Registered User
#1

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.

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.

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?

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.

Moderator
#5

Smythe said:

...
∂^2 f/∂t^2 = v^2 cos(x-vt) [edited to replace 'c' with 'v' like other equations]

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

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

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

which when divided by $v^2$ 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

$\displaystyle f(x - vt) + g(x + vt)$

is a solution. Specifically

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

and the left hand side is

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

Dividing the first result above by $v^2$, 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.

Registered User
#6

Michael Collins said:
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.

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

Michael Collins said:

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

$\displaystyle f(x - vt) + g(x + vt)$

is a solution. Specifically

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

and the left hand side is

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

Dividing the first result above by $v^2$, 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.

That's interesting. Thank you for that.