D'alembert derivation

[tom_rylex]

1. The problem statement, all variables and given/known data
I am looking at the derivation of the D'alembert equation, and I'm having trouble with understanding where the limits of integration come in.


2. Relevant equations
Given the 1-d wave equation:
u_{tt} = c^2u_{xx} , with the general solution u(x,t)= \theta(x-ct) + \psi(x+ct) and the initial conditions
u(x,0)=f(x) , u_t(x,0)=g(x)

Show that the solution is
u(x,t)=\frac{1}{2} \left[ f(x+ct) + f(x-ct) +\frac{1}{c}\int_{x-ct}^{x+ct} g(y)dy \right]


3. The attempt at a solution
If I take the second of the initial conditions, I get
-c\theta'(x)+c\phi'(x)=g(x)
-\theta(x)+\phi(x)=\frac{1}{c}\int g(x) ,

I guess I just don't understand where the limits of integration come from to yield
\frac{1}{c} \int_{-\infty}^x g(y) dy
on the right hand side.

[HallsofIvy]

1. The problem statement, all variables and given/known data
I am looking at the derivation of the D'alembert equation, and I'm having trouble with understanding where the limits of integration come in.


2. Relevant equations
Given the 1-d wave equation:
u_{tt} = c^2u_{xx} , with the general solution u(x,t)= \theta(x-ct) + \psi(x+ct) and the initial conditions
u(x,0)=f(x) , u_t(x,0)=g(x)

Show that the solution is
u(x,t)=\frac{1}{2} \left[ f(x+ct) + f(x-ct) +\frac{1}{c}\int_{x-ct}^{x+ct} g(y)dy \right]


3. The attempt at a solution
If I take the second of the initial conditions, I get
-c\theta'(x)+c\phi'(x)=g(x)
-\theta(x)+\phi(x)=\frac{1}{c}\int g(x) ,
You want "psi", \psi, not "phi", \phi!

I guess I just don't understand where the limits of integration come from to yield
\frac{1}{c} \int_{-\infty}^x g(y) dy
on the right hand side.
Okay, you have, correctly, -\theta(x)+ \psi(x)= (1/c)\int g(x)dx
If you add the first initial condition, \theta(x)+ \psi(x)= f(x), you get 2\psi(x)= f(x)+ (1/c)\int g(x)dx so \psi(x)= (1/2)f(x)+ (1/2c)\int g(x)dx. We can "fix" the undermined constant in that integral by choosing what every limits of integration are convenient, say 0 and x: \psi(x)= (1/2)f(x)+ (1/2c)\int_0^x g(y)dy+ C. Then, since, from your equation again, \theta(x)= \psi(x)- (1/c)\int g(x)dx, \theta(x)= (1/2)f(x)- (1/2c)\int_0^x g(y)dx+ C. Now, replace x with "x-ct" and "x+ ct" in \theta and \psi respectively:
u(x,t)= \theta(x-ct)+ \psi(x+ ct)= (1/2)f(x-ct)- \int_0^{x-ct}g(t)dt- C+ (1/2)f(x+ ct)+ \int_0^{x-ct}g(t)dt+ C

Notice that "-C" and "C" cancel while \int_0^{x-ct}g(y)dy= -\int_{x-ct}^0 g(y)dy so the two integrals combine as \int_0^{x+ ct}g(y) dy+ \int_{x-ct}^0 g(y)dy= \int_{x-ct}^{x+ct} g(y)dy.