i need heat transfer in sphere

[logistic_guy]

here is the question

Solve the heat equation $\displaystyle \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial r^2} + \frac{2}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial u}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 u}{\partial \phi^2}$

my attemb
i'm chemical engineering and i'm studying heat and mass transfer for spherical objects
i need to solve the heat equation and then to use solution to find the heat transfer on the surface of the sphere
it's not clear to me what to do and i think this is differential equation or something like that

 

[mario99]

i need to solve the heat equation and then to use solution to find the heat transfer on the surface of the sphere

How will you find the heat on the surface of the sphere without initial and boundary conditions?



Anyway, you can assume that there is a solution of this form:

$u(r,\theta,\phi,t) = \Psi(r,\theta,\phi) \ T(t)$

Continue.

 

[logistic_guy]

thank

How will you find the heat on the surface of the sphere without initial and boundary conditions?

i'll do my best

Anyway, you can assume that there is a solution of this form:

$u(r,\theta,\phi,t) = \Psi(r,\theta,\phi) \ T(t)$

Continue.

ok i assume, so what? it won't help me. i don't want assumption, i want to start with the correct solution

 

[mario99]

i'll do my best

What do you mean by you will do your best? I am saying that even if you were able to solve the PDE and get the general solution, you would not be able to find the heat on the surface of the sphere if there were no initial and boundary conditions!

ok i assume, so what? it won't help me. i don't want assumption, i want to start with the correct solution

How did you decide that my assumption would not lead to the correct solution when you did not even try it? Isn't the main idea of differential equations guessing a solution and trying to use it? And if it did not work, you would try something else!

All you have to do for now is just to substitute $\Psi(r,\theta,\phi) \ T(t)$ into the original PDE. For simplicity omit the independent variables and imagine that you have only $\Psi\ T$. What do you get?

FYI. Your PDE is just:

$\displaystyle \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}$

The PDE in post #1 looks scary just because it is written in spherical coordinate!

 

[khansaheb]

here is the question

Solve the heat equation $\displaystyle \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial r^2} + \frac{2}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial u}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 u}{\partial \phi^2}$

my attemb
i'm chemical engineering and i'm studying heat and mass transfer for spherical objects
i need to solve the heat equation and then to use solution to find the heat transfer on the surface of the sphere
it's not clear to me what to do and i think this is differential equation or something like that

Review the process you followed when you solved

1. temperature distribution in a rod (1 dimensional)
   
   
2. temperature distribution in a square plate (2 dimensional)
   
   
3. temperature distribution in a cube (3 dimensional)
   

This problem is very similar to the cube problem.

 

[logistic_guy]

What do you mean by you will do your best? I am saying that even if you were able to solve the PDE and get the general solution, you would not be able to find the heat on the surface of the sphere if there were no initial and boundary conditions!


How did you decide that my assumption would not lead to the correct solution when you did not even try it? Isn't the main idea of differential equations guessing a solution and trying to use it? And if it did not work, you would try something else!

All you have to do for now is just to substitute $\Psi(r,\theta,\phi) \ T(t)$ into the original PDE. For simplicity omit the independent variables and imagine that you have only $\Psi\ T$. What do you get?

FYI. Your PDE is just:

$\displaystyle \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}$

The PDE in post #1 looks scary just because it is written in spherical coordinate!

i'm confused. i don't know what to do

Review the process you followed when you solved

1. temperature distribution in a rod (1 dimensional)
   
   
2. temperature distribution in a square plate (2 dimensional)
   
   
3. temperature distribution in a cube (3 dimensional)

This problem is very similar to the cube problem.

i don't never solve this type of process. most of my heat work is on laplace equation

 

[mario99]

i'm confused. i don't know what to do

I will help you in doing the first step. Let me first write the original differential equation.

$\displaystyle \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial r^2} + \frac{2}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial u}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 u}{\partial \phi^2}$


When I plug $\Psi \ T$ in the orignal differential equation, I get this:

$\displaystyle \Psi\frac{\partial T}{\partial t} = T\left(\frac{\partial^2 \Psi}{\partial r^2} + \frac{2}{r}\frac{\partial \Psi}{\partial r} + \frac{1}{r^2}\frac{\partial^2 \Psi}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial \Psi}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 \Psi}{\partial \phi^2}\right)$

This shows that the separation of variables method works. We got (almost) two separated differential equations. And we can rearrange them so that each side has one of them.

$\displaystyle \frac{1}{T}\frac{\partial T}{\partial t} = \frac{1}{\Psi}\left(\frac{\partial^2 \Psi}{\partial r^2} + \frac{2}{r}\frac{\partial \Psi}{\partial r} + \frac{1}{r^2}\frac{\partial^2 \Psi}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial \Psi}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 \Psi}{\partial \phi^2}\right)$

Now we have two fully separated differential equations. And the set up above means that they are equal to a constant. Let us call this constant $-\lambda$.

$\displaystyle \frac{1}{T}\frac{\partial T}{\partial t} = \frac{1}{\Psi}\left(\frac{\partial^2 \Psi}{\partial r^2} + \frac{2}{r}\frac{\partial \Psi}{\partial r} + \frac{1}{r^2}\frac{\partial^2 \Psi}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial \Psi}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 \Psi}{\partial \phi^2}\right) = -\lambda$

Now we can write each differential equation alone.

$\displaystyle \frac{\partial T}{\partial t} + \lambda T = 0$

$\displaystyle \frac{\partial^2 \Psi}{\partial r^2} + \frac{2}{r}\frac{\partial \Psi}{\partial r} + \frac{1}{r^2}\frac{\partial^2 \Psi}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial \Psi}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 \Psi}{\partial \phi^2} + \Psi \lambda = 0$

The second equation is called the Helmholtz equation. There is a theorem states that if all types of the boundary conditions of the Helmholtz equation are smooth and homogenous, then the eigenvalues $\lambda \geq 0$. The proof of the theorem is not our concern. And all types of the boundary conditions meant to be Dirichlet, Neumann, and Robin boundary conditions.

At this point, all you have to do is just to repeat the process by assuming there is a solution to the Helmholtz equation of this form:

$\Psi(r,\theta,\phi) = R(r) \ Y(\theta, \phi)$

Check and see. Does separation of variables work again?

 

[logistic_guy]

I will help you in doing the first step. Let me first write the original differential equation.

$\displaystyle \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial r^2} + \frac{2}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial u}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 u}{\partial \phi^2}$


When I plug $\Psi \ T$ in the orignal differential equation, I get this:

$\displaystyle \Psi\frac{\partial T}{\partial t} = T\left(\frac{\partial^2 \Psi}{\partial r^2} + \frac{2}{r}\frac{\partial \Psi}{\partial r} + \frac{1}{r^2}\frac{\partial^2 \Psi}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial \Psi}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 \Psi}{\partial \phi^2}\right)$

This shows that the separation of variables method works. We got (almost) two separated differential equations. And we can rearrange them so that each side has one of them.

$\displaystyle \frac{1}{T}\frac{\partial T}{\partial t} = \frac{1}{\Psi}\left(\frac{\partial^2 \Psi}{\partial r^2} + \frac{2}{r}\frac{\partial \Psi}{\partial r} + \frac{1}{r^2}\frac{\partial^2 \Psi}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial \Psi}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 \Psi}{\partial \phi^2}\right)$

Now we have two fully separated differential equations. And the set up above means that they are equal to a constant. Let us call this constant $-\lambda$.

$\displaystyle \frac{1}{T}\frac{\partial T}{\partial t} = \frac{1}{\Psi}\left(\frac{\partial^2 \Psi}{\partial r^2} + \frac{2}{r}\frac{\partial \Psi}{\partial r} + \frac{1}{r^2}\frac{\partial^2 \Psi}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial \Psi}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 \Psi}{\partial \phi^2}\right) = -\lambda$

Now we can write each differential equation alone.

$\displaystyle \frac{\partial T}{\partial t} + \lambda T = 0$

$\displaystyle \frac{\partial^2 \Psi}{\partial r^2} + \frac{2}{r}\frac{\partial \Psi}{\partial r} + \frac{1}{r^2}\frac{\partial^2 \Psi}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial \Psi}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 \Psi}{\partial \phi^2} + \Psi \lambda = 0$

The second equation is called the Helmholtz equation. There is a theorem states that if all types of the boundary conditions of the Helmholtz equation are smooth and homogenous, then the eigenvalues $\lambda \geq 0$. The proof of the theorem is not our concern. And all types of the boundary conditions meant to be Dirichlet, Neumann, and Robin boundary conditions.

At this point, all you have to do is just to repeat the process by assuming there is a solution to the Helmholtz equation of this form:

$\Psi(r,\theta,\phi) = R(r) \ Y(\theta, \phi)$

Check and see. Does separation of variables work again?

thank

At this point, all you have to do is just to repeat the process by assuming there is a solution to the Helmholtz equation of this form:

$\Psi(r,\theta,\phi) = R(r) \ Y(\theta, \phi)$

Check and see. Does separation of variables work again?

you mean i remove $\displaystyle \Psi$ and place $\displaystyle RY$?

like this?

$\displaystyle \frac{\partial^2 RY}{\partial r^2} + \frac{2}{r}\frac{\partial RY}{\partial r} + \frac{1}{r^2}\frac{\partial^2 RY}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial RY}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 RY}{\partial \phi^2} + RY \lambda = 0$

 

[khansaheb]

thank


you mean i remove $\displaystyle \Psi$ and place $\displaystyle RY$?

like this?

$\displaystyle \frac{\partial^2 RY}{\partial r^2} + \frac{2}{r}\frac{\partial RY}{\partial r} + \frac{1}{r^2}\frac{\partial^2 RY}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial RY}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 RY}{\partial \phi^2} + RY \lambda = 0$

YES....

What is the next step .....

continue till you are explicitly stuck ....​

let us know where you are stuck....​

 

[logistic_guy]

YES....

What is the next step .....

continue till you are explicitly stuck ....​

let us know where you are stuck....​

i'm stuck. what should i do with this equation

 

[mario99]

$\displaystyle \frac{\partial^2 RY}{\partial r^2} + \frac{2}{r}\frac{\partial RY}{\partial r} + \frac{1}{r^2}\frac{\partial^2 RY}{\partial \theta^2} + \frac{\cos \theta}{r^2\sin \theta}\frac{\partial RY}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 RY}{\partial \phi^2} + RY \lambda = 0$

This step was good. But you did not completely follow post #7.

The idea is to have two separated equations, one on the left and one on the right. If you could do that, you would let them equal to a constant, say $\eta$.

You have to analyze term by term, for example the first terms is:

$\displaystyle \frac{\partial^2 RY}{\partial r^2}$

The variable in the denominator is $r$. Which one of the two functions, $R \ \text{and } \ Y$ is a function of $r$? From post #7, we know that it is $R(r)$, so you keep it and the other you pull it out like this:

$\displaystyle Y\frac{\partial^2 R}{\partial r^2}$

This means that the first term treats $Y$ as a constant and treats $R$ as a variable. If you keep doing this to all terms, you will be able to recognize the two separated equations.