Understanding PDEs: Evaluating a Solution to the 1-Dimensional Wave Equation

[ZedCar]

Homework Statement

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) = exp(x-ivt) a possible solution?

Homework Equations

The Attempt at a Solution

∂^2 f/∂x^2 = exp(x-ivt)

and

∂f/∂t = -iv exp(x-ivt)

Possible if v = -i

 

[Ray Vickson]

Homework Statement

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) = exp(x-ivt) a possible solution?

Homework Equations

The Attempt at a Solution

∂^2 f/∂x^2 = exp(x-ivt)

and

∂f/∂t = -ic exp(x-ivt)

Possible if v = -i

You need to compute \partial^2 f/\partial t^2, \text{ not just } \partial f/\partial t. Anyway: what is "c"? The PDE does not have "c" in it, nor does your f.

RGV

 

[ZedCar]

Anyway: what is "c"? The PDE does not have "c" in it, nor does your f.

Sorry, c should have been v. I've corrected it now.

 

[ZedCar]

You need to compute \partial^2 f/\partial t^2, \text{ not just } \partial f/\partial t.

RGV

So when I obtain the 2nd partial differentiation for 't' I obtain;

-v^2 exp(x-ivt)

So I assume this is not a possible solution since

exp(x-ivt) ≠ -exp(x-ivt)