Applying Initial Conditions (14.4) - D’Alembert’s Solution of Wave Equation

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Session 1: Understanding Initial Conditions

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SarahInstructor

Today, we are discussing how to apply initial conditions to D'Alembert’s solution of the wave equation. Can anyone remind me what these initial conditions represent?

Noah

They represent the initial state of the wave at time zero, right? Like its position and velocity?

SarahInstructor

Exactly! We typically have two initial conditions: the initial displacement, $u(x, 0) = \phi(x)$ , and the initial velocity, $\frac{\partial u}{\partial t}(x, 0) = \psi(x)$ . These functions help define how the wave behaves as it propagates.

Isabella

So, if we set these up correctly, we can figure out what the wave looks like at any time t?

SarahInstructor

Correct! This sets the stage for deriving our solution to the wave equation.

SarahInstructor

Can anyone summarize why we differentiate the initial conditions?

Akash

Differentiating helps us establish the relationship between position and motion of the wave!

SarahInstructor

Excellent! Let's move on to how we can formulate this into a full solution of the wave equation.

Session 2: Deriving Functions from Initial Conditions

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RobertInstructor

Now that we understand the initial conditions, let's learn how to derive the functions $f(x)$ and $g(x)$ . After substituting $t = 0$ in D’Alembert’s solution, what do we derive?

Ananya

We get $u(x, 0) = f(x) + g(x) = \phi(x)$ !

RobertInstructor

Right! And we also have the velocity condition, which after applying provides another equation involving the derivatives of $f$ and $g$ . Can someone express that?

Noah

It becomes $c[f'(x) - g'(x)] = \psi(x)$ !

RobertInstructor

Perfect! This tells us how the shapes of our functions relate to the initial velocity of the wave.

Isabella

So we can find $f'(x)$ by rearranging, right?

RobertInstructor

Absolutely! This fundamental manipulation is key to solving our wave equation.

Session 3: Finalizing D'Alembert's Solution

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SarahInstructor

So after deriving functions $f$ and $g$ , what does our final formula for the wave solution look like?

Ananya

It is $u(x, t) = \frac{1}{2}{[\phi(x + ct) + \phi(x - ct)]} + \frac{1}{2c}\int \psi(s) ds$ !

SarahInstructor

Excellent! What do the two terms in this equation represent?

Akash

The first term is the wave's displacement propagating to the left and right, and the second term accounts for initial velocity!

SarahInstructor

Exactly! Understanding this helps us interpret the wave behavior in physical systems effectively.

Noah

So if we know the initial conditions, we can use this complete formula to describe the wave at any time!

SarahInstructor

That's correct! Remember, applying initial conditions is crucial for real-world applications of wave equations in physics.

Session 4: Example Problem Solving

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RobertInstructor

Let's apply what we’ve learned to an example problem. Consider, we have the wave equation with initial conditions: $u(x, 0) = \sin x$ and $\frac{\partial u}{\partial t}(x, 0) = 0$ . How can we start?

Isabella

First, we identify $\phi(x) = \sin x$ and $\psi(x) = 0$ .

RobertInstructor

Correct! Next, how do we set up our D'Alembert's solution with these initial conditions?

Akash

We would substitute into the formula. So it becomes $u(x, t) = \frac{1}{2}[\sin(x + 2t) + \sin(x - 2t)]$ since the wave speed $c = 2$ .

RobertInstructor

Exactly! And what's the final simplified form of the wave function?

Ananya

After applying the identity for sine, it is $u(x, t) = \sin x \cos(2t)$ .

RobertInstructor

Fantastic! This shows how initial conditions directly impact the wave's displacement over time.

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Initial Conditions Overview

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Suppose initial conditions are:

∂𝑢 𝑢(𝑥,0) = 𝜙(𝑥), (𝑥,0) = 𝜓(𝑥) ∂𝑡

Detailed Explanation

In this chunk, we set the initial conditions for our wave equation. The function 𝑢(𝑥,0) represents the initial displacement of the wave at any position 𝑥 when time 𝑡 is 0. It is defined by another function 𝜙(𝑥). Additionally, 𝜓(𝑥) denotes the initial velocity of the wave at time zero, which is represented by the derivative of 𝑢 with respect to time, ∂𝑢/∂𝑡.

Examples & Analogies

Think of a guitar string that is plucked. The initial displacement of the string (how far it is from its rest position) is similar to 𝜙(𝑥), while the speed at which the string moves immediately after being plucked is analogous to 𝜓(𝑥). At the moment of plucking, you can observe both the position where the string is displaced and how fast it is moving.

D'Alembert's Solution Application

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Then from D'Alembert’s solution: At 𝑡 = 0: • 𝑢(𝑥,0) = 𝑓(𝑥)+𝑔(𝑥) = 𝜙(𝑥) → (1) • ∂𝑢 (𝑥,0) = 𝑐[𝑓′(𝑥)−𝑔′(𝑥)]= 𝜓(𝑥) → (2)

Detailed Explanation

Here, we apply the initial conditions directly within the framework of D'Alembert's solution for the wave equation. At time 𝑡 = 0, we know that u's value can be decomposed into two arbitrary functions, 𝑓(𝑥) and 𝑔(𝑥). The first point (1) shows the relation between these functions and the initial displacement 𝜙(𝑥). The second point (2) uses the velocity condition to relate the derivatives of these functions to the initial velocity 𝜓(𝑥). This helps to establish a system of equations that we can solve.

Examples & Analogies

Imagine the initial: when you pull a rubber band and let it go—the shape of the band when you release it is the 𝜙(𝑥), and the speed at which the band snaps back to its original shape is like 𝜓(𝑥). The functions 𝑓 and 𝑔 are the underlying functions that describe how the rubber band behaves after release, representing the way it’s affected by initial conditions.

Finding Functions from Initial Conditions

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From (1): 𝑓(𝑥)+ 𝑔(𝑥) = 𝜙(𝑥) ⇒ 𝑔(𝑥) = 𝜙(𝑥)−𝑓(𝑥) Differentiate and substitute into (2): 𝑐[𝑓′(𝑥)− (𝜙′(𝑥)−𝑓′(𝑥))] = 𝜓(𝑥) ⇒ 2𝑐𝑓′(𝑥)= 𝜓(𝑥)+ 𝑐𝜙′(𝑥)⇒ 𝑓′(𝑥) 1 1 = 𝜓(𝑥)+ 𝜙′(𝑥) 2𝑐 2

Detailed Explanation

In this section, we derive function 𝑔(𝑥) using the relationship established in (1) by isolating it. We then differentiate the equation for the velocity condition before substituting this result into the second equation from the initial conditions. This leads to a new equation for the derivative of 𝑓(𝑥), which can then be further manipulated to isolate 𝑓′(𝑥). This step is crucial as it allows us to express 𝑓′(𝑥) in terms of known functions, which can then be integrated to find 𝑓 and subsequently 𝑔.

Examples & Analogies

Returning to our rubber band analogy, discovering 𝑓 and 𝑔 is like determining the exact shape of the rubber band once it's released based on its starting point and velocity. If we know how far we pulled it back and how fast it snaps back, we can calculate every position it takes on its way back to its original position.

Integration to Find Functions

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Integrate to get 𝑓(𝑥), then use 𝑔(𝑥) = 𝜙(𝑥)−𝑓(𝑥) to find 𝑔(𝑥).

Detailed Explanation

Now, we take the expression we found for 𝑓′(𝑥) and integrate it to find 𝑓(𝑥). This gives us the exact functional form of the displacement related to 𝑓. Once we have that, it is straightforward to find 𝑔(𝑥) by substitutively using the relation we derived earlier, where 𝑔(𝑥) is defined as the difference between 𝜙(𝑥) and 𝑓(𝑥). This step is the finalization of the component functions required for the complete D'Alembert solution.

Examples & Analogies

Continuing with the rubber band metaphor, once you calculate how the rubber band moves back into shape (represented by 𝑓(𝑥)), you can also easily determine how much of that initial displacement remains and flows back into the overall wave motion (represented by 𝑔(𝑥)). Essentially, it’s about combining what we know about the initial conditions to predict the entire movement of the wave thereafter.

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14.4. Applying Initial Conditions

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