Let $0<a<b$ and impose zero Dirichlet boundary conditions at $a,b$. With the convention $L_{x}G(x,s)=\delta (x-s)$ for the Dirac delta distribution, the Green's function is continuous and its first derivative has jump one. For $n\ne 0$ choose $\ell (x)=x^{-n}-a^{-n}$ and $r(x)=x^{-n}-b^{-n}$. Their Wronskian is $n(a^{-n}-b^{-n})s^{-n-1}$, soFor $n=0$, take the continuous limit or use $\ell =\log (x/a)$ and $r=\log (x/b)$ to obtainThe solution is $u(x)={\int }_a^{b}G(x,s)f(s)ds$. These kernels use ordinary measure $ds$. The factor $s^{n+1}$ comes from the nonconstant Wronskian and must not be discarded by confusing the original operator with its multiplied divergence form $(x^{n+1}{u}^{\prime }{)}^{\prime }$.

For the exterior of the unit disc in two dimensions, with the sign convention ${\Delta }_{x}G=+\delta (x-y)$, this is the Dirichlet Green function that vanishes on the circle and is bounded at infinity. The method of images places an opposite logarithmic source at the inverse point $y/|y{|}^2$. The constant $-(2\pi {)}^{-1}\log |y|$ enforces the zero boundary value. The inversion in a circle identity $|x-y|=|y||x-y/|y{|}^2|$ for $|x|=1$ verifies that condition directly.

The outward normal of the exterior domain is $-x$ on the unit circle. Its normal derivative gives the positive exterior Poisson kernelFor bounded harmonic functions this kernel integrates boundary data to the interior value. Boundedness at infinity is essential to uniqueness: without it, a multiple of $\log |y|$ can be added without changing the circle's Dirichlet boundary condition.

For the three-dimensional ball $|x|<a$, the displayed Dirichlet Green function has convention ${\Delta }_x{G}_D=+\delta (x-y)$. Its method of images uses the exterior inverse point $y^{\ast }=a^{2}y/|y{|}^2$. On $|x|=a$, the distance identity $|x-y^{\ast }|=(a/|y|)|x-y|$ makes the boundary value zero. The image lies outside the ball and therefore adds no source singularity to the domain. At a central source the weighted image term tends to $1/a$, giving $G_D(x,0)=-(4\pi |x|{)}^{-1}+(4\pi a{)}^{-1}$. Spherical inversion with the required radial weight is the Kelvin transform.

For a three-dimensional slab $0<x_1<L$ and source $y=(y_1,y_2,y_3)$, define image positions $y_n^{+}=(y_1+2nL,y_2,y_3)$ and $y_n^{-}=(-y_1+2nL,y_2,y_3)$. The displayed method of images series uses convention ${\Delta }_x{G}_D=+\delta (x-y)$. Reflection in $x_1=0$ pairs the plus image $n$ with the minus image $-n$, while reflection in $x_1=L$ pairs it with the minus image $1-n$. Both boundary values therefore vanish. Only $y_0^{+}=y$ lies in the slab. Each bracket is $O(n^{-2})$, giving absolute convergence of the paired series away from sources. This selects the usual Dirichlet Green function that decays at large transverse separation.

For $D=d/dx$ and $L_a=(D-a{)}^2+b^2$ on $(0,L)$, with $b>0$ and $\sin (bL)\ne 0$, the substitution $y=e^{ax}v$ gives $L_{a}y=e^{ax}(v^{\prime \prime }+b^{2}v)$. Here $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. The displayed Green function vanishes at the endpoints, is continuous at $x=\xi$, and its first derivative has jump one, so $L_{a,x}G=\delta (x-\xi )$. It is the usual sine Dirichlet Green function conjugated by the exponential factor. At $bL\in \pi \mathbb{Z}\setminus \{0\}$ there is a homogeneous zero mode, so the inverse kernel does not exist without a solvability condition and a normalization.

For $\kappa \ge 0$ in the upper half-space, let $r^{\prime \ast }$ be the reflection of the interior source $r^{\prime }$ across its boundary plane. The displayed decaying Green function solves $(\Delta -{\kappa }^2)G_D=\delta (r-r^{\prime })$ with zero Dirichlet boundary condition. The direct term supplies the source; the image source lies outside the domain and its opposite sign makes the two terms cancel on the plane. The minus sign relative to the positive static Yukawa Green function comes from the operator convention. The case $\kappa =0$ is the Coulomb limit. An infinity condition is needed for uniqueness on an unbounded domain.

With the convention ${\Delta }_z{G}_D=\delta (z-\zeta )$, the displayed Dirichlet Green function vanishes on the unit circle. Indeed $|z-\zeta |=|1-z\bar{\zeta }|$ when $|z|=1$. For $\zeta \ne 0$, the second logarithm corresponds to an opposite image source at $\zeta /|\zeta {|}^2$, outside the unit disc; it is a harmonic function throughout the interior. At $\zeta =0$ the formula becomes $(2\pi {)}^{-1}\log |z|$. Its outward normal derivative in the boundary variable is $(2\pi {)}^{-1}(1-|\zeta {|}^2)/|z-\zeta {|}^2$, the Poisson kernel divided by $2\pi$. Green's second identity therefore gives the Poisson integral on the unit disk.

For $L=d^2/d{x}^2+4x^{-1}d/dx-4x^{-2}$ on $(0,1)$, regularity at zero selects the homogeneous solution $u=x$, while the right Dirichlet boundary condition selects $v=x^{-4}-x$. Their Wronskian is $-5x^{-4}$. Thus the displayed Green function, with $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$, is continuous, has unit first-derivative jump and satisfies $L_{x}G=\delta (x-\xi )$. For bounded forcing $f$, $y(x)={\int }_0^{1}G(x,\xi )f(\xi )d\xi$ is the unique solution continuous at zero with zero endpoint values. The singular homogeneous branch $x^{-4}$ is excluded by that regularity. The kernel uses ordinary source measure $d\xi$, not the measure for the multiplied self-adjoint operator.

A Dirac comb is a periodic train of Dirac delta distributions. With Fourier transform convention $\hat{f}(q)=\int f(x)e^{-iqx}dx$ and period $D>0$,The identity holds as a tempered distribution. Truncating the real-space train by a finite window convolves the reciprocal train with the transform of that window, broadening its peaks. Convolving the train with one slit's transmission instead multiplies the reciprocal peaks by the slit's field transform, the basic construction of a diffraction grating pattern.

A light-emitting diode is a semiconductor diode producing electroluminescence under electric current. Forward bias injects electrons and electron holes; radiative electron-hole recombination emits photons. A double heterostructure uses band offsets to retain both charge carriers in an active layer. A sufficiently thin layer is a quantum well.^[ref]

The optical and electrical questions are different: photon energy sets a color scale, while light-emitting diode efficiency determines how much supplied energy escapes as light. The blue light-emitting diode made compact white light-emitting diodes and full-color displays practical. A modulated light-emitting diode can also supply the optical signal in a chopped-light lock-in experiment.

A rectifier converts an alternating input into a unidirectional output. Diode orientation selects which direction of electric current is transmitted.^[ref]

The New Jersey Institute of Technology laboratory uses a silicon 1N4148 diode. For its forward current-voltage characteristic, the supply operates in current-source mode; a benchtop voltmeter independently reads the diode electric potential difference. Small electric current steps resolve the steep branch. For reverse measurements, a sensitive ammeter sits in series; the handout suggests a $10$–$100\mathrm{k\Omega }$ series resistor. Reversing the applied polarity supplies the comparison. Meter placement must avoid counting the voltmeter's own electric current, and the zero reading may have an offset. The same diode is then measured with an Agilent U2722A source measure unit for graphical comparison.^[ref]

The prescribed analysis uses a logarithmic scale for forward electric current to infer $I_S$ and $n$ from the straight region. Large-electric current departures can indicate series electrical resistance. The handout cautions that the accessible range may not determine it precisely; meter loading also matters. This is a measurement procedure, without a published set of device readings.^[ref]

The reason for the logarithmic scale follows directly from the Shockley diode equation. If $I_D\gg I_S$ and series electrical resistance is negligible,where $I_0>0$ is a fixed reference electric current making the logarithm dimensionless. The slope is $1/(n{V}_T)$, so extracting $n$ needs the junction temperature. Fitting only an appropriate interval matters: a straight-looking subset cannot establish the model at all electric currents.

A sufficiently large reverse electric potential difference can produce a rapidly increasing reverse electric current. Both the ideal diode model and the unextended Shockley diode equation omit this behavior.^[ref]

For a p-n junction at fixed temperature, the Shockley diode equation models its electric current asHere $I_S>0$ is the diode saturation current, $n>0$ the diode ideality factor, $V_T$ the thermal voltage, $k_B$ the Boltzmann constant, and $q>0$ the elementary charge.^[ref]

For an original model example, choose $I_S=1\mathrm{nA}$, $n=2$, and $T=300\mathrm{K}$. Using $k_B=1.380649\times 10^{-23}J/K$ and $q=1.602176634\times 10^{-19}\mathrm{C}$ gives $V_T=25.8520\mathrm{mV}$. Then $V_D=0.600\mathrm{V}$ gives $I_D=0.109591\mathrm{mA}$, whereas $V_D=0.800\mathrm{V}$ gives $I_D=5.24450\mathrm{mA}$. These are predictions of the declared model. There is no universal turn-on electric potential difference: solving the Shockley diode equation for the value needed at a chosen electric current givesAt $I_D\gg I_S$, increasing $V_D$ by $0.100\mathrm{V}$ multiplies the electric current by approximately $e^{0.100/(n{V}_T)}=6.91772$. At $V_D=-0.300\mathrm{V}$ the same model gives $I_D=-0.996979\mathrm{nA}$. Its reverse branch approaches $-I_S$; it cannot describe diode reverse breakdown.

A constructed extension adds internal series electrical resistance $R_s$ to a junction whose electric potential difference is $V_j$. Applying the Kirchhoff cycle law givesFor $R_s=20\Omega$ and the same $I_S,n,T$, sustaining $I_D=5.24450\mathrm{mA}$ needs $V_D=0.904890\mathrm{V}$: $0.800\mathrm{V}$ across the junction plus $0.104890\mathrm{V}$ across its internal electrical resistance. The derivativeshows why the current-voltage characteristic becomes nearly a straight line when the internal series electrical resistance dominates. This is still a fixed-temperature model; self-heating can change $I_S$, $n$ and $V_T$. Real diodes also have leakage and charge-storage effects omitted here.^[ref]

The ideal diode model sets the conducting electric potential difference to zero and the blocking electric current to zero. With the stated orientation,^[ref]

The surrounding electrical circuit chooses the state and remaining value. The two branches meet at $V_D=I_D=0$. Their product vanishes, so the ideal diode dissipates no power: $P_D=V_D{I}_D=0$.

The ideal diode model omits forward drop, reverse leakage, diode reverse breakdown and switching delay.

The terminal of a diode at which positive forward electric current leaves. The bar in the diode symbol identifies this terminal.

The terminal of a diode at which positive forward electric current enters, with the orientation used in the ideal diode model.

For positive diffusivity $D(x)$, a diffusion equation with potential-driven drift has diffusive flux $J=-D(c_x+\beta V^{\prime }c)$. On a finite interval with zero flux at both endpoints, the total amount is conserved. The substitution $c=e^{-\lambda t}{e}^{-\beta V}y$ converts the decay eigenvalue problem to the Sturm-Liouville problemThe Rayleigh quotient is $\lambda =\int p|y^{\prime }{|}^2/\int w|y{|}^2\ge 0$. Its zero eigenfunction is constant, giving the stationary solution $c_{\mathrm{eq}}=C{e}^{-\beta V}$. This is a spatial Boltzmann distribution; $D$ changes relaxation rates but cancels out of the zero-flux equilibrium profile. For a sufficiently regular positive $D$ and bounded $V$, the constant eigenfunction is the unique zero mode up to scale.

These Robin boundary conditions give the spatial Sturm-Liouville operator $-d^2/d{x}^2$ one negative eigenvalue, $-1$, with eigenfunction $e^x$. The other eigenvalues are $n^2{\pi }^2$, with eigenfunctions $V_n=n\pi \cos (n\pi x)+\sin (n\pi x)$, $n\ge 1$; there is no zero eigenvalue. Hence the diffusion equation has a growing contribution $c_0{e}^{t+x}$ in addition to decaying modes.

There is no contradiction with ordinary diffusive smoothing. The outward normal derivative is $-u_x$ at the left endpoint and $u_x$ at the right endpoint, so imposing the same coordinate derivative at both ends does not impose dissipative outward-normal data at both ends. The energy identity is $\frac{1}{2}{\partial }_t\int u^2=[u{u}_x{]}_0^1-\int u_x^2$, whose boundary term is $u(1{)}^2-u(0{)}^2$ here and can supply energy. Orthogonal projection of initial data $f$ gives $c_0=2{\int }_0^{1}f{e}^{x}dx/(e^2-1)$; it vanishes precisely when the growing mode is absent.

The reflected boundary conditions $u_x(0,t)=-u(0,t)$ and $u_x(1,t)=-u(1,t)$ have the same spectrum, because reflection $x\mapsto 1-x$ reverses the coordinate derivative. Here the growing eigenfunction is $e^{-x}$ and the decaying functions are $W_n=n\pi \cos (n\pi x)-\sin (n\pi x)$, givingFor square-integrable initial data $f$, orthogonal projection givesThe norms are $(1-e^{-2})/2$ and $(n^2{\pi }^2+1)/2$. The spatial Sturm-Liouville problem has no zero eigenvalue and exactly one negative eigenvalue for $-D^2$, namely $-1$. The energy boundary term now reads $u(0{)}^2-u(1{)}^2$, so the left endpoint can supply energy. The growing mode disappears exactly when ${\int }_0^1{e}^{-x}f(x)dx=0$.

For constant $\nu$, variation of this positive functional gives $({\partial }_t+\nu {\nabla }^2)({\partial }_t-\nu {\nabla }^2)h=0$, or $h_{tt}-{\nu }^2{\nabla }^{4}h=0$. The diffusion equation $h_t=\nu {\nabla }^{2}h$ gives zero-residual minima, but is only a subset of the stationary equations of the squared functional. Each nonzero spatial Fourier mode has both growing and decaying temporal solutions in the stationary equation. Boundary data decide which stationary paths are admissible. In two spatial dimensions, the bulk action is invariant under the anisotropic scale invariance $x\mapsto bx$, $t\mapsto b^{2}t$, $h\mapsto h$.