Problem 75
Question
Traveling waves (for example, water waves or electromagnetic waves) exhibit periodic motion in both time and position. In one dimension (for example, a wave on a string) wave motion is governed by the one-dimensional wave equation $$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u(x, t)\) is the height or displacement of the wave surface at position \(x\) and time \(t,\) and \(c\) is the constant speed of the wave. Show that the following functions are solutions of the wave equation. \(u(x, t)=A f(x+c t)+B g(x-c t),\) where \(A\) and \(B\) are constants and \(f\) and \(g\) are twice differentiable functions of one variable
Step-by-Step Solution
Verified Answer
Question: Verify that the given solution \(u(x, t) = Af(x+ct) + Bg(x-ct)\) indeed satisfies the one-dimensional wave equation, \(\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}\).
Answer: After finding the second order partial derivatives of \(u(x, t)\) with respect to time and position and substituting them into the wave equation, we see that the equality holds. Therefore, the given solution \(u(x, t) = Af(x+ct) + Bg(x-ct)\) satisfies the one-dimensional wave equation.
1Step 1: Find \(\frac{\partial u}{\partial t}\)
To find the first partial derivative of \(u(x, t)\) with respect to time \(t\), we can use the chain rule:
\(\frac{\partial u}{\partial t} = A \frac{\partial f(x+ct)}{\partial t} + B \frac{\partial g(x-ct)}{\partial t}\).
Now, we further apply the chain rule on both terms:
\(\frac{\partial u}{\partial t} = Ac f'(x+ct) - Bc g'(x-ct)\).
2Step 2: Find \(\frac{\partial^2 u}{\partial t^2}\)
Now, we compute the second partial derivative of \(u(x, t)\) with respect to time \(t\) by differentiating the expression from Step 1 with respect to \(t\) again. Applying the chain rule, we get:
\(\frac{\partial^2 u}{\partial t^2} = Ac^2 f''(x+ct) + Bc^2 g''(x-ct)\).
3Step 3: Find \(\frac{\partial u}{\partial x}\)
Next, we find the first partial derivative of \(u(x, t)\) with respect to position \(x\). Using the chain rule, we have:
\(\frac{\partial u}{\partial x} = A f'(x+ct) + B g'(x-ct)\).
4Step 4: Find \(\frac{\partial^2 u}{\partial x^2}\)
Finally, we compute the second partial derivative of \(u(x, t)\) with respect to position \(x\) by differentiating the expression from Step 3 with respect to \(x\) again. Applying the chain rule, we get:
\(\frac{\partial^2 u}{\partial x^2} = A f''(x+ct) + B g''(x-ct)\).
5Step 5: Verify that the given solution satisfies the wave equation
Now we have the second order partial derivatives of \(u(x, t)\). We can verify whether the given solution satisfies the wave equation by substituting these partial derivatives into the wave equation:
\(\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}\).
Substitute the expressions for the partial derivatives found in steps 2 and 4:
\(Ac^2 f''(x+ct) + Bc^2 g''(x-ct) = c^2 (A f''(x+ct) + B g''(x-ct))\).
We can now observe that both sides of the equation are equal, meaning that the given solution \(u(x, t) = Af(x+ct) + Bg(x-ct)\) indeed satisfies the one-dimensional wave equation.
Key Concepts
Partial DerivativesChain RuleDifferentiable FunctionsOne-Dimensional Waves
Partial Derivatives
Partial derivatives play a vital role in understanding how functions change with respect to one variable while keeping others constant. In the case of the wave equation, we use partial derivatives to find out how the wave function changes over time, keeping the position constant, and vice versa. When working on the wave equation, we take derivatives of the function \( u(x, t) = Af(x+ct) + Bg(x-ct) \). We start by finding the first partial derivative with respect to \( t \), which involves differentiating with respect to time:
- The result is \( \frac{\partial u}{\partial t} = Ac f'(x+ct) - Bc g'(x-ct) \), which tells us how rapidly the wave changes over time.
- The expression becomes \( \frac{\partial^2 u}{\partial t^2} = Ac^2 f''(x+ct) + Bc^2 g''(x-ct) \).
Chain Rule
The Chain Rule is an essential tool in calculus when dealing with the differentiation of composite functions. In the context of the wave equation, where we have functions nested inside other functions (such as \( f(x+ct) \) and \( g(x-ct) \)), the Chain Rule helps us handle these derivatives.Here’s a quick breakdown of how it works for the wave equation:
- When differentiating \( f(x+ct) \), we treat \( x+ct \) as a single variable and multiply by the derivative of its inside function, giving us \( c \cdot f'(x+ct) \).
- For \( g(x-ct) \), similarly, we get \( -c \cdot g'(x-ct) \).
Differentiable Functions
To prove that the given functions are solutions to the wave equation, we assume they are differentiable functions. A function being differentiable means that it is smooth enough to have derivatives.For the wave equation solution, the functions \( f \) and \( g \) need to be twice differentiable:
- This means we must be able to compute both the first and second derivatives of these functions, like \( f'(x+ct) \) and \( f''(x+ct) \).
- Being differentiable is crucial as it ensures that the partial derivative computations will not break due to discontinuities or undefined behaviors.
One-Dimensional Waves
One-dimensional waves, such as a wave moving along a string, are beautifully described by the wave equation. The quintessential form of this equation, \( \frac{\partial^{2} u}{\partial t^{2}} = c^{2} \frac{\partial^{2} u}{\partial x^{2}} \), encapsulates how waves evolve over time and space.In this setting:
- \( u(x, t) \) represents the displacement of the wave at any given point \( x \) over time \( t \).
- The constant \( c \) is the speed of the wave which indicates how fast the wave propagates through the medium.
- \( Af(x+ct) \) describes the wave moving in the positive \( x \)-direction.
- \( Bg(x-ct) \) describes the wave moving in the negative \( x \)-direction.
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