From Math 2220 Class 22 Double Integrals Double Integral Dr. - - PowerPoint PPT Presentation

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From Math 2220 Class 22 V1cc IFT Why the IFT? From Math 2220 Class 22 Double Integrals Double Integral Dr. Allen Back Problems Oct. 17, 2014 Inverse Function Theorem From Math 2220 Class 22 V1cc IFT The identity function Id is the

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

From Math 2220 Class 22

Dr. Allen Back
Oct. 17, 2014

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

The identity function Id is the function defined by f(x)=x. If we want to indicate a domain (e.g. x ∈ Rn) we might write IdRn.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

The inverse function theorem says that for a C 1 function f : U ⊂ Rn → Rn with f (p0) = q0, then one can locally uniquely solve (for p in terms of q with p near p0 and q near q0) the equation q = f (p) if the derivative Df (p0) is invertible.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

Under these circumstances, the solution g(q) = p will be a (differentiable) local inverse to f ; i.e. there are neighborhoods U0 of p0 and V0 of q0 so that

1 f : U0 ⊂ Rn → Rn has range f (U0) = V0. 2 g : V0 ⊂ Rn → Rn has range g(V0) = U0. 3 f (g(q)) = q for all q ∈ V0. (i.e f ◦ g = IdV0.) 4 g(f (p)) = p for all p ∈ U0. (i.e g ◦ f = IdU0.)

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

So f and g are giving a one-to-one correspondence between points of U0 and V0; to each point q of V0 there is a unique point p of U0 with f (p) = q and g(q) = p. And vice versa; to each point p of U0, . . .

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

Moreover the derivative of g is the inverse of the derivative of f , as the chain rule applied to f ◦ g = Id shows must be the case if a differentiable inverse exists. Dg(f (p)) = [Df (p)]−1 .

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

Moreover the derivative of g is the inverse of the derivative of f , as the chain rule applied to f ◦ g = Id shows must be the case if a differentiable inverse exists. Dg(f (p)) = [Df (p)]−1 . The example of inverse functions f (x) = x3 g(x) =

√x shows that even when Df fails to be invertible (f ′(0) = 0 here), a continuous inverse CAN still possibly exist; it just can’t be differentiable. (g′(0) does not exist.)

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

A Basic Inverse Function Theorem Problem

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

Implicit Function Theorem (1 Constraint): Suppose f : U ⊂ R3 → R is a C 1 function and p0 = (x0, y0, z0) is a point on the level set f (x, y, z) = C where ∂f ∂z

(x0,y0,z0)

= 0. Then we can locally solve for z in terms of x and y near p0. Namely there exists a C 1 function g(x, y) with domain a neighborhood of (x0, y0) so that g(x0, y0) = z0 and f (x, y, g(x, y)) = C.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

One way to remember this condition is to realize that the “worst thing that can can happen” in trying to solve f (x, y, z) = C for z is for z to not really appear in the equation; e.g. f (x, y, z) = x2 + y2 = 1 CANNOT be solved for z. Notice here fz = 0; i.e. the hypothesis of fz = 0 is ruling out this extreme obstruction to being able to solve for z. The conclusion of the implicit function theorem is saying the nonvanishing of this partial derivative together with a point to get started with is enough to guarantee a locally defined implicit function.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

A Basic Implicit Function Theorem Problem

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

In cases where the implicit function theorem applies, one can use implicit differentiation to compute the partials of the implicitly defined function g. There is a formula one can memorize, but I suggest you just think of implicit diff as the result of applying e.g. ∂ ∂x to the equation f (x, y, z(x, y)) = C.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

For example if we think of z as a function of x in f (x, y, z) = x2 + y2 + z2 + z9 = 4 near (x, y, z) = (1, 1, 1) (so we are really thinking x2 + y2 + [z(x, y)]2 + [z(x, y)]9 = 4), we obtain 2x + 2zzx + 9z8zx = 0 and zx = − 2x 2z + 9z8 .

r zx(1, 1) = 2

11.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

Note that the partial derivative assumed not to vanish in the implicit function theorem hypothesis is exactly the one you divide by in implicit differentiation.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

In the case of a system of m equations (where we can hope to solve for m variables z1, . . . , zm) F1(x1, . . . , xn, z1, . . . , zm) = C1 . . . = . . . Fm(x1, . . . , xn, z1, . . . , zm) = Cm the general implicit function theorem says this is locally possible near a point satisfying the system if the matrix of partials  

∂F1 ∂z1

. . .

∂F1 ∂zm

. . . . . . . . .

∂Fm ∂z1

. . .

∂Fm ∂zm

  is invertible. (One possibility is to check its determinant is nonzero.)

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

Two Constraint Implicit Function Theorem Problem

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

Another Two Constraint Implicit Function Theorem Problem

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem

y2 + z2 = 1 x2 + z2 = 1

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Why the IFT?

Existence of solutions to f (p) = q for p near p0 and q near q0 = f (p0) may be established by showing that Newton’s method with initial choice p0 converges to a solution.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Why the IFT?

We want f (p) = q or f (p) − f (p0) = q − q0. (Df (p0)) (p − p0) ∼ q − f (p0) Set p1 = (Df (p0))−1 (q − f (p0))+ p0 (Df (p1)) (p − p1) ∼ q − f (p1) Set p2 = (Df (p1))−1 (q − f (p1))+ p1 . . . ∼ . . . Set pn = . . .+ . . .

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Why the IFT?

Uniqueness of solutions to f (p) = q for p near p0 and q near q0 = f (p0) may be established for a C 2 function using Taylor’s formula with remainder. For example f (p) − f (r) − Df (p)(r − p) ≤ Cp − r2 ⇒ f (p) − f (r) ≥ 1 2Df (p)(r − p) for p and r close enough implying f (p) = f (r) when p = r in this neighborhood.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Why the IFT?

The implicit function theorem for the case f (x, y, z(x, y)) = C may be established by applying the inverse function theorem to the system u = x v = y w = f (x, y, z) and extracting z(x, y) from the local inverse which arises.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals

A regular partition P of the interval [a, b] × [c, d].

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals

A Riemann Sum for a bounded function f : [a, b] × [c, d] → R. SP =

n

m

f (cij)∆xi∆yj where cij ∈ [xi−1, xi] × [yj−1, yj]. (Regular partitions have all the xi (resp. yj) equally spaced.)

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals

A Riemann Sum Picture

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals

The Riemann Sum is sum of these (signed) volumes.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals

A finer partition.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals

We say a function is integrable if the Riemann sums settle down to a limit L as the mesh diameter goes to zero. If so we write

f dA = L where R denotes the rectangle [a, b] × [c, d].

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals

Density of Foxes in England

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals

Density of Foxes in England

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals

The example f (x, y) = 1 if x is rational

therwise

shows that not all functions are integrable.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals

Theorem: If f : R → R is continuous, then f is integrable; i.e.

R f dA exists.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals

We actually compute double integrals by a sequence of 1 dimensional integrals using Fubini’s Theorem: Theorem: If f is continuous then

f dA = d

c

b

a

f (x, y) dx

(or b

a

d

c

f (x, y) dy

dx.)

At the Riemann sum level this comes down to organizing your sums over all rectangles by first adding up contributions form each row of rectangles and then adding up the row totals. But there are subtleties to this!

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals

Defining double integrals for regions more general than a rectangle:

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integral Problems

Setup the integral of f (x, y) = ex over the region bounded by the curves y = 2x and y = x3 + x. (or just the area . . . .)

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integral Problems

Interchange the order of integration in 1 2x x2 dy dx.

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From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems