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Multiplication by the integrating factor $e^{2x}$ givesHence the Sturm-Liouville problem is

Yes. Fubini's theorem and the triangle inequality giveThus $T$ is Lipschitz continuous with respect to $d_1$ as well.

Let $\Vert K{\Vert }_{\infty }={\max }_{[0,1{]}^2}|K|$. For $f,g\in C([0,1])$,Taking the supremum over $x$ proves that the integral operator $T$ is Lipschitz continuous, hence continuous, in the uniform norm.

It remains to see that $Tf$ is continuous. Since $K$ is uniformly continuous on the compact square and $f$ is bounded,as $x^{\prime }\to x$. Therefore $T:C([0,1])\to C([0,1])$ is well defined and continuous.

Starting with $A_1=A$, for $k=1,\dots ,n-2$ choose a Householder reflection $H_k$ that acts only on coordinates $k+1,\dots ,n$ and maps the tail of column $k$,to a multiple of its first coordinate vector. SetThis zeros all entries in column $k$ below its first subdiagonal. Because $H_k$ fixes the first $k$ coordinates, it preserves the zeros created in earlier columns. After $n-2$ steps,is an upper Hessenberg matrix.

Each $H_k$ is orthogonal and symmetric. Ifthen $Q$ is orthogonal andPart (a) shows that each similarity update costs $O(n^2)$ arithmetic operations, and there are $O(n)$ updates. The total cost is therefore

For a nonzero vector $v\in {\mathbb{R}}^n$, a Householder reflection isIt is immediately symmetric. If $P=v{v}^T/(v^{T}v)$, then $P^2=P$, soHence $H$ is also an orthogonal matrix, with $H^{-1}=H$.

The similarity transformation can be expanded asComputing $v^{T}A$, $Av$, and the scalar $v^{T}Av$ costs $O(n^2)$ arithmetic operations, after which the remaining updates are outer products and scalar multiples, also costing $O(n^2)$. Thus $HA{H}^{-1}$ can be formed in $O(n^2)$ operations rather than by two general $O(n^3)$ matrix multiplications.

Successive reflections in the two grounded planes require four charges:Their total charge and electric dipole moment vanish. Applying the second-order expansion from part (a), the terms proportional to $a^2{x}^2$, $b^2{y}^2$, and $a^2+b^2$ also cancel, while the mixed terms add. ThusThis is an electric quadrupole potential.

The method of images replaces the earthed plane by an image charge $-q$ at $(-a,0,0)$. The two Coulomb potentials cancel at $x=0$, so uniqueness for the Dirichlet problem makes the resulting field the physical field in $x>0$. With $\mathbf{r}=(x,y,z)$ and ${\mathbf{e}}_x=(1,0,0)$,and

The electric multipole expansion from part (a) gives, for $r\gg a$,Thus the leading field is that of an electric dipole.

On the plane,Taking the plane normal to be $+{\mathbf{e}}_x$ and using polar coordinates,With the outward normal of the region $x>0$, the sign is reversed. This is consistent with Gauss's law: all electric flux from the real charge terminates on the grounded conductor.

The electrostatic boundary condition gives the induced surface charge densityand its integral isIn the plane $z=0$, the field lines leave the positive charge, meet the conductor normally, and are the right-half-plane portions of the field lines joining the real charge to its negative image.

WriteUsing the Taylor seriesand retaining terms through second order in $|\mathbf{y}|/|\mathbf{x}|$ givesThis is the beginning of the multipole expansion.

PutThe time-independent Schrödinger equation has the formsThe first region contains the incident and reflected waves, while the final region contains only the transmitted wave.

Continuity of $\psi$ and ${\psi }^{\prime }$ at both edges of the finite square well gives the standard transmission coefficientWhen $V_0=3E$, one has $K=2k$, and thereforeHence the transmission probability is

For a one-dimensional wavefunction,The probability continuity equation isFor a stationary state $\psi (x,t)=\varphi (x)e^{-iEt/\hslash }$, the probability density is independent of time, so ${\partial }_{x}j=0$ and the probability current is constant in space.

For the plane waveone obtainsThis is a momentum eigenstate with momentum $p=\hslash k$ and, when it satisfies the free Schrödinger equation, energy $E={\hslash }^2{k}^2/(2m)$. Its constant density and current describe a spatially uniform beam carrying probability in the sign of $k$. It is not a normalizable wavefunction, so it represents an idealized state rather than a localized particle.

By the superposition principle, interchange $x$ and $y$ in the solution from part (c) and add the two solutions:The first term supplies the required data on the positive $x$-axis and vanishes on the positive $y$-axis; the second does the reverse. Explicitly,It is harmonic in the quarter-plane, has the stated boundary values, and decays in both unbounded directions.

Extend the boundary data oddly to the whole real axis. The extended function is precisely the function $f$ from part (a). Taking the Fourier transform in $x$, the Laplace equation becomesDecay as $y\to \infty$ selectsPart (b), together with the convolution theorem, says that the inverse transform is convolution with the Poisson kernelFor $x,y>0$ this givesThe odd extension enforces $u(0,y)=0$, while the Poisson integral takes the prescribed boundary values at every continuity point and decays at infinity.

Because $e^{-\lambda |k|}$ is an even function, Fourier inversion reduces to a cosine integral:Therefore

Direct evaluation of the integral givesThe continuous value at $k=0$ is zero, in agreement with the vanishing integral of the odd function $f$. Thus