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

In summary, the conversation discusses how to find the function L when given the first order PDE: $ (\frac{\partial L}{\partial x})^2 - (\frac{\partial L}{\partial y})^2 = -1$. The author initially wrote L = y + ax^2 + .. but after further discussion, it was determined that L = ax + by + c is a solution, as long as a^2 - b^2 = -1. The conversation also mentions the possibility of other solutions, but due to a lack of expertise in PDEs, it is uncertain.
  • #1
PhyAmateur
105
2
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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  • #2
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.
 
  • #3
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?
 
  • #4
I can't edit the title :(, I would really appreciate if anyone from the mentors could edit my title!
 
  • #5
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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FAQ: Solving 1st Order PDE: Finding L with Ax^2 Form

What is a first-order partial differential equation (PDE)?

A first-order partial differential equation (PDE) is a mathematical equation that involves derivatives of a function with respect to multiple independent variables. It is called "first-order" because the highest derivative involved is of first order.

What is the "Ax^2" form of a first-order PDE?

The "Ax^2" form of a first-order PDE refers to a specific form of the equation where the coefficient of the highest derivative term is a function of only one of the independent variables, typically denoted as x. This form is useful for solving certain types of first-order PDEs.

How do you determine the value of L in solving a first-order PDE using the Ax^2 form?

In order to determine the value of L, you must first substitute the given values of the independent variables into the equation in Ax^2 form. This will result in a simpler equation with only one unknown variable, which can then be solved for L.

Can the Ax^2 form be used to solve all first-order PDEs?

No, the Ax^2 form is only applicable to certain types of first-order PDEs, specifically those where the coefficient of the highest derivative term is a function of only one independent variable. Other forms and methods may be needed to solve other types of first-order PDEs.

What are some practical applications of solving first-order PDEs using the Ax^2 form?

The Ax^2 form is commonly used in fields such as physics and engineering to model and solve problems involving heat transfer, fluid flow, and other physical phenomena. It can also be applied in economics and finance to analyze and predict market trends and behaviors.

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