THE D'Alembert Solution for the 1D Wave Equation

[tom_rylex]

Homework Statement

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.

Homework Equations

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

Show that the solution is
[tex] 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] [/tex]

The Attempt at a Solution

If I take the second of the initial conditions, I get
[tex] -c\theta'(x)+c\phi'(x)=g(x) [/tex]
[tex] -\theta(x)+\phi(x)=\frac{1}{c}\int g(x) [/tex],

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

 

[HallsofIvy]

Homework Statement

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.

Homework Equations

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

Show that the solution is
[tex] 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] [/tex]

The Attempt at a Solution

If I take the second of the initial conditions, I get
[tex] -c\theta'(x)+c\phi'(x)=g(x) [/tex]
[tex] -\theta(x)+\phi(x)=\frac{1}{c}\int g(x) [/tex],

You want "psi", [itex]\psi[/itex], not "phi", [itex]\phi[/itex]!

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

Okay, you have, correctly, [itex]-\theta(x)+ \psi(x)= (1/c)\int g(x)dx[/itex]
If you add the first initial condition, [itex]\theta(x)+ \psi(x)= f(x)[/itex], you get [itex]2\psi(x)= f(x)+ (1/c)\int g(x)dx[/itex] so [itex]\psi(x)= (1/2)f(x)+ (1/2c)\int g(x)dx[/itex]. We can "fix" the undermined constant in that integral by choosing what every limits of integration are convenient, say 0 and x: [itex]\psi(x)= (1/2)f(x)+ (1/2c)\int_0^x g(y)dy+ C[/itex]. Then, since, from your equation again, [itex]\theta(x)= \psi(x)- (1/c)\int g(x)dx[/itex], [itex]\theta(x)= (1/2)f(x)- (1/2c)\int_0^x g(y)dx+ C[/itex]. Now, replace x with "x-ct" and "x+ ct" in [itex]\theta[/itex] and [itex]\psi[/itex] respectively:
[tex]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[/tex]

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