# Solution of the Wave Equation

• Apr 28th 2011, 04:30 AM
Phugoid
Solution of the Wave Equation
Hi,

I'm having a bit of trouble understanding D'Alembert's solution to the wave equation.

The Problem

$\displaystyle \displaystyle \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}$

Or, in short notation:

$\displaystyle \displaystyle u_{tt} = c^2 u_{xx}$

Find $\displaystyle \displaystyle u(x,t)$ subject to the initial conditions:

$\displaystyle \displaystyle u(x,0) = f(x)$

$\displaystyle \displaystyle u_t(x,0) = g(x)$

The Solution (the part I understand)

The first stage is to express the equation in characteristic coordinates, which, once performed, gives:

$\displaystyle \displaystyle u_{\xi \eta} = 0$

Here, $\displaystyle \displaystyle \xi = x + ct$ and $\displaystyle \displaystyle \eta = x-ct$.

Integrating the equation w.r.t both variables gives us the simple general solution:

$\displaystyle \displaystyle u(\xi, \eta) = F_1(\xi)+F_2(\eta)$

Or, back in cartesian coordinates:

$\displaystyle \displaystyle u(x,t) = F_1(x+ct) + F_2(x-ct)$ (1)

Now, differentiating this, we get:

$\displaystyle \displaystyle u_t(x,t) = cF_1'(x+ct) - cF_2'(x-ct)$ (2)

Applying the initial conditions to equations (1) and (2), we get:

$\displaystyle \displaystyle u(x,0) = F_1(x) +F_2(x) = f(x)$ (3)

$\displaystyle \displaystyle u_t(x,0) = cF_1'(x) - cF_2'(x) = g(x)$ (4)

The Solution (part I don't understand)

Now, the book I am using then makes the following statement:

If we integrate equation (4), we get:

$\displaystyle \displaystyle \int_a^x g(\tau)d\tau = cF_1(x) - cF_2(x)$

To be honest, this is the only part I don't understand. Where do the other terms which results from the integration disappear to? i.e., by my working, I would get:

$\displaystyle \displaystyle \int_a^x g(\tau)d\tau = \int_a^x (cF_1'(\tau) - cF_2'(\tau)) d\tau$

$\displaystyle \displaystyle = \biggg(cF_1(\tau)-cF_2(\tau)\biggg)^x_a$

$\displaystyle \displaystyle = cF_1(x) - cF_1(a) -cF_2(x) + cF_2(a)$

How do they manage to conclude that $\displaystyle \displaystyle cF_2(a) - cF_1(a) = 0$??

• Apr 28th 2011, 05:35 AM
bugatti79
Quote:

Originally Posted by Phugoid
If we integrate equation (4), we get:

$\displaystyle \displaystyle \int_a^x g(\tau)d\tau = cF_1(x) - cF_2(x)$

I believe what the book has done from equation 4 is integrated the middle terms and the RHS

$\displaystyle cF_1(x) - cF_2(x)=\int_a^x g( \tau) d \tau + F_1(0)-F_2(0)$

where the ast two terms are constants of integration which are a function of t and additionally we know u_t (x,0) hence the constants go to zero.

Can you continue on?
• Apr 28th 2011, 06:31 AM
Phugoid
Quote:

Originally Posted by bugatti79
I believe what the book has done from equation 4 is integrated the middle terms and the RHS

$\displaystyle cF_1(x) - cF_2(x)=\int_a^x g( \tau) d \tau + F_1(0)-F_2(0)$

where the ast two terms are constants of integration which are a function of t and additionally we know u_t (x,0) hence the constants go to zero.

Can you continue on?

Sorry, I don't fully understand what you mean.

Could you elaborate?
• Apr 28th 2011, 07:01 AM
bugatti79
From $\displaystyle \displaystyle u(x,t) = F_1(x+ct) + F_2(x-ct)$ (1) you can see that F1 and F2 are both functions of x and t. Therefore when integrating equation 4 wrt to x the resulting constants of integration must be a function of t.

I have attached my version using your initial conditions. This should help you find the general solutio u(x,t)
• Apr 28th 2011, 02:04 PM
Phugoid
Quote:

Originally Posted by bugatti79
From $\displaystyle \displaystyle u(x,t) = F_1(x+ct) + F_2(x-ct)$ (1) you can see that F1 and F2 are both functions of x and t. Therefore when integrating equation 4 wrt to x the resulting constants of integration must be a function of t.

I have attached my version using your initial conditions. This should help you find the general solutio u(x,t)

Ah, I see... so the constants don't necessarily disappear as might be suggested by the book, but eventually cancel out at a later stage in the derivation.

Thanks for that, cleared a lot up.