Answer Exercise 3.2.37 for the Sobolev H^1 inner product ⟨f, g⟩=∫0^1[f(x) g(x)+f^'(x) g^'(x)] d

Answer Exercise 3.2.37 for the Sobolev H^1 inner product ⟨f,...

Key Concepts

Induced Norm (Energy Norm)

The norm induced by the Sobolev inner product, commonly referred to as the energy norm, is defined by the square root of the sum of the L² norm of the function and the L² norm of its derivative. This norm is crucial in the analysis of boundary value problems and in the finite element method, as it naturally measures both the amplitude and the variation of a function.

Sobolev Inner Product

The Sobolev inner product combines the L² inner product of a function with that of its weak derivative, typically in the form ?f, g? = ?[f(x)g(x) + f?(x)g?(x)]dx. This definition reflects both the function's magnitude and its smoothness, providing a Hilbert space structure that is particularly well-suited for variational formulations and the study of partial differential equations.

Sobolev Space H^1

The Sobolev space H^1 is the space of functions defined on a domain (typically a subset of ?^n) that, together with their first (weak) derivatives, are square-integrable. This space broadens the classical notion of differentiability by allowing functions whose derivatives exist in the weak sense, which is essential for many variational and PDE methods.

Weak Derivative

A weak derivative extends the concept of a classical derivative to functions that may not be differentiable in the usual sense. Instead of requiring pointwise differentiability, the weak derivative is defined through an integration by parts formula with smooth test functions, making it a powerful tool in functional analysis and Sobolev spaces.

Transcript

00:02 In problem 41, the inner product between two functions is defined as the definite integral from minus 1 to 1 for the multiplication of these two functions.

00:11 Here, the two functions are defined as x and g and e use of out of x.

00:19 For party a, it's required to find the value of the inner product between f and g.

00:28 It equals by definition the definite integral from minus 1 to 1 for 1.

00:33 Because it's f of x multiplied by e to the power of x because it's re of x d x we can evaluate this integral by birds then it equals u multiplied by v minus the integral from minus 1 to 1 for v d u by defining u as x and dv as as e as e to power of x d x then v is just e to the word of x by integrating from minus 1 to 1 and u is defined as x then it equals x multiplied by e to the 1 of x integrates from minus 1 to 1 minus sorry we substitute from minus one to one minus integral from minus one to one for v which is defined as e to the bar of x d u d u is d x let's substitute for the first term we substitute by x equals 1 it's e multiplied by sorry it's e minus we substitute by the second term it's minus 1 e to the bar of minus x minus 1 minus the integral of e to the bar of x is e to the bar of x we substitute from minus 1 to 1 then it's e plus e to the bar of minus 1 minus 1 minus e minus e to the bar of minus 1 we can find that it's 2 divided by e this is for part a let's move to part b it's required to find the magnitude of the function f by definition it's the square root of the inner product between f and itself then it's the square root of the definite integral from minus 1 to 1 for f for f which is x multiplied by f d x then it's the square root of the integral of x squared the integral of x squared is x squared is x cubed divided by 3 we substitute for minus 1 to 1.

04:01 It equals the square root we substitute by x equals 1 first minus minus 1 third.

04:10 It's square root of 6 divided by 3...

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