MHB Find the general solution of the problem

mathmari
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Hey! :o

Find the solution of the problem $$u_{tt}(x, t)-u_{xx}(x, t)=0, 0<x<\pi, t>0 \tag {*} \\ u(0, t)=0, t>0 \\ u_x(\pi ,t)=-u_{tt}(\pi ,t), t>0$$

I have done the following:

We are looking for solutions of the form $$u(x, t)=X(x) \cdot T(t)$$

$$u(0, t)=X(0) \cdot T(t)=0 \Rightarrow X(0)=0 \\ X'(\pi ) \cdot T(t)+X(\pi ) \cdot T''(t)=0 \Rightarrow X'(\pi)+X(\pi )\frac{T''(t)}{T(t)}=0$$

$$(*) \Rightarrow X(x) \cdot T''(t)-X''(x) \cdot T(t)=0 \\ \Rightarrow \frac{X(x) \cdot T''(t)}{X(x) \cdot T(t)}-\frac{X''(x) \cdot T(t)}{X(x) \cdot T(t)}=0 \\ \Rightarrow \frac{T''(t)}{T(t)}=\frac{X''(x)}{X(x)}=-\lambda$$

So, we get the following two problems:
$$\left.\begin{matrix}
X''(x)+\lambda X(x)=0, 0<x<\pi \\
X(0)=0 \\
X'(\pi )-\lambda X(\pi )=0
\end{matrix}\right\}(1)
$$

$$\left.\begin{matrix}
T''(t)+\lambda T(t)=0, t>0
\end{matrix}\right\}(2)$$

For the problem $(1)$ we do the following:

The characteristic polynomial is $d^2+\lambda=0$.
  • $\lambda <0$ :

    $X(x)=c_1 \sinh (\sqrt{-\lambda} x)+c_2 \cosh (\sqrt{-\lambda}x)$

    Using the initial values we get that $X(x)=0$, trivial solution.
  • $\lambda=0$ :

    $X(x)=c_1 x+c_2$

    Using the initial values we get that $X(x)=0$, trivial solution.
  • $\lambda >0$ :

    $X(x)=c_1 cos (\sqrt{\lambda}x)+c_2 \sin (\sqrt{\lambda}x)$

    $X(0)=0 \Rightarrow c_1=0 \Rightarrow X(x)=c_2=\sin (\sqrt{\lambda}x)$

    $X'(\pi )-\lambda X(\pi )=0 \Rightarrow \tan (\sqrt{\lambda} \pi )=\frac{1}{\sqrt{\lambda}}$

That means that the eigenvalue problem $(1)$ has only positive eigenvalues $0<\lambda_1 < \lambda_2 < \dots < \lambda_k < \dots $ .

The graph of $\tan (y \cdot \pi)$ and $\frac{1}{y}$ is the following: https://www.wolframalpha.com/input/?i=plot%5Btan%28y*%28pi%29%29%2C1%2Fy%2C+%7By%2C0%2C20%7D%5D

So, the $\tan (y \cdot \pi )$ has a period of $\pi$ and in each period it has exactly 1 intersection with $\frac 1 y$.
Since there are a countable number of periods of the tangent, that means that the number of solutions is also countable.

The eigenfunctions are $\sin (\sqrt{\lambda_k} x)$.

For the problem $(2)$ we have the following:

$$T''(t)+\lambda T(t)=0 \Rightarrow T_k(t)=C_1 \sin (\sqrt{\lambda_k} t)+C_2 \cos (\sqrt{\lambda_k} t)$$

The eigenfunctions are $\sin (\sqrt{\lambda_k} t)$, $\cos (\sqrt{\lambda_k} t)$.

So, the general solution is the following:

$$u(x, t)=\sum_{k=1}^{\infty}(a_k \cos (\sqrt{\lambda_k} t)+b_k \sin (\sqrt{\lambda_k} t)) \sin (\sqrt{\lambda_k} x)$$

Is this correct?? (Wondering)
 
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mathmari said:
Hey! :o

Find the solution of the problem $$u_{tt}(x, t)-u_{xx}(x, t)=0, 0<x<\pi, t>0 \tag {*} \\ u(0, t)=0, t>0 \\ u_x(\pi ,t)=-u_{tt}(\pi ,t), t>0$$

Did you type this out properly? Judging by your working, the PDE appears to be $\displaystyle \begin{align*} u_t \left( x, t \right) - u_{xx} \left( x, t \right) = 0 \end{align*}$...
 
It is as I wrote it in the post #1.

So, is the methodology I applied wrong?? (Wondering)

Should I solve the problem in an other way?? (Wondering)
 
mathmari said:
It is as I wrote it in the post #1.

So, is the methodology I applied wrong?? (Wondering)

Should I solve the problem in an other way?? (Wondering)

Why have you only differentiated your t function once then?
 
It should be twice differentiated... I edited my initial post... But how can the last condition $u_x(\pi,t)=-u_{tt}(\pi,t)$ be satisfied by the $u$ that I found?? At the one side of the equation we will have cos and at the other side we will have sin, or not?? (Wondering)
 
There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...

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