# D'Alembert's Solution

1. Apr 12, 2007

### DanielO_o

I have to solve the 1-dimensional wave equation using D'Alembert's solution.

I know the solution has the forum 1/2[f(x-ct) + f(x+ct)] + 1/2C*integral(g(s)ds)

and i have two intial conditions f(x) and g(x).... do i just plug in (x-ct) where i see x etc...?

also what is g(s) in the integral?

Thanks

Dan

2. Apr 12, 2007

### HallsofIvy

You've written this very confusingly! You have f used in at least two different ways and g used in two different ways.

What you are giving as the "D'Alembert solution" is too specialized.

The general D'Alembert solution to the wave equation is $\phi (x,t)= F(x+ ct)+ G(x- ct)$ where F and G can be any two twice differentiable functions of a single variable. Then $\phi_t(x,t)= cF'(x+ct)- cG'(x- ct)$. If one initial condition is $\phi_t(x,0)= 0$, then it follows that $F'(x)= G'(x)$ so F and G differ by a constant- in that case, $\phi (x,t)= (1/2)(F(x+ct)+ F(x-ct))+ constant$.

But that is not the case here. (It is also confusing to say just "initial values f(x) and g(x)" since functions are not "conditions". I assume you mean $\phi (x,0)= f(x)$ and $\phi_t(x,0)= g(x)$.)

Then you have $\phi (x,0)= F(x)+ G(x)= f(x)$ and $\phi_t(x,0)= cF'(x)- cG'(x)= g(x)$. Solve those two equations for F and G.

3. Apr 12, 2007

### DanielO_o

Yes, sorry to be confusing but your assumptions were right...

The initial conditions are #(x,0) = e^(-x^2) and #'(x,0) = 2cxe^(-x^2)...

I wasn't sure how to use these to get to the solution but it seems like i need to solve those two equations.

Thanks!

4. Apr 12, 2007

### DanielO_o

I have got the answer #(x,t) = e^-((x-ct)^2)

Am i right?

5. Apr 12, 2007

### HallsofIvy

Well, it's easy to check, isn't it? Calculate #xx and #tt and see if it satisfies the differential equation. Is #(x,0)= e^(-x^2)?
Is #_t(x,0)= -2cxe^(-x^2)?