Laplace and Divergence theorem

In summary, the conversation discusses using the Divergence theorem to determine an alternate formula for the integral of u multiplied by the Laplacian of u, and then using this formula to prove the uniqueness of solutions for Laplace's equation. The plan is to show that if u is given on the boundary, then u can be represented as v-w, where v and w are arbitrary vectors, and then prove that v and w must be equal, thus proving the uniqueness of u. However, there may be a mistake in the textbook and further clarification is needed regarding the given boundary conditions for u.
  • #1
draco193
7
0

Homework Statement



Use Divergence theorem to determine an alternate formula for [tex]\int\int u \nabla^2 u dx dy dz[/tex] Then use this to prove laplaces equation [tex]\nabla^2 u = 0[/tex] is unique. u is given on the boundary.

Homework Equations



[tex]u \nabla^2 u = \nabla * (u \nabla u) -(\nabla u)^2[/tex]

The Attempt at a Solution



Using Divergence theorem, I get that the new equation should be [tex]\oint (u \nabla u) *n -\oint (\nabla u)^2[/tex] where n is the normal vector.

I wanted to make sure that I had applied this correctly before moving onto the next part.

My plan for the next part would then be to say that u=v-w, where v and w are arbitrary vectors, and show that v =w to show uniqueness of u.
 
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  • #2
Purely as a TeXnical note: The dot product operator can be written \cdot, and you can get double and triple integral symbols which are properly spaced using \iint and \iiint. Also, differentials in integrals look better if you put a thin space \, before them. So your first equation might be set [tex]\iint u \, \nabla^2 u \,dx \,dy \,dz [/tex].
 
  • #3
draco193 said:
Use Divergence theorem to determine an alternate formula for [tex]\int\int u \nabla^2 u dx dy dz[/tex] Then use this to prove laplaces equation [tex]\nabla^2 u = 0[/tex] is unique. u is given on the boundary.

Please post the original problem statement word for word... If you only specify [itex]u[/itex] on the boundary of a finite region, the solution to [itex]\nabla^2 u = 0[/itex] is not unique.

Consider a simple counter example: [itex]u_1=1[/itex] and [itex] u_2=\frac{r}{R}[/itex]...both satisfy Laplace's equation, and both have the same value on the boundary of the sphere [itex]r \leq R[/itex].
 
  • #4
@ystael: Thanks. The formatting of the Latex is new stuff to me. :smile:

@gabba: That is the exact problem. I think the idea is to prove that [tex]u=v-w=0[/tex], and then by the min max principle, u is the only solution.

As an update, the professor said there was a mistake in the book, so we could just treat it as a double integral (not a double with respect to three variables). I also realized that there is no reason to apply divergence to the [tex](\nabla u)^2[/tex] term, so my answer for the first part should read

[tex]\oint (u\nabla u)\cdot n-\iint (\nabla u)^2[/tex]
 
  • #5
draco193 said:
@gabba: That is the exact problem.

You can't prove something that isn't true. Specifying [itex]u[/itex] alone on the boundary will not generate a unique solution (Laplace's equation is a 2nd order DE, and hence requires at least two boundary conditions to be uniquely solved for a given region). If the problem also tells you that [itex]\mathbf{\nabla}u[/itex] is specified on the boundary, then you can use the hint by assuming [itex]u_1[/itex] and [itex]u_2[/itex] are both solutions that satisfy the same boundary conditions and then looking at [itex]u_3 \equiv u_2 - u_1[/itex].

Is this problem from a textbook? If so, what text and what is the problem number?
 
  • #6
gabbagabbahey said:
You can't prove something that isn't true. Specifying [itex]u[/itex] alone on the boundary will not generate a unique solution (Laplace's equation is a 2nd order DE, and hence requires at least two boundary conditions to be uniquely solved for a given region). If the problem also tells you that [itex]\mathbf{\nabla}u[/itex] is specified on the boundary, then you can use the hint by assuming [itex]u_1[/itex] and [itex]u_2[/itex] are both solutions that satisfy the same boundary conditions and then looking at [itex]u_3 \equiv u_2 - u_1[/itex].

Is this problem from a textbook? If so, what text and what is the problem number?

It is a textbook problem. Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 4th Edition Haberman #2.5.12 a and b.
 

Related to Laplace and Divergence theorem

What is Laplace's theorem?

Laplace's theorem, also known as the divergence theorem or Gauss's theorem, states that the flux of a vector field through a closed surface is equal to the volume integral of the divergence of the vector field over the region enclosed by the surface.

What is the significance of Laplace's theorem?

Laplace's theorem is significant because it provides a relationship between the surface and volume integrals of a vector field, allowing for the calculation of one from the other. It is also a fundamental tool in the study of vector calculus and has many applications in physics and engineering.

What is the difference between Laplace's theorem and the divergence theorem?

Laplace's theorem and the divergence theorem are two names for the same mathematical concept. They both refer to the relationship between the surface and volume integrals of a vector field, but they are named after different mathematicians who contributed to its development.

How is Laplace's theorem used in real-world applications?

Laplace's theorem has many applications in physics and engineering, including fluid dynamics, electromagnetism, and heat transfer. It is used to calculate flux through surfaces, which is important in understanding the flow of fluids and the distribution of heat and electrical charge.

What are some common misconceptions about Laplace's theorem?

One common misconception is that Laplace's theorem only applies to three-dimensional vector fields. In reality, it can be extended to any number of dimensions. Another misconception is that it only applies to closed surfaces, when in fact it can also be applied to open surfaces with some modifications. Additionally, Laplace's theorem is sometimes confused with Laplace's equation, which is a separate mathematical concept.

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