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[?] Solve a 2D heat equation inside a circular domain?

I am trying to solve the Heat Equation in 2D for a circular domain and I used the example attached, however, for some reason I do not get any answer from it, and in principle, it seems that I am following the same steps as in the original document from wolfram tutorials. Any help will be much appreciated. I am using version 11.1.1

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POSTED BY: David Quesada
6 Replies

If the region is becoming irregular, like the one, in the picture attached, is there a way in Mathematica to create irregular shapes by composing, subtracting images?enter image description here

POSTED BY: David Quesada

I am trying to solve the heat equation inside and outside a region, boundary conditions are based on exchange of heat across the boundary separating both regions. I see NDSolveValue accepts BVP for a single function as well as a description of the region for that specific function, how could I implement the solution with NDSolveValue inside a circle and outside and to match both solutions at the boundary with Mathematica?

POSTED BY: David Quesada
Posted 7 years ago

NDSolve would return a rule which could be used like this. But in this case you are using NDSolveValue, which returns the solution as an Interpolating Function. So this is for direct use. This has not provided a rule for u, or even a value for u. All it has done is assign a value to bvpdisk, which is the interpolating function.

POSTED BY: David Keith

thank you once again, got it.

POSTED BY: David Quesada
Posted 7 years ago

Your NDSolveValue assigns the solution to bvpdisk, but then you plot u[t,x,y}.

Plot3D[bvpdisk[0.2, x, y], {x, y} \[Element] omega, Mesh -> All]

enter image description here

POSTED BY: David Keith

thank you so much, it worked now, however raised then another question, let say I use Plot3D[Evaluate[u[t,x,y] /. bvpdisk], ....] it should give a similar answer also, and for some reason it does not ,,, could you please help me to understand the reason?

POSTED BY: David Quesada
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