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How are these results obtained in d'Alembert's Solution to the $1+1$ wave equation?

After getting, $\frac{1}{c}\psi(c)=f'(x)-g'(x)$ we integrate from $a$ to $x$ and we get

$$f(x)-g(x)=\frac{1}{c}\int_{a}^{x}\psi(s)\,ds.$$ My question is: Why do we integrate it from $a$ to $x$?

I think we require answer in variable $x$, that's why we integrate..

Bernard
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Goldy
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  • $a$ is an arbitrary parameter, corresponding to the integration constant. This is similar to the usual 'indefinite integral' notation, but more coherent: $$\frac{1}{c} \int \psi(x) dx +C$$ – Yuriy S Feb 12 '18 at 17:53

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The parameter $a$ is arbitrary. However, the author was a bit careless with an intermediate step.

Note that from $f'(x)-g'(x)=\frac1c\psi(x)$ we have for any constant $a$

$$f(x)-g(x) -(f(a)-g(a))=\frac1c \int_a^x \psi(x')\,dx'\tag1$$

Then, since $f(x)+g(x)=\phi(x)$ we see that

$$f(x)=\frac12\phi(x)+\frac1{2c}\int_a^x \psi(x')\,dx'+f(a)-g(a)\tag 2$$

and

$$g(x)=\frac12 \phi(x)-\frac1{2c}\int_a^x \psi(x')\,dx'-(f(a)-g(a))\tag 3$$

In the referenced development, the author omitted the term $f(a)-g(a)$ from the expressions for $f(x)$ and $g(x)$. However, these terms cancel in the final answer.

Finally, from $u(x,t)=f(x+ct)+g(x-ct)$ we find that

$$u(x,t)=\frac12\left(\phi(x+ct)+\phi(x-ct)\right)+\frac1{2c}\int_{x-ct}^{x+ct}\psi(x')\,dx'$$

and the term $f(a)-g(a)$ is absent as expected.

Mark Viola
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  • Thanks for details... but why do you integrate it from $a$ to $x$ ? Can't we write $f(x)-g(x)=\frac{1}{c}\int{\psi(x),dx}+K$. – Goldy Feb 12 '18 at 18:21
  • @Goldy One can write $f(x)-g(x)$ in terms of an arbitrary antiderivative of $\psi$. But then, how does that lead to the coveted solution? – Mark Viola Feb 12 '18 at 19:10