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Not sure how to set up the Laplacian/Poisson Equation

Use NDSolve to solve a PDEWhy does NDSolveValue giving crappy results?Solution of Poisson equation with two regionsNDSolve for unsteady Taylor-Goldstein equation, can't use MethodOfLinesSolving Partial Differential Equations using NDSolveConfusion with Neumann and DirichletSystem of nonlinear PDE 2D (Reaction-Diffusion type) with periodic boundary conditionFluid Flow PDE with Mass ConservationThe rectangular region composed of two triangular regions contains a pde connecting the bc of the first and the second kindPoisson equation with pure Neumann boundary conditions

2

$begingroup$

As stated, I am having trouble trying to set up a Laplacian/Poisson Equation. I have boundary conditions with this too, and I've tried using the DirichletCondition function, but I don't know what I'm doing there either. (I have almost zero Mathematica experience, and the Wolfram site's help is just as confusing to me as the program.)

Laplacian[V[x, y], {x, y} == 0;



 V[x, 0] == 0;

 V[x, 0.05] == 1;

 V[0, y] == 0;

 V[0.1, y] == 0;]



Plot[{x, -0.25, 0.25}, {y, -0.15, 0.15}]

While I am getting that plot to appear, it's not even close to what I need. What I'm needing is the solution to appear within the region 0 ≤ x ≤ 0.1 and 0 ≤ y ≤ 0.05, as stated by the boundary conditions (rectangular). It's supposed to be a distribution type of plot, kind of like elevation contour graphs and similar.

And for now, the PDE I'm solving is equal to 0, so once I get that done, how do I set up the PDE when it's not equal to 0 (Poisson's Equation)? I would think that since Laplacian is a function, I can't use it anymore since the PDE isn't equal to 0 anymore.

Help is greatly appreciated!

differential-equations

share|improve this question

asked 3 hours ago

LtGenSpartanLtGenSpartan

132

New contributor

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$endgroup$

add a comment | 

2

$begingroup$

As stated, I am having trouble trying to set up a Laplacian/Poisson Equation. I have boundary conditions with this too, and I've tried using the DirichletCondition function, but I don't know what I'm doing there either. (I have almost zero Mathematica experience, and the Wolfram site's help is just as confusing to me as the program.)

Laplacian[V[x, y], {x, y} == 0;



 V[x, 0] == 0;

 V[x, 0.05] == 1;

 V[0, y] == 0;

 V[0.1, y] == 0;]



Plot[{x, -0.25, 0.25}, {y, -0.15, 0.15}]

While I am getting that plot to appear, it's not even close to what I need. What I'm needing is the solution to appear within the region 0 ≤ x ≤ 0.1 and 0 ≤ y ≤ 0.05, as stated by the boundary conditions (rectangular). It's supposed to be a distribution type of plot, kind of like elevation contour graphs and similar.

And for now, the PDE I'm solving is equal to 0, so once I get that done, how do I set up the PDE when it's not equal to 0 (Poisson's Equation)? I would think that since Laplacian is a function, I can't use it anymore since the PDE isn't equal to 0 anymore.

Help is greatly appreciated!

differential-equations

share|improve this question

asked 3 hours ago

LtGenSpartanLtGenSpartan

132

New contributor

LtGenSpartan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

As stated, I am having trouble trying to set up a Laplacian/Poisson Equation. I have boundary conditions with this too, and I've tried using the DirichletCondition function, but I don't know what I'm doing there either. (I have almost zero Mathematica experience, and the Wolfram site's help is just as confusing to me as the program.)

Laplacian[V[x, y], {x, y} == 0;



 V[x, 0] == 0;

 V[x, 0.05] == 1;

 V[0, y] == 0;

 V[0.1, y] == 0;]



Plot[{x, -0.25, 0.25}, {y, -0.15, 0.15}]

While I am getting that plot to appear, it's not even close to what I need. What I'm needing is the solution to appear within the region 0 ≤ x ≤ 0.1 and 0 ≤ y ≤ 0.05, as stated by the boundary conditions (rectangular). It's supposed to be a distribution type of plot, kind of like elevation contour graphs and similar.

And for now, the PDE I'm solving is equal to 0, so once I get that done, how do I set up the PDE when it's not equal to 0 (Poisson's Equation)? I would think that since Laplacian is a function, I can't use it anymore since the PDE isn't equal to 0 anymore.

Help is greatly appreciated!

differential-equations

share|improve this question

asked 3 hours ago

LtGenSpartanLtGenSpartan

132

New contributor

LtGenSpartan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

Laplacian[V[x, y], {x, y} == 0;



 V[x, 0] == 0;

 V[x, 0.05] == 1;

 V[0, y] == 0;

 V[0.1, y] == 0;]



Plot[{x, -0.25, 0.25}, {y, -0.15, 0.15}]

share|improve this question

asked 3 hours ago

LtGenSpartanLtGenSpartan

132

New contributor

LtGenSpartan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked 3 hours ago

LtGenSpartanLtGenSpartan

132

New contributor

LtGenSpartan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 hours ago

LtGenSpartanLtGenSpartan

132

New contributor

LtGenSpartan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 hours ago

LtGenSpartanLtGenSpartan

132

1 Answer
1

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oldest

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4

$begingroup$

Something like this?

PDE = D[V[x, y], x, x] + D[V[x, y], y, y];



BCs = {DirichletCondition[V[x, y] == 0, y == 0], 

   DirichletCondition[V[x, y] == 1, y == 0.05], 

   DirichletCondition[V[x, y] == 0, x == 0], 

   DirichletCondition[V[x, y] == 0, x == 0.1]};



ufun = NDSolveValue[{PDE == 0, BCs}, V, {x, 0, 0.1}, {y, 0, 0.05}];



ContourPlot[ufun[x, y], {x, 0, 0.1}, {y, 0, 0.05}] 

For Poisson equation replace PDE == 0 by PDE == f[x,y], where f[x,y] is an arbitrary function.

share|improve this answer

edited 2 hours ago

answered 3 hours ago

zhkzhk

9,32411433

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh wow, that's exactly what I needed! Also thanks for clearing up the DirichletCondition, the way you did it was much easier than what the Wolfram site had. To make sure I understand the syntax for PDE, i.e does that mean derivative of V with respect to x, and x again (to satisfy a squared partial)?
  $endgroup$
  – LtGenSpartan
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LtGenSpartan Yes!
  $endgroup$
  – zhk
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I actually have another question, is there a way to add legends, axes, etc. to this plot?
  $endgroup$
  – LtGenSpartan
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LtGenSpartan Of course you can. Check the documentations on Plot.
  $endgroup$
  – zhk
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LtGenSpartan - FrameLabel -> Automatic, PlotLegends -> Automatic
  $endgroup$
  – Bob Hanlon
  2 hours ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

Something like this?

PDE = D[V[x, y], x, x] + D[V[x, y], y, y];



BCs = {DirichletCondition[V[x, y] == 0, y == 0], 

   DirichletCondition[V[x, y] == 1, y == 0.05], 

   DirichletCondition[V[x, y] == 0, x == 0], 

   DirichletCondition[V[x, y] == 0, x == 0.1]};



ufun = NDSolveValue[{PDE == 0, BCs}, V, {x, 0, 0.1}, {y, 0, 0.05}];



ContourPlot[ufun[x, y], {x, 0, 0.1}, {y, 0, 0.05}] 

For Poisson equation replace PDE == 0 by PDE == f[x,y], where f[x,y] is an arbitrary function.

share|improve this answer

edited 2 hours ago

answered 3 hours ago

zhkzhk

9,32411433

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh wow, that's exactly what I needed! Also thanks for clearing up the DirichletCondition, the way you did it was much easier than what the Wolfram site had. To make sure I understand the syntax for PDE, i.e does that mean derivative of V with respect to x, and x again (to satisfy a squared partial)?
  $endgroup$
  – LtGenSpartan
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LtGenSpartan Yes!
  $endgroup$
  – zhk
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I actually have another question, is there a way to add legends, axes, etc. to this plot?
  $endgroup$
  – LtGenSpartan
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LtGenSpartan Of course you can. Check the documentations on Plot.
  $endgroup$
  – zhk
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LtGenSpartan - FrameLabel -> Automatic, PlotLegends -> Automatic
  $endgroup$
  – Bob Hanlon
  2 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Something like this?

PDE = D[V[x, y], x, x] + D[V[x, y], y, y];



BCs = {DirichletCondition[V[x, y] == 0, y == 0], 

   DirichletCondition[V[x, y] == 1, y == 0.05], 

   DirichletCondition[V[x, y] == 0, x == 0], 

   DirichletCondition[V[x, y] == 0, x == 0.1]};



ufun = NDSolveValue[{PDE == 0, BCs}, V, {x, 0, 0.1}, {y, 0, 0.05}];



ContourPlot[ufun[x, y], {x, 0, 0.1}, {y, 0, 0.05}] 

For Poisson equation replace PDE == 0 by PDE == f[x,y], where f[x,y] is an arbitrary function.

share|improve this answer

edited 2 hours ago

answered 3 hours ago

zhkzhk

9,32411433

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh wow, that's exactly what I needed! Also thanks for clearing up the DirichletCondition, the way you did it was much easier than what the Wolfram site had. To make sure I understand the syntax for PDE, i.e does that mean derivative of V with respect to x, and x again (to satisfy a squared partial)?
  $endgroup$
  – LtGenSpartan
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LtGenSpartan Yes!
  $endgroup$
  – zhk
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I actually have another question, is there a way to add legends, axes, etc. to this plot?
  $endgroup$
  – LtGenSpartan
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LtGenSpartan Of course you can. Check the documentations on Plot.
  $endgroup$
  – zhk
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LtGenSpartan - FrameLabel -> Automatic, PlotLegends -> Automatic
  $endgroup$
  – Bob Hanlon
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Something like this?

PDE = D[V[x, y], x, x] + D[V[x, y], y, y];



BCs = {DirichletCondition[V[x, y] == 0, y == 0], 

   DirichletCondition[V[x, y] == 1, y == 0.05], 

   DirichletCondition[V[x, y] == 0, x == 0], 

   DirichletCondition[V[x, y] == 0, x == 0.1]};



ufun = NDSolveValue[{PDE == 0, BCs}, V, {x, 0, 0.1}, {y, 0, 0.05}];



ContourPlot[ufun[x, y], {x, 0, 0.1}, {y, 0, 0.05}] 

For Poisson equation replace PDE == 0 by PDE == f[x,y], where f[x,y] is an arbitrary function.

share|improve this answer

edited 2 hours ago

answered 3 hours ago

zhkzhk

9,32411433

$endgroup$

PDE = D[V[x, y], x, x] + D[V[x, y], y, y];



BCs = {DirichletCondition[V[x, y] == 0, y == 0], 

   DirichletCondition[V[x, y] == 1, y == 0.05], 

   DirichletCondition[V[x, y] == 0, x == 0], 

   DirichletCondition[V[x, y] == 0, x == 0.1]};



ufun = NDSolveValue[{PDE == 0, BCs}, V, {x, 0, 0.1}, {y, 0, 0.05}];



ContourPlot[ufun[x, y], {x, 0, 0.1}, {y, 0, 0.05}] 

share|improve this answer

edited 2 hours ago

answered 3 hours ago

zhkzhk

9,32411433

answered 3 hours ago

zhkzhk

9,32411433

answered 3 hours ago

zhkzhk

9,32411433