I refer to the first post. The cap is not meant to be a hemisphere. In normalization, length dimensions were re-scaled by means of ##R## (the radius of the sphere from which the spherical cap is derived), this means that ##p##, the depth of the cap below the surface, can vary from 0 to 1 (0 = no...
I try to upload an image of the surface of the semi-infinite solid. The spherical cap is isotherm while the rest of the surface is adiabatic. The semi-infinite solid is beneath the surface. I apologize for the misunderstandings.
Returning to the spherical cap, in a spherical coordinate system centered at the "center" of the cap, the PDE is:
##\frac{\partial }{\partial r}\left(r^2\frac{\partial T}{\partial r}\right)+\frac{1}{\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial T}{\partial...
I apologize for the bad exposition of the problem. I will try to explain my thoughts with another example.
Let's suppose there is a heat point source of power ##q## in an infinite solid initially at temperature ##T_0##. This source melts part of the solid forming a spherical melt pool; the...
On the surface of a semi-infinite solid, a point heat source releases a power ##q##; apart from this, the surface of the solid is adiabatic. The heat melts the solid so that a molten pool forms and grows. Let's hypothesize that the pool temperature is homogeneously equal to the melting...
Dear Office_Shredder, you are right. Both functions go to zero as ##x->0##, or ##y->0##, or both go to zero. Then, it is obvious that, under the same condition, their difference goes to zero as well.
Actually, I was looking for a way to solve this integral:
##\int_0^{\infty } \frac{\exp...
Thank for your message.
Let's call ##f1## the function:
##
\pmb{\text{f1}=\frac{\text{Exp}\left[-\frac{1 }{2 \epsilon ^2} \left(\frac{\left(x-\epsilon ^2 t \right)^2+y^2}{t +1}\right)\right]}{\sqrt{t} (t +1)}}\\##
And ##f2## the function:
##\pmb{\text{f2}=\frac{\text{Exp}\left[-\frac{1 }{2...
You are full right! It was my fault. Unfortunately, I cannot write formulae correctly in this forum, so I did a picture of my integrals, but the resolution was very bad.
The correct integrals are:
and
.
I lost the minus in the conversion process.
You can interpret x and y as related by the...