Seperaion of Variables (PDE's)

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In summary, the conversation discusses finding a separable solution for T(r,θ) and deriving equations for R(r) and Θ(θ) using a separation constant that results in sinusoidal solutions for Θ(θ). The conversation also mentions showing that the equation for R(r) has solutions of the form R(r)=r^p and finding the general solution for Θ(θ). The solution for Θ(θ) is found to be Asinkθ+Bcoskθ using k^2 as the separation constant. The conversation also discusses the process of showing that R(r)=r^p satisfies the differential equation r^2 R''+rR'=k^2 R.
  • #1
twoscoops
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Homework Statement


dsdsd.jpg

Look for a seperable solution T(r,θ) = R(r)Θ(θ) and derive equations for R(r) and Θ(θ) choosing a separation constant that gives sinusoidal solutions for Θ(θ). Write down a general solution for Θ(θ) and show the equation for R(r) has solutions of the form R(r)=r^p.

Homework Equations


p is just a random letter which has no real meaning.

The Attempt at a Solution


I can get the solution of Θ(θ)=Asinkθ+Bcoskθ using k^2 as the separation constant. That leaves me with r2R''+ rR' = k2R but I am not sure what solution that gives, likewise for the general solution for Θ(θ). Any help will be much appreciated. Thanks
 
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  • #2
twoscoops said:

Homework Statement


dsdsd.jpg

Look for a seperable solution T(r,θ) = R(r)Θ(θ) and derive equations for R(r) and Θ(θ) choosing a separation constant that gives sinusoidal solutions for Θ(θ). Write down a general solution for Θ(θ) and show the equation for R(r) has solutions of the form R(r)=r^p.

When you are required to "show" a function satisfies a DE, then just substitute that function into the DE and see if it satisfies it. So you're asked to "show" that R(r)=r^p satisfies the DE:

[tex]r^2 R''+rR'=k^2 R[/tex]

so when you substitute R(r)=r^p into that, what must the relationship between p and k be so that it satisfies the DE?
 
  • #3
jackmell said:
When you are required to "show" a function satisfies a DE, then just substitute that function into the DE and see if it satisfies it. So you're asked to "show" that R(r)=r^p satisfies the DE:

[tex]r^2 R''+rR'=k^2 R[/tex]

so when you substitute R(r)=r^p into that, what must the relationship between p and k be so that it satisfies the DE?

[itex]r^2 R''+rR' - k^2 R= 0[/itex] is Euler equation and solution is [itex]r^n \hbox { or } r^p[/itex] what ever which way you want to call it.
 
  • #4
okay thanks for clearing that up, but what about the general solution for Θ(θ)? still not sure on that...
 
  • #5
You said you already found it in your original post. What specifically are you stuck on?
 

What is Separation of Variables in PDEs?

Separation of variables is a method used to solve partial differential equations (PDEs) by breaking down the solution into simpler parts that can be solved separately.

How does Separation of Variables work?

The method of Separation of Variables involves assuming that the solution to the PDE can be written as a product of two functions, each of which depends on only one variable. The PDE is then rewritten as a set of ordinary differential equations (ODEs) involving each of these functions. These ODEs can then be solved separately and the solutions are combined to obtain the final solution to the PDE.

What types of PDEs can be solved using Separation of Variables?

Separation of Variables can be used to solve linear, homogeneous PDEs with constant coefficients. It is most commonly used for second-order PDEs, but can also be applied to higher-order PDEs with some modifications.

What are the advantages of using Separation of Variables to solve PDEs?

One of the main advantages of Separation of Variables is that it can be used to solve a wide range of PDEs, including some that cannot be solved using other methods. It also provides a systematic approach to solving PDEs and can often lead to simple and elegant solutions.

Are there any limitations to using Separation of Variables for PDEs?

Although Separation of Variables can be a powerful method for solving PDEs, it does have some limitations. It can only be used for linear, homogeneous PDEs with constant coefficients, and it may not always be possible to find a solution using this method. In addition, the solutions obtained using Separation of Variables may not be valid for all values of the variables, and may need to be checked for consistency.

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