Solving a Second-Order Wave Equation

[KFC]

Hi there,
I am solving a second-order wave equation of the following form:

$v^2\dfrac{\partial^2 y}{\partial x^2} = \dfrac{\partial^2y}{\partial t^2}$

where $t$ is time; $y$ is the wave function. I learn the wave function by assuming the system is linear and has a linear solution. The book tells to try a solution of sinusoidal wave so I get the following solution

$y = A\sin(kx - \omega t)$

M question is why the wave equation has to be a second-order PDE but not a first-order PDE. I saw some explanation online, it is stated that for a physical system, we need to know both initial position and initial speed so to learn the time evolution of the system. This makes sense to me. However, besides this points, is it anything to do with the time reversibility to explain why we need a second-order PDE instead?

 

[DarkBabylon]

$v^2\dfrac{\partial^2 y}{\partial t^2} = \dfrac{\partial^2y}{\partial t^2}$

$y = A\sin(kx - \omega t)$

Would like to point out that as is, the equation can only makes sense if v= +/-1, and the solution wouldn't work. It should be:
$v^2\dfrac{\partial^2 y}{\partial x^2} = \dfrac{\partial^2y}{\partial t^2}$

 

[KFC]

Would like to point out that as is, the equation can only makes sense if v= +/-1, and the solution wouldn't work. It should be:
$v^2\dfrac{\partial^2 y}{\partial x^2} = \dfrac{\partial^2y}{\partial t^2}$

Yes, it is a typo there. I just get it fixed. Thanks for that.

 

[DarkBabylon]

As for why it HAS to be a second order PDE:
In ordinary differential equations you have first order and second order. It is safe to assume that the solution would always be in the form of:
y(x)=A*exp(α*x)
The first order would return that α is what multiplies the derivative.
The second order however becomes something that I personally find comfort with as it is a middle school level maths from here until further notice:
aα²+bα+c=0
For the homogeneous solution, where {a,b,c} are parameters. That is a quadratic equation so we can solve it using this following formula:
α_1,2=[-b±sqrt(b²-4ac)]/(2a)
Notice we have a square root. This would return us several kinds of roots:
One repeated root where the solution is:
y(x)=A*exp(αx)+Bx*exp(αx)
Two different real roots where the solution is:
y(x)=A*exp(α₁*x)+B*exp(α₂*x)
Or two complimentary complex roots where the solution is:
y(x)=A*exp((λ+μi)x)+B*exp((λ-μi)x)
where λ is the solution for -b/(2a) and μ is the solution for sqrt(b²-4ac)/(2a).
However these are complex numbers. Through some formulation and Euler's formula you can then end up with this:
y(x)=A*exp(λx)*cos(μx)+B*exp(λx)*sin(μx)
or if you don't like two trigonometric functions for whatever reason:
y(x)=C*exp(λx)*cos(μx+φ)
where φ is the phase, C is the amplitude, A is a parameter and B is also a parameter, all dependent on the initial conditions. i is the imaginary unit.
The sinusoidal solution can't appear for first order differential because α is a real number, if and only if the multiplier of the derivative is a real number. If the multiplier of the derivative is a complex number then α must also be complex however the solution will return a function that belongs to the complex plane.
As for why in this particular equation it must be sinusoidal I trust you can, with this information, arrive at the answer yourself.

I would like to also point out it is not just a sinusoidal function solution solves this. Any wave form can solve this equation, be it square, triangle, saw-tooth etc. but you must use their Fourier representation to do that, which incidentally involves the trigonometric functions sine and cosine and if you work with complex numbers, exponent (must have an imaginary unit in the exponent) as well.

 

[DarkBabylon]

Can't edit on the phone.
Where I said:
"μ is the solution for sqrt(b2-4ac)/(2a)."
It should be μi, excuse me for the inaccuracy.

 

[KFC]

Can't edit on the phone.
Where I said:
"μ is the solution for sqrt(b2-4ac)/(2a)."
It should be μi, excuse me for the inaccuracy.

Thanks a lot for your explanation. I agree the reason to have a second order PDE for the wave equation in terms of mathematics. But what I want to know more is why physically it should be 2nd order and if it is any thing to do with the time reversibility. In physical system, there is 1st order PDE to describe some system as well. I just don't understand why wave equation is not 1st order or 3rd order or even 4rd order but just 2nd order in terms of physics.

 

[olivermsun]

The order has nothing to do with time reversibility. (The heat equation is also 2nd order.)

The 2nd order comes about because the wave equation is relating the acceleration (2nd derivative of position with respect to time) of a piece of string to the net tension force exerted on it (essentially the 2nd derivative of position with respect to space).

 

[nashed]

You should look up a derivation of the wave equation then it would make more sense, essentially what they do is they take a system of N coupled oscillators where they connect all oscillators in a line such that they are all evenly spaced (assume that the line's length is L, then the distance between each couple is L/N), they write the movement equations for the system and then take the limit of N going to infinity, the result of that limit is the wave equation, so basically the second derivative nature of the wave equation is a result of the movement equations for the coupled oscillators.

 

[KFC]

Thanks all above. Yes, by looking at the force, it makes sense to have 2nd order PDE.