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:
$$u_{tt} = c^2u_{xx}$$, 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
$$u(x,t)=\frac{1}{2} \left[ f(x+ct) + f(x-ct) +\frac{1}{c}\int_{x-ct}^{x+ct} g(y)dy \right]$$

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]

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:
$$u_{tt} = c^2u_{xx}$$, 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
$$u(x,t)=\frac{1}{2} \left[ f(x+ct) + f(x-ct) +\frac{1}{c}\int_{x-ct}^{x+ct} g(y)dy \right]$$

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 $\int_0^{x+ ct}g(y) dy+ \int_{x-ct}^0 g(y)dy= \int_{x-ct}^{x+ct} g(y)dy$.