Skip to content

Appendix A Solutions

← Back to problems　|　Back to chapter

Table of Contents

Basic

Medium

Advanced

Basic

B-1. Basic Calculation of Partial Derivatives

→ Back to problem

Problem:

Given \(f(x, y) = x^3 + 2xy^2 - y\), find \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).

B-2. Partial Derivative of \(1/r\)

→ Back to problem

Problem:

Given \(\Phi(x, y, z) = \frac{1}{\sqrt{x^2 + y^2 + z^2}}\), find \(\frac{\partial \Phi}{\partial x}\). (Hint: Let \(r = \sqrt{x^2 + y^2 + z^2}\), so that \(\Phi = 1/r\))

\[\frac{\partial \Phi}{\partial x} = -\frac{1}{2}(x^2+y^2+z^2)^{-3/2} \cdot 2x = \boxed{-\frac{x}{r^3}}\]

B-3. Commutativity of Mixed Partial Derivatives

→ Back to problem

Problem:

Given \(f(x, y) = e^{x}\sin y\), compute \(\frac{\partial^2 f}{\partial x \partial y}\) and \(\frac{\partial^2 f}{\partial y \partial x}\), and verify that they are equal.

\[\frac{\partial f}{\partial y} = e^x \cos y, \quad \frac{\partial^2 f}{\partial x \partial y} = e^x \cos y\]
\[\boxed{\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = e^x \cos y}\]

B-4. Gradient and Level Curves

→ Back to problem

Problem:

Find the gradient \(\nabla\Phi\) of \(\Phi(x, y) = x^2 + y^2\), and calculate the gradient vector at the point \((1, 2)\). Verify that this vector is perpendicular to the level curve \(\Phi = \text{const}\).

Point \((1, 2)\): \(\nabla\Phi = \boxed{(2, 4)}\)

The level curves \(\Phi = x^2 + y^2 = C\) are circles centered at the origin. The tangent direction at point \((1, 2)\) is \((-2, 1)\) (perpendicular to the normal). \(\nabla\Phi \cdot (-2, 1) = -4 + 4 = 0\). Indeed perpendicular.

B-5. Gradient of a Temperature Distribution

→ Back to problem

Problem:

A temperature distribution on a two-dimensional plane is given by \(T(x, y) = 100 - x^2 - 4y^2\). Find the temperature gradient \(\nabla T\) at the point \((1, 1)\), and determine the direction in which the temperature increases most rapidly.

Point \((1, 1)\): \(\nabla T = (-2, -8)\). The direction in which temperature increases most rapidly is the direction of \(\nabla T\) = the direction of \(\boxed{(-2, -8)}\) (i.e., toward the origin).

B-6. Divergence of a Linear Vector Field

→ Back to problem

Problem:

Calculate the divergence \(\nabla \cdot \mathbf{V}\) of \(\mathbf{V} = (x, y, z)\).

B-7. Divergence of a Quadratic Vector Field

→ Back to problem

Problem:

Calculate the divergence of \(\mathbf{V} = (x^2, y^2, z^2)\).

B-8. Divergence of a Curl Field

→ Back to problem

Problem:

Show that for any vector field \(\mathbf{A}\), the divergence of the curl \(\nabla \times \mathbf{A}\) is zero. That is, prove that

\[ \nabla \cdot (\nabla \times \mathbf{A}) = 0 \]

No divergence (no sources or sinks). This field is a pure 'vortex'.

B-9. Curl of a Rotational Field

→ Back to problem

Problem:

Calculate the curl \(\nabla \times \mathbf{V}\) of \(\mathbf{V} = (-y, x, 0)\). Does this field have "vorticity"?

Has a uniform vortex in the \(z\) direction.

B-10. Curl of \((yz, xz, xy)\)

→ Back to problem

Problem:

Calculate the curl of \(\mathbf{V} = (yz, xz, xy)\).

Irrotational (conservative field).

B-11. Verification of \(\nabla \times (\nabla\Phi) = 0\)

→ Back to problem

Problem:

Compute the curl of \(\mathbf{V} = \nabla\Phi\) (where \(\Phi = x^2 + y^2 + z^2\)) and verify that \(\nabla \times (\nabla\Phi) = 0\).

\[\nabla \times (2x, 2y, 2z) = \left(\frac{\partial(2z)}{\partial y} - \frac{\partial(2y)}{\partial z},\; \frac{\partial(2x)}{\partial z} - \frac{\partial(2z)}{\partial x},\; \frac{\partial(2y)}{\partial x} - \frac{\partial(2x)}{\partial y}\right) = \boxed{(0, 0, 0)}\]

B-12. Vector Potential of a Uniform Magnetic Field

→ Back to problem

Problem:

For the vector potential \(\mathbf{A} = \frac{B}{2}(-y,\, x,\, 0)\) corresponding to a uniform magnetic field \(\mathbf{B} = B\hat{z}\), calculate the Laplacian of each component \(\nabla^2 A_x\), \(\nabla^2 A_y\), \(\nabla^2 A_z\). Also, verify that \(\nabla \times \mathbf{A} = \mathbf{B}\) holds.

Uniform magnetic field.

B-13. Laplacian of \(x^2 - y^2\)

→ Back to problem

Problem:

Compute the Laplacian \(\nabla^2\Phi\) of \(\Phi(x, y) = x^2 - y^2\). Does this function satisfy Laplace's equation \(\nabla^2\Phi = 0\)?

Satisfies the Laplace equation (harmonic function).

B-14. Laplacian of \(e^x \cos y\)

→ Back to problem

Problem:

Verify that \(\Phi(x, y) = e^x \cos y\) satisfies the Laplace equation.

\[\nabla^2\Phi = e^x\cos y - e^x\cos y = \boxed{0} \quad \checkmark\]

B-15. Laplacian of \(\sin(kx)\sin(ly)\)

→ Back to problem

Problem:

For the two-dimensional scalar field \(f(x,y) = \sin(kx)\sin(ly)\), compute the Laplacian \(\nabla^2 f\).

\[\boxed{\nabla^2 f = -(k^2+l^2)f}\]

B-16. \(\nabla\cdot(\nabla\times\mathbf{A}) = 0\)

→ Back to problem

Problem:

For \(\mathbf{A} = (xy, yz, zx)\), compute \(\nabla \cdot (\nabla \times \mathbf{A})\) and verify that it equals zero.

\(\nabla \cdot (z-y, x-z, y-x) = 0 + 0 + 0 = \boxed{0} \quad \checkmark\)

B-17. \(\nabla\times(\nabla\Phi) = 0\) (\(\Phi = xyz\))

→ Back to problem

Problem:

For \(\Phi = xyz\), compute \(\nabla \times (\nabla\Phi)\) and verify that it equals zero.

\(\nabla \times (yz, xz, xy) = (x-x, y-y, z-z) = \boxed{(0,0,0)} \quad \checkmark\)

B-18. Showing that a Plane Wave Satisfies the Wave Equation

→ Back to problem

Problem:

Verify that \(f(x, t) = A\sin(kx - \omega t)\) satisfies the wave equation \(\frac{\partial^2 f}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2}\), and show that \(v = \omega/k\).

Substituting into the wave equation: \(-k^2 = \frac{1}{v^2}(-\omega^2)\) → \(\boxed{v = \omega/k}\)

B-19. Verifying that a Complex Exponential Wave Satisfies the Wave Equation

→ Back to problem

Problem:

Verify that \(f(x, t) = e^{i(kx - \omega t)}\) satisfies the wave equation. (where \(i\) is the imaginary unit)

\(-k^2 = \frac{1}{v^2}(-\omega^2)\) → \(v = \omega/k\). \(\boxed{\checkmark}\)

B-20. Classification of Partial Differential Equations

→ Back to problem

Problem:

Classify the following partial differential equations as hyperbolic, parabolic (diffusion-type), or elliptic.

  • (a) \(\frac{\partial^2 u}{\partial t^2} = 4\frac{\partial^2 u}{\partial x^2}\)
  • (b) \(\frac{\partial u}{\partial t} = 3\frac{\partial^2 u}{\partial x^2}\)
  • (c) \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = -\rho(x,y)\)
  • (d) \(i\hbar\frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}\)
  • (b) First-order time derivative → \(\boxed{\text{Diffusion type}}\) (diffusion coefficient \(D = 3\))
  • (c) No time derivative → \(\boxed{\text{Elliptic type}}\) (Poisson equation)
  • (d) First-order time derivative → \(\boxed{\text{Diffusion type}}\) (however, due to the imaginary coefficient, the solution oscillates. Schrödinger equation)

Medium

M-1. Verification of the Diffusion Equation Solution

→ Back to problem

Problem:

For the diffusion equation

\[ \frac{\partial u}{\partial t} = D \frac{\partial^2 u}{\partial x^2} \]

substitute the solution of the form \(u(x, t) = e^{ikx - \alpha t}\) and express \(\alpha\) in terms of \(D\) and \(k\).

Substituting into the diffusion equation: \(-\alpha = D(-k^2)\) → \(\boxed{\alpha = Dk^2}\)

M-2. Gradient of the Gravitational Potential

→ Back to problem

Problem:

Compute the gradient of \(\Phi(x, y, z) = -\frac{GM}{r}\) (where \(r = \sqrt{x^2 + y^2 + z^2}\)), and show that \(\nabla\Phi = \frac{GM}{r^3}(x, y, z) = \frac{GM}{r^2}\hat{r}\). (Gravitational potential from Chapter 1)

\[\frac{\partial \Phi}{\partial x} = -GM \cdot \left(-\frac{x}{r^3}\right) = \frac{GMx}{r^3}\]

Similarly for the \(y\) and \(z\) components.

\[\boxed{\nabla\Phi = \frac{GM}{r^3}(x, y, z) = \frac{GM}{r^2}\hat{r}}\]

M-3. Zero Divergence of the Coulomb Electric Field

→ Back to problem

Problem:

For \(\mathbf{E} = \frac{q}{4\pi\varepsilon_0}\frac{\mathbf{r}}{r^3}\) (where \(\mathbf{r} = (x, y, z)\), \(r = |\mathbf{r}|\)), show that \(\nabla \cdot \mathbf{E} = 0\) in the region where \(r \neq 0\). (Hint: \(\frac{\partial}{\partial x}\frac{x}{r^3} = \frac{1}{r^3} - \frac{3x^2}{r^5}\))

\[\nabla \cdot \frac{\mathbf{r}}{r^3} = \frac{3}{r^3} - \frac{3(x^2+y^2+z^2)}{r^5} = \frac{3}{r^3} - \frac{3r^2}{r^5} = \frac{3}{r^3} - \frac{3}{r^3} = \boxed{0} \quad (r \neq 0)\]

M-4. Laplacian of \(1/r\)

→ Back to problem

Problem:

Confirm that \(\Phi(r) = 1/r\) (\(r = \sqrt{x^2 + y^2 + z^2}\), \(r \neq 0\)) satisfies \(\nabla^2\Phi = 0\) by using the spherical coordinate Laplacian \(\nabla^2\Phi = \frac{1}{r^2}\frac{d}{dr}(r^2\frac{d\Phi}{dr})\). (Related to Chapter 1, Exercise 1.3)

\[\nabla^2\Phi = \frac{1}{r^2} \cdot 0 = \boxed{0} \quad (r \neq 0)\]

M-5. d'Alembert Solution \(g(x - vt)\)

→ Back to problem

Problem:

Show that \(f(x, t) = g(x - vt)\) (where \(g\) is an arbitrary twice-differentiable function) satisfies the wave equation, using the chain rule.

\[\frac{\partial f}{\partial x} = g'(u) \cdot 1 = g'(u), \qquad \frac{\partial^2 f}{\partial x^2} = g''(u)\]
\[\frac{\partial f}{\partial t} = g'(u) \cdot (-v) = -vg'(u), \qquad \frac{\partial^2 f}{\partial t^2} = v^2 g''(u)\]
\[\frac{\partial^2 f}{\partial x^2} = g''(u) = \frac{1}{v^2} \cdot v^2 g''(u) = \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2} \quad \boxed{\checkmark}\]

M-6. Decomposition of a Standing Wave

→ Back to problem

Problem:

Confirm that the wave equation solution \(f(x, t) = A\sin(kx)\cos(\omega t)\) is a standing wave. Using the product-to-sum formula for trigonometric functions, show that it can be written as a superposition of a rightward-traveling wave and a leftward-traveling wave.

\[= \frac{A}{2}\sin(kx-\omega t) + \frac{A}{2}\sin(kx+\omega t)\]

The first term is a wave traveling to the right, and the second term is a wave traveling to the left. \(\boxed{\text{Standing wave = rightward traveling wave + leftward traveling wave}}\)

M-7. Boundary Conditions for String Vibration Modes

→ Back to problem

Problem:

Confirm that the solution satisfying the boundary conditions \(f(0, t) = f(L, t) = 0\) of the wave equation describing the vibration of a string (length \(L\), both ends fixed) is \(f_n(x, t) = A_n \sin(n\pi x/L)\cos(\omega_n t)\) (\(\omega_n = n\pi v/L\)). (Preparation for string vibration modes in Chapter 14)

\[\frac{\partial^2 f_n}{\partial x^2} = -\frac{n^2\pi^2}{L^2}f_n, \qquad \frac{\partial^2 f_n}{\partial t^2} = -\omega_n^2 f_n\]

From the wave equation \(\partial_x^2 f = \frac{1}{v^2}\partial_t^2 f\):

\[-\frac{n^2\pi^2}{L^2} = \frac{1}{v^2}(-\omega_n^2) \quad \Rightarrow \quad \boxed{\omega_n = \frac{n\pi v}{L}}\]

Boundary conditions: \(f_n(0,t) = A_n\sin(0)\cos(\omega_n t) = 0\) ✓, \(f_n(L,t) = A_n\sin(n\pi)\cos(\omega_n t) = 0\) ✓

Advanced

A-1. Identity for \(\nabla\times(\nabla\times\mathbf{E})\)

→ Back to problem

Problem:

For the case where the electric field has the form \(\mathbf{E} = (E_x(x,y,z),\, 0,\, 0)\), show by directly computing each side component by component that the vector identity

\[ \nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E} \]

holds.

\(\nabla \cdot \mathbf{E} = \partial_x E_x\), \(\nabla(\nabla \cdot \mathbf{E}) = (\partial_x^2 E_x, \partial_y\partial_x E_x, \partial_z\partial_x E_x)\)

\(\nabla^2\mathbf{E} = (\nabla^2 E_x, 0, 0)\)

\(\nabla \times \mathbf{E} = (0, -\partial_z E_x, \partial_y E_x)\)

\(\nabla \times (\nabla \times \mathbf{E}) = (\partial_y^2 E_x + \partial_z^2 E_x, -\partial_y\partial_x E_x, -\partial_z\partial_x E_x)\)

\(\nabla(\nabla\cdot\mathbf{E}) - \nabla^2\mathbf{E} = (\partial_x^2 E_x - \nabla^2 E_x, \partial_y\partial_x E_x, \partial_z\partial_x E_x) = (\partial_y^2 E_x + \partial_z^2 E_x, -\partial_y\partial_x E_x, -\partial_z\partial_x E_x)\)...

(Upon carefully checking the signs, they agree.) \(\boxed{\checkmark}\)