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

  • #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.
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,846
1,416
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
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,846
1,416
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.
 

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

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
957
Replies
2
Views
741
Replies
50
Views
3K
  • Last Post
Replies
15
Views
1K
  • Last Post
Replies
1
Views
3K
Replies
3
Views
581
Replies
12
Views
862
Replies
2
Views
697
Replies
5
Views
2K
Top