I assumed they were, but this problem seems to suggest otherwise. In spherical coordinates we are to solve (grad^2) T = 0, which has an easy general solution of T = -c_1/r + c_2. However, the prescribed boundary conditions require that dT/dr = B at r = 0, and T(a) = T_a. This requires that c_1 = B*r^2.
I was under the impression that constants of integration are never variables. Nevertheless, I see no other way to satisfy the boundary conditions. Any help?
I don’t think that’s a monomial serving as a constant; to me, it just looks like the solution at the boundary condition.
It’s been a long time since vector calculus, but isn’t gradient an operation on a scalar field that gives a vector field as its result? Then grad^2 would be grad(grad(scalar field)) which is trying to operate gradient on a vector field.
Or does grad^2 T mean the divergence of the gradient of T?
I likely confused my nomenclature. What is meant is merely the second derivative operation, which I now recall is called the del operator, here operated in terms of spherical coordinates; in this case I ignored the dependence on the the phi and theta variables and concentrated on the r, or radius variable.
Taking the equation I provided to be true, what do you Dopers say of the validity of obtained constant of integration.
When calculating an indefinite integral (or solving a differential equation) you get a family of functions for answer. Each function from the family has a different value for the integration constant.
When you calculate a definite integral (or impose boundary conditions to your differential equation) you choose one particular function from the family, the one whose integration constant has a particular value.
Boundary conditions are exactly the things that tell you what the “constants of integration” are equal to.
If I tell you that:
dy/dx = 2x
you can tell me that
y = x[sup]2[/sup] + c
where c is any value that doesn’t depend on y or x (since all such values are valid solution to the original differential equation). But, if I give you a boundary condition also:
dy/dx = 2x, and y(0)=1
then only one value for c works, and the solution is:
y=x[sup]2[/sup] + 1 .
Generally, for every differentiation that you undo by an integration, youl introduce one unknown quantity. But, for every boundary condition you are given, you can constrain one unknown. So if you have a second-order differential equation and two boundary conditions, you will not have any unknown “constants” at the end.
Based on the OP, the question does seem to be about the Laplacian in spherical coordinates (for which the solution is C1/r+C2 when there is point symmetry). The differential equation would be (1/r^2)d(r^2dT/dr)/dr=0, not d^2(T)/dr^2=0.
I don’t think you can have del^2(T)=0 and dT/dr=B at r=0, unless B is zero.
If the question is related to the steady state temperature distribution in a sphere of radius “a” when heat is generated at the center at a certain rate and surface is at temperature T_a, then consider a sphere with a tiny cavity in the cetner of radius e. The boundary conditions would then be T(a)=T_a and the boundary condition at r=e, would be 4pie^2*dT/dr = B, where B would be related to thermal conductivity of the sphere and how fast heat is being generated at the center. Then take limit as e approaches zero. The factor of e^2, which comes from the surface area of the tiny cavity takes care of problems with r=0 in the denominator. In this case, del^2(T) would be zero everywhere except at r=0.
Thanks, this answered my question great. The problem is in fact related to the temperature gradient; however, this is merely an academic exercise so I likely don’t need to perform the limit you described - seems the answer is merely the condition is only met where B = 0.
Since the question’s answered I think jokes are permitted.
Two mathematicians are discussing public innumeracy when one needs to go to the lavatory. The other hurriedly tips the cocktail waitress and asks her to memorize “x cubed over three.” The other guy comes back and they arrange a $20 bet; they call the waitress over. “What’s the integral of x squared?” She replies “x cubed over three.” The guy wins $20.
As she’s walking away the waitress looks over her shoulder, sneering at the guy like he’s a moron. “Plus a constant!” she says.
Ahh, wonderful. I hereby sanction a formal thread hijack, where all those who partake must contribute their math jokes. Seeing that we are trying to build something here, I require that these jokes be entirely original.
What is the most mystical property of infinity?
It’s over nine-thousand!