1. Hey!! Find the solution of the problem $$u_t(x, t)-u_{xx}(x, t)=0, 0<x<1, t>0 \tag {*} \\ u(0, t)=0, t>0 \\ u_x(1,t)+u_t(1,t)=0, t>0$$

I have done the following:

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

$$u(0, t)=X(0) \cdot T(t)=0 \Rightarrow X(0)=0 \\ X'(1) \cdot T(t)+X(1) \cdot T'(t)=0 \Rightarrow \frac{X'(1)}{X(1)}=-\frac{T'(t)}{T(t)}$$

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

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

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

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.
• $\lambda >0$ :

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

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

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

That means that the eigenvalue problem $(1)$ has only positive eigenvalues $0<\lambda_1 < \lambda_2 < \dots < \lambda_k < \dots$ that are the positive roots of the equation $\tan \sqrt{x}=\frac{1}{\sqrt{x}}$.

Is this correct??

Why can we say that the number of the eigenvalues is countable ?? How can we show that $$\lim_{k \rightarrow +\infty} \frac{\sqrt{\lambda_k}}{k \pi}=1$$ ??   Reply With Quote

2.

3. Hi!!!  Originally Posted by mathmari Is this correct??
Yep.  Quote:
Why can we say that the number of the eigenvalues is countable ?? Take a look at of $\tan y$ and $\frac 1 y$.

The tangent 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.  Quote:
How can we show that $$\lim_{k \rightarrow +\infty} \frac{\sqrt{\lambda_k}}{k \pi}=1$$ ?? In the same graph, you can see that each successive intersection of $\tan y$ and $\frac 1 y$ is closer and closer to where $\tan y$ intersects the axis. That is at $k\pi$.
So those intersections, that correspond to $\sqrt {\lambda_k}$, approach $k\pi$.
Thus:
$$\lim_{k\to \infty} \frac{\sqrt {\lambda_k}}{k\pi} = 1$$   Reply With Quote

I understand!!  Originally Posted by I like Serena Take a look at of $\tan y$ and $\frac 1 y$.

The tangent 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.
When we have an other problem, and we find that the eigenvalues are the positive roots of the equation $\tan (\sqrt{\lambda } \pi)=\frac{1}{\sqrt{\lambda}}$, is the argument the same?? I have also an other question... Originally Posted by mathmari $$u(0, t)=X(0) \cdot T(t)=0 \Rightarrow X(0)=0 \\ X'(1) \cdot T(t)+X(1) \cdot T'(t)=0 \Rightarrow \frac{X'(1)}{X(1)}=-\frac{T'(t)}{T(t)}$$
To divide with $X(1)$ are we sure that it isn't zero??   Reply With Quote

5. Originally Posted by mathmari I understand!! Good!  Quote:
When we have an other problem, and we find that the eigenvalues are the positive roots of the equation $\tan (\sqrt{\lambda } \pi)=\frac{1}{\sqrt{\lambda}}$, is the argument the same?? Yes. In this case the limit would be $\displaystyle \lim_{k\to\infty} \frac{\sqrt{\lambda_k}}{k} = 1$ Quote:
I have also an other question...

To divide with $X(1)$ are we sure that it isn't zero?? No, we can't be sure.   Reply With Quote Originally Posted by I like Serena No, we can't be sure. So, can't we do that or do we have to suppose that $X(1) \neq 0$ ??   Reply With Quote

7. Originally Posted by mathmari So, can't we do that or do we have to suppose that $X(1) \neq 0$ ?? Easiest is to avoid dividing by $X(1)$. Otherwise we should distinguish two cases. One where $X(1) \neq 0$ and one where $X(1) = 0$.   Reply With Quote Originally Posted by I like Serena Easiest is to avoid dividing by $X(1)$. Otherwise we should distinguish two cases. One where $X(1) \neq 0$ and one where $X(1) = 0$. How could we avoid dividing by $X(1)$ ??   Reply With Quote

9. Originally Posted by mathmari How could we avoid dividing by $X(1)$ ?? When do you need to?   Reply With Quote Originally Posted by I like Serena When do you need to? We have the problem $$u_t(x, t)-u_{xx}(x, t)=0, 0<x<1, t>0 \tag {*} \\ u(0, t)=0, t>0 \\ u_x(1,t)+u_t(1,t)=0, t>0$$

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

$$u(0, t)=X(0) \cdot T(t)=0 \Rightarrow X(0)=0 \\ u_x(1,t)+u_t(1,t)=0 \Rightarrow X'(1) \cdot T(t)+X(1) \cdot T'(t)=0 \Rightarrow \frac{X'(1)}{X(1)}=-\frac{T'(t)}{T(t)}$$

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

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

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

If we don't divide by $X(1)$ how else could we use the condition $u_x(1,t)+u_t(1,t)=0 \Rightarrow X'(1) \cdot T(t)+X(1) \cdot T'(t)=0$ ??   Reply With Quote

11. Originally Posted by mathmari If we don't divide by $X(1)$ how else could we use the condition $u_x(1,t)+u_t(1,t)=0 \Rightarrow X'(1) \cdot T(t)+X(1) \cdot T'(t)=0$ ?? Substitute $T'(t)=-\lambda T(t)$? Afterwards, we can divide by $T(t)$.
If we assume that the solution for $u$ is non-trivial, $T(t)$ must be non-zero for at least one value of $t$.   Reply With Quote

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