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Use the mixed Dirichlet-Neumann modes on an interval. To justify their Fourier series, extend $y$ evenly across $x=1$ to $[0,2]$, then oddly across $x=0$, and repeat with period four. The condition $y(0)=0$ makes the odd extension continuous, and $y^{\prime }(1)=0$ makes the even reflection continuously differentiable. The resulting periodic function has a convergent Fourier sine series. Reflection about $x=1$ eliminates the even sine harmonics, leavingAlternatively these values follow directly from $\sin {\lambda }_{n}0=0$ and ${\lambda }_n\cos {\lambda }_n=0$. Fourier orthogonality givesThe extension argument also supplies completeness, not merely orthogonality, of this Fourier sine basis.

Expand the forcing as $x\cos \pi x=\sum b_n\sin {\lambda }_{n}x$. Product-to-sum and integration by parts giveEach mode of the differential operator has eigenvalue $-({\lambda }_n^2+{\alpha }^2)$. Hence, for real $\alpha$, the solution isThe coefficients are $O(n^{-4})$, so the series and its first two differentiated series converge uniformly. Therefore it solves the equation and the boundary conditions term by term. Uniqueness follows by multiplying the homogeneous equation by its real solution and integrating: the boundary terms vanish and ${\int }_0^1[(y^{\prime }{)}^2+{\alpha }^2{y}^2],dx=0$, forcing $y=0$.

For fixed real $\kappa$, consider initial data $u_0\in L^2(0,1)$ and a solution continuous into $L^2$ at time zero, with homogeneous Dirichlet boundary conditions for positive times. Existence, uniqueness and continuous dependence in this norm give Hadamard well-posedness.

To construct a solution, put $v(x,t)=e^{\kappa x}u(x,t)$. Substitution givesLet $b_j=2{\int }_0^1{e}^{\kappa x}{u}_0(x)\sin (j\pi x)dx$. The Fourier sine series solution isMultiplication by $e^{\pm \kappa x}$ is bounded on $L^2(0,1)$. Completeness of the sine basis gives convergence to $u_0$ in $L^2$ at time zero, while the exponential decay gives smoothness for positive times and the required equation and boundary values. For a classical solution continuous up to the endpoints at time zero, the usual initial-boundary compatibility conditions are additionally required.

For the energy method, integration by parts givesBoth boundary terms vanish. The same identity applies to the difference of two solutions, proving uniqueness and continuous dependence. The Poincaré inequality further giveswith the identical estimate for differences of initial data. This Dirichlet convection-diffusion contraction holds for every real $\kappa$, and its energy constant is independent of the drift strength.

For small transverse displacement, the wave equation on a string iswhere $T$ is the tension and $\rho$ is mass per unit length. Using separation of variables, put $y=X(x)F(t)$, the ratios $X^{\prime \prime }/X=F^{\prime \prime }/(c^{2}F)$ equal a constant. The fixed-end conditions $X(0)=X(L)=0$ exclude nontrivial solutions for nonnegative separation constants; for $-{\kappa }^2$ they give $X=\sin (\kappa x)$ and $\kappa =n\pi /L$, $n\ge 1$. The associated angular frequencies are ${\omega }_n=n\pi c/L$. Thus the general superposition of separated normal modes isThe string is initially straight and at rest before the impulse, so $A_n=0$. Assume the hammer interval lies inside $[0,L]$. The imposed velocity $g(x)=y_t(x,0)$ has Fourier sine series coefficientTherefore $B_n=g_n/{\omega }_n$, giving the amplitude of every excited normal mode.

The string energy is $\frac{\rho }{2}{\int }_0^L(y_t^2+c^2{y}_x^2)dx$. At the impulse it is $E_{\mathrm{total}}=\rho a{v}^2/2$. Orthogonality of the sine modes gives $E_n=\rho L{g}_n^2/4$ initially; each mode subsequently exchanges kinetic and elastic energy while retaining that total. Thus the modal energy of a localized velocity impulse on a string has fractionsFor $l=L/3$ and $a=L/10$, a mode is absent exactly when $\sin (n\pi /3)=0$ or $\sin (n\pi /20)=0$. Hence the unexcited modes are precisely

The Fourier sine series coefficients of $x$ on $(0,\pi )$ areby integration by parts. The odd periodic extension is piecewise smooth, so the Fourier series converges to $x$ at every interior point. HenceTo justify integration from zero, the Fourier sine series partial sums converge to $x$ in L2 norm. By the Cauchy-Schwarz inequality, integration from zero to any $x\in [0,\pi ]$ has error at most $\sqrt{\pi }$ times that L2 norm error, uniformly in $x$. Integrating the partial sums and taking the limit therefore givesThe integrated series is absolutely and uniformly convergent, since its terms are bounded by $4/n^2$. Thus the resulting Fourier cosine series hasIndependently, the constant coefficient isComparison yields the requested exact sum

For the disc, put $u(r,\phi )=R_0(r)\Theta (\phi )$. The polar Laplace equation separates aswhere periodicity requires an integer $n\ge 0$. The radial ansatz $R_0=r^{\alpha }$ gives $\alpha =\pm n$; for $n=0$ the solutions are a constant and $\log r$. Regularity at the origin excludes negative powers and the logarithm.

Writing the boundary data as a Fourier serieswith $a_0=(2\pi {)}^{-1}{\int }_0^{2\pi }{u}_D$, $a_n={\pi }^{-1}{\int }_0^{2\pi }{u}_D\cos n\phi$, and the analogous formula for $b_n$, givesEquivalently it is the Poisson integral on the unit disk with kernelFor continuous boundary data this integral converges to the prescribed value at every boundary point, is smooth and harmonic inside, and uniqueness follows from the maximum principle for harmonic functions.

For the square, superpose four solutions, each carrying one side's data and zero data on the other three sides. Set $k_n=n\pi /a$ andSeparation of variables gives the explicit sumEach summand has zero Laplacian and the required zero values on three open sides. The Fourier sine series supplies the fourth side. For regular continuous data with matching values at the corners, the harmonic extension also takes the common corner values; termwise boundary evaluation of the sine series at the corners is not valid in general. For incompatible corner data there is no continuous solution on the closed square, but the displayed harmonic solution still attains the prescribed data on each open side.

Put $u=\epsilon v+O({\epsilon }^2)$. The linearization at the zero solution isThe appropriate Fourier sine series isOrthogonality of the modes givesThus each mode grows or decays exponentially with its own eigenvalue, and the largest growth rate is the $n=1$ value. Strict linear stability analysis, meaning decay of every mode, is equivalent toAt equality the first mode is neutral in the linearization; the nonlinear terms decide its subsequent behavior. Above this length the first mode grows, giving linear instability.

Integration by parts gives the Fourier sine series coefficientsHenceThe odd periodic extension is a sawtooth, equal to $x$ on $(-\pi ,\pi )$ and repeated every $2\pi$; its jumps at odd multiples of $\pi$ have midpoint value zero in the Fourier series. In particular the sine series equals zero, rather than the specified endpoint value $\pi$, at $x=\pi$. The even periodic extension is the continuous triangular wave equal to $|x|$ on $[-\pi ,\pi ]$. Their sketches are below.

The Parseval identity for a Hilbertian basis for this sine expansion gives

Take real $\kappa$ and use the L2 norm on the spatial interval. Existence, uniqueness and continuous dependence are the three requirements of Hadamard well-posedness. An energy method supplies the decisive estimate. For a smooth solution with homogeneous Dirichlet boundary conditions, integration by parts givesThe drift contributes only a boundary term, which vanishes. The Poincaré inequality $\Vert u_x{\Vert }_2^2\ge {\pi }^2\Vert u{\Vert }_2^2$ further givesApply the same argument to the difference of two solutions to obtain uniqueness and continuous dependence on the initial data.

For existence, use the Dirichlet gauge transform for constant drift: $w=e^{\kappa x/2}u$ satisfies $w_t=w_{xx}-{\kappa }^{2}w/4$ with zero boundary values. Expanding $w_0$ in its Fourier sine series givesFor $u_0\in L^2(0,1)$ the series defines a solution continuous in $L^2$ down to $t=0$ and smooth for positive time; multiplication by the fixed bounded exponentials preserves this interpretation. Its energy estimate follows by approximation with smooth initial data. For a classical solution at the initial corners, require the usual smoothness and boundary compatibility instead. The problem is well posed in $L^2$, with a contraction estimate independent of the initial data.

Fix the convention $\Delta q-m^{2}q=0$, $m>0$, and let $c_m=m^2/4$. With $z=x+iy$, two families of modified Helmholtz adjoint plane waves areBoth satisfy $\Delta E_{\pm }=m^2{E}_{\pm }$, since the products of their $z$ and $\bar{z}$ exponents are $m^2/4$. They obey $E_{-}(z,k)=\overline{E_{+}(z,\bar{k})}$. Green second identity gives the two conjugate global relations for the modified Helmholtz equationFor real boundary traces, ${\mathcal{G}}_{-}(k)=\overline{{\mathcal{G}}_{+}(\bar{k})}$. Thus the second identity is the conjugate spectral companion, rather than an unrelated extra boundary condition. In complex differential form the first relation is equivalentlyIts integrand is a closed one-form: differentiating its coefficients gives $2E_{+}(q_{z\bar{z}}-c_{m}q)=0$. This supplies a direct derivation of the global relation as well as its outward-normal sign convention.

Write the known Dirichlet boundary condition on side $j$ as $f_j(s)$ and the unknown outward normal derivative as $h_j(s)$, with $-1<s<1$. For any linear combination $v$ of the adjoint waves, the global relation isExpanding the $h_j$ in a finite basis and enforcing these identities at selected collocation points for a global relation gives a linear system with a known right side. The square permits especially simple paired tests.

Let $p_n=n\pi /2$, ${\eta }_n=\sqrt{m^2+p_n^2}$, and ${\phi }_n(s)=\sin [p_n(s+1)]$ for $n\ge 1$. Define four real adjoint solutionsThey satisfy the modified Helmholtz equation because ${\eta }_n^2-p_n^2=m^2$. Each is a linear combination of two adjoint exponentials, hence is obtained from the two spectral global relations. More explicitly, put ${\rho }_n^{\pm }=({\eta }_n\pm p_n)/2$, so ${\rho }_n^{+}{\rho }_n^{-}=c_m$. For the $E_{+}$ convention above, the required spectral pairs areAppropriate phase-weighted differences produce $\sin [p_n(s+1)]$. For example the top test is $e^{-{\eta }_n}[e^{i{p}_n}{E}_{+}({\rho }_n^{-})-e^{-i{p}_n}{E}_{+}({\rho }_n^{+})]/(2i)$. The conjugate relation ensures a real system for real data. This is sine collocation of square modified Helmholtz global relations.

The functions ${\phi }_n$ are an orthonormal Fourier sine basis on $[-1,1]$. Set $h_{j,n}={\int }_{-1}^1{h}_j{\phi }_{n}ds$ and compute the known quantity $R_{j,n}={\int }_{\partial \Omega }f{\partial }_n{v}_{j,n}ds$ by numerical integration. Since ${\phi }_n(\pm 1)=0$, each adjoint test vanishes on both adjacent sides. On its own side it equals ${\phi }_n$, and on the opposite side it equals ${\delta }_n{\phi }_n$, where ${\delta }_n=e^{-2{\eta }_n}$. Thus the square modified Helmholtz Dirichlet-to-Neumann coefficients obeyEach block is inverted explicitly: for opposite sides $j,j^{\prime }$,Compute these coefficients for $1\le n\le N$ and reconstruct $h_j^{(N)}={\sum }_{n=1}^N{h}_{j,n}{\phi }_n$. This is a semi-analytical scheme: the spectral basis integrals and two-by-two inverses are explicit, while the known boundary integrals are evaluated numerically. Increase $N$ and the quadrature resolution until the desired convergence is observed; compatible smooth side data have convergent normal-trace expansions, while corner singularities require the usual weaker trace interpretation and more careful quadrature.

The requested weak interaction between sides is particularly clear: the adjacent-side unknown traces contribute exactly zero, and the opposite-side coefficient is $e^{-2\sqrt{m^2+(n\pi /2{)}^2}}\to 0$. The own-side coefficient stays equal to one. Thus diagonal dominance of paired square global-relation collocation is genuine after pairing; individual uncombined exponential samples need not have the same conditioning. The eigenvalues of each block are $1\pm {\delta }_n$, so its condition number is $(1+{\delta }_n)/(1-{\delta }_n)$.

There is also a direct localization check for an individual adjoint wave. On side $j$ with outward unit normal $n_j$ and tangent $t_j$, the normalized test $e^{\eta (n_j\cdot r-1)+ip{t}_j\cdot r}$ has $\eta =\sqrt{m^2+p^2}$. Its opposite-side magnitude is $e^{-2\eta }$, and its magnitude integrates to at most $1/\eta$ along either adjacent side. This side localization of modified Helmholtz plane waves explains why suitable large spectral parameters suppress remote-side effects even before the exact sine cancellation.

Finally the interior solution can be evaluated from the computed traces using the two-dimensional modified Helmholtz fundamental solutionHere $K_0$ is the Modified Bessel function of the second kind, and $(-\Delta +m^2){\Gamma }_m=\delta$. Replace $h$ by its computed finite Fourier sine series and use numerical integration; at interior points the boundary kernels are smooth. The sign follows from Green second identity with this fundamental-solution convention. This completes the numerical integration of the boundary value problem with a Dirichlet boundary condition, rather than stopping at an equation for its unknown boundary derivatives.

Choose continuous boundary values equal to $\sin x$ on the lower side and zero on the other three sides. These values agree at all four corners, so they define a continuous function on the boundary of the rectangle. Any solution would satisfy the zero vertical-side conditions of part (a), whose Fourier sine series forces its lower boundary trace to be zero. This contradicts the prescribed value $\sin x$ for $0<x<\pi$. Hence these continuous Dirichlet data admit no solution in $C^0(\overline{R})\cap C^2(R)$.

Label the sides by $L,R,B,T$, and parametrize the two vertical sides by $s=y\in [-1,1]$, the horizontal sides by $s=x\in [-1,1]$. Their outward normal derivatives are $q_L=-u_x(-1,s)$, $q_R=u_x(1,s)$, $q_B=-u_y(s,-1)$ and $q_T=u_y(s,1)$. Let $f_j$ denote the prescribed side values. Assume compatible, sufficiently regular boundary traces; the square's corners have zero arclength measure and do not require separate normal values.

For sine collocation of square modified Helmholtz global relations, use Legendre polynomials for the known Dirichlet boundary condition and a Fourier sine series for the unknown normal derivative:The known coefficients are $d_{j\ell }=(2\ell +1){\int }_{-1}^1{f}_j{P}_{\ell }ds/2$. The sine functions have ${\int }_{-1}^1{\varphi }_m{\varphi }_{n}ds={\delta }_{mn}$ and form a complete basis in $L^2(-1,1)$. Choosing them for the normal derivative does not impose zero flux at a corner: the expansion is an $L^2$ representation, and endpoint values are not determined by it. If preserving corner values of the approximated Dirichlet trace is necessary, subtract its endpoint-interpolating line before polynomial approximation, and add that line back. All subsequent known-data integrals can alternatively be evaluated with the exact $f_j$.

Define the entire function basis transformsAt $\tau =\pm i{p}_m$ the quotient is evaluated by its removable limit or by the defining integral. For example, ${\mathsf{P}}_0=2\sinh \tau /\tau$, and polynomial expansion of $P_{\ell }$ expresses every ${\mathsf{P}}_{\ell }$ in derivatives of ${\mathsf{P}}_0$. PutFor compactness write $A=A(\lambda )$ and $B=B(\lambda )$. Substituting these expansions into the first global relation for a linear boundary value problem givesThe signs are fixed by the outward normal vectors, rather than by a choice of traversal direction. The second approximate global relation for a linear boundary value problem replaces $A$ by $-A$, leaving $B$ unchanged:Here $\simeq 0$ records omitted boundary-expansion tails. In a finite spectral method the selected equations are set equal to zero to solve for the $4N$ unknown real coefficients $c_{jm}$. For real data the two families obey the same complex conjugation relation as their exact counterparts.

A direct elimination of pressure gives the same result as taking a double curl of the momentum equation. Incompressibility gives ${\nabla }^{2}p=\mathrm{Ra}\mathrm{Pr}{\partial }_z\theta$. Apply ${\nabla }^2$ to the vertical momentum equation and subtract this pressure derivative to obtainHere ${\nabla }_h^2={\partial }_x^2+{\partial }_y^2$ is the horizontal Laplacian. At $z=0,1$, impermeability gives $w=0$ and fixed temperature gives $\theta =0$. The stress-free boundary condition gives ${\partial }_{z}u={\partial }_{z}v=0$, since horizontal derivatives of the boundary value $w=0$ vanish. Differentiating incompressibility once in $z$ therefore givesFor a horizontal Fourier mode $X=e^{i(k_{x}x+k_{y}y)}$, ${\nabla }_h^{2}X=-{\lambda }^{2}X$ with real $\lambda =\sqrt{k_x^2+k_y^2}$. More generally the nonnegative spectrum of $-{\nabla }_h^2$ follows by integration by parts for periodic or square-integrable horizontal disturbances. Substituting the separated normal mode gives, with $D=d/dz$,The boundary conditions admit a complete Fourier sine series in $z$. For a nonzero horizontal wavenumber, take $W=W_n\sin (n\pi z)$, $\Theta ={\Theta }_n\sin (n\pi z)$, and set $a_n^2={\lambda }^2+n^2{\pi }^2$. The two amplitude equations areThus the stress-free convection growth-rate polynomial isIts discriminant is $(\mathrm{Pr}-1{)}^2{a}_n^4+4\mathrm{Ra}\mathrm{Pr}{\lambda }^2/a_n^2$, so both growth rates are real for $\mathrm{Ra}>0$. For $\mathrm{Ra}<0$, its two coefficients after the leading term are positive: their sum gives ${\sigma }_1+{\sigma }_2=-(1+\mathrm{Pr})a_n^2<0$, and their product is $\mathrm{Pr}(a_n^4-\mathrm{Ra}{\lambda }^2/a_n^2)>0$. Real roots are both negative; nonreal roots have real part $-(1+\mathrm{Pr})a_n^2/2<0$. Hence stable heating from above damps every mode. Horizontally uniform thermal modes and uncoupled horizontal viscous modes are also purely damped; impermeability and incompressibility exclude a horizontally uniform nonzero $w$.

Because the unstable-side roots are real, the onset is stationary, an exchange of stabilities in stress-free convection. Setting $\sigma =0$ gives the free-slip convection neutral curveWith $x={\lambda }^2$, its derivative has the sign of $2x-n^2{\pi }^2$. Its minimum is at $x=n^2{\pi }^2/2$, with value $27n^4{\pi }^4/4$. Minimizing also over $n\ge 1$ gives the critical values

Separation of variables with $u=X(x)Y(y)$ and the three homogeneous boundary conditions gives $X_n=\sin (n\pi x)$ and $Y_n=\sinh (n\pi y)$. The boundary value at $y=1$ contains only the third Fourier sine series mode. ThereforeThis harmonic function has all four prescribed boundary values.

Inside the infinite square well, the stationary Schrödinger equation for unit mass isIts normalized eigenstates and energies areThe general normalized time-dependent solution is

Immediately after removing the barrier, the normalized wavefunction isIts Fourier sine series coefficients in the full-well basis areThey areThe Born rule therefore gives

Expand $g$ in its Fourier sine seriesBy separation of variables, the decaying solution of Laplace equation in each half-strip has $n$th mode proportional to $e^{-k_n|x|}\sin (k_{n}y)$. Continuity at $x=0$ and integration of the equation across the Dirac delta function require the normal-derivative jumpThe jump of $C_n{e}^{-k_n|x|}$ is $-2k_n{C}_n$, so

For $g(y)=\delta (y-c)$, its Fourier coefficients areThusTranslation in $x$ gives the Dirichlet Green function for an infinite strip with source $(b,c)$: