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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 $d^2+\lambda=0$.

  • $\lambda<0$:

    General solution: $X(x)=c_1 e^{\sqrt{-\lambda }x}+c_2e^{-\sqrt{-\lambda}x}$

$$X(0)=0 \Rightarrow c_1+c_2=0 \Rightarrow c_1=-c_2$$

$$X(1)=c_1e^{\sqrt{-\lambda}}+c_2e^{-\sqrt{-\lambda}}=c_1(e^{\sqrt{-\lambda}}-e^{-\sqrt{-\lambda}})$$

$$X'(x)=\sqrt{-\lambda}c_1e^{\sqrt{-\lambda }x}-\sqrt{-\lambda}c_2 e^{-\sqrt{-\lambda}x} \\ X'(1)=\sqrt{-\lambda}c_1e^{\sqrt{-\lambda }}-\sqrt{-\lambda}c_2 e^{-\sqrt{-\lambda}}=\sqrt{-\lambda}c_1(e^{-\lambda}+e^{-\sqrt{-\lambda}}$$

$$X'(1)-\lambda X(1)=0 \Rightarrow \sqrt{-\lambda}c_1(e^{\sqrt{-\lambda}}+e^{-\sqrt{-\lambda}})-\lambda c_1(e^{\sqrt{-\lambda}}-e^{-\sqrt{-\lambda}})=0 \Rightarrow c_1 [ e^{\sqrt{-\lambda}}(\sqrt{-\lambda}-\lambda)+e^{-\sqrt{-\lambda}}(\sqrt{-\lambda}+\lambda)]=0$$

How could we continue to show that for $\lambda <0$ we get the trivial solution $X(x)=0$ ??

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EDIT:

  • $\lambda <0$ :

    $X(x)=c_1 \sinh (\sqrt{-\lambda} x)+c_2 \cosh (\sqrt{-\lambda}x)$

    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 $$\lim_{k \rightarrow +\infty} \frac{\sqrt{\lambda_k}}{k \pi}=1$$ ??

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It may be easier in this case to take the general solution as $$ X(x)=c_1\sinh \mu x+c_2\cosh \mu x,\quad \mu:=\sqrt{-\lambda}. $$ Then the first boundary condition implies that $c_2=0$ and the second is that either $$ c_1=0 $$ or $$ \tanh \mu =-\frac{1}{\mu}. $$ But the last equation has no solutions, hence $c_1=0$.

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  • $\begingroup$ How do we know that the equation $$\tanh \mu=-\frac{1}{\mu}$$ has no solutions ?? @Artem $\endgroup$ – Mary Star Apr 6 '15 at 1:42
  • $\begingroup$ I have edited my initial post... Could you take a look at it ?? @Artem $\endgroup$ – Mary Star Apr 6 '15 at 2:59
  • $\begingroup$ @MaryStar It looks fine to me. To answer your two last questions try to sketch the graphs of $\tan h$ and $1/h$. $\endgroup$ – Artem Apr 6 '15 at 3:14
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Using an integrating factor, we can show that the solution to $$ X''+\lambda X=0\tag{1} $$ for $\lambda\gt0$, is $$ X(x)=a\cos(\sqrt\lambda\,x)+b\sin(\sqrt\lambda\,x)\tag{2} $$ for $\lambda=0$, is $$ X(x)=a+bx\tag{3} $$ and for $\lambda\lt0$, is $$ X(x)=a\cosh(\sqrt{-\lambda}\,x)+b\sinh(\sqrt{-\lambda}\,x)\tag{4} $$ Furthermore, we also have $$ X(0)=0\tag{5} $$ and $$ X'(1)-\lambda X(1)=0\tag{6} $$


If $\lambda\gt0$, $(2)$ and $(5)$ say that $a=0$. Then $(2)$ and $(6)$ say that $$ b(\sqrt\lambda\cos(\sqrt\lambda)-\lambda\sin(\sqrt\lambda))=0\tag{7} $$ if $b=0$, then $X(x)=0$, otherwise, we need $$ \tan(\sqrt\lambda)=\frac1{\sqrt\lambda}\tag{8} $$ which has an infinite set of solutions, the smallest being $\lambda=0.7401738843949670422$.


If $\lambda=0$, $(3)$ and $(5)$ say that $a=0$. Then $(3)$ and $(6)$ say that $b=0$. Therefore, $X(x)=0$.


If $\lambda\lt0$, $(4)$ and $(5)$ say that $a=0$. Then $(4)$ and $(6)$ say that $$ b(\sqrt{-\lambda}\cosh(\sqrt{-\lambda})-\lambda\sinh(\sqrt{-\lambda}))=0\tag{9} $$ if $b=0$, then $X(x)=0$, otherwise, we need $$ \tanh(\sqrt{-\lambda})=-\frac1{\sqrt{-\lambda}}\tag{10} $$ which has no solutions since $\tanh(x)\gt0$ when $x\gt0$ and $\tanh(x)\lt0$ for $x\lt0$.


Thus, as you have added since I started writing this up, there is a sequence of solutions with $\lambda\gt0$, where $\lambda$ satisfies $(8)$.

As $\lambda$ gets larger, $\frac1{\sqrt\lambda}\to0$. The places were $\tan(x)$ is positive and near $0$ are slightly greater than multiples of $\pi$. Thus, $\sqrt\lambda$ must be slightly greater than $k\pi$ for some non-negative $k\in\mathbb{Z}$. Certainly, we can label each such $\lambda$ as $\lambda_k$. In that case, we have $$ \lim_{k\to\infty}\frac{\sqrt{\lambda_k}}{k\pi}=1\tag{11} $$

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  • $\begingroup$ To divide with $X(1)$ are we sure that it isn't zero?? $$X'(1) \cdot T(t)+X(1) \cdot T'(t)=0 \Rightarrow \frac{X'(1)}{X(1)}=-\frac{T'(t)}{T(t)}$$ ?? @robjohn $\endgroup$ – Mary Star Apr 6 '15 at 17:28
  • $\begingroup$ Indeed. I am not sure why you are bringing this up. Are you worried that we get that $a=0$ and $b=0$ imply that $X(x)=0$? Well, $X(x)=0$ is a solution, so we don't need to worry about the division by $0$. However, if you want to worry about it, you can assume that $X(x)\ne0$ for some $x$ and then when we derive $X(x)=0$, it is a contradiction; that is $b\ne0$. $\endgroup$ – robjohn Apr 6 '15 at 17:58
  • $\begingroup$ Ok... When we have an other problem and we end up having $$ \tanh(\sqrt{-\lambda}\pi)=-\frac1{\sqrt{-\lambda}}\tag{10} $$ would we say again that it has no solutions since $\tanh(x)\gt0$ when $x\gt0$ and $\tanh(x)\lt0$ for $x\lt0$ ?? @robjohn $\endgroup$ – Mary Star Apr 6 '15 at 23:27
  • $\begingroup$ Yes. That is just what I wrote. To be inclusive, I wanted to also knock out the negative $\sqrt{-\lambda}$, too. If we use $\sqrt{-\lambda}\gt0$, the left hand side is positive and the right hand side is negative. If we use $\sqrt{-\lambda}\lt0$, the left hand side is negative and the right hand side is positive. $\endgroup$ – robjohn Apr 6 '15 at 23:56
  • $\begingroup$ So can we write it in that way although at tan we have $\sqrt{-\lambda }\cdot \pi $ and at the other side of the equation just $\sqrt{-\lambda}$ ?? @robjohn $\endgroup$ – Mary Star Apr 7 '15 at 0:04

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