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Example of a Dirichlet Problem



Special thanks to Professor A. David Wunsch

A Dirichlet problem deals with functions that are harmonic and analytic within a domain.

Definitions:

Complex Number – Denoted as z in this document, a complex number has a real part and an imaginary part. The real part is denoted x and the imaginary part is denoted y. For example, 3+i4 is drawn in the complex plane as a point that is 3 over on the x-axis and 4 up on the y-axis. The y-axis is known as the imaginary axis and the x-axis is known as the real axis. See the complex variables document on this web site for more.

Domain – [simple definition] – As it applies here a domain is an area that is unbounded within the complex plane. For example, [pic] states that the real part of Z (the x in x+iy) is less than 3. That domain would include all points in the complex x-y plane where x is less than three (the domain will NOT include the border line where x=3.

Analyticity in a Domain – Given a domain as specified above, a function is analytic in that domain if at every point in the domain, a derivative exists, not only at the point in question, but at every other point around that point. [synonyms: holomorphic, regular]

Harmonic – Functions that satisfy LaPlace’s equation in a domain are said to be harmonic in that domain. Here is LaPlace’s equation for the function f([pic] where [pic] is complex.

Laplace’s Equation:

[pic]

Example of a harmonic function:

[pic]

Some facts about Harmonic Functions –

If a harmonic function is analytic in a domain, then its real and imaginary parts are both analytic in the domain as well.

If you have a real harmonic function in a domain, there exists an imaginary harmonic function that is related to this real function. This imaginary function is called the harmonic conjugate.

In the real (yet idealized) world, examples of harmonic function include heat conduction within a medium and also water flow and electrostatic charge within a medium.

Dirichlet Defined Through Example

To put a more physical spin on things, here is a typical situation where Dirichlet comes into play. This is just a description – our solved example will be a different situation.

The goal is to find a function that is harmonic within a domain when all that is known is the value of the function on the boundary of a domain. For example, imagine a long cylinder with a radius of 1 inch filled with a heat conducting material. By painstakingly measuring the temperature on the outside of this cylinder, you have determined that you can specify the temperature at any point by using the following formula:

[pic]

The radius is always one on the surface of the cylinder and since it is a long cylinder, in the ideal world, the temperature is the same along the whole cylinder at each value of the angle theta.

In the real non-idealized world, such problems are solved using numerical analysis on the computer, and are approximations. For this paper, we will use an easier surface and assume ideal conditions. The method will simplify data manipulations, but will not sully the point behind this problem.

Preparations for Our Simplified Dirichlet Example

In order to solve a Dirichlet problem, we need to derive the Poisson Integral Formula, which we will do in this section. This is an extension of the Cauchy Integral Formula. Once derived, we will be able to use it to determine the value of the function at every point within a bounded domain, provided we know the value on the boundary. (See the complex variable documentation on this web site for a discussion of the Cauchy Integral Theorem. It will merely be stated here below.

Cauchy Integral Theorem

[pic]

We are working in the w complex plane, and Z is a point inside a circle of radius R.

[pic]

We define Z1 as follows:

[pic]

Z1 = X1+iY1

Z bar (in the denominator) is the conjugate of Z, or X1-iY1.

The magnitude of Z is the same as the magnitude of Z bar. ([pic])

The argument ([pic]is the same for both Z and Z1 and with Z1 defined as it is, the magnitude is indeed greater than R, so it is outside the circle. To show this more clearly,

[pic]

Since [pic]

So, if we use Z1 in the Cauchy Integral Formula, the singularity is outside the circle, the function is analytic within and on the boundary, and the Cauchy-Goursat theorem tells us that the value of the formula is zero.

[pic]

To get a relationship between both Z and Z1, we subtract both Cauchy integrals

[pic]

Simplifying

[pic]

• Note the bar over the Z in the two denominators.

We integrate around the circular contour, so it seems it would be wise to convert the above equation to polar coordinates. The following transformation formulas are utilized.

For any value ’w’ on the boundary R: [pic]

For any value within the boundary, we’ll let [pic]

So the conjugate of Z is equal to [pic]

The derivative of w is [pic] so [pic]

[pic]

Simplifying

[pic]

Multiply the denominator

[pic]

Multiply both Numerator and Denominator by –r/R

[pic]

Multiply both Numerator and Denominator by [pic]

[pic]

Making use of the identity

[pic]

[pic]

The final step is to allow [pic] to be complex and separate out both real and imaginary parts.

[pic]

[pic]

By making these substitutions, the REAL part becomes the POISSON INTEGRAL FORMULA for the interior of a circle. There is a corresponding IMAGINARY part as well.

The real part is:

[pic]

This leads us to our real world simplified example. The equation basically says that if we know the values of U on the circumference of the circle, then for a harmonic function, we can find the value at any point within the circle.

Real World Example Using the POISSON INTEGRAL FORMULA

So we will be looking at a long electrically conducting tube of radius 1 (units not important, but you can say it is in inches if it is easier to picture). The tube is filled uniformly with a dielectric substance. Dielectric means that the substance can maintain an electric field but does not conduct. Dry air is a dielectric as is glass.

This conducting tube is slit down the side on both sides and both sections are very close to each other. One part of the tube is at a potential of 1 Volt, and the other part is at a potential of -1 volt.

|[pic] | |

| |[pic] is the angle from 0 to the black line in the |

| |second quadrant |

| | |

| |[pic] is the angle from 0 to the first black line in the|

| |first quadrant. |

| | |

| |The red circle is the top of the tube with small slits |

| |at the 0 degree and 180 degree points. |

| | |

| |The top section is at +1 volt and the bottom section is |

| |at -1 volt. |

Our goal is to find out what the voltage is at any point within the dielectric material.

The electrostatic potential field is known to be a harmonic function, so we can use Poisson’s Integral Formula to help us find the voltage we are looking for.

Using the angles as shown in the figure above, we can define [pic] in this way:

|Top section, V=1 |0 < [pic] < [pic] |

|Bottom Section, V=-1 |[pic] < [pic] < 2[pic] |

So with R=1, we need to sum up the two sections of the pipe. [pic] for the top half and -1 for the bottom half. The limits of integration will be from 0 to pi for the first half and pi to 2 pi for the second half.

[pic]

In order to avoid complete insanity, we will try to make these integrals look like something we may find in an integral table.

We will let the following identities hold:

[pic]

[pic]

[pic]

This allows us to re-write the above equation: (we can move [pic] outside the integral because it does not vary with x)

[pic]

[pic] is available in a table of integrals. The solution shown below is valid for [pic].

[pic]

Since [pic] and [pic], and since the radius can never be less than 0, it is obvious that the conditions shown are met

[pic]

See Appendix A if you are interested in how this integral is arrived at.

Now we need to substitute back in for a and b (and x) to get this in terms of the radius r and [pic] and [pic].

[pic]

Reducing and Ranging

[pic]

This can be simplified slightly by realizing that [pic] where n is any integer.

[pic]

One thing to note here is the multi-valued nature of the tan and arctan functions. These operations are inverses of each other so you expect the following to hold:

[pic]

In reality, since [pic], the following is true:

[pic]

Note that n=0 when evaluated between [pic]

So, look at each term in our function and remember we are evaluating between 0 and pi, and then pi and 2*pi. Plugging this into the first term for theta tells us that we will be evaluating this tangent within the [pic] range. The other two terms we must account for the n[pic] factor.

In order to calculate out what these factors may be, we note that due to symmetry, the voltage at the center of this tube is 0 volts and r=0. Let’s rewrite the formula taking this into account.

[pic]

Since tan and arctan are odd functions, [pic] we can do the following:

[pic]

And combine like terms, bringing the factor of 2 outside the brackets

[pic]

Let’s evaluate this when theta goes from 0 to pi (where the edge of the tube is 1 volt. This means that the first term tangent goes from pi/2 to –pi/2. So when the first term is evaluated, it returns pi/2 – theta/2 and the second term returns theta/2.

[pic]

Reducing

[pic]

[pic]

It is apparent that m+n must equal -1 in this case where theta is between 0 and pi.

Now, let’s do the same for when theta goes from pi to 2 pi (where the edge of the tube has -1 Volt applied. In this case, the second term theta/2 will go from pi/2 to pi. When this term is evaluated, we get theta/2 – pi. The first term is not affected, because with the range of pi to 2 pi, we end up with an evaluation range of 0 to –pi/2.

[pic]

Reducing

[pic]

So m+n = 1 in this case.

Since tan and arctan are odd functions, [pic] we can do the following:

[pic]

This is our function. We use the minus pi/2 when evaluating theta from 0 to pi and the plus pi/2 when evaluating between pi and 2 pi.

You can comment or correct anything here by emailing me at steve@.

Here is a plot of our function for radii of .5 and .9 inches. As expected, the closer we get to the boundary, the closer to either 1 or -1 volt we get.

Radius of .5 inches

[pic]

Radius of .9 inches

[pic]

Appendix A – How to actually do the integration we found in the table

Here’s what we found in the table:

[pic]

Identities used in this solution (These can be proven – there are several pages online that show this)

[pic]

[pic]

[pic]

[pic]

To solve the integral, Let

[pic]

So

[pic]

(Because [pic])

Rearranging

[pic]

Or

[pic]

Using another identity

[pic]

Substituting into the original integral

[pic]

Pulling out the constants

[pic]

Let

[pic]

So our integral now becomes

[pic]

Factor out the constant [pic]

[pic]

Let

[pic]

So

[pic]

Now it is a matter of solving

[pic]

Using our integral identity, this becomes

[pic]

Substituting

[pic]

The integral is (Between the limits 0 to PI and PI to 2 PI)

[pic]

Time to reinsert the value we assigned to c and t

[pic]

Appendix 2 Tan and Arctan Reference Plots

Tan(x)

[pic]

Arctan(x)

[pic]

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