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everyone!! Could you please help me with a question?

Using separation of variables one can derive the general solution of the wave equation as can be seen in the following picture:

General solution

There is a solution to the following problem, which I do not fully understand. It is probably some basic maths that I have overlooked but I still cannot figure it out.

Problem and solution

In the general solution one can see that the coefficients $b_n$ and $c_n$ are computed using the two integrals in which the initial conditions $f(x)$ and $g(x)$ are multiplied with $sin(\frac{n\pi}{L}x)$. In the solution, one simply reads off the coefficients of the sine series in the initial conditions without multiplying it with the $sin(n\pi x/L)$ and integrating. My question is why one may simply take the coefficients from the IC? Also, I do not understand why the last term in the solution is $sin(3x)$ instead of $sin(\frac{3x}{2})$ because L = 2*pi.

euler132
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  • Having links to images in a question are never a good idea, particularly when they're not even rotated so it's easier to read! – thesmallprint Jan 16 '18 at 20:16
  • Sorry. I have rotated them. Did not notice in the beginning. The site does not allow me to post images without the links. It says I do not have enough reputation or smth. – euler132 Jan 16 '18 at 20:23

1 Answers1

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Alright, I think I understand the solution now. One simply uses the general solution and compares it to the initial condition in the following way:

The general solution of the wave equation is $$u(x,t)=\sum_{n=1}^\infty sin(\frac{n\pi}{L}x)((b_n cos(\frac{n\pi c}{L}t)+(\frac{c_n L}{n \pi c} )sin(\frac{n\pi c}{L}t)$$

We plug in the $0$ as in the initial condition, at the same time equating it to the actual initial condition, and get $$u(x,0)=\sum_{n=1}^\infty sin(\frac{n\pi}{L}x)((b_n cos(0)+(\frac{c_n L}{n \pi c} )sin(0))$$ $$=\sum_{n=1}^\infty b_n sin(\frac{n\pi}{L}x)=sin(\frac{x}{2})+sin(\frac{5x}{2})$$ Therefore, $b_n$ is equal to 1. If we compare $\frac{n\pi x}{L}$ to $\frac{x}{2}$, with L=2*pi, as in the boundary and the initial conditions, we get n=1. So, $b_1=1$, and, with $c=\sqrt2$, the first term of the solution is $$1\times sin(\frac{x}{2})\times cos(\frac{\sqrt2}{2}t)$$ One can apply the same principle to compute the second $b_n$ coefficient and the $c_n$ coefficient, and then the other two terms of the solution to this PDE. For the $c_n$ coefficient one simply uses $u_t(x,0)$ instead of $u(x,0)$.

euler132
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