MHB Find the general solution of the problem

[mathmari]

Hey!

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)

 

[Prove It]

Hey!

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 $\displaystyle \begin{align*} u_t \left( x, t \right) - u_{xx} \left( x, t \right) = 0 \end{align*}$...

 

[mathmari]

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)

 

[Prove It]

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?

 

[mathmari]

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)