
MHB Master
#1
April 6th, 2015,
09:03
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 $$\lim_{k \rightarrow +\infty} \frac{\sqrt{\lambda_k}}{k \pi}=1$$ ??

April 6th, 2015 09:03
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MHB Seeker
#2
April 6th, 2015,
11:19

MHB Master
#3
April 6th, 2015,
13:53
Thread Author
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??

MHB Seeker
#4
April 6th, 2015,
14:01
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.

MHB Master
#5
April 6th, 2015,
14:06
Thread Author
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 $X(1) \neq 0$ ??

MHB Seeker
#6
April 6th, 2015,
14:10

MHB Master
#7
April 6th, 2015,
14:13
Thread Author
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 $X(1)$ ??

MHB Seeker
#8
April 6th, 2015,
14:16
Originally Posted by
mathmari
How could we avoid dividing by $X(1)$ ??
When do you need to?

MHB Master
#9
April 6th, 2015,
14:31
Thread Author
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 $u_x(1,t)+u_t(1,t)=0 \Rightarrow X'(1) \cdot T(t)+X(1) \cdot T'(t)=0$ ??

MHB Seeker
#10
April 6th, 2015,
14:42