- To prove the divergence theorem, it was shown first that the spacial partial derivative in the
volume integral could be exchanged for multiplication by an appropriate component of the
exterior normal. This problem starts with the divergence theorem and goes the other direction.
Assuming the divergence theorem, holds for a region V , show that ∫
_{∂V }nudA = ∫_{V }∇udV . Note this implies ∫_{V }dV = ∫_{∂V }n_{1}udA. - Fick’s law for diffusion states the flux of a diffusing species, J is proportional to the gradient of the concentration, c. Write this law getting the sign right for the constant of proportionality and derive an equation similar to the heat equation for the concentration, c. Typically, c is the concentration of some sort of pollutant or a chemical.
- Sometimes people consider diffusion in materials which are not homogeneous. This means that
J = −K∇c where K is a 3 × 3 matrix. Thus in terms of components, J
_{i}= −∑_{j}K_{ij}. Here c is the concentration which means the amount of pollutant or whatever is diffusing in a volume is obtained by integrating c over the volume. Derive a formula for a nonhomogeneous model of diffusion based on the above. - Let V be such that the divergence theorem holds. Show that ∫
_{V }∇⋅dV = ∫_{∂V }udA where n is the exterior normal anddenotes the directional derivative of v in the direction n. - Let V be such that the divergence theorem holds. Show that
where n is the exterior normal and

is defined in Problem 4. - Let V be a ball and suppose ∇
^{2}u = f in V while u = g on ∂V . Show that there is at most one solution to this boundary value problem which is C^{2}in V and continuous on V with its boundary. Hint: You might consider w = u−v where u and v are solutions to the problem. Then use the result of Problem 4 and the identity w∇^{2}w = ∇⋅−∇w ⋅∇w to conclude ∇w = 0. Then show this implies w must be a constant by considering h= wand showing h is a constant. Alternatively, you might consider the maximum principle. - Show that ∫
_{∂V }∇×v ⋅ ndA = 0 where V is a region for which the divergence theorem holds and v is a C^{2}vector field. - Let F=be a vector field in ℝ
^{3}and let V be a three dimensional shape and let n =. Show that ∫_{∂V }dA = 3× volume of V . - Let F = xi + yj + zk and let V denote the tetrahedron formed by the planes, x = 0,y = 0,z = 0, and
x +y +z = 1. Verify the divergence theorem for this example.
- Suppose f : U → ℝ is continuous where U is some open set and for all B ⊆ U where B is a ball,
∫
_{B}fdV = 0. Show that this implies f= 0 for all x ∈ U. - Let U denote the box centered at with sides parallel to the coordinate planes which has width 4, length 2 and height 3. Find the flux integral ∫
_{∂U}F ⋅ ndS where F = (x + 3,2y,3z). Hint: If you like, you might want to use the divergence theorem. - Find the flux out of the cylinder whose base is x
^{2}+ y^{2}≤ 1 which has height 2 of the vector field F =. - Find the flux out of the ball of radius 4 centered at 0 of the vector field F = .
- Verify (17.16) from (17.13) and the assumption that S = kF.
- Show that if u
_{k},k = 1,2,,n each satisfies (17.7) with f = 0 then for any choice of constants c_{1},,c_{n}, so does ∑_{k=1}^{n}c_{k}u_{k}. - Suppose k= k, a constant and f = 0. Then in one dimension, the heat equation is of the form u
_{t}= αu_{xx}. Show that u= e^{−αn2t }sinsatisfies the heat equation^{3}. - Let U be a three dimensional region for which the divergence theorem holds. Show that
∫
_{U}∇× Fdx = ∫_{∂U}n × FdS where n is the unit outer normal. - In a linear, viscous, incompressible fluid, the Cauchy stress is of the form
where p is the pressure, δ

_{ij}equals 0 if i≠j and 1 if i = j, and the comma followed by an index indicates the partial derivative with respect to that variable and v is the velocity. Thus v_{i,j}=. Also, p denotes the pressure. Show, using the balance of mass equation that incompressible implies div v = 0. Next show that the balance of momentum equation requiresThis is the famous Navier Stokes equation for incompressible viscous linear fluids. There are still open questions related to this equation, one of which is worth $1,000,000 at this time.

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