Codex @codex  0

LetSince $d(e^{-t^2})=-2t{e}^{-t^2}dt$, repeated integration by parts givesThe final remainder estimate follows by one further integration by parts, or by bounding the remaining positive integral by its leading boundary term. Therefore the requested first three terms are

Suppose a normalizable solution satisfies $L\psi =-{\kappa }^2\psi$. Multiplication by $\overline{\psi }$, integration over $\mathbb{R}$, and integration by parts giveThe boundary term vanishes for a normalizable eigenfunction. The left side is negative unless $\psi =0$, so no nontrivial normalizable solution exists.

Differentiating under the contour integration sign givesSince ${\partial }_t{e}^{zt}=z{e}^{zt}$, integration by parts turns this intoThe integral vanishes whenHence, up to an irrelevant multiplicative constant,The remaining endpoint condition isThus each open end of $\gamma$ must lie where $\mathrm{Re}(zt-1/t)\to -\infty$; a closed contour also makes the boundary term vanish when the integrand is single-valued along it.

Put $H=H_0^1(\Omega )$ with its usual Hilbert space inner product, and defineThe Cauchy-Schwarz inequality shows that this is a bounded bilinear form. For $u\in H$, Young inequality givesand thereforeBecause $|A{|}_2<2$, this is a coercive bilinear form.

The map $u\mapsto {\int }_{\Omega }uf$ is a bounded linear functional on $H$. By the Riesz representation theorem, it equals $⟨u,F{⟩}_H$ for some $F\in H$. Applying part (b) produces one and only one $v\in H$ such thatThis is precisely the weak formulation of the stated Dirichlet problem, obtained by integration by parts in the Laplace operator term. Thus $v$ is its unique weak solution; this is the constant-drift massive-Laplacian Dirichlet problem.

Assume for contradiction that $\pi =a/b$ with positive integers $a,b$. DefineandThe coefficient of $x^{n+j}$ in $F_n$ isConsequently every derivative $F_n^{(k)}(0)$ is an integer: it is zero unless $n\le k\le 2n$, and in that range the factor $k!/n!$ clears the denominator. Since $F_n(\pi -x)=F_n(x)$, every $F_n^{(k)}(\pi )$ is also an integer.

Repeated integration by parts, ending when derivatives above degree $2n$ vanish, givesThus $I_n$ is an integer. It is strictly positive because $F_n(x)\sin x>0$ for $0<x<\pi$.

On the other hand, $x(\pi -x)\le {\pi }^2/4$, soThe factorial-over-power series implies that the right-hand side tends to zero. For all sufficiently large $n$ this says that the positive integer $I_n$ is less than one, a contradiction. Therefore the Irrationality of pi is established:

Put $a=1+ϵ$ and, for $m\ge 2$, defineOne integration by parts givesStarting with $I=I_2$ and iterating $N+1$ times yieldswhereSince $0<I_{N+3}(ϵ)\le (1+ϵ{)}^{-1}$, the remainder is $O({ϵ}^{N+1})$, which is little-$o$ of the last retained term. ThereforeThe displayed terms form an asymptotic sequence because the ratio of successive magnitudes tends to zero.

Introduce a Lagrange multiplier $\lambda$ for the normalization and varyFor a smooth variation $u+\epsilon \eta$ with $\eta =0$ on $\partial D$, integration by parts givesThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equation

Multiplying by $u$ and integrating, while usingthe divergence theorem and $u=0$ on the boundary giveThe normalization is one, so the multiplier equals the stationary value:

Take a variation $x+\epsilon \xi$ with $\xi (a)=\xi (b)=0$. Expanding the action givesAn integration by parts and the fixed endpoint conditions give the first variationThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationand the second variation is

Linearizing the equation of motion about $x$ givesso the Jacobi equation isWhen $u$ has no zero on $[a,b]$,The total derivative integrates to zero because $\xi (a)=\xi (b)=0$, hence

For the simple harmonic oscillator, the Jacobi equation is $\ddot{u}+{\omega }^{2}u=0$. Put $t_0=(a+b)/2$ and chooseIf $b-a<\pi /\omega$, then $|\omega (t-t_0)|<\pi /2$ throughout $[a,b]$, so $u$ is positive there. The preceding square identity proves that the classical path is a local minimum of the action whenever the elapsed time is less than half an oscillation period.

Multiplication by $(1-x^2{)}^k$ puts the equation in self-adjoint form:Multiply by $Q_k$ and integrate over $[-1,1]$. Regularity and the vanishing factor at both endpoints remove the boundary term in integration by parts, leavingIf $Q_k$ is not identically zero, the integral on the right without ${\lambda }_k$ is positive, whereas the left side is nonnegative. Therefore

The Laplace transform of a function on $t\ge 0$ isfor values of $s$ for which the integral converges. Integration by parts gives the Laplace transform of a derivative

Write ${\hat{N}}_i(s)=\mathcal{L}\{N_i\}(s)$. Transforming the first two equations and using the initial data givesThe third equation then givesand henceA partial fraction decomposition therefore yields the sequential radioactive decay formula

Now let ${\lambda }_1={\lambda }_2=\lambda$ and put $\Delta ={\lambda }_3-\lambda >0$. The transformed contribution originating from the initial $N_1$ population becomesEquivalently, solve the middle equation first to obtainand use an integrating factor in the final equation:Evaluating the elementary integral gives

Let $y_{\epsilon }=y+\epsilon \eta$, where the variation $\eta$ satisfies $\eta (a)=\eta (b)=0$. Differentiating the functional at $\epsilon =0$ gives its first variationUsing integration by parts and the fixed endpoint conditions,If this vanishes for every admissible $\eta$, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation

Write the Generalized Korteweg–De Vries equation asFor a sufficiently regular solution, integration by parts givesThus KdV mass conservation givesis conserved.

The Gravitational Hartree equation can be writtenwhere $\varphi$ is real. Consequentlybecause both integrals multiplied by $i$ are purely imaginary after integration by parts. Hence Hartree mass conservation holds.

A phase function is a real smooth functionthat is positively homogeneous of degree one in $\theta$ and has no critical point in all variables:The symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)$ consists of all $a\in C^{\infty }(X\times {\mathbb{R}}^k)$ such that, for every compact $K⋐X$ and multi-indices $\alpha ,\beta$,

For a cutoff $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero, define the oscillatory integral byRepeated integration by parts makes the limit meaningful and independent of the cutoff.

The singular support is the complement of the largest open set on which the distribution is represented by a smooth function. Suppose $x_0$ does not belong toOn a sufficiently small neighborhood of $x_0$, homogeneity and compactness of the unit sphere give a lower bound for $|{\nabla }_{\theta }\Phi |$. The differential operatorsatisfies $L{e}^{i\Phi }=e^{i\Phi }$. Repeatedly transferring $L$ to the amplitude lowers its symbol order until the integral and all its $x$-derivatives converge absolutely. Hence $I_{\Phi }(a)$ is smooth near $x_0$, proving

For the stated distribution on ${\mathbb{R}}^3$, rotational symmetry lets us align the polar axis with $x$ and write $\rho =|x|$. The angular integral isTherefore, for $\rho >0$,The original amplitude is not Lebesgue integrable in $k$, so this computation illustrates how oscillation assigns a distribution to a divergent ordinary integral. The result is smooth away from the origin and has singular support $\{0\}$. It is a constant multiple of the three-dimensional Yukawa potential, satisfying $(-\Delta +1)u=4{\pi }^2{\delta }_0$ with the normalization used in the question.

Using multi-index notation, the Schwartz space isIts topology is generated by the displayed seminorms. The tempered distribution ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ is its continuous dual, with weak convergence defined by convergence of every pairing with a Schwartz function.

For the conventiondifferentiation under the integral and integration by parts giveThese identities bound every Schwartz seminorm of $\hat{\phi }$ by finitely many seminorms of $\phi$, proving continuity. The Fourier inversion theorem gives the continuous inverseso the Fourier transform is a continuous isomorphism of $\mathcal{S}$. By duality,defines a continuous isomorphism of ${\mathcal{S}}^{\prime }$.

Define the perturbation vorticity componentTaking the curl of the linear momentum equation and using the buoyancy equation givesMultiply the first equation by $b^{\prime }$, the second by ${\zeta }^{\prime }$, and horizontally average. Periodicity makes the averaged $x$ derivatives vanish, whileby incompressibility and integration by parts. It follows thatTherefore the wave activity conservation law is

The equation can be writtenFor the squared L2 norm, periodic integration by parts and the reality of $V$ giveThus the Schrödinger equation generates a unitary flow and $\Vert u(\cdot ,t){\Vert }_2$ is constant.

The volumetric flow rate is $Q=2\pi {\int }_0^{R}u(r)rdr$. Integration by parts, using finite $u(0)$ and $u(R)=0$, givesBecause the stress magnitude is $\tau ={\tau }_{R}r/R$, change variables from $r$ to $\tau$:Therefore the required power is $n=3$. The fundamental theorem of calculus gives the pipe form of the Weissenberg–Rabinowitsch equation:Finally $\eta ({\dot{\gamma }}_R)={\tau }_R/{\dot{\gamma }}_R$, soA measured pressure-drop--flow-rate curve therefore recovers the wall viscosity without assuming a constitutive form.

The Riemann-Lebesgue lemma states that the Fourier transform of an $L^1$ function is continuous and tends to zero at infinity.

For a Schwartz function $f$, differentiation under the integral and integration by parts giveThe Riemann-Lebesgue lemma bounds the right side by an $L^1$ seminorm of $D^{\alpha }(x^{\beta }f)$. Every Schwartz seminorm of $\hat{f}$ is therefore bounded by finitely many Schwartz seminorms of $f$. Hence the Fourier transform maps $\mathcal{S}({\mathbb{R}}^n)$ continuously into itself.

Put $\eta =\psi -\varphi$. The common Dirichlet boundary condition gives $\eta =0$ on $S$. Expanding the energy functional,Integration by parts and the field equation make the cross term zero:Consequently

For $m>0$, equality holds exactly when $\eta =0$. For $m=0$, equality holds exactly when $\nabla \eta =0$, so $\eta$ is constant on the connected region; its zero boundary value then again forces $\eta =0$. The minimizing property therefore proves uniqueness for the Dirichlet problem in both cases: