PDE Existence/Uniqueness Question

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In summary, if you know the function and the value of σ at a given point, you can find V in principle.
  • #1
thegreenlaser
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My equation is:

[tex] \left(\mathbf{\nabla}\sigma\right)\cdot\left(\mathbf{\nabla}V\right) + \sigma\nabla^2V = 0[/tex]

If I'm given V(r) on the boundary of some volume, and I know σ(r) inside the volume, is there a unique solution V(r) inside that volume for any arbitrary (well-behaved) function σ(r)?

I suspect the answer is yes, but I've never taken a formal PDE class, so I wanted to double-check.

Edit: just so it's clear, I don't need to know how to solve for V, I just need to know that it's possible to find V in principle.
 
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  • #2
thegreenlaser said:
My equation is:

[tex] \left(\mathbf{\nabla}\sigma\right)\cdot\left(\mathbf{\nabla}V\right) + \sigma\nabla^2V = 0[/tex]

If I'm given V(r) on the boundary of some volume, and I know σ(r) inside the volume, is there a unique solution V(r) inside that volume for any arbitrary (well-behaved) function σ(r)?

I suspect the answer is yes, but I've never taken a formal PDE class, so I wanted to double-check.

Edit: just so it's clear, I don't need to know how to solve for V, I just need to know that it's possible to find V in principle.

You have [tex]
\left(\mathbf{\nabla}\sigma\right)\cdot\left(\mathbf{\nabla}V\right) + \sigma\nabla^2V
= \nabla \cdot ( \sigma \nabla V) = 0
[/tex] which if [itex]\sigma[/itex] is given is a linear PDE for [itex]V[/itex]. Hence if [itex]V_1[/itex] and [itex]V_2[/itex] satisfy this PDE in a volume [itex]U[/itex] with [itex]V_1 = V_2[/itex] on [itex]\partial U[/itex], then [itex]V = V_1 - V_2[/itex] satisfies the PDE with [itex]V = 0[/itex] on [itex]\partial U[/itex].

Hence [tex]
0 = \int_U V \nabla \cdot (\sigma \nabla V)\,dU =
\int_U \nabla \cdot( V \sigma \nabla V) - \sigma (\nabla V) \cdot (\nabla V)\,dU \\
= \int_{\partial U} V \sigma \frac{\partial V}{\partial n}\,dS - \int_U \sigma \|\nabla V\|^2\,dU
[/tex] and as [itex]V[/itex] vanishes on [itex]\partial U[/itex] we have [tex]
\int_U \sigma \|\nabla V\|^2\,dU = 0.
[/tex] Hence if [itex]\sigma[/itex] is everywhere strictly positive (or strictly negative) we can conclude that [itex]\|\nabla V\|^2 = 0[/itex] everywhere in [itex]U[/itex], from which it follows that [itex]V[/itex] is constant on [itex]U[/itex], and in view of the boundary conditions [itex]V = 0[/itex]. Hence [itex]V_1 = V_2[/itex].

It follows that if [itex]\sigma[/itex] is everywhere strictly positive or everywhere strictly negative and a solution exists, then that solution is unique.
 
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  • #3
Perfect, thanks!

I forgot, I do actually know that [itex]\sigma \geq 0 [/itex], and I could probably restrict it to [itex]\sigma > 0 [/itex].
 

1. What is the PDE Existence/Uniqueness Question?

The PDE Existence/Uniqueness Question is a fundamental question in the field of partial differential equations (PDEs) that seeks to determine whether a given PDE has a unique solution or if multiple solutions exist. This question is important in understanding the behavior and properties of PDEs, which are used to model a wide range of physical phenomena.

2. Why is the PDE Existence/Uniqueness Question important?

The PDE Existence/Uniqueness Question is important because it helps us determine the well-posedness of PDEs, which is a measure of the stability and accuracy of their solutions. A well-posed PDE has a unique solution that is continuous and depends continuously on the initial/boundary conditions and the parameters of the equation. This is crucial for making accurate predictions and conclusions based on PDE models.

3. What are some techniques used to prove PDE existence/uniqueness?

There are several techniques used to prove the existence and uniqueness of solutions to PDEs. These include the method of characteristics, separation of variables, and energy methods. Other approaches such as the maximum principle, the method of continuation, and the Galerkin method can also be used depending on the specific type of PDE and its properties.

4. Can a PDE have multiple solutions?

Yes, a PDE can have multiple solutions. This typically occurs when the PDE is not well-posed, meaning it lacks the necessary conditions for a unique solution. In this case, there may be infinitely many solutions or a finite number of solutions to the PDE. The existence of multiple solutions can also be caused by the presence of symmetries in the PDE, leading to different solutions that satisfy the same set of boundary/initial conditions.

5. Are there any open problems related to the PDE Existence/Uniqueness Question?

Yes, there are still many open problems related to the PDE Existence/Uniqueness Question. These include the existence and uniqueness of solutions to certain types of nonlinear PDEs, the well-posedness of PDEs in infinite dimensions, and the behavior of solutions to PDEs with singularities. These open problems continue to be an active area of research in the field of PDEs.

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