# Proving solution of wave equation

1. Nov 7, 2006

### tandoorichicken

Hello everyone.

I'm asked in a problem to prove that a given general solution is valid for the wave equation
$$\nabla^2 p - \frac{1}{c_0^2} \frac{\partial^2 p}{\partial t^2} = 0$$.

The given solution was
$$p(x, t) = A_1 f_1 (x - c_0 t) + A_2 f_2 (x + c_0 t)$$.

I just need a check of work here. I plugged in the solution and calculated out for a 1-dimensional wave equation, but I might have made some assumptions along the way that are un-kosher.

Last edited: Nov 7, 2006
2. Nov 7, 2006

### tandoorichicken

Here is the work:

$$\nabla^2 p - \frac{1}{c_0^2}\frac{\partial^2 p}{\partial t^2} = \frac{\partial^2 p}{\partial x^2} - \frac{1}{c_0^2}\frac{\partial^2 p}{\partial t^2} = 0$$

$$\frac{\partial^2 p}{\partial x^2} = A_1 \frac{\partial^2 f_1}{\partial x^2} + A_2 \frac{\partial^2 f_2}{\partial x^2}$$
$$\frac{\partial^2 p}{\partial t^2} = A_1 c_0^2\frac{\partial^2 f_1}{\partial t^2} + A_2 c_0^2\frac{\partial^2 f_2}{\partial t^2}$$

$$\frac{\partial^2 p}{\partial x^2} - \frac{1}{c_0^2}\frac{\partial^2 p}{\partial t^2} = A_1 \frac{\partial^2 f_1}{\partial x^2} + A_2 \frac{\partial^2 f_2}{\partial x^2} - \frac{1}{c_0^2} (A_1 c_0^2\frac{\partial^2 f_1}{\partial t^2} + A_2 c_0^2\frac{\partial^2 f_2}{\partial t^2}) = 0$$

$$A_1 (\frac{\partial^2 f_1}{\partial x^2} - \frac{\partial^2 f_1}{\partial t^2}) + A_2 (\frac{\partial^2 f_2}{\partial x^2} - \frac{\partial^2 f_2}{\partial t^2}) = 0$$

At this point I assumed that since this was the LINEAR wave equation, this meant that spatial position varied linearly with time, so some equation could be written $x = kt + b$ where k and b are arbitrary constants.

$$\frac{\partial^2 f}{\partial t^2} = \frac{\partial^2 f}{\partial x^2}\frac{d^2 x}{d t^2} = 0$$

Likewise, while it doesn't make sense physically, one could make the mathematically sound statement $t = mx + d$ where m and d are arbitrary constants.

$$\frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 f}{\partial t^2}\frac{d^2 t}{d x^2} = 0$$

Therefore the above equations reduce to 0.

Q.E.D?

Last edited: Nov 7, 2006
3. Nov 7, 2006

### Daverz

The assumption doesn't make sense to me. Spatial position of what? How does you assumption work for a known solution like exp[i(x - c0 t)]?

I was thinking that you might be able to do this by transforming to new variables

$\phi_\pm = x \pm c_0 t$

4. Nov 7, 2006

### tandoorichicken

Sorry, completely forgot the context

This is for ultrasonic imaging. The function p(x,t) is a function of position (depth) in tissue and time.

5. Nov 10, 2006

### physics girl phd

I'm just not sure your problem is 1-D, or that you just made that assumption... Things I am just concerned about -- are your givens A1, A2 constants or vectors (like wih E-fields, they would be indicative of polarizations) and is x one-dimensional... or a vector (which would often be indicated by r, but not always)? These thoughts complicate the things, as now the spacial derivatives are more interesting...

6. Nov 10, 2006

### dextercioby

The nabla is for more than one spatial coordinate. Let's assume for simplicity that the "x" in the solution means that the problem is essentially one-dimensional.

Then you might redo the differentiations using the chain rule properly.

E.g.

$$\frac{\partial p}{\partial x}=\frac{dp}{d\left(x-c_{0}t\right)}\frac{\partial \left(x-c_{0}t\right)}{\partial x} +\frac{dp}{d\left(x+c_{0}t\right)}\frac{\partial \left(x+c_{0}t\right)}{\partial x}$$

and using the expression the problem offers, you finally get

$$\frac{\partial p}{\partial x}=A_{1}\frac{df_{1}}{d\left(x-c_{0}t\right)}\frac{\partial \left(x-c_{0}t\right)}{\partial x}+A_{2}\frac{df_{2}}{d\left(x+c_{0}t\right)}\frac{\partial \left(x+c_{0}t\right)}{\partial x}=...$$

Daniel.