2

The D'Alembert solution of the wave equation $$u_{tt}-c^2u_{xx}=0, x \in \mathbb{R}, t>0$$ $$u(x,0)=\phi(x), x \in \mathbb{R}$$ $$u_t(x,0)=\psi(x), x \in \mathbb{R}$$

is the following:

$$u(x,t)=\frac{1}{2}\begin{bmatrix} \phi(x-ct)+\phi(x+ct) \end{bmatrix}+\frac{1}{2c} \int_{x-ct}^{x+ct} \psi(\tau)d \tau$$

I am looking at the proof of this solution, which is the following:

The general solution of the wave equation is given from: $\displaystyle{u(x,t)=f(x+ct)+g(x-ct)} \ \ \ \ \ (1)$

$\displaystyle{u(x,0)=f(x)+g(x)=\phi(x)} \ \ \ \ \ \ (2)$

$(1): \displaystyle{u_t(x,t)=c(f'(x+ct)-g'(x-ct))} \ \ \ \ \ (3)$

$\displaystyle{u_t(x,0)=c(f'(x)-g'(x))=\psi(x)} \ \ \ \ \ \ (4)$

$(2): \displaystyle{f'(x)+g'(x)=\phi'(x)} \ \ \ \ \ (5)$

From the relations $(4)$ and $(5)$ we get:

$$f'(x)=\frac{1}{2}\begin{bmatrix} \phi'(x)+\frac{\psi(x)}{c} \end{bmatrix}$$ $$g'(x)=\frac{1}{2}\begin{bmatrix} \phi'(x)-\frac{\psi(x)}{c} \end{bmatrix}$$

Integrating these equations we get:

$$f(s)=\frac{1}{2}\begin{bmatrix} \phi(s)+\frac{1}{c} \int_0^s \psi(\tau)d \tau \end{bmatrix}+a$$ $$g(s)=\frac{1}{2}\begin{bmatrix} \phi(s)-\frac{1}{c} \int_0^s \psi(\tau)d \tau \end{bmatrix}+b$$

Why are the limits of the integral $0$ and $s$, and not $-\infty$ and $+\infty$??

Mary Star
  • 13,956
  • 3
    These are antiderivatives (indefinite integrals), not definite integrals. Antiderivatives are functions, definite integrals are just numbers. – MPW Jun 13 '14 at 23:05
  • 2
    If you're doing differential equations, it's important you remember calculus and how to integrate: for any function $f$ smooth on $\Bbb R$, for all $L$ there exists an $a$ such that $f(s)=\int_L^s f'(\tau)d\tau+a$. Indeed, $a=f(L)$. One probably doesn't want to set $L=-\infty$, because the integral might diverge, and one also obviously doesn't want to set the upper bound to be $+\infty$, because then the left side is a function of $s$ and the right side is a constant. – anon Jun 13 '14 at 23:10
  • 2
    (Continuing) For example, if you know $f'(x)=2x$ $f(0)=0$, then you can write $f(s)=\int_0^s2x;dx=\left.x^2\right|_0^s = s^2$. – MPW Jun 13 '14 at 23:10
  • Ok!!! I understand!!! Thank you both!!! :-) – Mary Star Jun 19 '14 at 01:20

1 Answers1

2

This is the fundamental theorem of calculus.

Under reasonable conditions, ${d\over dx}\int_a^xf(x')dx' = f(x)$.

Think about what happens as you vary $x$ to $x+\delta x$. The area under the curve gets an extra little sliver with a width of $\delta x$ and a height of $f(x)$. So the rate of change of this integral with respect to the right endpoint is the height of the integrand at the endpoint.

It would be bizarre if the integrand were between fixed limits. You'd be saying that the derivative of something that didn't depend on $s$ depends on $s$.

Dan Piponi
  • 4,406