Iderown
Registered User

#2

There is a very wide range of solutions for the 1-D wave equation. Any expression of the form f(x - nt) + g(x + nt) is a solution. Obviously, not all are waves.

The further requirement for a wavelike solution is to choose initial/boundary conditions which are appropriate to a wavelike solution.

Smythe
Registered User

#3

Thanks Iderown.

So do you mean I am correct that f(x, t) = cos(x-vt) + (x-vt) is *not* a solution?

Iderown
Registered User

#4

It is a solution. Just not a fully wavelike solution. The h(x,t) = x - vt part is not wavelike. You can easily check that this h(x,t) satisfies the wave equation.

The wave equation is linear - that means that sums of solutions are also solutions. The initial/boundary conditions are an important feature in the solution of partial differential equations.

Michael Collins
Moderator

#5

There's a minus missing here for the second derivative with respect to time. You obviously know the chain rule since you realised that the 'v' must come outside the cos( ) function, so it was probably just a sign slip up.

So you should have

[latex] \displaystyle \frac{\partial^2 \psi}{\partial t^2} = \frac{\partial}{\partial t} \Big(-\sin(x-vt)\cdot(-v)\Big) = \frac{\partial}{\partial t} \Big(v\sin(x-vt)\Big) = v\cos(x-vt)\cdot(-v)=-v^2\cos(x-vt)[/latex]

which when divided by [latex] v^2 [/latex] then will equal the second derivative with respect to x.

What Iderown says is quite useful too, and in fact it's easy to show that any function of the form

[latex] \displaystyle f(x - vt) + g(x + vt) [/latex]

is a solution. Specifically

[latex] \displaystyle \frac{\partial^2}{\partial t^2} \Big(f(x - vt) + g(x + vt)\Big)=f''(x-vt)\cdot(-v)\cdot(-v)+g''(x+vt)\cdot(v)\cdot(v) = v^2f''(x-vt) + v^2g''(x+vt)[/latex]

and the left hand side is

[latex] \displaystyle \frac{\partial^2}{\partial x^2} \Big(f(x - vt) + g(x + vt)\Big)=f''(x-vt)+g''(x+vt).[/latex]

Dividing the first result above by [latex] v^2 [/latex], according to the 1-D wave equation, gives the last result above. So any function of the given form must be a solution (provided all the relevent derivatives exist, of course). Also as Iderown says, initial and boundary conditions must be met for a particular solution to a given problem.

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Smythe
Registered User

#6

Thanks. Don't know how I made that mistake.

That's interesting. Thank you for that.

I'm just trying to get an understanding of answering PDEs, so wanted to ask what you thought of my answer to this question.

The one-dimensional wave equation is given by the first equation shown in this link;

http://mathworld.wolfram.com/WaveEquation1-Dimensional.html

where Ψ = f(x, t)

Is f(x, t) = cos(x-vt) + (x-vt) a possible solution?

∂^2 f/∂x^2 = -cos(x-vt)

and

∂^2 f/∂t^2 = c^2 cos(x-ct)

So, since -cos(x-vt) ≠ cos(x-vt) I assume this is not a solution.