Finding Eigenvalues for the Heat Equation: A Step-by-Step Approach

In summary, the conversation involves finding solutions for the partial differential equation $u_t(x,t) - u_{xx}(x,t) = 0$ with boundary conditions $u(0,t) = 0$ and $u_x(1,t) + u_t(1,t) = 0$. The solutions are of the form $u(x,t) = X(x) \cdot T(t)$ and the two problems that arise are $X''(x) + \lambda X(x) = 0$ and $T'(t) + \lambda T(t) = 0$. The eigenvalues for the problem are the positive roots of the equation $\tan (\sqrt{\lambda} \pi) = \frac{
  • #1
mathmari
Gold Member
MHB
5,049
7
Hey! :eek:

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)
 
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  • #2
Hi! (Wave)

mathmari said:
Is this correct??

Yep. (Smile)

Why can we say that the number of the eigenvalues is countable ?? (Wondering)

Take a look at this plot 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. (Thinking)
How can we show that $$\lim_{k \rightarrow +\infty} \frac{\sqrt{\lambda_k}}{k \pi}=1$$ ?? (Wondering)

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:
$$\lim_{k\to \infty} \frac{\sqrt {\lambda_k}}{k\pi} = 1$$
(Wasntme)
 
  • #3
I understand! (Sun)
I like Serena said:
Take a look at this plot 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?? (Wondering)
I have also an other question...

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

To divide with $X(1)$ are we sure that it isn't zero?? (Wondering)
 
  • #4
mathmari said:
I understand! (Sun)

Good! (Happy)


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?? (Wondering)

Yes. (Nod)
In this case the limit would be \(\displaystyle \lim_{k\to\infty} \frac{\sqrt{\lambda_k}}{k} = 1\)
I have also an other question...

To divide with $X(1)$ are we sure that it isn't zero?? (Wondering)

No, we can't be sure. (Shake)
 
  • #5
I like Serena said:
No, we can't be sure. (Shake)

So, can't we do that or do we have to suppose that $X(1) \neq 0$ ?? (Wondering)
 
  • #6
mathmari said:
So, can't we do that or do we have to suppose that $X(1) \neq 0$ ?? (Wondering)

Easiest is to avoid dividing by $X(1)$. (Wasntme)
Otherwise we should distinguish two cases. One where $X(1) \neq 0$ and one where $X(1) = 0$. (Nerd)
 
  • #7
I like Serena said:
Easiest is to avoid dividing by $X(1)$. (Wasntme)
Otherwise we should distinguish two cases. One where $X(1) \neq 0$ and one where $X(1) = 0$. (Nerd)

How could we avoid dividing by $X(1)$ ?? (Wondering)
 
  • #8
mathmari said:
How could we avoid dividing by $X(1)$ ?? (Wondering)

When do you need to? (Thinking)
 
  • #9
I like Serena said:
When do you need to? (Thinking)

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 $u_x(1,t)+u_t(1,t)=0 \Rightarrow X'(1) \cdot T(t)+X(1) \cdot T'(t)=0$ ?? (Wondering)
 
  • #10
mathmari said:
If we don't divide by $X(1)$ how else could we use the condition $u_x(1,t)+u_t(1,t)=0 \Rightarrow X'(1) \cdot T(t)+X(1) \cdot T'(t)=0$ ?? (Wondering)

Substitute $T'(t)=-\lambda T(t)$? (Thinking)

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 $t$. (Wasntme)
 
  • #11
I like Serena said:
Substitute $T'(t)=-\lambda T(t)$? (Thinking)

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 $t$. (Wasntme)

Ok... I see... (Emo)
I like Serena said:
Take a look at this plot 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. (Thinking)

mathmari said:
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?? (Wondering)

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 do we say the following ?? (Wondering)

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. How do we know that there are a countable number of periods of the tangent ?? (Wondering)
 
Last edited by a moderator:
  • #12
mathmari said:
So do we say the following ?? (Wondering)

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.

Sounds fine to me. (Smile)

How do we know that there are a countable number of periods of the tangent ?? (Wondering)

Each period on the positive axis is identified by $k\pi$ where $k \in \mathbb N$.
In other words, we have a bijective mapping from each period to $\mathbb N$.
That means we have a countably infinite number of periods. (Wasntme)
 
  • #13
I understand! Thank you very much! (flower)
 

FAQ: Finding Eigenvalues for the Heat Equation: A Step-by-Step Approach

1. What are eigenvalues?

Eigenvalues are a set of numbers that represent the characteristics or properties of a linear transformation or a matrix. They are used in various fields of mathematics and science, such as physics, engineering, and computer graphics, to analyze and understand the behavior of systems.

2. How do eigenvalues relate to eigenvectors?

Eigenvalues and eigenvectors are closely related. An eigenvector is a vector that, when multiplied by a matrix, only changes in magnitude, not in direction. The corresponding eigenvalue is the factor by which the eigenvector is scaled. In other words, eigenvectors are the directions in which a matrix only stretches or compresses, and eigenvalues are the amounts by which it stretches or compresses.

3. What do eigenvalues tell us about a matrix?

Eigenvalues provide valuable information about a matrix, such as its size, shape, and behavior. They can tell us how a matrix will affect the space it operates on and whether it will stretch or compress the vectors within that space. Additionally, the eigenvalues of a matrix can also indicate its stability, convergence, and other important properties.

4. How are eigenvalues calculated?

Eigenvalues can be calculated using various methods, such as the characteristic polynomial, eigenvalue decomposition, or iterative methods. The most common approach is the characteristic polynomial method, in which the eigenvalues are found by solving the characteristic equation det(A - λI) = 0, where A is the matrix and λ is the eigenvalue. The eigenvectors can then be found by solving the corresponding system of equations.

5. What are some real-world applications of eigenvalues?

Eigenvalues have numerous applications in different fields of science and engineering. For example, in physics, they are used to study the properties of quantum mechanics and particle physics. In engineering, they are used to analyze the stability and behavior of mechanical systems, electrical circuits, and control systems. In computer graphics, eigenvalues are used in image processing and 3D modeling to understand and manipulate visual data. They are also commonly used in data analysis, signal processing, and machine learning.

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