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Laplacian in Spherical Shell

by urista
Tags: laplacian, shell, spherical
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urista
#1
May2-04, 04:28 AM
P: 11
I have a problem on my homework that is really confusing. I need to solve the partial differential equation in a spherical shell with inner radius = a and outer radius=b: (Laplacian u)=1 in spherical coordinates. The boundary conditions are u=0 on the inner radius r=a, and du/dr=0 on outer radius r=b. The part that is really confusing is that the (Laplacian u)=1 not zero which doesn't make since to me physically. I can't use seperation of variables because the PDE is inhomogeneous. I am really confused, would someone help me please?
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arildno
#2
May2-04, 06:56 AM
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1. This seems to me as a linearized "displacement equation", i.e. something gotten from stationary elasticity theory, where maximal displacement occurs at radius b, no displacement at r=a.

2. That the Laplacian to a function is non-zero may very well be physically meaningful;
for example for stationary flow of a viscous fluid are governed by such equations
(It all depends on the problem at hand)

3. Are you sure this is a PARTIAL differential equation in u???
The boundary conditions given are solely dependent on the radius value; no conditions for u's behaviour has been given at specific angle values!!
Hence, I believe you should treat u=u(r), i.e an ordinary differential equation.
(I am quite certain that u has the physical interpretation of being radial displacement).

4. What you now need to solve the problem, is the expression of the Laplacian operator in spherical coordinates where the differentiations of u with respect to the angular variables should be set to zero.

Good luck!
urista
#3
May2-04, 12:02 PM
P: 11
Thank you; it became evident that the boundary conditions are independent of the angles which means it reduces to an ODE in r. I appreciate your help and let you know what I get for a result.

urista
#4
May2-04, 01:07 PM
P: 11
Laplacian in Spherical Shell

OK, I got an answer u=b^3/(3*r)-(a^3+2*b^3)/(6*a)+r^2/6 which was completely valid to satisfy the ODE and the boundary conditions. The second part of the question is asking to let the inner radius "a" go to zero in the above solution and to interpret the result. Obviously the limit doesn't exist as a goes to zero, so I went back to the general solution u=c1/r+c2+r^2/6 before applying the boundary conditions and eliminated the constant c1 in order to bound the solution. The problem is that the boundary condition on the outer radius has become invalid du/dr=0 on r=b because I tried applying it and it was meaningless. What went wrong? Is it because physically now we have a solid sphere and it doesn't make sense to have a constant displacement on the outside without fixing the center?
arildno
#5
May2-04, 01:41 PM
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From what I can see the original Laplacian problem is:
1/r^(2)*d/dr(r^(2)du/dr)=1 (Right?)

This gives:
r^(2)du/dr=1/3r^(3)+C1,
or
du/dr=1/3*r+C1/r^(2).

By regularity demand, we must have du/dr<inf-->C1=0

And we gain:
u(r)=1/6r^(2)+C2

I think we are in full agreement up to this point.
However, the introduction of the regularity demand at r=0 has made our problem have 3 boundary conditions rather than 2! One of the original ones have to go..

Let us examine the condition du/dr=0?
What does this imply physically?
I suggest that this means, that we are to have maximal displacement at r=b.
This is then in accordance with the other original boundary equation u(a)=0,
that is, no displacement is to occur at r=a.
(For example, we might think that for a not equal to 0, the model is with a core (0<r<a) that is much more rigid than the exterior layer (a<r<b))

With the interpretation that the "real" boundary condition at r=b is that maximal displacement is to be found there , the problem is resolved by requiring u(0)=0; i.e C2=0
The maximal displacement will still be found at r=b, since we now have:
u(r)=1/6r^(2)

This is how I interpret it, anyways..
urista
#6
May2-04, 02:08 PM
P: 11
so with u(r)=1/6 r^2 we'll get du/dr=1/3 r, how could this be equal to zero at r=b?
arildno
#7
May2-04, 02:24 PM
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It isn't, because when the regularity demand at a=0 appears, one of the two original boundary conditions have to go,
You cannot fulfill all 3: (regularity at r=0, u(0)=0, du/dr=0 at r=b)
In particular, note that the requirement du/dr=0 at r=b CANNOT be fulfilled when the regularity demand appears, hence, you have to let it go.

The integral of the Laplacian yields:
du/dr=1/3r+C1/r^(2).
You cannot fullfill the condition at r=b if C1 is equal to 0
urista
#8
May2-04, 02:33 PM
P: 11
Thank you arilando, that was great help. Do you think you might have time later on to give me some input on a 1D wave equation? I won't bother you now!
urista
#9
May2-04, 05:57 PM
P: 11
I have a wave equation Ytt=c^2 Yxx - g where g is a constant. The boundary conditions are Y(0,t)=Y(L,t)=0 with initial conditions Y(x,0)=0 and Yt(x,0)=0 I tried to solve it by Laplace transfoming the PDE in time and everything worked fine until I got to the point where I had to inverse the transform but things got ugly. Obviously, I have a nonhomogenous PDE with homogeneous boundary conditions. I was going to expand everything in terms of the related eigenfunctions sin(n Pi x/L)but it's not right to expand the constant g in terms of eigenfunctions. I can't do seperation of variables because the PDE is inhomogeneous. What's the trick here to get me started without using integral transforms?


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