# Preface

This tutorial was made solely for the purpose of education and it was designed
for students taking Applied Math 0340. It is primarily for students who
have some experience using *Mathematica*. If you have never used
*Mathematica* before and would like to learn more of the basics for this computer algebra system, it is strongly recommended looking at the APMA
0330 tutorial. As a friendly reminder, don't forget to clear variables in use and/or the kernel. The *Mathematica* commands in this tutorial are all written in **bold black font**, while *Mathematica* output is in regular fonts.

Finally, you can copy and paste all commands into your *Mathematica* notebook, change the parameters, and run them because the tutorial is under the terms of the GNU General Public License
(GPL). You, as the user, are free to use the scripts for your needs to learn the *Mathematica* program, and have
the right to distribute and refer to this tutorial, as long as
this tutorial is accredited appropriately. The tutorial accompanies the
textbook *Applied Differential Equations.
The Primary Course* by Vladimir Dobrushkin, CRC Press, 2015; http://www.crcpress.com/product/isbn/9781439851043

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Introduction to Linear Algebra

## Glossary

# Laplace equation in a wedge

Recall that both the real and imaginary parts of an analytic function satisfy Laplace’s equation in two dimension. Suppose the region of interest is defined by the angular wedge

*W*= 0 ≤ θ ≤ α. Consider the analytic function

*m*for some integer

*m*, then

*f*(

*z*) is analytic everywhere. If α is an arbitrary real number, then

*f*(

*z*) may have a branch point at the origin, but we may choose the branch cut so that

*f*(

*z*) is still analytic in our region everywhere except at the origin. In fact the function \( f_n (z) = z^{n\pi /\alpha} \) for any integer

*n*has the same nice properties. Then its real part \( u(t, \theta ) = \Re f(z) = r^{\pi /\alpha} \cos \frac{\pi\theta}{\alpha} \) and imaginary part \( v(t, \theta ) = \Im f(z) = r^{\pi /\alpha} \sin \frac{\pi\theta}{\alpha} \) are both solutions of the Laplace's equation:

*W*with opening angle α and conducting boundaries at potential

*V*

_{0}is described by the complex potential

*r*.

Since the origin is included within our wedge-shaped region *W*, the sum
is over positive *n* only, so that the potential remains finite. Then the
potential near the origin (small *r*) is dominated by the first (*n*
= 1) term,and the field near the origin has components:

FullSimplify[%]

As one might expect, the behavior of solution near *r* = 0 has to be
restricted:

*c*

_{0}and

*c*

_{1}are some constants. The constant

*c*

_{0}cannot be chosen arbitrary. (It is analogous to the so-called blockage coefficient in other potential flows.) The above condition on behavior of harmonic function near the corner point is called the

**wedge condition**.

For instance, if we consider the Neumann boundary conditions on the two sides of the wedge,

*f*is a specified function. Evidently, the solution of the problem,

*u*, is not unique because we can always add

*c*

_{0}+

*c*

_{1}log

*r*, where

*c*

_{0}and

*c*

_{1}are arbitrary constants.

**Example: **
Consider Laplace's equation in a corner:

text1 = Graphics[ Text[Style["\[CapitalDelta] u = 0", FontSize -> 16], {0.6, 0.2}]];

text2 = Graphics[Text[Style["u = 0", FontSize -> 16], {0.4, 0.4}]]; text3 = Graphics[Text[Style["u = 0", FontSize -> 16], {0.6, -0.05}]]; text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.18, 0.35}]]; txt0 = Graphics[Text[Style["0", FontSize -> 16], {-0.03, 0.0}]]; txtr = Graphics[Text[Style["r = a", FontSize -> 16], {1.1, 0.0}]]; Show[w, text1, text2, text3, text4, txt0, txtr]

*u*(

*r,θ*) =

*R*(

*r*) ⋅ Θ(&theta). Substitution of this product into the Laplace equation yields

_{0}= a+ bθ must be identically zero to satisfy the homogeneous boundary condition.

Therefore, we get a discrete sequence of positive eigenvalues

*R*(

*r*). If we choose

*R*(

*r*) to be a power function

*R*(

*r*) =

*r*

^{k}, we get

*k*:

*b*

_{n}= 0. Adding all nontrivial solution, we obtain

*r = a*, we have to determine the values of coefficients a

_{n}from the equation

**Example: **
Consider the Dirichlet problem for Laplace's equation in a corner:

*u*

_{0}and

*u*

_{α}are given numbers.

text1 = Graphics[ Text[Style["\[CapitalDelta] u = 0", FontSize -> 16], {0.6, 0.2}]];

text2 = Graphics[ Text[Style[ "u(r,\[Alpha]) = \!\(\*SubscriptBox[\(u\), \(\[Alpha]\)]\)", FontSize -> 16], {0.3, 0.4}]]; text3 = Graphics[ Text[Style["u(r,0) = \!\(\*SubscriptBox[\(u\), \(0\)]\)0", FontSize -> 16], {0.6, -0.05}]]; text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.18, 0.35}]]; txt0 = Graphics[Text[Style["0", FontSize -> 16], {-0.03, 0.0}]]; txtr = Graphics[Text[Style["r = a", FontSize -> 16], {1.1, 0.0}]]; Show[w, text1, text2, text3, text4, txt0, txtr]

*u*(

*r,θ*) as a sum of two functions:

*w*can be used any function that satisfies the prescribed boundary conditions: \( w(r,0) = u_0 \) and \( w(r,\alpha ) = u_{\alpha} . \) Then for function

*v*(

*r,θ*) we get the following boundary value problem

**Example: **.

text1 = Graphics[ Text[Style["\[CapitalDelta] u = 0", FontSize -> 16], {0.6, 0.15}]];

text2 = Graphics[ Text[Style[ "\!\(\*SubscriptBox[\(u\), \(\[Theta]\)]\)(r,\[Alpha]) = 0", FontSize -> 16], {0.3, 0.4}]];

text3 = Graphics[ Text[Style["\!\(\*SubscriptBox[\(u\), \(\[Theta]\)]\)(r,0) = 0", FontSize -> 16], {0.6, -0.05}]];

text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.18, 0.35}]]; txt0 = Graphics[Text[Style["0", FontSize -> 16], {-0.03, 0.0}]]; txtr = Graphics[Text[Style["r = a", FontSize -> 16], {1.1, 0.0}]]; Show[w, text1, text2, text3, text4, txt0, txtr]

**Example: **
Consider the boundary value problem for the Laplace equation in circular domain:

text1 = Graphics[ Text[Style["\[CapitalDelta] u = 0", FontSize -> 16, FontWeight -> "Bold"], {0.4, 1.0}]];

text2 = Graphics[ Text[Style["u(r,\[Beta]) = 0", FontSize -> 16], {-0.3, 0.74}]];

text3 = Graphics[ Text[Style["u(r,\[Alpha]) = 0", FontSize -> 16], {1.0, 0.75}]];

text4 = Graphics[ Text[Style["u(b,\[Theta]) = f(\[Theta])", FontSize -> 16], {0.8, 1.4}]];

text5 = Graphics[ Text[Style["u(a,\[Theta]) = g(\[Theta])", FontSize -> 16], {0.3, 0.45}]];

Show[w, text1, text2, text3, text4, text5]

**Example: **
Consider the boundary value problem for Laplace's equation

text1 = Graphics[ Text[Style["\[CapitalDelta] u = 0", FontSize -> 16, FontWeight -> "Bold"], {-0.05, 0.45}]];

text2 = Graphics[ Text[Style["\!\(\*SubscriptBox[\(u\), \(\[Theta]\)]\)(r,0) = 0", FontSize -> 16], {0.5, -0.15}]];

text3 = Graphics[ Text[Style["u(r,\[Pi]) = 0", FontSize -> 16], {-0.5, -0.1}]];

text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.0, 0.8}]];

p = Graphics[{PointSize[Medium], Point[{0, 0}]}];

rec = Graphics[{Brown, Rectangle[{1, -0.05}, {0, 0}]}];

Show[w, text1, text2, text3, text4, p, rec]

Substituting the assumed form *u*(*r,θ*) = *R*(*r*)
⋅ Θ(&theta) into Laplace's equation, and separating variables, we
obtain two differential equations

*r*=

*a*, we have to choose coefficients

*c*n so that

**Example: **
Consider the boundary value problem for Laplace's equation

p = Graphics[{PointSize[Medium], Point[{0, 0}]}];

p = Graphics[{PointSize[Medium], Point[{0, 0}]}];

text1 = Graphics[ Text[Style["\[CapitalDelta] u = 0", FontSize -> 16, FontWeight -> "Bold"], {-0.05, 0.45}]];

text2 = Graphics[ Text[Style["u(r,0+0) = \!\(\*SubscriptBox[\(u\), \(0\)]\)(r)", FontSize -> 16], {0.5, 0.1}]];

text3 = Graphics[ Text[Style["u(r,2\[Pi]-0) = \!\(\*SubscriptBox[\(u\), \(1\)]\)(r)", FontSize -> 16], {0.5, -0.1}]];

text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.0, 0.8}]];

Show[w, text1, text2, text3, text4, p, rec]

If the boundary conditions on the crack are the same, *u*_{0} =
*u*_{1}, the solution can be obtained from our first two examples
by taking the limit \( \lim_{\alpha \to 2\pi} u(r, \theta ) . \) When they are not the same, the problem becomes very hard to solve.
■

# Laplace equation in a corner

- Peirce, A., Circular domains, 2018.

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