Energy methods derive some sort of system “energy” from a partial differential equation. That energy may then be used to derive such things as existence and/or uniqueness of the solution, and whether it depends continuously on the data.
Consider the following very general Poisson equation problem:
The objective in this subsection is to show that solutions to this problem are unique as long as and do not have opposite sign. Assuming that the problem has a solution in the first place, there is only one.
To prove that, it must be proved that any two solutions and of the same problem cannot be different. That they must be the same. In other words, it must be proved that the difference between any two solutions and is zero.
The proof starts with deriving the equations satisfied by the
difference . Since by assumption and are both
solutions of the same problem:
If you subtract these two problems from each other, and replace
by , you get
The difference between two solutions of a problem always satisfies the homogeneous problem.
Now we must prove that is zero. That will mean that the difference between and is zero. And that will in turn imply that and must be the same; different solutions are not possible.
To prove that is zero, the trick is to multiply the Laplace equation
for by and integrate over the entire domain :
Now consider some special cases, starting with the simplest one:
So must be zero everywhere, and that means that must equal some constant everywhere. That includes that must be on the boundary. But on the boundary, so must be zero. So is everywhere zero. So the difference between any two solutions and of the original problem must be everywhere zero. That means that the solutions must be equal, different solutions are not possible. So the solution is unique.
So solutions of the Neumann problem are not unique. However, different solutions differ only by a constant, nothing more than that.
(1.20) |
To do that, note that in that case the final integral must again be zero. The final integral cannot be negative because it is an integral of a square. And the final integral cannot be positive; otherwise the first term in the sum above would have to be negative to make their sum zero. And the first term cannot be negative because it too is an integral of a square.
Since the second integral is zero, must again be some constant . And since is nonzero at at least some point, (otherwise it would be the pure Neumann problem), it follows from the boundary condition that the constant is zero. So like the Dirichlet problem, the solution is unique.
In that case, the first, surface, integral in (1.19) is a sum over all these parts. As long as you can show that the combined integral is zero or negative, must be zero and the solution is unique. (Except that in the pure Neumann case, the solution is indeterminate by a constant.)
(1.21) |
Show that the Poisson equation
Using the arguments given in the text, uniqueness can not be shown for the Poisson equation
For the heat equation, similar arguments can be made as for the Laplace equation. This subsection briefly indicates the general lines.
Like for the Laplace equation in the previous subsection, the
difference between any two solutions of a heat equation
problem must satisfy the homogenous problem. That problem is here
Multiply the partial differential equation by and
integrate like for the Laplace equation to get
(1.22) |
Now consider conditions like those for the Laplace equation; Dirichlet
or Neumann boundary conditions, or mixed boundary boundary conditions
where and have the same sign. For those the final two terms
cannot by negative. So the first term cannot be positive. So the
“energy integral”
If the difference between two solutions is always zero, different solutions are not possible. Solutions are unique.
Under the same conditions on the boundary conditions, you can see that one condition for properly posedness is satisfied: small changes in the initial conditions produce small changes in the solution. To see this, allow a nonzero initial condition for in the arguments above. The energy integral for at a later time is still never larger than the energy integral of the initial condition for . And the energy integral is a measure for the magnitude of : if you divide the energy integral by half the volume of the domain and take a square root, it gives the root mean square value of .
For the wave equation, similar arguments can be made as for the Laplace and heat equations. This subsection briefly indicates the general lines.
Like in the previous two subsections, the difference
between any two solutions of a wave equation problem must satisfy the
homogenous problem. That problem is here
Multiply by and integrate over as before using
the divergence theorem to get
(1.23) |
Now consider conditions like those for the Laplace equation; Dirichlet
or Neumann boundary conditions, or mixed boundary boundary conditions
where and have the same sign. Under such conditions the surface
integral term cannot be negative. So the time derivative of the
“energy integral”
If the difference between two solutions is always zero, different solutions are not possible. Solutions are unique.
It may be noted that physically, the first term in the energy equation is typically kinetic energy, and the second potential energy. If you generalize the problem for to still have homogeneous boundary conditions, but inhomogeneous initial conditions, you can derive energy conservation. In particular for homogeneous Dirichlet or Neumann boundary conditions, the total energy is preserved. For mixed boundary conditions where and have the same sign, the energy can only decrease. If and have opposite sign, the problem is unstable in the sense that the energy will increase. You can also use this to show that small changes in the initial conditions produce small changes in the solution for appropriate boundary conditions.