1st order PDE, seems easy but still confusing

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    Confusing Pde
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Discussion Overview

The discussion revolves around a first-order partial differential equation (PDE) related to a scalar field q(x,y,t) in a physics context. Participants explore potential solutions and initial conditions, specifically addressing how to satisfy the conditions q(x,y,t=0) = 1 and q(x,y,t=0) = delta(x,y).

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant presents a PDE and two tentative solutions, questioning their ability to satisfy the initial conditions.
  • Another participant points out the absence of 't' in the proposed solutions, suggesting a possible substitution error with 's'.
  • A later reply indicates that the original poster has resolved their confusion by employing the Method of Characteristics.
  • Another participant suggests substituting q(x,y,t) with a product of functions and separating the equation into two ordinary differential equations (ODEs) to find a general solution.
  • One participant proposes a specific general solution involving an arbitrary function C(x,y) and provides a form of the solution that satisfies the delta function initial condition.

Areas of Agreement / Disagreement

The discussion includes multiple competing views on how to approach the PDE and its solutions. There is no consensus on the best method or the correctness of the proposed solutions.

Contextual Notes

Participants express uncertainty regarding the initial conditions and the form of the solutions. The dependence on the parameter 'a' and the nature of the arbitrary function C(x,y) are also noted as potential limitations.

Who May Find This Useful

Readers interested in partial differential equations, methods of characteristics, and initial value problems in physics may find this discussion relevant.

Irid
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Hello,
I am doing some physics and I end up with this PDE:

\frac{\partial q(x,y,t)}{\partial t} = -(x^2 + y^2)q(x,y,t) + ax\frac{\partial q(x,y,t)}{\partial y}
where q(x,y,t) is the scalar field to determine and a is a parameter. I need to consider two types of initial conditions: q(x,y,t=0) = 1; and q(x,y,t=0) = delta(x,y).

I have found two tentative solutions:

q(x,y,t) = C\exp \left(-\lambda s + \frac{y}{ax}(x^2 + y^2/3 - \lambda) \right)
where lambda is any (?) number. Another solution is

q(x,y,t) = C\exp \left(-sx^2 + -y^3/3ax\right)

They both seem to satisfy the PDE, but I can't make them satisfy the required initial condition (either 1 or delta function). Any ideas or experience with this kind of equations?
 
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I don't see a 't' in either solution. Did you substitute it with s?
 
jaytech said:
I don't see a 't' in either solution. Did you substitute it with s?

Oops, sorry, I did.
Actually I think I figured out how to find the solution. It's by the Method of Characteristics. No more help needed...
 
Substitute q(x,y,t)=X(x)Y(y)T(t) and treat x and X(x) as constants and then try to separate the equation to two ordinary ones.Then solve those two ODEs.The general answer is the linear combination of the answers to the ODEs but the coefficients will depend on x.
 
The general solution to your PDE is

q(x,y,t) =C(x,axt+y)\exp(\frac{xy}{a}+\frac{y^3}{3ax}),

where C(x,y) is an arbitrary function.

So

q(x,y,t) =δ(x,axt+y)\exp(-\frac{t(a^2t^2x^2+3atxy+3x^2+3y^2)}{3})

is the solution with

q(x,y,0) =δ(x,y).
 

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