Class Contents

• 1: Tohoku Univ.
• 1.1: Specialized Education
• 1.1.1: Exercise in Mathematics and Physics Ⅱ
• 1.1.1.1: Chapter 25 Solutions
• 1.1.1.2: Chapter 26 Solutions
• 1.1.2: Mathematics Ⅱ
• 1.1.2.1: 4 - Fourier Transform
• 1.1.2.2: 5 - PDE
• 1.2: General Education

1 - Tohoku Univ.

1.1 - Specialized Education

1.1.1 - Exercise in Mathematics and Physics Ⅱ

1.1.1.1 - Chapter 25 Solutions

Chapter 25 Solutions

1. Complex Trigonometric Equation

Problem

Find all complex numbers \(z\) which satisfy the equation \(\sin(z) = 5\).

Solution

We use the exponential definition of the sine function for a complex variable \(z = x + iy\):

Set this equal to 5:

To solve this, we make a substitution. Let \(w = e^{iz}\). Then \(e^{-iz} = \frac{1}{w}\). The equation becomes:

Multiply the entire equation by \(w\) to eliminate the fraction:

Rearrange this into a standard quadratic equation \(aw^2 + bw + c = 0\):

We solve for \(w\) using the quadratic formula \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Now we have two possible values for \(w\). We must solve for \(z\) by substituting back \(w = e^{iz}\), which means \(iz = \log(w)\). Remember that the complex logarithm is multi-valued: \(\log(w) = \ln|w| + i(\arg(w) + 2n\pi)\) for \(n \in \mathbb{Z}\).

Case 1: \(w = (5 + 2\sqrt{6})i\)

\(|w| = 5 + 2\sqrt{6}\). The argument \(\arg(w)\) is \(\frac{\pi}{2}\) since it’s on the positive imaginary axis.

Multiply by \(\frac{1}{i} = -i\):

Case 2: \(w = (5 - 2\sqrt{6})i\)

\(|w| = 5 - 2\sqrt{6}\). The argument \(\arg(w)\) is also \(\frac{\pi}{2}\).

Multiply by \(\frac{1}{i} = -i\):

Note that \(\ln(5 - 2\sqrt{6}) = \ln\left(\frac{(5 - 2\sqrt{6})(5 + 2\sqrt{6})}{5 + 2\sqrt{6}}\right) = \ln\left(\frac{1}{5 + 2\sqrt{6}}\right) = -\ln(5 + 2\sqrt{6})\). So, this second case gives \(z = i \ln(5 + 2\sqrt{6}) + \left(\frac{\pi}{2} + 2n\pi\right)\).

We can combine both cases into a single expression:

4. Cauchy’s Integral Formula / Residue Theorem

Problem

For the path integral \(\oint_C \frac{z}{(z - 3)(z + 1 + i)} dz\), calculate its values at \(|z| = 1\) and \(|z| = 4\).

Solution

The integrand is \(f(z) = \frac{z}{(z - 3)(z - (-1 - i))}\). The function has two simple poles at \(z_1 = 3\) and \(z_2 = -1 - i\).

Case 1: C is the circle \(|z| = 1\)

We check if the poles lie inside the contour \(C\).

• For \(z_1 = 3\): \(|z_1| = |3| = 3\). Since \(3 > 1\), this pole is outside the contour.
• For \(z_2 = -1 - i\): \(|z_2| = |-1 - i| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2} \approx 1.414\). Since \(\sqrt{2} > 1\), this pole is also outside the contour.

Since the function \(f(z)\) is analytic everywhere inside and on the contour \(|z|=1\), by Cauchy’s Integral Theorem, the value of the integral is 0.

Case 2: C is the circle \(|z| = 4\)

We check if the poles lie inside this new contour.

• For \(z_1 = 3\): \(|z_1| = 3\). Since \(3 < 4\), this pole is inside the contour.
• For \(z_2 = -1 - i\): \(|z_2| = \sqrt{2}\). Since \(\sqrt{2} < 4\), this pole is also inside the contour.

Since both poles are inside, we use the Residue Theorem. The integral is \(2\pi i\) times the sum of the residues at the enclosed poles.

• Residue at \(z_1 = 3\):
  $$\text{Res}(f, 3) = \lim_{z\to 3} (z-3)f(z) = \lim_{z\to 3} \frac{z}{z+1+i} = \frac{3}{4+i}$$
  $$= \frac{3(4-i)}{(4+i)(4-i)} = \frac{12-3i}{16-i^2} = \frac{12-3i}{17}$$
• Residue at \(z_2 = -1 - i\):
  $$\text{Res}(f, -1-i) = \lim_{z\to -1-i} (z-(-1-i))f(z) = \lim_{z\to -1-i} \frac{z}{z-3} = \frac{-1-i}{-1-i-3} = \frac{-(1+i)}{-(4+i)} = \frac{1+i}{4+i}$$
  $$= \frac{(1+i)(4-i)}{(4+i)(4-i)} = \frac{4-i+4i-i^2}{17} = \frac{5+3i}{17}$$
• Sum of residues:
  $$\frac{12 - 3i}{17} + \frac{5 + 3i}{17} = \frac{17}{17} = 1$$

The value of the integral is \(2\pi i \times (\text{Sum of residues}) = 2\pi i \times 1 = \mathbf{2\pi i}\).

5. Path Integral on the Unit Circle

Problem

For the path integral \(\oint_C \frac{d\theta}{z + \cos\theta}\) calculate its value at \(|z| = 1\) (where \(z = e^{i\theta}, 0 \le \theta \le 2\pi\)).

Solution

We convert the entire integral into the \(z\) domain. The contour \(C\) is the unit circle.

• \(z = e^{i\theta}\)
• \(dz = i e^{i\theta} d\theta = iz d\theta \implies d\theta = \frac{dz}{iz}\)
• \(\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2} = \frac{z + 1/z}{2}\)

Substitute these into the integral:

Simplify the expression inside the integral:

Now we use the Residue Theorem. The poles are the roots of \(3z^2 + 1 = 0\), which means \(z^2 = -1/3\), so \(z = \pm \frac{i}{\sqrt{3}}\).

Both poles, \(z_1 = \frac{i}{\sqrt{3}}\) and \(z_2 = -\frac{i}{\sqrt{3}}\), have a magnitude of \(\frac{1}{\sqrt{3}} \approx 0.577\), which is less than 1. Therefore, both poles are inside the unit circle contour.

Let \(g(z) = \frac{1}{3z^2+1} = \frac{1}{3(z - i/\sqrt{3})(z + i/\sqrt{3})}\).

• Residue at \(z_1 = i/\sqrt{3}\):
  $$\text{Res}(g, i/\sqrt{3}) = \lim_{z\to i/\sqrt{3}} (z - i/\sqrt{3})g(z) = \lim_{z\to i/\sqrt{3}} \frac{1}{3(z + i/\sqrt{3})} = \frac{1}{3(2i/\sqrt{3})} = \frac{1}{2i\sqrt{3}}$$
• Residue at \(z_2 = -i/\sqrt{3}\):
  $$\text{Res}(g, -i/\sqrt{3}) = \lim_{z\to -i/\sqrt{3}} (z + i/\sqrt{3})g(z) = \lim_{z\to -i/\sqrt{3}} \frac{1}{3(z - i/\sqrt{3})} = \frac{1}{3(-2i/\sqrt{3})} = -\frac{1}{2i\sqrt{3}}$$
• Sum of residues:
  $$\frac{1}{2i\sqrt{3}} - \frac{1}{2i\sqrt{3}} = 0$$

The value of the integral is \(\frac{2}{i} \times (2\pi i \times \text{Sum of residues}) = \frac{2}{i} \times (2\pi i \times 0) = \mathbf{0}\).

6. Real Integral using Contour Integration

Problem

Calculate the integral \(\int_{-\infty}^{\infty} \frac{x \sin(x)}{x^2 + a^2} dx\).

Solution

We use contour integration. Consider the complex function:

The integral we want is the imaginary part of the integral of \(f(z)\) along the real axis:

We integrate \(f(z)\) over a semi-circular contour \(C\) in the upper half-plane, consisting of the real axis from \(-R\) to \(R\) and a semi-circular arc \(\Gamma\) of radius \(R\). As \(R \to \infty\):

By Jordan’s Lemma, the integral over the arc \(\Gamma\) goes to zero as \(R \to \infty\). Thus:

Now we find the value of the contour integral using the Residue Theorem. The poles of \(f(z) = \frac{z e^{iz}}{(z - ia)(z + ia)}\) are at \(z = \pm ia\). Assuming \(a > 0\), only the pole at \(z = ia\) is inside our contour in the upper half-plane.

• Residue at \(z = ia\):
  $$\text{Res}(f, ia) = \lim_{z\to ia} (z-ia)f(z) = \lim_{z\to ia} \frac{ze^{iz}}{z+ia}$$
  $$= \frac{(ia)e^{i(ia)}}{ia+ia} = \frac{iae^{-a}}{2ia} = \frac{e^{-a}}{2}$$

The contour integral is \(2\pi i\) times the sum of residues inside \(C\):

Equating this with the real integral:

By equating the imaginary parts of both sides, we get our result:

1.1.1.2 - Chapter 26 Solutions

Chapter 26 Solutions

1. Solving ODEs with Complex Functions

Problem

Solve the following differential equations using complex functions:

Solution

We can combine these two real equations into a single complex differential equation by defining \(Z(t) = x(t) + i y(t)\). The forcing function becomes \(\cos(4t) + i \sin(4t) = e^{i4t}\).

The complex ODE is:

We seek a particular (steady-state) solution of the form \(Z_p = A e^{i4t}\), where \(A\) is a complex constant. We find its derivatives:

Substitute these into the complex ODE:

Divide by \(e^{i4t}\):

Now, we rationalize the complex number \(A\):

Now we find the complex solution \(Z_p\):

The solutions for \(x(t)\) and \(y(t)\) are the real and imaginary parts of \(Z_p\), respectively. \(x(t) = \text{Re}(Z_p)\) = \(\frac{3}{85}\sin(4t) - \frac{7}{170}\cos(4t)\) \(y(t) = \text{Im}(Z_p)\) = \(-\frac{7}{170}\sin(4t) - \frac{3}{85}\cos(4t)\)

2. Complex Potential Function

Problem

Given a complex function \(f(z) = z + \frac{a^2}{z}\):

1. Find the real part \(\Phi(r, \theta)\) and the imaginary part \(\Psi(r, \theta)\) of \(f(z)\).
2. Show \(\left. \frac{\partial\Phi}{\partial r} \right|_{r=a} = 0\).

Solution

(1) Real and Imaginary Parts

We express \(z\) in polar coordinates as \(z = r e^{i\theta} = r(\cos\theta + i \sin\theta)\). Substitute this into \(f(z)\):

For the second term, we use \(\frac{1}{e^{i\theta}} = e^{-i\theta} = \cos\theta - i \sin\theta\):

Now, we group the real and imaginary terms:

The real part \(\Phi\) (the velocity potential) and the imaginary part \(\Psi\) (the stream function) are: \(\Phi(r, \theta) = \left(r + \frac{a^2}{r}\right)\cos\theta\) \(\Psi(r, \theta) = \left(r - \frac{a^2}{r}\right)\sin\theta\)

(2) Show the Boundary Condition

We need to calculate the partial derivative of \(\Phi\) with respect to \(r\) and then evaluate it at \(r=a\).

First, find \(\frac{\partial\Phi}{\partial r}\):

Since \(\cos\theta\) is treated as a constant with respect to \(r\):

Now, evaluate this derivative at \(r = a\):

This confirms that the radial velocity \(u_r\) is zero on the circle \(r=a\), which is the correct boundary condition for non-penetrating flow around a cylinder. ✅

3. Boundary Condition for Flow in a Corner

Problem

Regarding the flow shown in Figure 26-3, find the boundary condition corresponding to eq. (26-10) with respect to \(\Psi\).

Solution

Figure 26-3 typically depicts potential flow within a 90-degree corner, where the boundaries are the positive x-axis and the positive y-axis.

• Physical Meaning of \(\Psi\): The imaginary part of the complex potential, \(\Psi\), is the stream function. By definition, lines where \(\Psi\) is constant are streamlines, which are the paths that fluid particles follow.

• Boundary Condition on a Solid Wall: In ideal fluid flow, there can be no flow across (i.e., through) a solid boundary. This means that any solid boundary must itself be a streamline.

• Applying to the Figure: The walls in the figure are the positive x-axis (\(y=0, x>0\)) and the positive y-axis (\(x=0, y>0\)). Since these are solid boundaries, they must be streamlines. Therefore, the stream function \(\Psi\) must be constant along these axes.

By convention, the constant value for one of the boundaries is chosen to be zero. For the flow in a corner bounded by the positive x and y axes, the boundary condition for the stream function \(\Psi\) is:

\(\Psi = 0\) on the positive x-axis and the positive y-axis.

This ensures that the axes themselves represent the path of the fluid particles adjacent to the walls, fulfilling the no-flow-through condition.

1.1.2 - Mathematics Ⅱ

1.1.2.1 - 4 - Fourier Transform

Fourier Transform and Integral Evaluation

(1-1) Find the Fourier transform of \(f(x) = e^{-|x|}\)

We define the Fourier transform of a function \(f(x)\) as:

For the function \(f(x) = e^{-|x|}\), we split the integral based on the definition of the absolute value:

Now, we evaluate each integral:

Combining the fractions, we get:

So, the Fourier transform is: \(\hat{f}(k) = \frac{2}{1+k^2}\).

(1-2) Evaluate the integral \(\int_{-\infty}^{\infty} \frac{1}{(1+x^2)^2}dx\)

We will use Plancherel’s Theorem (also known as Parseval’s identity for Fourier transforms), which states:

First, we evaluate the left-hand side (LHS) with our original function \(f(x) = e^{-|x|}\):

Next, we evaluate the right-hand side (RHS) using the Fourier transform \(\hat{f}(k) = \frac{2}{1+k^2}\):

By equating the LHS and RHS (\(1 = \frac{2}{\pi} \int \dots\)), we can solve for the integral. Replacing the dummy variable \(k\) with \(x\), we find:

The value of the integral is \(\frac{\pi}{2}\).

Fourier Cosine Transform and Integral Evaluation

(2-1) Find the Fourier cosine transform of the given function

The function is a triangular pulse defined as:

Since \(f(x)\) is an even function, we can find its Fourier cosine transform using the definition:

For our function, this becomes:

The first integral is:

For the second integral, we use integration by parts (\(\int u dv = uv - \int v du\)) with \(u=x\) and \(dv = \cos(\omega x)dx\):

Combining everything:

Using the half-angle identity \(1-\cos(2\theta) = 2\sin^2(\theta)\), the final transform is:

The Fourier cosine transform is \(\hat{f}_c(\omega) = \left(\frac{\sin\omega}{\omega}\right)^2\).

(2-2) Evaluate the integral \(\int_{0}^{\infty} \left(\frac{\sin x}{x}\right)^2 dx\)

We use the inverse Fourier cosine transform:

Substituting our result and evaluating at \(x=0\):

From the function’s definition, we know that \(f(0) = 1 - \frac{|0|}{2} = 1\). Equating the two expressions for \(f(0)\):

Solving for the integral and replacing the dummy variable \(\omega\) with \(x\) gives:

The value of the integral is \(\frac{\pi}{2}\).

1.1.2.2 - 5 - PDE

Here is a detailed solution to the heat conduction problem.

(1) Obtain two ODEs by the method of separating variables

We are given the partial differential equation (PDE):

We assume a solution of the form \(u(x,t) = X(x)T(t)\). Substituting this into the PDE gives:

To separate the variables, we divide by \(\alpha X(x)T(t)\):

Since the left side depends only on \(t\) and the right side depends only on \(x\), they must both be equal to a constant. We call this the separation constant, \(k\). This leads to two ordinary differential equations (ODEs):

• Time-dependent ODE: \(\frac{T’(t)}{\alpha T(t)} = k \implies T’(t) - \alpha k T(t) = 0\)
• Space-dependent ODE: \(\frac{X’’(x)}{X(x)} = k \implies X’’(x) - k X(x) = 0\)

(2) Show that the separation constant k must be negative

We analyze the spatial ODE \(X’’(x) - kX(x) = 0\) along with the boundary conditions. The boundary conditions for \(u(x,t)\) are \(u_x(0,t) = 0\) and \(u_x(2,t) = 0\). In terms of \(X(x)\), these become \(X’(0) = 0\) and \(X’(2) = 0\).

• Case 1: \(k > 0\). Let \(k = \lambda^2\) where \(\lambda > 0\). The ODE is \(X’’(x) - \lambda^2 X(x) = 0\). The general solution is \(X(x) = C_1 e^{\lambda x} + C_2 e^{-\lambda x}\). The derivative is \(X’(x) = C_1 \lambda e^{\lambda x} - C_2 \lambda e^{-\lambda x}\). Applying the boundary conditions:

  • \(X’(0) = C_1\lambda - C_2\lambda = \lambda(C_1 - C_2) = 0 \implies C_1 = C_2\).
  • \(X’(2) = C_1\lambda e^{2\lambda} - C_2\lambda e^{-2\lambda} = C_1\lambda(e^{2\lambda} - e^{-2\lambda}) = 0\). Since \(\lambda > 0\), the term \((e^{2\lambda} - e^{-2\lambda})\) is not zero. Thus, we must have \(C_1 = 0\), which implies \(C_2 = 0\). This gives \(X(x) = 0\), the trivial solution, which is not of interest.

• Case 2: \(k = 0\). The ODE is \(X’’(x) = 0\). The general solution is \(X(x) = C_1 x + C_2\). The derivative is \(X’(x) = C_1\). Applying the boundary conditions:

  • \(X’(0) = C_1 = 0\).
  • \(X’(2) = C_1 = 0\). This requires \(C_1 = 0\), but \(C_2\) can be any constant. So, \(X(x) = C_2\) is a non-trivial solution. This constant solution is physically important as it relates to the steady-state temperature.

• Case 3: \(k < 0\). Let \(k = -\lambda^2\) where \(\lambda > 0\). The ODE is \(X’’(x) + \lambda^2 X(x) = 0\). The general solution is \(X(x) = C_1\cos(\lambda x) + C_2\sin(\lambda x)\). The derivative is \(X’(x) = -C_1\lambda\sin(\lambda x) + C_2\lambda\cos(\lambda x)\). Applying the boundary conditions:

  • \(X’(0) = -C_1\lambda(0) + C_2\lambda(1) = C_2\lambda = 0 \implies C_2 = 0\).
  • With \(C_2=0\), the second condition becomes \(X’(2) = -C_1\lambda\sin(2\lambda) = 0\). To have a non-trivial solution, we need \(C_1 \ne 0\), which means we must have \(\sin(2\lambda) = 0\). This occurs when \(2\lambda = n\pi\) for \(n = 1, 2, 3, \dots\).

Therefore, to obtain non-trivial, time-decaying solutions of interest, the separation constant \(k\) must be negative. The case \(k=0\) gives a constant steady-state solution.

(3) Find u(x, t) for the initial condition f(x) = 100

From our analysis, the eigenvalues are \(\lambda_n = \frac{n\pi}{2}\), which gives \(k_n = -\lambda_n^2 = -\left(\frac{n\pi}{2}\right)^2\) for \(n=1, 2, \dots\). The case \(k=0\) corresponds to an eigenvalue \(\lambda_0 = 0\). The general solution is a superposition of all possible solutions:

We apply the initial condition \(u(x,0) = f(x) = 100\):

This is the Fourier cosine series for the function \(f(x) = 100\). By inspection, we can see that the function is already a constant. The constant term \(A_0\) must be 100, and all other coefficients \(A_n\) for \(n \ge 1\) must be 0.

Alternatively, using the formula for Fourier cosine coefficients on an interval \([0, L]\) with \(L=2\):

The solution is therefore just the constant term:

This makes physical sense: if the bar starts at a uniform temperature and its ends are insulated, the temperature will not change.

(4) Find u(x, t) for the initial condition f(x) = x

We use the same general solution form and apply the initial condition \(u(x,0) = f(x) = x\):

This requires finding the Fourier cosine series for \(f(x) = x\) on the interval \([0,2]\). The coefficients are:

This means \(A_n=0\) for even \(n\), and \(A_n = -\frac{8}{n^2\pi^2}\) for odd \(n\). Substituting these coefficients into the general solution for \(u(x,t)\):

To write this as a single sum, we can use the index \(k\) where \(n = 2k-1\):

(5) Draw u(x, t) at different times

We will plot the partial sum of the solution from Q(4) up to \(n<10\) (which includes \(n=1,3,5,7,9\), so the sum runs from \(k=1\) to \(k=5\)). We set \(\alpha=1\). The function to plot is:

Description of the graphs:

• t = 0.0: The exponential term is 1. The graph is a Fourier cosine series approximation of the initial condition \(f(x) = x\). It will be a line starting near \((0,0)\) and rising to near \((2,2)\).
• t = 0.2: The terms with higher frequencies (larger \(k\)) decay faster due to the \((2k-1)^2\) term in the exponent. The graph becomes a smoother curve, still rising but flatter than the initial line.
• t = 0.7: The decay is much more significant. The curve is now much flatter and getting closer to the steady-state temperature.
• t = 2.0: All exponential terms in the sum are extremely small. The graph is now almost a flat horizontal line at the steady-state temperature.

The steady-state solution is the value as \(t \to \infty\). In this case, all exponential terms go to zero, and \(u(x,t) \to A_0 = 1\). This value represents the average of the initial temperature over the bar: \(\frac{1}{2}\int_0^2 x,dx = 1\).

The evolution of the temperature is shown in the plot. The temperature profile starts as the line \(u=x\), and as time progresses, heat redistributes within the insulated bar, causing the temperature to even out until it reaches a uniform steady state of \(u=1\).

Data for Plotting in Excel:

1.2 - General Education