Solving Partial Differential Equations with UsubX and Variable Transformation

  • Thread starter schattenjaeger
  • Start date
  • Tags
    Partial
In summary, the problem is meant to teach us that the SOV method might not always give the most general solution, but if you use it and get a solution that works, it is valid.
  • #1
schattenjaeger
178
0
UsubX = means partial of u with respect to x, for my halfasses notation...

so I have x^2*Usubx - Usuby + U = 0, I'm sposed to solve by a variable transformation thingy, and by separation of variables. Doing the transformation I get U=const*e^(1/x) which is a solution, by my own checking. Yay

Now at this point I noted if that's the general solution(which I think that method finds)then there IS no separated solution, and indeed despite my careful efforts whatever I get using SOV doesn't check out right. So assuming I'm right and didn't just muck it up, is this problem meant to teach us the potential failing of the SOV method? If so is there anyway to tell earlier it won't work besides doing it and checking?

And a semi-related question for clarification, if you use SOV(and get a solution that works) the solution is not guaranteed to be the most general solution, rather the most general separated solution?
 
Physics news on Phys.org
  • #2
Whoops, impressive what a minus sign will do, eh?

so I did get a more general solution, cool. So doing it by the variable transformation does NOT get you the most general solution? Or something?
 
  • #3
it's late but I see what I may've done, I didn't go through and solve the resulting ODEs setting the separation paramater thing equal to 0, a negative const. and a positive const.

if upon doing that I see that I get some manner of trivial or no solutions for two cases, and one case that ends up as the e^(1/x) solution, I shall be very much reassured about how all this works. If not, do explain

Edit: But my answer I did get using SOV worked when I checked it. So whatever, I'll figure it out later
 
Last edited:
  • #4
schattenjaeger said:
UsubX = means partial of u with respect to x, for my halfasses notation...

so I have x^2*Usubx - Usuby + U = 0, I'm sposed to solve by a variable transformation thingy, and by separation of variables. Doing the transformation I get U=const*e^(1/x) which is a solution, by my own checking. Yay

Now at this point I noted if that's the general solution(which I think that method finds)then there IS no separated solution, and indeed despite my careful efforts whatever I get using SOV doesn't check out right. So assuming I'm right and didn't just muck it up, is this problem meant to teach us the potential failing of the SOV method? If so is there anyway to tell earlier it won't work besides doing it and checking?

And a semi-related question for clarification, if you use SOV(and get a solution that works) the solution is not guaranteed to be the most general solution, rather the most general separated solution?

Of course, [itex]U= Ce^{-\frac{1}{x}}[/itex] is NOT the most general slolution. Remember that the most general solutions to partial differential equations have involve functions as the "constants" of integration. Since you haven't shown exactly what you have done, it is hard to say what you are doing. Separating U= X(x)Y(y) gives the two equations [itex]X'= \frac{\lambdaX}{x^2}[/itex] ( where [itex]\lambda[/itex] is an arbitrary constant), which has general solution [itex]X(x)= Ce^{\frac{-\lambda}{x}}[/itex]and [itex]Y'= \lambdaY+ 1[/itex] which has solution [itex]Y(x)= Ce^{(1+\lambda)y}[/itex].

The general "separated" solution is of the form [itex]U(x,y)= Ce^{\frac{-\lambda}{x}}e^{(1+\lambda)y}[/itex].

What values [itex]\lambda[/itex] can take depends on the boundary conditions.

Of course, not every solution to the original equation is of that form- but every solution can be written as a (possibly infinite) linear combination of them.
 
Last edited by a moderator:
  • #5
ok, thanks! I had a lightbulb over head moment in class today. I forgot that when you integrate partial diff. eqs that you don't get a constant, whether a function of the other variable, and in doing so it all works out happily ever after. And stuff
 

1. What are partial differential equations?

Partial differential equations (PDEs) are mathematical equations that involve multiple independent variables and their partial derivatives. They are used to describe complex systems in various fields such as physics, engineering, and economics.

2. What is the purpose of using UsubX and variable transformation to solve PDEs?

UsubX and variable transformation are techniques used to simplify and solve PDEs. They involve substituting the original variables with new ones and transforming the equation into a simpler form that is easier to solve.

3. How does UsubX work in solving PDEs?

UsubX involves substituting a new variable u for one of the original variables in the PDE. This new variable is chosen in a way that simplifies the equation and makes it easier to solve. The transformed equation is then solved using standard techniques, and the solution is then converted back to the original variables.

4. What are the benefits of using variable transformation in solving PDEs?

Variable transformation allows for the simplification of complex PDEs, making them easier to solve. It also helps to find a general solution that can be applied to a range of similar problems. Additionally, it can reveal symmetries and patterns in the solution that may not have been apparent in the original equation.

5. Are there any limitations to using UsubX and variable transformation in solving PDEs?

UsubX and variable transformation may not always work for every PDE. The choice of the new variable and the transformation itself can be challenging and require advanced mathematical knowledge. Additionally, the transformed equation may be more difficult to interpret physically, making it necessary to convert back to the original variables to gain insight into the problem.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
833
  • Calculus and Beyond Homework Help
Replies
1
Views
442
  • Calculus and Beyond Homework Help
Replies
11
Views
742
  • Calculus and Beyond Homework Help
Replies
6
Views
389
Replies
5
Views
966
  • Calculus and Beyond Homework Help
Replies
5
Views
267
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
3K
  • Calculus and Beyond Homework Help
Replies
2
Views
967
  • Calculus and Beyond Homework Help
Replies
3
Views
968
Back
Top