxj + c − bi xi v. n i=1 Note that here we have to assume that bi ∈ C 1( ¯U ). However, what really interests us is the adjoint bilinear form, which is simply given by B†[v... |
solution u ∈ H 1 0 (U ) to Lγu = Lu + γu = f, u|∂U = 0. Moreover, we have the bound uH 1(U ) ≤ Cf L2(U ) (which gives uniqueness). Thus, we can set L−1 0 (U ) is a bounded linear map. Composing with the inclusion L2(U ), we get a compact endomorphism of L2(U ). Now suppose u ∈ H 1 γ f to be this u, and then L−1 γ 0 is... |
). (ii) As above, we know that the non-zero solution u. There are two things to show. First, we have to show that v − K †v = 0 iff v is a weak solution to L†v = 0, v|∂U = 0. Next, we need to show that h = L−1 γ f ∈ (N ∗)⊥ iff f ∈ (N ∗)⊥. For the first part, we want to show that v ∈ ker(I − K †) iff B†[v, u] = B[u, v] = 0 ... |
trivial kernel. For µ ≤ γ, (L + µ)u = 0 may or may not have a non-trivial solution, but we know this satisfies the Fredholm alternative, since L + µ is still an elliptic operator. Rewriting (L + µ)u = 0 as Lu = −µu, we are essentially considering eigenvalues of L. Of course, L is not a bounded linear operator, so our u... |
ormal basis of eigenvectors. 48 4 Elliptic boundary value problems III Analysis of PDEs From this, it follows easily that Theorem (Spectrum of L). (i) There exists a countable set Σ ⊆ R such that there is a non-trivial solution to Lu = λu iff λ ∈ Σ. (ii) If Σ is infinite, then Σ = {λk}∞ k=1, the values of an increasing s... |
an orthonormal basis {wk}∞ 0 (U ) an eigenfunction of L with eigenvalue λk. k=1 of L2(U ) with wk ∈ H 1 Proof. Note that positivity implies c ≥ 0. So the inverse L−1 : L2(U ) → L2(U ) exists and is a compact operator. We are done if we can show that L−1 is self-adjoint. This is trivial, since for any f, g, we have (L−... |
= D2uL2(Rn). So we have deduced that D2uL2(Rn) = ∆uL2(Rn). This is of course not a very useful result, because we have a priori assumed that u and f are C∞, while what we want to prove that u is, for example, in H 2(u). However, the fact that we can control the H 2 norm if we assumed that u ∈ H 2(U ) gives us some str... |
)v dx U k w)∆h k(wv) = (τ h ∆h kv + (∆h kw)v, where τ h k w(x) = w(x + hek). Theorem (Interior regularity). Suppose L is uniformly elliptic on an open set U ⊆ Rn, and assume aij ∈ C 1(U ), bi, c ∈ L∞(U ) and f ∈ L2(U ). Suppose further that u ∈ H 1(U ) is such that B[u, v] = (f, v)L2(U ) (†) for all v ∈ H 1 0 (U ). The... |
that Ru ∈ L2(U ), this tells us u ∈ H 2 loc(U ). Moreover, on V U, 51 4 Elliptic boundary value problems III Analysis of PDEs – We can control uH 2(V ) by f − RuL2(V ) and uL2(V ) (by theorem). – We can control f − RuL2(V ) by f L2(V ), uL2(V ) and DuL2(V ). – By G˚arding’s inequality, we can control DuL2(V ) by uL2(V... |
k(aijuxi )(ζ 2∆h ku)xj dx (τ h k aij∆h kuxi + (∆h kaij)uxi)(ζ 2∆h kuxj + 2ζζxj ∆h ku) dx where A1 = i,j U i,j U A2 = ξ2(τ h k aij)(∆h kuxi)(∆h kuxj ) dx (∆h kaij)uxiζ 2∆h kuxj + 2ζζxj ∆h ku(τ h k aij∆h kuxi + (∆h kaij)uxi) dx. By uniform ellipticity, we can bound A1 ≥ θ U ξ2|∆h kDu|2 dx. This is what we want to be sma... |
|2 dx − C W |Du|2 dx. U This is promising. It now suffices to bound (f, v) from above. By Young’s inequality, |(f, v)| ≤ |f ||∆−h k (ζ 2∆h ku)| dx ≤ C ≤ ε ≤ ε |f ||D(ζ 2∆h ku)| dx |D(ζ 2∆h ku)|2 dx + C |f |2 dx |ζ 2∆h kDu|2 dx + C(f 2 L2(U ) + Du2 L2(U )) Setting ε = θ 4, we get U ζ 2|∆h kDu|2 dx ≤ C(f 2 L2(W ) + Du2 L2(... |
53 4 Elliptic boundary value problems III Analysis of PDEs that u ∈ H 3 loc(U ). Of course, some bookkeeping has to be done if we were to do this properly, since we need to write everything in weak form. However, this is not particularly hard, and the details are left as an exercise. Theorem (Elliptic regularity). If ... |
and Most of the proof in the previous proof goes through, as long as we restrict to v = −∆−h k (ζ 2∆h ku) with k = n, since all the translations keep us within U, and hence are well-defined. Thus, we control all second derivatives of the form DkDiu, where k ∈ {1,..., n − 1} and i ∈ {1,..., n}. The only remaining second... |
�) = n+1 i,j=1 aij(y)ξiξj has signature (+, −, −,...) for all y. That is to say, after perhaps changing basis, at each point we can write q(ξ) = λ2 n+1ξ2 n+1 − n i=1 i ξ2 λ2 i with λi > 0. It turns out not to be too helpful to treat this equation at this generality. We would like to pick out a direction that correspond... |
if we take L = −∆, then we obtain the wave equation. Let’s see what we can do with it. Example. Start with the equation utt − ∆u = 0. Multiply by ut and integrate over Ut to obtain 0 = uttut − ut∆u dx dt Ut Ut Ut t ∂ ∂t Σt−Σ0 u2 t − ∇ · (utDu) + Dut · Du dx dt u2 t + |Du|2 − ∇ · (utDu) t + |Du|2 u2 dx − ut ∂∗Ut dx dt ... |
0 ∂U aijuxj v dS dt. Using the boundary conditions, we find that UT f v dx dt = UT −utvt + aijuxivxj + biuxiv + cuv dx dt − Σ0 ψv dx. (†) Conversely, suppose u ∈ C 2( ¯UT ) satisfies (†) for all such v, and u|Σ0 = ψ and u|∂∗UT = 0. Then by first testing on v ∈ C∞ c (UT ), reversing the integration by parts tells us 0 = (... |
of energy. There could be some exponential growth in the energy, so want to suppress this. t 57 5 Hyperbolic equations III Analysis of PDEs Then this function belongs to H 1(UT ), v = 0 on ΣT ∪ ∂∗UT, and vt = −e−λtu. Using the fact that u is a weak solution, we have UT utue−λt − vtxj vxieλt + i biuxiv + (c − 1)uv − vv... |
integral must vanish. In particular, the integral of u2eλt = 0. So u = 0. We now want to prove the existence of weak solutions. While we didn’t need to assume much regularity in the uniqueness result, since we are going to subtract the boundary conditions off anyway, we expect that we need more regularity to prove exis... |
k)xj + biuN xi ϕk + cuN ϕk dx = (f, ϕk)L2(U ). (∗) uk(0) = (ψ, ϕk)L2(U ) ˙uk(0) = (ψ, ϕk)L2(U ). Notice that if we have a genuine solution u that can be written as a finite sum of the ϕk(x), then these must be satisfied. This is a system of ODEs for the functions uk(t), and the RHS is uniformly C 1 in t and linear in the... |
˙uN e−λt. Integrating in time, and estimating as before, for λ sufficiently large, we get 1 2 Στ ( ˙uN )2 + |DuN |2 dx + Uτ ( ˙uN )2 + |DuN |2 + (uN )2 dx dt ≤ C(ψ2 H 1(U ) + ψ2 L2(U ) + f 2 UT ). This, in particular, tells us uN is bounded in H 1(UT ), Since uN (0) = N N large enough, we have n=1(ψ, ϕk)ϕk, we know this... |
dt − ψv dx Σ0 = UT f v dx dt. So ut satisfies the identity required for u to be a weak solution. Now for k = 1,..., M, the map w ∈ H 1(UT ) → linear map, since the trace is bounded in L2. So we conclude that wϕk dx is a bounded Σ0 Σ0 uϕk dx = lim N→∞ Σ0 uN ϕk dx = (ψ, ϕk)L2(H). Since this is true for all ϕk, it follows... |
we know that utt is controlled in L2, then we can use the elliptic estimate to gain control on the second-order spatial derivatives of u. So uH 2(Σt) ≤ C(∆uL2(Σt)) = CuttL2(Σt). So we control all second-derivatives of u in terms of the data. Theorem. If aij, bi, c ∈ C 2(UT ) and ∂U ∈ C 2, then for ψ ∈ H 2(U ) and ψ ∈ ... |
2(Σ0) + f L2(UT ) + ftL2(UT ). 61 5 Hyperbolic equations III Analysis of PDEs We know N uN t |t=0 = (ψ, ϕk)L2(U )ϕk. k=1 Since ϕk are a basis for H 1, we have uN t H 1(Σ0) ≤ ψH 1(Σ0). To control uN −∆. From the fact that tt, let us assume for convenience that in fact ϕk are the eigenfunctions (¨uN, ϕk)L2(U ) + Σt i,j a... |
the equation as holding pointwise almost everywhere by undoing the integration by parts that gave us the definition of the weak solution. The initial conditions can also be understood in a trace sense. Returning to the case ψ ∈ H 1 0 (U ) and ψ ∈ L2(U ), by approximating in H 2(U ), by approximating in H 2(U ), H 1 0 (... |
. The “physicist’s version” of the wave equation involves a constant c, and says ¨u − c2∆x = 0. This constant c is the speed of propagation. This tells us in the wave equation, information propagates at a speed of at most c. We can see this very concretely in the 1-dimensional wave equation, where d’Alembert wrote down... |
to show that if u|Σ0 = 0 if ψ|S0 = ψ|S0 = 0 and f |D = 0. We take as test function v(t, x) = τ (x) t 0 e−λsu(s, x) ds (t, x) ∈ D (t, x) ∈ D. One checks that this is in H 1(UT ), and v = 0 on ΣT ∪ ∂∗UT with vxi = τxie−λτ u(x, τ ) + vt = −e−λtu(x, t). τ (x) t e−λsuxi(x, s) ds Plugging these into the definition of a weak ... |
0,γ( ¯U ), 20 C k,δ boundary, 26 C k,γ( ¯U ), 20 H k(U ), 22 H k 0 (U ), 22 Lp space, 21 Lp(U ), 21 W k,p(U ), 22 W k,p (U ), 22 0 a priori estimates, 56 analytic real, 10 associated eigenvector, 48 autonomous ODE, 9 boundary value problem, 37 Cauchy problem, 9 Cauchy–Kovalevskaya theorem for ODEs, 10 for PDEs, 13 cha... |
–Lindelof theorem, 9 Poincar´e’s inequality, 40 point spectrum, 48 Poisson’s equation, 5 positive operator, 49 principal symbol, 55 quasi-linear PDE, 7 real analytic, 10 real analytic hypersurface, 16 Rellich–Kondrachov theorem, 44 resolvent set, 48 Ricci flow, 6 Schr¨odinger’s equation, 6 Schwartz notation, 7 self-adjo... |
at some examples showing how this result can be applied. Example 0.7: Graphs. The three graphs each is a deformation retract of a disk with two holes, but we can also deduce this are homotopy equivalent since from the collapsing criterion above since collapsing the middle edge of the first and third graphs produces the... |
Some Underlying Geometric Notions X/A and X/B are homotopy equivalent to X. The space X/A is the quotient S 2/S 0, the sphere with two points identified, and X/B is S 1 ∨ S 2. Hence S 2/S 0 and S 1 ∨ S 2 are homotopy equivalent, a fact which may not be entirely obvious at first glance. Example 0.9. Let X be the union of... |
suspension X is slightly simpler than SX (X ∨ Y ) = Σ Σ Σ Σ is in its CW structure. In SX there are two 0 cells (the two suspension points) and an (n + 1) cell en × (0, 1) for each n cell en of X, whereas in and an (n + 1) cell for each n cell of X other than the 0 cell x0. X there is a single 0 cell Σ The reduced sus... |
n−1 the mapping cone Cf is the space obtained from Y by attaching an n cell via f : S n−1→Y. A mapping cone Cf can also be viewed as the quotient Mf /X of the mapping cylinder Mf with the subspace X = X × {0} collapsed to a point. If one varies an attaching map f by a homotopy ft, one gets a family of spaces whose sha... |
֓X. For we have X/A = (X∪CA)/CA ≃ X∪CA since CA is a contractible subcomplex of X ∪ CA. Example 0.14. If (X, A) is a CW pair and A is contractible in X, that is, the inclusion A֓ X is homotopic to a constant map, then X/A ≃ X ∨ SA. Namely, by the previous example we have X/A ≃ X ∪ CA, and then since A is contractible i... |
× I→X × {0} ∪ A× I, so X × {0} ∪ A× I is a retract of X × I. The converse is equally easy when A is closed in X. Then any two maps X × {0}→Y and A× I→Y that agree on A× {0} combine to give a map X × {0} ∪ A× I→Y which is continuous since it is continuous on the closed sets X × {0} and A× I. By composing this map X × {... |
exists a map f : B→A and a homeomorphism h : Mf →N with h || A ∪ B = 11. Mapping cylinder neighborhoods like this occur fairly often. For example, the thick let- ters discussed at the beginning of the chapter provide such neighborhoods of the thin letters, regarded as subspaces of the plane. To verify the homotopy ext... |
× I onto X n × {0} ∪ (X n−1 ∪ An)× I during the t interval [1/2n+1, 1/2n], this infinite concatenation of homotopies is a deformation retraction of X × I onto X × {0} ∪ A× I. There is no problem with continuity of this deformation retraction at t = 0 since it is continuous on X n × I, being stationary there during the ... |
slightly more refined version of one of our earlier criteria for homotopy equivalence, is the following: Proposition 0.18. If (X1, A) is a CW pair and we have attaching maps f, g : A→X0 that are homotopic, then X0 ⊔f X1 ≃ X0 ⊔g X1 rel X0. Here the definition of W ≃ Z rel Y for pairs (W, Y ) and (Z, Y ) is that there are... |
֓ X. ⊔⊓ Corollary 0.21. A map f : X→Y is a homotopy equivalence iff X is a deformation retract of the mapping cylinder Mf. Hence, two spaces X and Y are homotopy equivalent iff there is a third space containing both X and Y as deformation retracts. The Homotopy Extension Property Chapter 0 17 Proof: In the diagram at the... |
|| A = 11 and gt = ht on A, || A starts and ends with the identity, and its second half simply rethe homotopy kt traces its first half, that is, kt = k1−t on A. We will define a ‘homotopy of homotopies’ ktu : A→X by means of the figure at the right showing the parameter domain I × I for the pairs (t, u), with the t axis ... |
1. Construct an explicit deformation retraction of the torus with one point deleted onto a graph consisting of two circles intersecting in a point, namely, longitude and meridian circles of the torus. 2. Construct an explicit deformation retraction of Rn − {0} onto S n−1. (a) Show that the composition of homotopy equi... |
. Fill in the details in the following construction from [Edwards 1999] of a compact space Y ⊂ R3 with the same properties as the space Y in Exercise 6, that is, Y is contractible but does not deformation retract to any point. To begin, let X be the union of an infinite se- quence of cones on the Cantor set arranged end... |
s ≤ 1, of X onto A, where continuity means that the map X × I × I→X sending (x, s, t) to r s t and r 1 t (x) is continuous. 14. Given positive integers v, e, and f satisfying v − e + f = 2, construct a cell structure on S 2 having v 0 cells, e 1 cells, and f 2 cells. 15. Enumerate all the subcomplexes of S ∞, with the... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.