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Let's assume a solution I found to second order PDE using separation of variables for the wave equation.

The solution is expressed in series: $$u(x,t) = C_0 +\sum_{n=1}^{\infty}C_n \cos(\frac{n\pi t}{2})\cos(\frac{nx}{2}) $$

I found the cofficents $C_n$, and $\sum_{n=1}^{\infty} C_n $ converges. I need to show that this solution is not classical solution of the problem.

I guess I have to show that the at least one of the second order derivatives of $u : u_{xx}, u_{tt} $ does not exist. I'm not sure I know how to do that ? and I wonder what is the meaning of a non-classical solution ?

$C_0 = 0.25, C_n = 0$ for any even $n \in N$
$C_n = \frac{16}{\pi n(16-n^2)}$ for $ n = 1,5,9,\ldots$
$C_n = \frac{-16}{\pi n(16-n^2)}$ for $ n = 3,7,11,\ldots$

the original problem: $$ u_{tt} = \pi^2u_{xx} ; 0 < x < 2\pi,t>0 $$ $$ u(x,0) = f(x), u_t(x,0) = 0; 0 \leq x \leq 2\pi$$ $$u_x(0,t)=u_x(2\pi,t) = 0; t > 0$$ where $f(x) = sin^2(x)$ between $0$ and $\pi$, and $0$ between $\pi$ and $2\pi$

d_e
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  • thanks. but with respect to this question, I don't know how to show that the derivative does not exist. I'm not even sure I solved correctly, and my solution is indeed non-classical. – d_e Nov 30 '15 at 21:38
  • posted. clearly , I might have mistakes in my solution :( – d_e Nov 30 '15 at 22:04
  • thanks for the hint. I wonder if it is ok to say that $u(x,4) = f(x)$ so because $f''(\pi)$ does not exist so $u_{xx}(\pi,4)$ does not exist either ? How can I see that $u_{xx}(\pi,4)$ does not exist directly from the series ? – d_e Dec 01 '15 at 00:00
  • thanks. you right. if you want to post your comments as answer I'll select it. – d_e Dec 01 '15 at 09:05

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A solution is deemed to be a classical solution if it's differentiable as many times as needed to be able to evaluate it in the PDE (see e.g. this and this question for more about this). For your $u$ not to be a classical solution of the wave-equation this means that we need to show that $u_{xx}$ and $u_{tt}$ fails to exist at some point $(x,t) \in(0,2\pi) \times (0,\infty)$.

The solution to the wave-equation in this case, with zero initial velocity $u_t(x,0) = 0$, can be written

$$u(x,t) = \frac{f(x+\pi t) + f(x-\pi t)}{2}$$

where $f(x) = u(x,0)$ is the initial data. The initial data $f(x) = \sin^2(x)\theta(\pi-x)$ for $x\in [0,2\pi]$ has a kink at $x=\pi$ as

$$\lim_{x\to\pi^-} f''(x) = 2,~~~~~\lim_{x\to\pi^+} f''(x) = 0$$

and it follows (in particular) that $u_{xx}(x,t)$ does not exist at any point where $x \pm \pi t = \pi$.

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Winther
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