1st order PDE, seems easy but still confusing

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The discussion centers on solving the first-order partial differential equation (PDE) given by \(\frac{\partial q(x,y,t)}{\partial t} = -(x^2 + y^2)q(x,y,t) + ax\frac{\partial q(x,y,t)}{\partial y}\). Two initial conditions are considered: \(q(x,y,t=0) = 1\) and \(q(x,y,t=0) = \delta(x,y)\). The participant initially proposed two solutions but struggled to satisfy the initial conditions. Ultimately, the Method of Characteristics was identified as the effective approach, leading to the general solution \(q(x,y,t) = C(x,axt+y)\exp(\frac{xy}{a}+\frac{y^3}{3ax})\), which satisfies the delta function initial condition.

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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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