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For causal functions the convolution is $(f\ast g)(t)={\int }_0^{t}f(\tau )g(t-\tau )d\tau$. In a half-plane of absolute convergence, changing variables $u=t-\tau$ in its Laplace transform and using Fubini's theorem givesThis proves the convolution theorem for Laplace transforms. Transforming the differential equation gives $(s^2+1)Y(s)=sy(0)+y^{\prime }(0)+F(s)$. The direct transform pairs $\mathcal{L}(\cos t)=s/(s^2+1)$ and $\mathcal{L}(\sin t)=1/(s^2+1)$ then giveFor a continuous forcing, call the last term $w(t)$. The Leibniz integral rule gives $w^{\prime }={\int }_0^{t}f(\tau )\cos (t-\tau )d\tau$ and $w^{\prime \prime }=f(t)-w(t)$. Also $w(0)=w^{\prime }(0)=0$. Thus both the original equation and the initial values hold.

For an impulse at positive time $T$, the Dirac delta evaluates the convolution, yieldingHere $H$ is the Heaviside step function. The displacement remains continuous at $T$, while its derivative jumps upward by one, exactly as required by integrating the differential equation across the unit impulse.

The residue theorem states that a positively oriented simple closed contour enclosing isolated poles and none on the contour has integral $2\pi i$ times the sum of their residues, provided the function is holomorphic elsewhere inside. Near a simple zero of $\psi$, $\psi (z)={\psi }^{\prime }(a)(z-a)+O((z-a{)}^2)$. HenceFor $\cot z=\cos z/\sin z$, this gives $\boxed{{\mathrm{Res}}_{z=k\pi }\cot z=1}$ for every integer $k$.

To evaluate the real integral, integrate $F(z)=\mathrm{Log}z/(z^2+a^2)$ around an upper semicircle, with a small upper semicircular indentation excluding zero. Use the branch real on the positive axis and with argument between zero and $\pi$ in the upper half-plane. The large arc tends to zero as $O(\ln R/R)$ and the small arc as $O(ϵ|\ln ϵ|)$. The only enclosed pole is $ia$, whose contribution isOn the negative real axis the upper boundary value is $\mathrm{Log}(-x)=\ln x+i\pi$. The two real segments therefore contributeComparing real parts yieldsThe integral is absolutely convergent at both zero and infinity, so no principal-value interpretation is needed.

Use Green's second identity in the source coordinate:Take $u({\mathbf{x}}^{\prime })=\psi ({\mathbf{x}}^{\prime })$ and $v({\mathbf{x}}^{\prime })=-1/[4\pi |\mathbf{x}-{\mathbf{x}}^{\prime }|]$, the fundamental solution of the Laplace equation satisfying ${\Delta }^{\prime }v=\delta ({\mathbf{x}}^{\prime }-\mathbf{x})$. With localized charge and the decaying potential, the boundary integral tends to zero on expanding spheres. The volume identity then gives $\psi (\mathbf{x})+(1/{ϵ}_0)\int vqd{V}^{\prime }=0$, hence the Coulomb potentialConvergence of this integral and a condition fixing the potential at infinity are needed: mere pointwise decay of $q$ is not enough, and a harmonic function could otherwise be added. A bounded, compactly supported charge distribution is a sufficient setting for this derivation.

For the disk put $\sigma =Q/(\pi a^2)$. On its axis, every ring of radius $\rho$ is at distance $\sqrt{{\rho }^2+z^2}$, soFor $z>0$ subtract $z$; for $z<0$ add $z$. This is the electrostatic potential of a uniformly charged disk with zero potential at infinity. The absolute value is essential for reflection symmetry and the correct behavior on the lower axis.

For variations $y+ϵ\eta$ with $\eta (a)=\eta (b)=0$, integration by parts gives $\delta I={\int }_a^b[f_y-d(f_{y^{\prime }})/dx]\eta dx$. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationFor the ramp, use conservation of energy from rest at the rim. The vertical drop is $R\cos \theta$, so $v=\sqrt{2gR\cos \theta }$. The spherical surface arc length is $ds=R\sqrt{d{\theta }^2+{\sin }^2\theta d{\varphi }^2}$. On a monotone segment described by $\varphi (\theta )$, the travel time is consequentlyFor increasing $\theta$ this has the printed limits ${\theta }_0$ to ${\theta }_1$; for a descent, reverse the decreasing angular limits to keep time positive. The azimuth is a cyclic coordinate of this integrand. Its conserved derivative gives the first integral for the brachistochrone on a hemispherical ramp:Solving for its square yieldsFor a regular stationary path with finite $C$, this tends to zero as the rim $\theta \to \pi /2$ is approached. The ratio of azimuthal to meridional displacements is $\sin \theta {\varphi }^{\prime }$, which also tends to zero. Hence the initial direction is along a meridian toward the bottom. This conclusion concerns the initial heading; the later route can turn before reaching the bottom when its rim endpoints require a nonzero azimuthal displacement.

A real orthogonal matrix satisfies $Q^{T}Q=I$; a complex unitary matrix satisfies $U^{†}U=I$. Their columns are orthonormal in the corresponding real or complex inner product.

Adopt the convention that a complex inner product is linear in its second argument, conjugate symmetric, and positive definite: $⟨u,u⟩\ge 0$, with equality exactly when $u=0$. Conjugate symmetry and second-argument linearity then imply antilinearity in the first:For the specified real symmetric matrix use $b=a^2$. Its upper block is the outer product $(1,a{)}^T(1,a)$, so it has one zero eigenvalue and one eigenvalue $1+a^2$. A complete orthonormal eigenbasis isIf $c=1+b$, the last two eigenvalues coincide and their eigenspace is the span of $v_{+},v_c$; every nonzero vector in that span is an eigenvector. Since $c>0$, the zero eigenspace remains one-dimensional.

With constant positive thermal diffusivity $\nu$, no internal heat source and one spatial coordinate, the heat equation isSeparation of variables makes each spatial mode decay by $e^{-\nu k^{2}t}$. The endpoint conditions determine whether the initial triangular profile is expanded in a Fourier sine series or a Fourier cosine series.

For example, impose homogeneous Dirichlet boundary conditions $y(a)=y(b)=0$, and use variations vanishing there. Homogeneous Neumann boundary conditions $p{y}^{\prime }=0$ are another choice removing the boundary terms. Put $N[y]={\int }_a^b(p{y}^{\prime 2}+q{y}^2)dx$, $D[y]={\int }_a^{b}w{y}^{2}dx>0$, and $I=N/D$. At a stationary function let $\lambda =I[y]$. The quotient rule gives $\delta I=(\delta N-\lambda \delta D)/D$. After integration by parts, stationarity under every allowed variation givesso the fundamental lemma of the calculus of variations yields the stated Sturm-Liouville problem. Conversely an eigenfunction with those boundary conditions has $N=\lambda D$ and makes the quotient stationary. The lowest eigenvalue minimizes the Rayleigh quotient under the usual self-adjoint, lower-bounded hypotheses; higher eigenvalues are stationary values with orthogonality constraints when minimizing successively.

The hydrogen equation becomes $-(r^2{\psi }^{\prime }{)}^{\prime }-2r\psi =2E{r}^2\psi$. Its radial energy quotient is thereforeRegularity at zero and decay at infinity remove the boundary term $r^2\psi {\psi }^{\prime }$. The condition $\psi (0)=1$ fixes an arbitrary amplitude for these trial functions; it is not their three-dimensional probability normalization, which cancels from the quotient.

For $\psi =e^{-\alpha r}$, elementary exponential integrals give $\int r^2{\psi }^{2}dr=1/(4{\alpha }^3)$, $\int r^2{\psi }^{\prime 2}dr=1/(4\alpha )$, and $\int r{\psi }^{2}dr=1/(4{\alpha }^2)$. HenceFor the compactly supported linear trial for hydrogen, the norm, gradient and Coulomb integrals are ${\beta }^3/30$, $\beta /3$, and ${\beta }^2/12$, respectively. ThusThe derivative changes from negative to positive at four, so this is the minimum over positive $\beta$. The quantum variational principle gives upper bounds on the ground-state energy; the exponential trial is better, and in fact $e^{-r}$ satisfies the exact differential equation at $E=-1/2$. It is the nodeless hydrogen ground state in the given units. The linear trial is admissible for the energy quadratic form despite its derivative jump at $r=\beta$; it need not possess a classical second derivative at that point.