Let β∈ℝ. Suppose f, g: ℝ →ℝ are differentiable functions such that f^'=βg, g^'=-βf, f(0)=0,

Let β∈ℝ. Suppose f, g: ℝ →ℝ are differentiable functions...

Key Concepts

Energy Method (Lyapunov Function Technique)

This is a method where a carefully chosen function – often involving the sum of squares of deviations – is used to study the stability or uniqueness of solutions. By differentiating this function and showing that its derivative is zero, one can conclude that the function remains constant over time. Combined with initial conditions, this establishes that the deviation from the expected solution is zero, hence confirming the correctness and uniqueness of the answer.

System of Linear Differential Equations

This concept involves analyzing multiple related functions whose derivatives are expressed as linear combinations of each other. In such systems, the evolution of each function is intertwined, often leading to common solution structures. Recognizing the linearity in the relationship allows the application of techniques from linear algebra and differential equations to find general solutions.

Initial Value Problem and Uniqueness

Setting specific values at an initial point, like f(0) = 0 and g(0) = 1, turns a differential equation into an initial value problem (IVP). The theory behind IVPs ensures that if the system of differential equations meets certain conditions, there will be a unique solution that satisfies these initial conditions. This concept guarantees that the particular solution identified is the only one possible for the problem.

Harmonic Oscillator and Trigonometric Functions

The nature of the derivatives, where each function’s derivative is proportional to the other function with a possible sign change, leads to a second-order differential equation characteristic of a harmonic oscillator. The solutions to such equations are naturally expressed in terms of sine and cosine functions, which inherently satisfy the required derivative relationships. Recognizing this connection simplifies the process of solving the system.

Transcript

00:01 All right, we have a nice little proof here where beta is between 0 and 1 exclusive, or strictly between 0 and 1, and we're going to prove that f is not differentiable at 0 if f satisfies these conditions where the absolute value of f is greater than equal to the absolute value of x, the b the b power, and f of 0 equals 0.

00:29 So this is what we're given.

00:30 This is an information that's useful to us.

00:34 So let's go through us.

00:37 So for a function to be differentiable, so we're going to look at for it to be deferential to show that it's not differentiable.

00:45 So for it to be differentiable, the point has to exist, the function has to exist at that point, which it does right here, is told f of 0 equals 0.

00:57 So that satisfies that part of the definition.

01:00 But that leaves one part unfulfilled, and that is that the limit, as h approaches 0 of f of x plus h of x all over h, that this limit needs to exist for it to be differentiable.

01:27 So we want to show that this limit does not exist.

01:30 So that's what we're going to do.

01:36 So the first thing we're going to do is we're going to take our x value of 0 and substitute it in for x here.

01:43 So let's go ahead and do that.

01:45 Let's use the magic of computers to help us.

01:55 Look that, neat.

01:56 And then i'm going to erase these x's and put in a zero.

02:04 Erase that x put in a zero.

02:06 And then we, that's nice because then this term goes away because f of zero is zero.

02:11 So we can get rid of that.

02:13 And now we have this, f of zero plus h.

02:17 So this just becomes f of h.

02:24 And that's gone.

02:25 So we go.

02:26 So we've simplified it to that, which is great.

02:28 So unfortunately, this does not show that it's differentiable or not because there's no way to, for sure, way to get this h out of the denominator here.

02:39 But that's where this function comes in, this absolute value of x to the b.

02:45 So let's play with that one.

02:48 Because this thing has to be greater than or equal to that.

02:55 The limit of that function...

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