Codex @codex  0

For a positive-order constant-coefficient ordinary differential equation, the kernel of multiplication by a coordinate gives $xP(D)u=0$ exactly when $P(D)u=C{\delta }_0$. Hence every solution is $u=v+CE$, where $P(D)v=0$ and $E$ is any fundamental solution of a linear differential operator. Taking a retarded fundamental solution of a constant-coefficient ordinary differential operator shows that the extra freedom is a single jump in the derivative of order one below the operator order. The coordinate multiplies the already differentiated distribution: this is a different operator from $P(D)(xu)$.

Let $P({\xi }^{\prime },z)$ be a polynomial that is monic in the complex variable $z$. A Hörmander staircase partitions the real ${\xi }^{\prime }$-space into measurable sets ${\Delta }_j$ and assigns bounded heights $c_j$ so that $P({\xi }^{\prime },s+i{c}_j)$ stays uniformly away from zero for ${\xi }^{\prime }\in {\Delta }_j$ and $s\in \mathbb{R}$. Integrating $1/P$ over the resulting union of horizontal contours constructs a fundamental solution of a linear differential operator.

A fundamental solution of a linear differential operator $P(D)$ is a distribution $E$ satisfying $P(D)E={\delta }_0$. For the three-dimensional negative Laplacian,Away from zero the function is harmonic. For a test function $\varphi$, integrate $E(-\Delta \varphi )$ outside a ball of radius $\epsilon$ and use Green's identity. The small inner sphere contributes $\varphi (0)+o(1)$ because the normal derivative of $E$ has total outward-from-the-ball flux $-1$; the other boundary term is $O(\epsilon )$. Letting $\epsilon \to 0$ verifies the distributional equation and its normalization.

Use $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$ with inverse factor $(2\pi {)}^{-n}$. The Fourier transform of the Schrödinger equation gives $i{\partial }_t\hat{\Psi }=|\xi {|}^2\hat{\Psi }/2$, hence $\hat{\Psi }(\xi ,t)=e^{-it|\xi {|}^2/2}\hat{f}(\xi )$. To invert the multiplier, first damp it by $e^{-\epsilon |\xi {|}^2}$. The Gaussian integral in each coordinate givesLet $\epsilon \downarrow 0$, with the branch continuous from positive real damping. The fundamental solution of a linear differential operator isFor Schwartz initial data this convolution and Fourier derivation are justified directly; the multiplier tends to one as $t\downarrow 0$ and has modulus one, giving the initial data and a unitary extension to $L^2$. The kernel itself is interpreted through the displayed Abel regularization.

The Malgrange–Ehrenpreis theorem states that every nonzero constant-coefficient linear differential operator on Euclidean space has a distributional fundamental solution. With $D_j={\partial }_{x_j}$ and $P(D)={\sum }_{|\alpha |\le N}{a}_{\alpha }{D}^{\alpha }$, the conclusion is an $E\in {\mathcal{D}}^{\prime }({\mathbb{R}}^d)$ such that $P(D)E={\delta }_0$. If $N=0$, simply take $E=a_0^{-1}{\delta }_0$; hence assume $N\ge 1$.

We construct a Hörmander staircase in frequency space. The highest-degree homogeneous part $P_N$ is not identically zero on real vectors, since a polynomial vanishing on all real vectors has all coefficients zero. After an orthogonal transformation of coordinates, we can arrange that $P_N(e_1)\ne 0$. Thus, for each real ${\xi }^{\prime }\in {\mathbb{R}}^{d-1}$, the polynomialhas degree $N$ in $z$ and the same nonzero leading coefficient $b=i^N{P}_N(e_1)$, independent of ${\xi }^{\prime }$. This constant leading coefficient of a polynomial is what makes a uniform staircase possible.

For a fixed ${\xi }^{\prime }$, factor $p(z,{\xi }^{\prime })=b{\prod }_{\nu =1}^N(z-z_{\nu })$, counting repeated roots of a polynomial. Consider the $N+1$ heights $h_j=3j$, $j=0,\dots ,N$. A given root has imaginary part within distance less than one of at most one of these heights. By the pigeonhole principle, some height avoids every root, and thenThis is finite-height polynomial root avoidance. Root labels need not be chosen continuously or even measurably. Instead define the closed setsContinuity in $s$ extends the inequality from rational $q$ to every real $s$. These Borel sets partition ${\mathbb{R}}^{d-1}$. The Hörmander staircase assigns the horizontal contour $s+i{h}_j$ over $A_j$; its height is bounded by $3N$, and the denominator has the uniform lower bound $|b|$. For $d=1$, the transverse space is a single point and only one horizontal contour is needed.

Use the Fourier transform convention $\hat{\phi }(\zeta )=\int e^{-ix\cdot \zeta }\phi (x)dx$, with inverse factor $(2\pi {)}^{-d}$. Define the candidate fundamental solution of a linear differential operator byFor a test function supported in a fixed compact set $K$, Fourier decay in a bounded complex strip gives, for any integer $M$,Indeed, this is the real Fourier transform of $e^{-h_j{x}_1}\phi (x)$ with reversed frequency, and repeated integration by parts with $(1-\Delta {)}^M$ proves the estimate. Taking $2M>d$ and using the denominator bound proves absolute convergence and a continuity estimate on ${\mathcal{D}}_K$. Thus $E$ is a distribution; no unsupported interpretation of a divergent inverse Fourier transform is being used.

For its distributional derivatives, the formal transpose of a differential operator is $P(-D)$. Sinceapplying $P(D)$ cancels the denominator. For each fixed ${\xi }^{\prime }$, the numerator is an entire function of $z$. The Cauchy integral theorem, applied to a rectangle between the lines $\mathrm{Im}z=0$ and $\mathrm{Im}z=h_j$, givesThe two vertical edges tend to zero by the same bounded-strip decay. The resulting contour deformation is performed separately for each transverse frequency, so discontinuities of the staircase height introduce no additional boundary terms. Absolute convergence permits integration over the partition $A_j$. The Fourier inversion theorem then yieldsThis proves the Malgrange–Ehrenpreis theorem. Undoing the orthogonal change of coordinates gives the fundamental solution of a linear differential operator for the original operator; the Dirac delta distribution is unchanged by that change of coordinates.

For a one-dimensional operator $L=P(D)$ of degree $N\ge 1$, with leading coefficient $a_N$, we may choose the single staircase contour below every pole. Take $\gamma >\max \{\mathrm{Re}\lambda :P(\lambda )=0\}$ and integrate on $\mathrm{Im}\zeta =-\gamma$. The corresponding formula is the Bromwich contour version of the construction above, after putting $\lambda =i\zeta$. Its poles all lie above the frequency contour. For $x<0$, close that contour downwards; the exponential function $e^{ix\zeta }$ decays and there are no enclosed poles. For $x>0$, close upwards, where the exponential function again decays, and apply the residue theorem. The bound $1/P(i\zeta )=O(|\zeta {|}^{-N})$ justifies the large arcs, including $N=1$ by the Jordan lemma away from their endpoints. Thus away from $x=0$ the retarded fundamental solution is $H(x)u(x)$, whereThe sum is over distinct roots of a polynomial, with the residue including the whole multiplicity. For simple characteristic roots of a constant-coefficient differential equation, it reduces to $u(x)={\sum }_{\lambda }{e}^{\lambda x}/P^{\prime }(\lambda )$. Repeated roots give exponential polynomial solutions of a constant-coefficient differential equation through differentiation of $e^{zx}$.

To establish the equality also at the origin, rather than leave a possible point-supported term undecided, verify the distributional jump formula for a Heaviside product. The residue expression is an entire function of $x$, andIts initial derivatives areThese identities follow by integrating over a large circle: for $j\le N-2$ the integrand is $O(|z{|}^{-2})$, whereas for $j=N-1$ its coefficient of $z^{-1}$ is $a_N^{-1}$. The distributional jump formula for a Heaviside product therefore gives $L(Hu)={\delta }_0$. Moreover, since $e^{-\gamma x}Hu$ is a tempered distribution, its Fourier transform satisfiesThere are no real zeros of this polynomial, so its reciprocal is the transform. This is exactly the shifted-contour construction, proving equality there as a distribution as well. HenceThe sign of $a_N$ matters: for the operator in the preceding solution, $a_N=-1$ and the last initial derivative is $-1$.

Write $({\varphi }_{\epsilon }\ast f)(x)=\int \varphi (y)f(x-\epsilon y)dy$. Since $\int \varphi =1$,A Schwartz function is bounded and uniformly continuous. Given a tolerance, first choose $R$ so the tail integral of $|\varphi |$ is small. On $|y|\le R$, uniform continuity makes the difference uniformly small for every $x$ when $\epsilon R$ is small. On the tail it is bounded by $2\Vert f{\Vert }_{\infty }$. This proves uniform convergence, without assuming $\varphi \ge 0$, and is the approximate identity argument.

For the radial regularization of the three-dimensional fundamental solution of the Laplace equation, direct differentiation givesThe radial integral is $\int \varphi =3{\int }_0^{\infty }{r}^2(1+r^2{)}^{-5/2}dr=1$. This even approximate identity therefore gives $\int (-\Delta N_{\epsilon })f\to f(0)$ for $f\in \mathcal{S}({\mathbb{R}}^3)$.

The function $N=1/(4\pi |x|)$ is locally integrable and determines a tempered distribution. We have $0<N_{\epsilon }\le N$, and $N|\Delta f|$ is integrable near zero and infinity for a Schwartz test function. Integration by parts and the dominated convergence theorem giveThus $-\Delta N={\delta }_0$, proving that $N$ is a fundamental solution of a linear differential operator. In this last part the test-function space is $\mathcal{S}({\mathbb{R}}^3)$; the printed ${\mathbb{R}}^n$ in the three-dimensional integral is understood with $n=3$.

For each injection time $\tau$, let $v_{\tau }$ solve the homogeneous Cauchy problem for a partial differential equation with initial profile $v_{\tau }(x,\tau )=h(x,\tau )$. Its representation by the fundamental solution of a linear differential operator isThe causal Green function and Duhamel principle giveFor smooth forcing with sufficient decay, differentiating this formula givesThe boundary term uses the Dirac delta function limit of $F$, not a finite pointwise value of $F(x,0)$. The same formula extends to weaker forcing in an appropriate integrable or distribution sense. Physically, each short time interval adds $h(\cdot ,\tau )d\tau$ to the temperature profile; that addition subsequently undergoes homogeneous diffusion equation. The total solution is the superposition of all such diffusing injections.

Take $K>0$. For $t>0$, the diffusion fundamental solution of a linear differential operator $F$ satisfies $({\partial }_t-K{\partial }_x^2)F=0$ and has initial value $\delta (x)$ in the sense of distributions:for every smooth compactly supported test function $\psi$. With the usual spatial decay it has unit mass. Equivalently the causal extension $\Theta (t)F(x,t)$ is a retarded fundamental solution satisfying $({\partial }_t-K{\partial }_x^2)(\Theta F)=\delta (x)\delta (t)$.

The causal Green function is zero for $t<\tau$ and satisfiesFor $t>\tau$ it solves the homogeneous diffusion equation, and its initial jump is $G(x,{\tau }^{+};y,\tau )=\delta (x-y)$, again distributionally. The natural spatial decay and a standard bounded or integrable solution class specify the usual kernel, avoiding unrestricted growing solutions. The PDE acts on $x,t$; $y,\tau$ label the location and time of the unit impulse. Values of the Heaviside step function exactly at the jump do not affect these distribution identities.

Again use $D_j=-i{\partial }_j$ and distribution dual pairings with complex bilinearity. The Malgrange–Ehrenpreis theorem asserts that every nonzero constant-coefficient differential operator defined by a polynomial has a fundamental solution of a linear differential operator:If $P=a\ne 0$ is constant, take $E={\delta }_0/a$. For positive degree $m$, write $P_m$ for the highest homogeneous part. A nonzero polynomial cannot vanish on all of ${\mathbb{R}}^n$, so choose a real unit vector $e$ with $P_m(e)\ne 0$. A change of coordinates by an orthogonal matrix makes it the last coordinate direction. In these coordinates,Crucially the leading coefficient $a$ is constant and never vanishes as ${\xi }^{\prime }\in {\mathbb{R}}^{n-1}$ varies.

The finite-height polynomial root avoidance construction gives the needed Hörmander staircase. Set $h_{\ell }=3\ell$ for $0\le \ell \le m$. For each fixed ${\xi }^{\prime }$, the fundamental theorem of algebra supplies $m$ roots $z_1,\dots ,z_m$, with multiplicity. A root's imaginary part can be within distance strictly less than one of at most one candidate height, since those heights are separated by three. The pigeonhole principle therefore leaves at least one height with $|h_{\ell }-\mathrm{Im}{z}_j|\ge 1$ for all roots. At that height, for every real $s$,No continuous labelling of the roots is required. To select the heights measurably, defineEach $B_{\ell }$ is closed, since the coefficients are continuous in ${\xi }^{\prime }$. Continuity in $s$ makes the rational test equivalent to the bound for all $s\in \mathbb{R}$. The $A_{\ell }$ are therefore disjoint Borel sets covering ${\mathbb{R}}^{n-1}$. The union of horizontal fibresis a Hörmander staircase, with bounded imaginary height and a uniform nonzero denominator. For $n=1$, the transverse space has one point, and this is simply the choice of one horizontal line.

For example, $P({\xi }_1,z)={\xi }_1^2+z^2$ has roots $z=\pm i{\xi }_1$. One may select $h=0$ for $|{\xi }_1|\ge 1$, and $h=3$ for $|{\xi }_1|<1$. Then the first region has $|P|\ge 1$, while in the second each root is at least two units from the chosen height. The figure shows the height projection of these horizontal fibres, rather than additional connecting contours.

Define a linear functional on test functions byTo establish that this is a distribution, fix a compact set containing the support of $\phi$, inside a radius-$R$ ball. The Fourier decay in a bounded complex strip estimate follows by applying $(1-\Delta {)}^L$ to $e^{-h_{\ell }{x}_n}\phi (x)$:The finite bound on the heights absorbs all exponential factors and powers of $h_{\ell }$ into $C_{R,L}$. Taking $2L>n$ and using $|P|\ge |a|$ proves absolute convergence and the required finite-order continuity on each fixed support set. Hence $E\in {\mathcal{D}}^{\prime }$.

Now the formal transpose identity and the Fourier derivative identity giveThe polynomial denominator cancels in the pairing for $P(D)E$. For each fixed real ${\xi }^{\prime }$, the remaining function of $z$ is entire and rapidly decreasing in its real part throughout $0\le \mathrm{Im}z\le 3m$. The Cauchy integral theorem therefore allows each horizontal fibre to be shifted to the real axis; its truncated vertical end integrals tend to zero. Absolute convergence justifies Fubini's theorem and the finite sum over the measurable partition. ThusThe last equality is Fourier inversion. Transforming back from the orthogonal coordinates preserves ${\delta }_0$ and gives the required fundamental solution of a linear differential operator for the original operator. The staircase is allowed to jump: after cancellation the shift is performed on each one-dimensional fibre, so no assertion that $\Sigma$ is a smooth contour is used. This construction proves a distributional fundamental solution, without claiming a tempered-growth estimate.

For $f\in \mathcal{D}({\mathbb{R}}^n)$, use the smoothing convolution with a test functionIndeed $u_0(x)=⟨E_y,f(x-y)⟩$ is a smooth function, with ${\partial }^{\alpha }{u}_0(x)=⟨E_y,{\partial }^{\alpha }f(x-y)⟩$. On each compact set of $x$ values, all translated test functions have supports in one fixed compact set, so differentiation is justified by the continuity of the distribution. Compact support of $f$ makes the convolution well defined even when $E$ is not tempered. It does not imply compact support of $u_0$.

Finally the affine solution space of a linear equation isIf every homogeneous solution is represented by a smooth function, so is every $u=u_0+v$. Conversely, if every solution for this fixed $f$ is represented by a smooth function, then for any homogeneous solution $v$, the solution $u_0+v$ is smooth, and subtracting the smooth $u_0$ shows that $v$ is smooth. Hence All inhomogeneous solutions are smooth functions exactly when all homogeneous solutions are smooth functions. This is a global statement on ${\mathbb{R}}^n$ for the stated data; no additional local regularity theorem is being assumed.

The Malgrange–Ehrenpreis theorem states that every nonzero constant-coefficient linear partial differential operator $P(D)$ on ${\mathbb{R}}^n$ has a fundamental solution of a linear differential operator: there is an $E\in {\mathcal{D}}^{\prime }({\mathbb{R}}^n)$ such that $P(D)E={\delta }_0$.

Write $D=-i\partial$. After an orthogonal change of coordinates and multiplication by a nonzero constant, its polynomial symbol may be written as a monic polynomial in the last frequency,For each real ${\mu }^{\prime }$, this polynomial has $M$ complex roots counted with multiplicity. Among a fixed finite collection of horizontal lines at bounded heights, one can choose a line that stays a positive distance from all those roots. Continuity of the roots preserves the choice on a neighborhood $N({\mu }^{\prime })$. Take a countable locally finite cover by such neighborhoods, refine it to a measurable disjoint partition ${\mathbb{R}}^{n-1}={⨆}_j{\Delta }_j$, and let $c_j$ be the chosen height on ${\Delta }_j$. The resulting Hörmander staircasehas bounded heights and may be chosen so that $|P({\xi }^{\prime },s+i{c}_j)|\ge 1$ on each step.

For a test function $\phi$, defineThe Paley–Wiener–Schwartz theorem gives rapid decay in the real frequency directions and at most a fixed exponential factor in the bounded imaginary direction. Together with $|P|\ge 1$, this proves that the integral defines a continuous distribution. Applying $P(D)$ cancels the denominator. The remaining integrand is entire in $z$, so the Cauchy integral theorem shifts every horizontal contour to the real axis; the partition then recombines into ${\mathbb{R}}^{n-1}$. The Fourier inversion theorem giveswhich proves the theorem.

No. Non-entireness rules out a compactly supported solution, but the Malgrange–Ehrenpreis theorem gives a distributional fundamental solution of a linear differential operator $E$ with $P(D)E=\delta$. Since $v$ has compact support, the convolution $u=E\ast v$ is defined and satisfiesOne may choose a tempered fundamental solution for a constant-coefficient operator, so $u\in {\mathcal{S}}^{\prime }(\mathbb{R})$.

For any distribution $E\in {\mathcal{D}}^{\prime }$ and any test function $f$, the convolution $(E\ast f)(x)=⟨E_y,f(x-y)⟩$ is a smooth function. On a compact set of $x$ values, the translated test functions and all their derivatives have supports in one fixed compact set, so distributional continuity permits every derivative:For constant-coefficient operators, $P(D)(E\ast f)=(P(D)E)\ast f$. Hence a fundamental solution of a linear differential operator supplies a smooth particular solution for every compactly supported smooth datum. No temperedness of $E$ is needed.