Wave Equation: Solutions and d'Alembert's Formula

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Questions and Answers

Which of the following is the expansion of $(a + b)^2$?

  • $a^2 - b^2$
  • $a^2 - 2ab + b^2$
  • $a^2 + b^2$
  • $a^2 + 2ab + b^2$ (correct)

What is the result of $(a + b)(a - b)$?

  • $a^2 + b^2$
  • $a^2 - 2ab + b^2$
  • $a^2 - b^2$ (correct)
  • $a^2 + 2ab + b^2$

What is the expansion of $(a - b)^2$?

  • $a^2 - b^2$
  • $a^2 + 2ab + b^2$
  • $a^2 - 2ab + b^2$ (correct)
  • $a^2 + b^2$

What is the simplified form of $(a + 1/a)(a - 1/a)$?

<p>$a^2 - 1/a^2$ (B)</p> Signup and view all the answers

What is the expansion of $(x + a)(x + b)$?

<p>$x^2 + (a + b)x + ab$ (B)</p> Signup and view all the answers

Expand $(a + b + c)^2$.

<p>$a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ (B)</p> Signup and view all the answers

What is $(a + b)^3$ equal to?

<p>$a^3 + b^3 + 3a^2b + 3ab^2$ (D)</p> Signup and view all the answers

Which expression equals $a^3 + b^3$?

<p>$(a + b)(a^2 - ab + b^2)$ (C)</p> Signup and view all the answers

If $a + b + c = 0$, what is the value of $a^3 + b^3 + c^3$?

<p>$3abc$ (B)</p> Signup and view all the answers

Flashcards

(a + b)² expansion

The square of a binomial (a + b) equals the square of 'a' plus twice the product of 'a' and 'b' plus the square of 'b': (a + b)² = a² + 2ab + b²

(a - b)² expansion

The square of a binomial (a - b) equals the square of 'a' minus twice the product of 'a' and 'b' plus the square of 'b': (a - b)² = a² - 2ab + b²

(a + b)(a - b) expansion

The product of (a + b) and (a - b) equals the difference of the squares of 'a' and 'b': (a + b)(a - b) = a² - b²

(a + b)² + (a - b)²

The sum of (a + b)² and (a - b)² equals twice the sum of the squares of 'a' and 'b': (a + b)² + (a - b)² = 2(a² + b²)

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(a + b)³ expansion

The cube of (a + b) equals a³ + 3a²b + 3ab² + b³

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(a - b)³ expansion

The cube of (a - b) equals a³ - 3a²b + 3ab² - b³

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(a + b)(a² - ab + b²)

(a + b) multiplied by (a² - ab + b²) equals a³ + b³

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(a - b)(a² + ab + b²)

(a - b) multiplied by (a² + ab + b²) equals a³ - b³

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a³ + b³ + c³ when a+b+c = 0

If a + b + c = 0, then a³ + b³ + c³ = 3abc

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Study Notes

  • Deals with the wave equation and methods to solve it

Physical Context of Wave Equation

  • Common situations include:
    • Small vibrations of a stretched string
    • Acoustics
    • Electromagnetism

Mathematical Context of Wave Equation

  • Wave Equation involves a scalar function of space and time: $u(x,t)$
  • Wave Equation: $\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2} $
    • $c$ represents the wave speed
    • It is a linear, second-order PDE (Partial Differential Equation)

Solution Methods for Wave Equations

  • Can be solved with:
    • Direct method, using d'Alembert's formula
    • Separation of variables, leading to Fourier series

d'Alembert's Formula Derivation

  • Define new variables: $\xi = x + ct$ and $\eta = x - ct$
    • Partial derivatives transform as follows: $\frac{\partial}{\partial x} = \frac{\partial}{\partial \xi} + \frac{\partial}{\partial \eta}$ and $\frac{\partial}{\partial t} = c\frac{\partial}{\partial \xi} - c\frac{\partial}{\partial \eta}$
    • Second-order partial derivatives transform as follows: $\frac{\partial^2}{\partial x^2} = (\frac{\partial}{\partial \xi} + \frac{\partial}{\partial \eta})^2 = \frac{\partial^2}{\partial \xi^2} + 2\frac{\partial^2}{\partial \xi \partial \eta} + \frac{\partial^2}{\partial \eta^2}$
    • Second-order partial derivatives transform as follows: $\frac{\partial^2}{\partial t^2} = c^2(\frac{\partial}{\partial \xi} - \frac{\partial}{\partial \eta})^2 = c^2(\frac{\partial^2}{\partial \xi^2} - 2\frac{\partial^2}{\partial \xi \partial \eta} + \frac{\partial^2}{\partial \eta^2})$
  • Wave equation transforms to: $\frac{\partial^2 u}{\partial \xi \partial \eta} = 0$
    • Integrating with respect to $\xi$ gives: $\frac{\partial u}{\partial \eta} = b(\eta)$
    • Integrating with respect to $\eta$ gives: $u(\xi, \eta) = \int b(\eta) d\eta + A(\xi) = B(\eta) + A(\xi)$
    • General solution: $u(x,t) = A(x + ct) + B(x - ct)$
  • The general solution Consists of two arbitrary functions
    • $A(x + ct)$ representing a wave moving to the left
    • $B(x - ct)$ representing a wave moving to the right

Finding Arbitrary Functions

  • Initial conditions are necessary:
    • Initial displacement: $u(x,0) = f(x)$
    • Initial velocity: $\frac{\partial u}{\partial t}(x,0) = g(x)$
  • From the initial displacement: $u(x,0) = A(x) + B(x) = f(x)$
  • The partial derivative with respect to time is: $\frac{\partial u}{\partial t}(x,t) = cA'(x + ct) - cB'(x - ct)$
  • Initial velocity condition yields: $\frac{\partial u}{\partial t}(x,0) = cA'(x) - cB'(x) = g(x)$
  • Rearranging, $A'(x) - B'(x) = \frac{1}{c}g(x)$
  • Integrating with respect to x: $A(x) - B(x) = \frac{1}{c}\int_0^x g(s) ds + k$
  • Solve the system of equations:
    • $A(x) + B(x) = f(x)$
    • $A(x) - B(x) = \frac{1}{c}\int_0^x g(s) ds + k$
  • Solving for A(x) and B(x):
    • $A(x) = \frac{1}{2}f(x) + \frac{1}{2c}\int_0^x g(s) ds + \frac{k}{2}$
    • $B(x) = \frac{1}{2}f(x) - \frac{1}{2c}\int_0^x g(s) ds - \frac{k}{2}$
  • Substituting back into the general solution $u(x,t) = A(x + ct) + B(x - ct)$
    • Substituting functions A and B, yields: $= \frac{1}{2}[f(x + ct) + f(x - ct)] + \frac{1}{2c}\int_0^{x+ct} g(s) ds - \frac{1}{2c}\int_0^{x-ct} g(s) ds$

d'Alembert's Formula

  • $u(x,t) = \frac{1}{2}[f(x + ct) + f(x - ct)] + \frac{1}{2c}\int_{x-ct}^{x+ct} g(s) ds$

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