- #1

mathmari

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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 ?? (Wondering)

How can we show that $$\lim_{k \rightarrow +\infty} \frac{\sqrt{\lambda_k}}{k \pi}=1$$ ?? (Wondering)