Solving 1st Order PDE: Finding L with Ax^2 Form

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    First order Pde
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Discussion Overview

The discussion revolves around solving a first-order partial differential equation (PDE) of the form $$ (\frac{\partial L}{\partial x})^2 - (\frac{\partial L}{\partial y})^2 = -1$$. Participants are exploring how to derive the function L, particularly in relation to a proposed solution involving a quadratic term in x.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Homework-related

Main Points Raised

  • One participant questions the correctness of the PDE's classification as first-order, suggesting it might be a second-order PDE instead, based on the title.
  • Another participant proposes that for the equation as stated, a linear form $$L = ax + by + c$$ could be a solution, provided that $$a^2 - b^2 = -1$$.
  • There is uncertainty regarding the proposed solution $$L = y + ax^2 + ..$$, with participants expressing confusion about how this form satisfies the original equation.
  • A participant reflects on their reasoning, suggesting that since the difference of the squares of the first partials is a constant, L should be first-degree in both x and y.
  • Some participants express a desire for clarification and assistance in editing the thread title to better reflect the discussion topic.

Areas of Agreement / Disagreement

Participants do not appear to reach consensus on the classification of the PDE or the validity of the proposed solutions, indicating multiple competing views remain.

Contextual Notes

There is a lack of clarity regarding the assumptions underlying the proposed solutions, particularly concerning the form of L and the classification of the PDE. The discussion also highlights potential confusion about the relationship between first-order and second-order PDEs.

PhyAmateur
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How to find L if the form is:

$$ (\frac{\partial L}{\partial x})^2 - (\frac{\partial L}{\partial y})^2 = -1$$

The author wrote, $$L = y + ax^2 + ..$$
but I didn't get how?
 
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PhyAmateur said:
How to find L if the form is:

$$ (\frac{\partial L}{\partial x})^2 - (\frac{\partial L}{\partial y})^2 = -1$$

The author wrote, $$L = y + ax^2 + ..$$
but I didn't get how?
This is actually a first order PDE, since both partials are first partials. Since you have "second order" in the title, I'm wondering if you have written the equation correctly.

Did you mean this equation?
$$\frac{\partial^2 L}{\partial x^2} - \frac{\partial^2 L}{\partial y^2} = -1 $$

For the equation as you wrote it, L = ax + by + c is a solution, with a2 - b2 = -1. I don't see how L = y + ax2 could be a solution.
 
Oh I have written the title wrongly! I will edit it right away. Can you please tell me how did you get L=ax+by+c?
 
I can't edit the title :(, I would really appreciate if anyone from the mentors could edit my title!
 
PhyAmateur said:
Oh I have written the title wrongly! I will edit it right away. Can you please tell me how did you get L=ax+by+c?
My approach was pretty simpleminded - it's been many years since I had a class on PDEs. My reasoning was that since the difference of the squares of the first partials was a constant (-1), then L(x, y) must be first-degree in both x and y, so that L(x, y) = ax + by + c.

You can check that this is a solution, provided that a2 - b2 = -1. If there are other solutions, my recollections from the class I had are too vague, and maybe someone else can weigh in.

PhyAmateur said:
I can't edit the title :(, I would really appreciate if anyone from the mentors could edit my title!
Done
 
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