Partial DE-separation of variables

In summary: If you have a negative constant that is negative then the left side will be positive and the right side will be negative. So you would get two more equations.
  • #1
emily600
2
0
Hi I'm having a bit of trouble with this question:

Use separation of variables to find all the possible separable solutions to the partial DE equation for u(x,y) given by
yux - 3x2 uy = 0
.I try u= X(x) Y(y)

ux = X'(x) Y(y)

uy = X(x) Y'
(y)


which gives y(X' Y)-3x2(X Y')

then I divide by (XY) and rearrange it into y[X'/X] = 3x2 [Y'/Y]

which is equal to the separationconstant K (I think)

y[X'/X] = 3x2 [Y'/Y] = Kand from here I get a bit lost. I try a positive constant, so I get two equations.

yX'/X = + K rearranged into {X'= XK/y}

3x2Y/Y = + K rearranged into {Y'= YK/3x2 }


and then I try a negative constant with more or less the same equations resulting, but I'm not sure what to do with them?

Is this how you find all the possible solutions of the pde (trying both a positive and a negative constant)?

I'm told I can check the solution by substitution but I'm not sure how. Any help would be really appreciated.

Thanks:confused::confused:.
 
Physics news on Phys.org
  • #2
emily600 said:
Hi I'm having a bit of trouble with this question:

Use separation of variables to find all the possible separable solutions to the partial DE equation for u(x,y) given by
yux - 3x2 uy = 0
.I try u= X(x) Y(y)

ux = X'(x) Y(y)

uy = X(x) Y'
(y)


which gives y(X' Y)-3x2(X Y')

then I divide by (XY) and rearrange it into y[X'/X] = 3x2 [Y'/Y]

which is equal to the separationconstant K (I think)

y[X'/X] = 3x2 [Y'/Y] = K
You would do better to rearrange it as $\mathbf{\dfrac{X'}{x^2X}} = \mathbf{\dfrac{3Y'}{yY}}$. The left side is then purely a function of $\mathbf{x}$ and the right side is then purely a function of $\mathbf{y}.$ You are then justified in putting both sides equal to a separation constant, and you have two separate ODEs for $\mathbf{x}$ and $\mathbf{y}$.
 
  • #3
Thanks that simplified things a little. For the positive constant I solved the 2 ODE's as:

View attachment 1591

And then put them together. If I do the same for a negative constant would that cover the possible solutions (assuming I can rearrange and solve them properly)?

Thanks:D
 

Attachments

  • ode.png
    ode.png
    8.7 KB · Views: 69

FAQ: Partial DE-separation of variables

What is partial DE-separation of variables?

Partial DE-separation of variables is a technique used to solve partial differential equations (PDEs). It involves breaking down a multi-variable PDE into several single-variable ordinary differential equations (ODEs) by assuming that the solution can be expressed as a product of separate functions of each variable.

Why is partial DE-separation of variables useful?

Partial DE-separation of variables is useful because it allows us to solve complex PDEs by breaking them down into simpler ODEs. This makes the solution process more manageable and allows us to apply well-known techniques for solving ODEs.

What types of PDEs can be solved using partial DE-separation of variables?

Partial DE-separation of variables can be used to solve linear, homogeneous PDEs with constant coefficients. Some examples include the heat equation, wave equation, and Laplace's equation.

What are the steps involved in partial DE-separation of variables?

The first step in partial DE-separation of variables is to write the PDE in its standard form. Then, we assume that the solution can be expressed as a product of separate functions of each variable. Next, we substitute this assumption into the PDE and separate the variables. This results in a system of ODEs, which can then be solved using standard techniques. Finally, we combine the solutions to obtain the complete solution to the PDE.

What are the limitations of partial DE-separation of variables?

Partial DE-separation of variables can only be used to solve certain types of PDEs, specifically linear, homogeneous PDEs with constant coefficients. It also may not always provide a complete solution, as some PDEs may have solutions that cannot be expressed in the assumed form. Additionally, the process can become quite complex for higher-order PDEs or those with non-constant coefficients.

Similar threads

Replies
3
Views
3K
Replies
3
Views
1K
Replies
6
Views
2K
Replies
4
Views
2K
Replies
3
Views
2K
Replies
11
Views
2K
Replies
15
Views
2K
Back
Top