Show that y(x,t)=y1(x,t)+y2(x,t) is a solution to the wave equation

In summary, you solved for y by taking the derivative of y with respect to x and t, and then using the principle of superposition to get that y=y_1+y_2.
  • #1
Duderonimous
63
1

Homework Statement


Let y[itex]_{1}[/itex](x,t)=Acos(k[itex]_{1}[/itex]x-ω[itex]_{1}[/itex]t) and y[itex]_{2}[/itex](x,t)=Acos(k[itex]_{2}[/itex]-ω[itex]_{2}[/itex]t) be two solutions to the wave equation
[itex]\frac{∂^{2}y}{∂x^{2}}[/itex]=[itex]\frac{1}{v^{2}}[/itex][itex]\frac{∂^{2}y}{∂t^{2}}[/itex]
for the same v. Show that y(x,t)=y[itex]_{1}[/itex](x,t)+y[itex]_{2}[/itex](x,t) is also a solution to the wave equation.

Homework Equations


Equations given in problem statement and

cos(a-b)=cosacosb+sinasinb

The Attempt at a Solution



I am pretty sure I put y[itex]_{1}[/itex](x,t)+y[itex]_{2}[/itex](x,t) in a form that does not involve addition like 2Acosasinb or something close to that. Then I take a second partial derivative with respect to both x and t and show that the ratio of ∂[itex]^{2}[/itex]y/∂t[itex]^{2}[/itex] over ∂[itex]^{2}[/itex]2/∂x[itex]^{2}[/itex] is equal to v[itex]^{2}[/itex].
I am hung up on the trig.

y(x,t)=y[itex]_{1}[/itex](x,t)+y[itex]_{2}[/itex](x,t)

y(x,t)=Acos(k[itex]_{1}[/itex]x-ω[itex]_{1}[/itex]t)+Acos(k[itex]_{2}[/itex]-ω[itex]_{2}[/itex]t)

y(x,t)=A(cosk[itex]_{1}[/itex]xcosω[itex]_{1}[/itex]t+sink[itex]_{1}[/itex]xsinω[itex]_{1}[/itex]t)+A(cosk[itex]_{2}[/itex]xcosω[itex]_{2}[/itex]t+sink[itex]_{2}[/itex]xsinω[itex]_{2}[/itex]t)

Cant figure out where to go from here. Please help! Thanks.
 
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  • #2
You're making it too complicated. Don't use any trig identities. You know that y1 and y2 are solutions. What does that mean mathematically?
 
  • #3
That their sum should also be a solution by the principle of superposition?
 
  • #4
Well, that's what you're supposed to show. :wink:

When you assume ##y_i## is a solution, you can say that
$$\frac{\partial^2}{\partial x^2}y_i(x,y) = \frac{1}{v^2}\frac{\partial^2}{\partial t^2} y_i(x,t),$$ where ##\frac{\omega_i}{k_i} = v##.

So now plug ##y=y_1+y_2## into the lefthand side of the wave equation. You get
$$\frac{\partial^2}{\partial x^2}y = \frac{\partial^2}{\partial x^2}(y_1+y_2)$$ which you can split into two terms because differentiation is a linear operation. What you want to do is keep manipulating the righthand side until you end up with
$$\frac{1}{v^2} \frac{\partial^2}{\partial t^2}y.$$
 
  • #5
[itex]\frac{\partial^2}{\partial x^2}y = \frac{\partial^2}{\partial x^2}(y_1+y_2)[/itex]
[itex]\frac{\partial^2}{\partial x^2}y = \frac{\partial^2 y_1}{\partial x^2} + \frac{\partial^2 y_2}{\partial x^2}[/itex]
[itex]\frac{\partial^2}{\partial x^2}y = \frac{1}{v^2}\frac{\partial^2 y_1}{\partial t^2}+\frac{1}{v^2}\frac{\partial^2 y_1}{\partial t^2}[/itex]
[itex]\frac{\partial^2}{\partial x^2}y = \frac{1}{v^2}(\frac{\partial^2 y_1}{\partial t^2}+\frac{\partial^2 y_2}{\partial t^2})[/itex]
[itex]\frac{\partial^2}{\partial x^2}y= \frac{1}{v^2}\frac{\partial^2}{\partial t^2}(y_1+y_2)[/itex]
[itex]\frac{\partial^2}{\partial x^2}y = \frac{1}{v^2}\frac{\partial^2}{\partial t^2}y[/itex]

Is that all I have to show since I already know that y is a solution to the wave equation and that it is assumed that is obeys that principle of superposition? Thanks again for your time.
 

1. What is the wave equation and why is it important?

The wave equation is a mathematical equation that describes the behavior of waves, such as sound and light, in a given medium. It is important because it allows us to predict and understand the behavior of these waves, which has numerous real-world applications in fields such as acoustics, optics, and engineering.

2. How do you show that y(x,t)=y1(x,t)+y2(x,t) is a solution to the wave equation?

To show that y(x,t)=y1(x,t)+y2(x,t) is a solution to the wave equation, we must substitute the given equation into the wave equation and show that it satisfies the equation. This involves taking the second derivative of y(x,t) with respect to both x and t, and then comparing it to the right-hand side of the wave equation. If they are equal, then y(x,t) is a solution to the wave equation.

3. What are the conditions for a solution to satisfy the wave equation?

A solution must satisfy the following conditions to be a solution to the wave equation:

  • It must be a function of both space (x) and time (t).
  • It must be continuous and differentiable.
  • It must satisfy the wave equation, which involves taking the second derivative of the function with respect to both x and t.

4. Are there other types of equations that govern wave behavior?

Yes, there are several other types of equations that govern wave behavior, such as the Schrödinger equation for quantum mechanics, the Navier-Stokes equation for fluid mechanics, and the Maxwell equations for electromagnetism. Each of these equations describes the behavior of waves in a specific medium or situation.

5. Can the wave equation be solved analytically?

Yes, the wave equation can be solved analytically for simple cases, such as a vibrating string or a sound wave in a uniform medium. However, for more complex cases, numerical methods are often used to approximate the solution. Additionally, the wave equation can also be solved using different techniques, such as separation of variables or Fourier transforms, depending on the specific problem at hand.

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