Solving Partial Differential Equations with UsubX and Variable Transformation

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Homework Help Overview

The discussion revolves around solving a partial differential equation (PDE) of the form x^2*Usubx - Usuby + U = 0. Participants explore methods such as variable transformation and separation of variables (SOV) to find solutions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the effectiveness of variable transformation and SOV in finding general solutions. Questions arise about the nature of solutions obtained through these methods and whether they represent the most general solutions. There is also consideration of the implications of potential failures of the SOV method.

Discussion Status

Some participants have made progress in understanding the relationship between the solutions found through different methods. There is acknowledgment of the need to clarify the conditions under which SOV may not yield the most general solution. The discussion reflects a mix of exploration and realization of conceptual nuances in solving PDEs.

Contextual Notes

Participants note the importance of boundary conditions in determining the values of constants in the solutions. There is also mention of the distinction between general solutions and specific forms derived from separation of variables.

schattenjaeger
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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?
 
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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?
 
it's late but I see what I may've done, I didn't go through and solve the resulting ODEs setting the separation parameter 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:
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.
 
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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
 

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