Find the general solution of the problem

In summary, the conversation discusses finding the solution of the PDE $u_{tt}(x,t)-u_{xx}(x,t)=0$, with boundary conditions $u(0,t)=0$ and $u_x(\pi,t)=-u_{tt}(\pi,t)$, using the method of separation of variables. The conversation also includes a discussion about the correctness of the methodology used and the satisfaction of the last boundary condition.
  • #1
mathmari
Gold Member
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Hey! :eek:

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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  • #2
mathmari said:
Hey! :eek:

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 $\displaystyle \begin{align*} u_t \left( x, t \right) - u_{xx} \left( x, t \right) = 0 \end{align*}$...
 
  • #3
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)
 
  • #4
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?
 
  • #5
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)
 

FAQ: Find the general solution of the problem

1. What is the general solution of a problem?

The general solution of a problem is a set of all possible solutions that satisfy the given conditions or constraints. It is a comprehensive solution that covers all cases and can be applied to different scenarios.

2. How do you find the general solution of a problem?

To find the general solution of a problem, you need to analyze the given information and apply relevant mathematical or scientific principles. This involves breaking down the problem into smaller parts, identifying patterns, and using equations or formulas to come up with a comprehensive solution.

3. Is the general solution always unique?

No, the general solution of a problem is not always unique. In some cases, there may be multiple solutions that satisfy the given conditions. However, the general solution should cover all possible cases and provide a comprehensive solution.

4. Can the general solution be applied to similar problems?

Yes, the general solution of a problem can be applied to similar problems as long as the conditions and constraints are similar. However, it is important to note that the general solution may need to be modified or adapted to fit the specific parameters of a similar problem.

5. Are there any limitations to finding the general solution of a problem?

Yes, there may be limitations to finding the general solution of a problem. Some problems may have complex or non-linear relationships, making it difficult to find a comprehensive solution. In these cases, approximate solutions or specific solutions for certain scenarios may be used instead of a general solution.

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