Properly posedness is really quite unique to partial differential equations. Ordinary differential equations can be hard to solve if they involve very different time scales. For example, that is an issue in many chemical reactions.
But for partial differential equations, “hard to solve” becomes “impossible to solve.” That happens even for apparently very simple linear partial differential equations with constant coefficients.
This section has a look at some of the issues involved.
In words, a properly posed problem in partial differential equations can be described as follows: it is a problem that has a unique solution that is physically reasonable.
Phrased more mathematically, a problem is properly posed if:
Requirement 3 above is the one that makes the solution physically reasonable. Physically, nothing is exactly known. There are always some errors in the data, however accurate they may be. If these negligible errors can produce a significant change in the solution, then all bets are off that the solution obtained is the right one.
You may wonder what “correspondingly small” in condition 3 really means. The true answer is that it varies. However, generally it is taken to mean that changes in the solution are no more than proportional to the changes in the data that cause these changes. And there must be some overall upper bound to the constant of proportionality that is independent of the details of the change in data.
That still requires that suitable measures of the magnitude of the changes in data and solution are defined. That however is beyond this discussion.
One thing should be emphasized. It is not partial differential equations that are properly or improperly posed. It is problems that are properly or improperly posed. Before you know what boundary and initial conditions are specified for your partial differential equation, you cannot say anything meaningful about properly posedness.
The following subsections give a few typical examples of how improperly posed problems arise. It illustrates that if you try to solve some partial differential equation numerically, you better know what sort of equation it is. Or you can get into major problems.
Show that the Dirichlet boundary-value problem for the Poisson equation on a finite domain,
Assuming that the Dirichlet boundary-value problem for the Laplace equation on a finite domain,
Repeat the previous two questions for the Dirichlet initial / boundary value problem for the heat equation,
This subsection will discuss an improperly posed problem involving the heat equation. Recall that the heat equation is an example of a parabolic equation.
Consider first a very standard properly posed problem for the heat equation. The problem is heat conduction in a bar. The unknown is the temperature. The ends of the bar are kept at zero temperature.
The below figure shows some computed temperature profiles in a bar at various times.
At the initial time the initial condition was assumed to be piecewise linear. There is then a singularity, a kink, at the center of the bar. A kink corresponds to a jump in the derivative. But the heat equation smooths away singularities. At any later time, even as small as 0.02 in the nondimensional units used in this problem, the temperature profile is perfectly smooth, with all derivatives continuous.
This problem was properly posed. Improperly posedness arises for the
“backward heat equation”. Physically the backward heat
equation is the heat equation solved backwards in time.
Mathematically, the backward heat equation takes the form
Note that the backward heat equation is equivalent to solving the normal heat equation forwards in time, but with a negative heat conduction coefficient. A negative heat conduction coefficient violates the second law of thermodynamics, so it is not really surprising that you get into trouble with the mathematics.
Suppose that you take the temperature profile at time in the figure above as the initial condition for the backward heat equation. Then you compute the solution of the backward heat equation up to . That should give you back the singular temperature profile at time zero. And it will, if you manage to do it exactly. The backward heat equation has a unique solution for the chosen initial condition in the interval from to .
But now suppose that you use the singular profile at time zero as the initial condition for the backward heat equation. Then you try again to compute the solution of the backward heat equation up to . It will not work. You will not be able to find a solution for any value of greater than zero. The reason is easy to understand. Suppose that you did find a solution at . Physically that would be a temperature distribution in the bar at time . But if a temperature distribution at time existed, even a singular one, then the temperature distribution would be smooth for all times greater than . The heat equation smooths away singularities. But the solution at times greater than is not nonsingular, because we know it is singular at time zero. So a solution at time can simply not exist.
Now let’s return to the problem that did have a unique solution. That was when we started the backward heat equation solution from the smooth temperature profile at . There is still a major physical problem with the solution. Physically, (and also in typical numerical solutions), the initial profile will not be exact to infinitely many digits at all locations. There will always some error. Suppose in particular that the actual profile has a very slight kink. It can be so tiny that you cannot see the error in the profile under a microscope if you plot it versus the exact one. But if there is a kink, there is no longer a solution to the backward wave equation. There cannot be a solution at earlier times if there is a singularity, however invisible the kink may be.
In physics your data, here being the temperature profile, are never truly exact. So you have no way of saying which one is the right answer, the unique solution or a complete lack of any solution at all. If the data have any imperfection, it is the latter.
The bottom line is that even though the backward heat equation can have unique solutions for some problems, these solutions are only meaningful if you have a problem that is mathematically exact. So the initial/boundary-value problem for the backward heat equation is improperly posed. Even though it may have unique solutions.
The pure initial-value problem is similarly improperly posed.
A typical improperly posed problem for the Laplace equation is shown in figure 1.4. Physically, it might correspond to heat conduction in a rectangular plate. For mathematical convenience, the horizontal size of the plate has been rescaled to length .
What is wrong in figure 1.4 is that both the temperature and the heat flow are specified at the lower boundary . That is wrong because the Laplace equation needs exactly one boundary condition at each point of the boundary, not two. It is also wrong that no boundary condition at all is given on the top boundary .
Mathematically speaking, you might say that figure 1.4 is an initial / boundary-value problem for the Laplace equation, with playing the part of time. And initial / boundary-value problems for the Laplace equation are not allowed. Note however that figure 1.4 would be perfectly fine if the partial differential equation was instead of . That would be a wave equation with a unit wave velocity, and initial/boundary-value problems are just what you want for wave equations. In short, a single sign in the partial differential equation makes all the difference.
The primary problem with initial/boundary-value problems for the Laplace equation is that they do not meet the third requirement for properly posedness. The effect of small changes in the data on the solution can be much larger than the small changes. For the example figure 1.4, that should be taken to mean that the solution inside the plate can be much larger than the given value of at the lower boundary.
That can be seen as follows. Consider the following type of solution:
Now note that the “data”, the values of , are no greater than 1 in magnitude. On the other hand, in the interior of the plate, can reach values up to . For any given value of that is a finite number. But there is no universal bound to it. You can make as large as you want by just taking large enough. The value of is only about 1.5, but is already about . In short, the size of can exceed the size of by any arbitrarily large factor.
Show that the given solution
Check the separation of variables solution to the above problem,
Can you immediately see that this separation of variables solution is probably no good?
Plot the separation of variables solution of the previous question for an example. Assume that the given boundary condition is a triangular profile:
Comment on whether a solution exists at and for .
This example should illustrate that typical improperly posed problems might have solutions if the data are perfectly smooth and their Taylor series have finite radii of convergence. But if there is a singularity, like the kink in the triangular profile, all bets are off.
You might know that if you talk about instability of ordinary differential equations, you wonder about what happens to the solution for infinite time. But in this problem you do not let the “time” coordinate go to infinity. The problem is not large , but large “wave number” . The large wave number problem is really unique to partial differential equations. (If you had a system of infinitely many ordinary differential equations, you might also run into it.)
Continuing the previous question, show analytically that for the supposed solution
Also show analytically that at the halfway point , the values that you get while summing increase monotonically to infinity.
Show that the Laplace equation
Explain the lack of solution in physical terms. To do so, consider this a steady heat conduction problem, with the temperature, and the gradient of the scaled heat flux.
Generalize the derivation to determine the requirement that
Show that if the Poisson equation
A typical improperly posed problem for the wave equation is shown in figure 1.5. Physically, it might correspond to transverse vibrations of a string over a finite time interval. For mathematical convenience, the length of the string has been rescaled to length . Also, the time has been rescaled to eliminate the wave speed from the wave equation. The scaled final time has been written as , again for mathematical convenience, but the value of can be anything.
What is wrong in figure 1.5 is that instead of specifying the initial position and velocity of the string, the initial and final position of the string are given. That is wrong because the wave equation is an evolution equation. It requires initial conditions, not final conditions.
Mathematically speaking, you might say that figure 1.5 is an boundary-value problem for the wave equation, with playing the part of a spatial coordinate. Boundary-value problems for the wave equation are not allowed. Note however that figure 1.5 would be perfectly fine if the partial differential equation was instead of . That would be the Laplace equation, and boundary-value problems are just what you want for the Laplace equation. In short, a single sign in the partial differential equation makes all the difference.
The given problem has special solutions of the form
The reason for the fact that figure 1.5 produces an
improperly posed problem depends on the value of . Consider first
the possibility that is a rational number. A rational number is a
number that can be written as
For such a rational , the solution to the problem figure 1.5 is not unique. The quickest way to see that is to take the function in the given problem zero. Then one solution to the problem is obviously . So any nonzero solution means that the solution is not unique. And you get a nonzero solution by taking or any whole multiple of in the special solutions given above.
Since the solution is not unique, the problem violates the second condition for properly posed problems.
Not all numbers are rational numbers, however. In fact, in some sense
there are infinitely many more irrational numbers than rational ones.
One simple example is . Irrational numbers can however be
approximated to arbitrary accuracy by rational ones. For example,
consider to 10 digits accuracy:
The same way, any irrational value is arbitrarily close to rational ones, and for rational ones the problem figure 1.5 is improperly posed. So surely you would not expect the boundary value problem to be properly posed for an irrational value. What happens in this case is that the criterion 3 for properly posedness is violated. Consider again the special solutions above. In the interior of the rectangle, the solution clearly has magnitude 1. (Or something comparable to 1, if you want to use an average magnitude as measure.) The data however have magnitude . So, if you can find values for that make arbitrarily small, you have shown that criterion 3 is violated. The magnitude of would be much larger than the magnitude of . Try solving the corresponding homework question only if you are really good in math.
Finally, you might argue that initial/boundary-value problems are not really required for the wave equation. Since appears in the wave equation exactly like does, surely you should be able to provide two “initial conditions” at instead of . And provide one “boundary condition” at and one “boundary condition” at . Physically that is not an initial/boundary-value problem; it might be called an initial/final/single-doubled-boundary-value problem.
You have a good point there. The stated problem is indeed properly
posed, as you say. However, that trick only works in
one-spatial dimension. For the wave equation in two spatial
Show that the given solution
How about twice that solution? Ten times? How about if ? How about if ? So how many solutions are there really to this single problem?
For the brave. Show without peeking at the solution that the problem for irrational is improperly posed by showing that you can make