Engine on the net that will solve PDEs

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Discussion Overview

The discussion revolves around finding an online engine capable of solving partial differential equations (PDEs), specifically a linear first-order PDE. Participants explore the nature of the PDE, initial conditions, and the validity of solutions provided by an external tool.

Discussion Character

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

Main Points Raised

  • One participant inquires about online engines for solving PDEs and specifies interest in a linear first-order PDE.
  • Another participant requests the specific form of the PDE to provide assistance.
  • A participant presents the PDE in question and the initial condition z=2 at the point (3,1).
  • Discussion includes the suggestion to solve the auxiliary equation related to the PDE, though uncertainty exists about incorporating the initial condition.
  • A participant shares a solution received from an online tool, questioning its readability and validity.
  • Concerns are raised about the appropriateness of the initial condition being a single point rather than a curve or line.
  • Further inquiry is made regarding how to determine the arbitrary function in the solution if the initial condition is indeed flawed.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the validity of the solution provided by the online engine and the appropriateness of the initial conditions. Multiple competing views about the nature of the initial condition and its implications for the solution remain unresolved.

Contextual Notes

There is a lack of consensus on whether the given initial condition is sufficient for solving the PDE, as well as on how to handle the arbitrary function in the solution.

coverband
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Hi

Does anybody know if there is an engine on the net that will solve PDEs.

I'm looking for the solution of a linear first order PDE

Thanks
 
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Could you post the form of the PDE?
 


yzx+xzy+z=y z=2 @ (3,1)
 


Is it not that we first solve the auxiliary equation ?
[tex]\frac{dx}{y} = \frac{dy}{x} = \frac{dz}{y - z}[/tex]

Let the two solutions be u=c1 and v=c2. Then the general solution is u=f(v).

But not sure how to include the given initial condition z=2 @ (3,1) .
 


Thank you for your input. I will consider this
 


I had sent the following to convode@riemann.physmath.fundp.ac.be

NMAX:=3;
SMAX:=4;
LPARTI:={};
ARGSTOP:=2;
DEPEND Z,X;
DEPEND Z,Y;
L2:={Z};
L3:={X,Y};
L4:={2,{3,1}};
L5:={};
L1:={Y*DF(Z,X)+X*DF(Z,Y)+Z-Y=0};
CONVODE(L1,L2,L3,L4,L5);

and received the reply for the solution as follows (if I retype it correctly)
[tex]z = (2x+2y)^{-1} (2\sqrt{- x^2 +y^2}f(- x^2 + y^2)+(y^2-x^2)\log(\frac{x+y}{\sqrt{-x^2 + y^2}})+x^2 +xy)[/tex]

where f is an arbitrary function.
Can we believe the solution given?
 


Thanks

Convode sends it back in a weird asss format in the email. Difficult to read no !?

I think the question might be faulty or at least the initial conditions. I don't know is it solvable by just being given the value of z at a POINT rather than along a curve/line
 
Last edited:


Never mind about that messy result given by convode. But it do gave something.

Let say there is a mistake giving a point. A curve or a line given instead as you said. How do we determine that arbitrary function?
 

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