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Differentiating $v=x^n{e}^{-x}$ gives $x{v}^{\prime }=(n-x)v$. Applying the Leibniz rule $n+1$ times to this identity givesOnly two terms survive on each side because the multiplying factors are linear. Rearranging yields $x{v}^{(n+2)}+(x+1)v^{(n+1)}+(n+1)v^{(n)}=0$. Since $v^{(n)}=n!e^{-x}{L}_n$, the next two derivatives are $n!e^{-x}(L_n^{\prime }-L_n)$ and $n!e^{-x}(L_n^{\prime \prime }-2L_n^{\prime }+L_n)$. Substituting and cancelling $n!e^{-x}$ givesThis is deriving the Laguerre equation from Rodrigues' formula, exhibiting $L_n$ as an eigenfunction of the Laguerre differential operator on polynomials.

Separate $\tilde{u}=f(t)g(x)$. On a nonzero product region,A decaying temporal factor requires $\lambda >0$, with $f=C{e}^{-\nu \lambda t}$. The zero spatial boundary values give $g=A\sin (\sqrt{\lambda }x)$ and $\sin \sqrt{\lambda }=0$, hence $\lambda =j^2{\pi }^2$ with $j=1,2,\dots$. Zero or negative separation parameters give no nonzero spatial mode with both boundary values zero.

The initial profile is already exactly the $j=n$ sine eigenfunction. Its coefficient therefore determines a single separated solution:The exponential tends to zero because $\nu >0$. More generally sine orthogonality would determine a sum of such modes, but all other coefficients vanish here. This is the transient of the heat equation with a single stationary sine forcing.

The equation as printed requires $v\ne 0$; its sign does not affect the result. Seek separated solutions $\psi =X(x)T(t)$. Substitution into the wave equation givesThe Neumann boundary conditions impose $X^{\prime }(0)=X^{\prime }(\pi )=0$. Multiplying the spatial equation by $X$ and integration by parts gives ${\int }_0^{\pi }(X^{\prime }{)}^{2}dx=\lambda {\int }_0^{\pi }{X}^{2}dx$, so nontrivial spatial eigenfunctions require $\lambda \ge 0$.

For $\lambda =0$, $X$ is constant and $T=A_0+B_{0}t$. For $\lambda =k^2>0$, $X=C\cos kx+D\sin kx$. The first boundary condition gives $D=0$, and the second gives $\sin k\pi =0$. Thus $k=n$ is a positive integer and the spatial eigenfunctions are $\cos nx$. Their time dependence is $A_n\cos (nvt)+B_n\sin (nvt)$. By superposition,The zero initial velocity sets $B_0=B_n=0$. The trigonometric identity gives the initial profile ${\cos }^2(x/2)=(1+\cos x)/2$, so only the constant mode and the first cosine mode occur. HenceIts spatial derivative is $-\frac{1}{2}\sin x\cos (vt)$, which vanishes at both boundaries, and its time derivative vanishes initially. Thus it satisfies all the required data.

Boundary form. For real functions and $\mathcal{L}y=-(p{y}^{\prime }{)}^{\prime }+qy$, integration by parts givesThe integral terms containing $p{y}_1^{\prime }{y}_2^{\prime }$ cancel, as do the $q$ terms. This is the Lagrange identity for a Sturm-Liouville operator. The boundary conditions must make this boundary form zero for every admissible pair. Several suitable choices are homogeneous Dirichlet boundary conditions at both endpoints; homogeneous Neumann boundary conditions at both endpoints; one of each; or separated homogeneous Robin boundary conditionswhere each fixed real pair $(\alpha ,\beta )$ is nonzero. At each endpoint the two boundary data vectors are then proportional, making their antisymmetric product zero. Another choice is periodic data $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$, which cancel the two endpoint contributions. These domains give the required self-adjoint differential operator in the regular setting.

Eigenfunctions. Set $x=\cos \theta$, $0\le \theta \le \pi$, and $Y(\theta )=y(\cos \theta )$. The chain rule givesThe Chebyshev differential equation therefore becomes $Y^{\prime \prime }+n^{2}Y=0$, with $Y(0)=Y(\pi )=0$. For a positive integer $n$, the first condition excludes the cosine term and the second is satisfied by $\sin (n\theta )$. ThusThere is no nonzero $n=0$ eigenfunction: the transformed solution is linear in $\theta$ and vanishes at both endpoints only if identically zero. Negative $n$ merely repeats a positive mode with the opposite sign, so it is excluded from the indexing of an independent eigenbasis.

Weight and normalization. Multiplying the differential equation by $-1/\sqrt{1-x^2}$ puts it into Sturm-Liouville form:Hence $p=\sqrt{1-x^2}$, $q=0$ and $w=(1-x^2{)}^{-1/2}$. Under $x=\cos \theta$, the measure $w(x)dx$ becomes $-d\theta$, soThereforeThese endpoints are singular because $p$ vanishes and $w$ diverges there, but the weight is integrable. Moreover $p{y}_n^{\prime }=-N_{n}n\cos (n\theta )$ remains finite, so the vanishing values of $y_n$ make the boundary form zero. These are the Dirichlet eigenfunctions of the Chebyshev differential operator, not the first-kind Chebyshev polynomials $\cos (n\arccos x)$, which have nonzero endpoint values.

Steady state. In the heat equation, time independence makes the second spatial derivative zero. The two endpoint temperatures therefore determine the linear profileInsulation is along the sides; the endpoint reservoirs can still exchange heat with the rod.

Transient solution. Subtract the steady profile before applying separation of variables: let $U=T-T_{0}x/l$. Then $U_t=\nu U_{xx}$, $U(0,t)=U(l,t)=0$ and $U(x,0)=-T_{0}x/l$ for $0<x<l$. Writing $U=X(x)\Theta (t)$ givesThe homogeneous Dirichlet boundary conditions exclude $\mu \le 0$ and select the eigenfunctions $\sin (n\pi x/l)$ with eigenvalues ${\mu }_n=(n\pi /l{)}^2$, $n\ge 1$. Their orthogonality gives the Fourier sine series coefficientsHere integration by parts uses ${\int }_0^{l}x\sin (n\pi x/l)dx=-l^2(-1{)}^n/(n\pi )$. ThusEvery summand satisfies the heat equation and the homogeneous endpoint conditions, while the steady profile supplies the prescribed endpoint values. The initial Fourier sine series equals $-T_{0}x/l$ in the open interval. At $(l,0)$ the initially cold rod and the suddenly imposed hot endpoint are incompatible; the formula describes the prescribed data for $t>0$ and the initial limit at interior points.

Heat absorption. Integrating in $x$ kills the even modes. With $a=\nu {\pi }^2/l^2$, the total heat relative to the initial state isConsequently the net rate of heat absorption isFor each $t\ge \epsilon >0$, any fixed number of differentiated modes is bounded by a polynomial in $n$ times $e^{-a\epsilon n^2}$. These bounds have finite sums, proving absolute convergence, uniform convergence on such time intervals and the validity of the integration and differentiation. In particular the displayed rate is finite for every $t>0$, positive when $T_0>0$, and tends to zero as equilibrium is approached. The same answer follows from the heat equation balance $H^{\prime }=\gamma \nu [T_x(l,t)-T_x(0,t)]$; this subtracts loss at the cold endpoint from input at the hot endpoint.

Mixed endpoint spectrum. For $y^{\prime \prime }+\lambda y=0$, integration against $y$ and integration by parts, using $y(0)=0,y^{\prime }(\pi /2)=0$, gives $\lambda \int y^2=\int (y^{\prime }{)}^2>0$ for every nonzero real eigenfunction. Thus $\lambda =k^2>0$ and the first condition makes $y=A\sin (kx)$. The second requires $\cos (k\pi /2)=0$, so $k=2n+1$, $n\ge 0$. Sincethe normalized eigenfunctions and eigenvalues areThe orthogonality also follows by applying integration by parts to two different eigenfunctions of the same self-adjoint differential operator.

Forced expansion. Let $k_n=2n+1$ and write $x=\sum f_n{u}_n$, $y=\sum c_n{u}_n$. The Fourier coefficients of the forcing arebecause the boundary cosine vanishes and $\sin (k_n\pi /2)=(-1{)}^n$. Applying $d^2/d{x}^2+\kappa$ to the eigenfunction expansion gives $(\kappa -k_n^2)c_n=f_n$. Since $\kappa$ is not an eigenvalue,The coefficients are $O(k_n^{-4})$, so the series and its first two differentiated series converge absolutely and uniformly on the interval. Each term satisfies the two endpoint conditions, and $y^{\prime \prime }+\kappa y$ equals the mixed-endpoint Fourier series of $x$. The forcing coefficients are $O(k_n^{-2})$, giving uniform convergence of that series as well. Uniqueness follows because the homogeneous problem has a nonzero solution only at one of the excluded eigenvalues.

Odd fourth-power sum. Set $\kappa =0$. Direct integration of $y^{\prime \prime }=x$, with the same endpoint conditions, givesAt $x=\pi /2$, the eigenfunction expansion instead gives $y(\pi /2)=-(4/\pi ){\sum }_{n\ge 0}{k}_n^{-4}$. Equating the two values yieldsAbsolute and uniform convergence justify evaluation at the endpoint.

Assume the regular real Sturm-Liouville problem, with $p,w>0$, ordered eigenvalues ${\lambda }_0<{\lambda }_1<\cdots$, and real orthonormal eigenfunctions in the weighted inner productThe Rayleigh quotient for an admissible trial function isThe second expression follows by integration by parts. It is the quantity used in the Rayleigh-Ritz method.

For $\tilde{y}=y_0+{\sum }_{n\ge 1}{a}_n{y}_n$, orthonormality eliminates every cross-term in the denominator:Applying $\mathcal{L}$ to the eigenfunction expansion and using the same orthogonality in the numerator givesHence, whenever the trial function has finite energy,Put $S=\sum a_n^2$ and $\Delta =\sum a_n^2({\lambda }_n-{\lambda }_0)$. The exact error and its expansion areIn particular, under a common small scaling $a_n=\epsilon b_n$ with fixed finite sums $\sum b_n^2$ and $\sum b_n^2({\lambda }_n-{\lambda }_0)$,The fourth-order remainder includes mixed products $a_n^2{a}_j^2$, not just individual fourth powers.

Error comparison and sign. Since $\Vert y_0{\Vert }_w=1$, the relative weighted norm error of the supplied trial function isFor ${\lambda }_0\ne 0$, the relative eigenvalue error is $\Delta /(|{\lambda }_0|(1+S))=O({\epsilon }^2)$. Thus the Rayleigh quotient has a second-order error for a first-order error in the trial eigenfunction, provided the energy-weighted coefficient sum is controlled. Small weighted norm error alone does not control arbitrarily high-frequency contamination, because ${\lambda }_n$ can grow without bound. If ${\lambda }_0=0$, an absolute error must replace the undefined relative one.

Finally every summand of $\Delta$ is nonnegative. Thuswith strict inequality whenever some $a_n\ne 0$. This is the upper-bound property of the Rayleigh-Ritz variational principle for the lowest eigenvalue.

Choose homogeneous Dirichlet boundary conditions, $y(a)=y(b)=0$, and nonzero admissible functions with finite $F$ and positive $G$. Variations obey the same zero endpoint values. At a stationary Rayleigh quotient, set $\lambda =F[y]/G[y]$. ThenThe boundary term vanishes, and the fundamental lemma of the calculus of variations gives the Sturm-Liouville equation $-(p{y}^{\prime }{)}^{\prime }+qy=\lambda wy$. Thus stationary functions are eigenfunctions. Conversely, multiply that equation by $y$ and integrate by parts to obtain $F=\lambda G$; its first variation then vanishes by the same calculation. This proves that stationary values are the corresponding eigenvalues. Homogeneous Neumann boundary conditions also make the displayed boundary terms vanish; general Robin boundary conditions can require a boundary contribution to the quadratic form.

For $p=1$, $q=0$, $w=x$ on $(0,\pi )$, the polynomial trial function $y=x(\pi -x)$ satisfies the endpoint conditions. Its Rayleigh quotient usesso the polynomial estimate is $\boxed{{\lambda }_{\mathrm{poly}}=20/{\pi }^3\simeq 0.64503}$. The trigonometric trial function $y=\sin x$ givesand hence $\boxed{{\lambda }_{\mathrm{trig}}=2/\pi \simeq 0.63662}$. Both are upper bounds on the lowest eigenvalue, by the Rayleigh-Ritz method; neither trial is an exact eigenfunction, so the inequalities are strict.

Their closeness has a useful explanation beyond their similar single-arch shapes. Both trial functions are symmetric under $x\mapsto \pi -x$, forcing ${\int }_0^{\pi }x{y}^{2}dx=(\pi /2){\int }_0^{\pi }{y}^{2}dx$. The reflection-symmetric trials for a linearly weighted Rayleigh quotient therefore reduce this problem to a constant multiple of the ordinary Dirichlet energy quotient. The sine is the best trial within this symmetric class, and the polynomial's value is only a factor $10/{\pi }^2\simeq 1.0132$ larger. The unrestricted lowest eigenfunction can be asymmetric because the weight $x$ is asymmetric.

Multiply the original equation by $p(x)=\exp ({\int }^{x}u(s)ds)>0$. Since $p^{\prime }=up$, it becomes $(p{y}^{\prime }{)}^{\prime }+qy+\lambda ry=0$ with $q=pv$, $r=pw>0$. This is Sturm-Liouville form. For two real eigenfunctions, multiply the equation for $y_n$ by $y_m$, subtract the equation for $y_m$ multiplied by $y_n$, and integrate. The Lagrange identity for a Sturm-Liouville operator givesEach separated homogeneous boundary condition makes the corresponding determinant zero, provided its pair of coefficients is not both zero. For distinct eigenvalues, this proves the orthogonality of Sturm-Liouville eigenfunctions.

For the constant drift, the drift-removing substitution for a Dirichlet Sturm-Liouville problem is $y=e^{-\alpha x}v$. It reduces the equation to $v^{\prime \prime }+\lambda v=0$ with $v(0)=v(\pi )=0$. Nonzero solutions occur exactly atThe zero and negative values give only the zero solution under both endpoint conditions. The self-adjoint expression is $(e^{2\alpha x}{y}^{\prime }{)}^{\prime }+{\alpha }^2{e}^{2\alpha x}y+\lambda e^{2\alpha x}y=0$. Hence the weight is $r=e^{2\alpha x}$, and for the displayed normalization,

For a nonzero real function, write $\Lambda =F/G$ with $G={\int }_a^{b}w{y}^{2}dx>0$. Multiplying the Sturm-Liouville equation by $y$ and integrating givesThus the stated vanishing boundary product proves $\lambda =\Lambda [y]$.

For the variational statement, the first variation isWith admissible boundary conditions that also make $[p{y}^{\prime }\eta {]}_a^b=0$ for every allowed variation, the fundamental lemma of the calculus of variations identifies stationary functions with the eigenfunctions and stationary values with their eigenvalues. Zero Dirichlet boundary conditions, zero Neumann boundary conditions, and matched periodic data give familiar examples. The single printed condition $[py{y}^{\prime }{]}_a^b=0$ proves the quotient identity but, by itself, does not specify the admissible variations needed for stationarity; the boundary conditions for variational Rayleigh quotients make this distinction explicit.

For the half-line equation, multiplication by $x^{n-1}$ gives $p=w=x^{n-1}$ and $q=x^{n+1}$. Use the regular radial boundary condition at zero and square-integrable decay at infinity, under which the proposed Gaussian is admissible. With $y=e^{a{x}^2}$, $a<0$, $y^{\prime }=2axy$, soThe ratio of the two Gaussian integrals is $n/(-4a)$, from a change of variables and the Gamma function recurrence. Differentiating gives ${\Lambda }^{\prime }(a)=-n[1-1/(4a^2)]$, so the minimum on $a<0$ occurs at $a=-1/2$. The Gaussian variational bound for a radial harmonic oscillator therefore givesHere the trial actually satisfies the differential equation with $\lambda =n$. Moreover, on the same regular decaying domain, integration by parts gives $F[y]-nG[y]={\int }_0^{\infty }{x}^{n-1}(y^{\prime }+xy{)}^{2}dx\ge 0$. Hence the estimate is exact, with $\boxed{{\lambda }_0=n}$. For $n=1$, regular radial data mean the Neumann boundary condition $y^{\prime }(0)=0$; imposing a zero Dirichlet boundary condition there would exclude this trial and give a different problem.

Multiplying by $y$ and integrating with the zero endpoint values first gives $\lambda {\int }_1^b{y}^{2}dx/x={\int }_1^{b}x(y^{\prime }{)}^{2}dx>0$ for a nonzero real eigenfunction; hence $\lambda >0$. Using the printed change of variables $\xi =\lambda \ln x$, write $y(x)=Y(\xi )$. Then $(x{y}^{\prime }{)}^{\prime }={\lambda }^2{Y}_{\xi \xi }/x$, so $Y_{\xi \xi }+Y/\lambda =0$. The first endpoint gives $Y=A\sin (\xi /\sqrt{\lambda })$, and the other gives $\sin (\sqrt{\lambda }\ln b)=0$. ThusThis is the logarithmic change of variable in a Sturm-Liouville problem. The measure $dx/x=d(\ln x)$ gives squared weighted norm $A_n^2\ln b/2$, so an orthonormal choice is $A_n=\sqrt{2/\ln b}$.

Multiply the equation for $y_n$ by $y_m$ and the equation for $y_m$ by $y_n$, then subtract. The common $h$ terms cancel, and integration by parts gives the Lagrange identity for a Sturm-Liouville operator:At each endpoint the two pairs $(y_n^{\prime },y_n)$ and $(y_m^{\prime },y_m)$ satisfy the same nontrivial homogeneous linear equation. They therefore lie on the same one-dimensional line, and their determinant vanishes. The boundary expression is zero. Since the eigenvalues are distinct, division gives the required orthogonality of Sturm-Liouville eigenfunctions:For complex-valued eigenfunctions of a real self-adjoint problem one may choose real eigenfunctions, or use the conjugated version as a weighted inner product.

Put $\Phi (r,\theta )=R(r)Y(\theta )$ in Laplace's equation, multiply by $r^2/(RY)$, and separate the two variables:With $s=\cos \theta$, the angular equation becomes $(1-s^2)Y_{ss}-2s{Y}_s+\lambda Y=0$. Requiring a smooth axisymmetric harmonic function at both poles selects $\lambda =n(n+1)$ and the Legendre polynomial $Y=P_n(s)$, for nonnegative integers $n$. The radial Cauchy-Euler equation has solutions $r^n$ and $r^{-n-1}$. Superposition over the angular eigenfunctions therefore givesThis expansion assumes regularity on each whole angular sphere; without it, singular angular solutions would also be possible. The boundary data in the shell involve only the $n=0,1$ modes. Define the spherical-shell harmonic interpolation weightsEach $p_n$ equals one at the inner boundary and zero at the outer one, and $q_n$ does the reverse. Hence the steady temperature isEach term is a harmonic function, and substitution verifies both prescribed boundary values. Uniqueness follows from the maximum principle for harmonic functions.

For nonzero $y$ with $G[y]>0$, the first variation of the Rayleigh quotient under $y\mapsto y+\epsilon h$ isChoose fixed admissible homogeneous boundary conditions that make the variation boundary term vanish, such as zero endpoint values or zero endpoint derivatives. Compactly supported $h$ then show that a stationary function satisfies the Sturm-Liouville equation with $\lambda =\Lambda [y]$. Conversely an eigenfunction satisfying those admissible conditions makes the variation zero. Multiplying its differential equation by $y$ and integrating by parts givesso the stipulated energy boundary term yields $\Lambda [y]=\lambda$. The displayed energy condition alone is not a full specification of the admissible variations; it must be paired with boundary conditions for this variational equivalence.

For the final Dirichlet problem take the trial function $y=\sin x$, with $p=1,q=0,w=x$. ThenThe Rayleigh-Ritz variational principle makes this an upper bound for the lowest eigenvalue. It is strictly above the true value: equality would require $\sin x$ itself to be a lowest eigenfunction, but $y^{\prime \prime }+\lambda xy=(-1+\lambda x)\sin x$ cannot vanish throughout $(0,\pi )$ for a constant $\lambda$. ThereforeAlthough the weight vanishes at the left endpoint, it is positive in the interior and its weighted quadratic norm is positive for every nonzero admissible continuous function; this does not obstruct the stated Dirichlet variational comparison.

Each spatial eigenfunction acquires the temporal factor $e^{-\kappa {\lambda }_{mn}t}$ under the heat equation. Expanding the initial constant using the preceding coefficients gives the cooling of an initially uniform square:For $t>0$ the exponential factors justify differentiating the series and enforce the zero boundary temperature. As $t\downarrow 0$, it tends to $T_0$ at every interior point and in L2 space; the instantaneous boundary cooling is incompatible with continuity at the initial boundary corners, which is expected for this idealization.

The slowest decaying nonzero mode is $(1,1)$, givingat each fixed interior point when $T_0\ne 0$. The next contributing modes have $(m,n)=(1,3),(3,1)$ and decay as $e^{-10\kappa {\pi }^{2}t/{\ell }^2}$, making the leading-mode approximation exponentially accurate.

Use the weighted inner product $⟨f,g{⟩}_w={\int }_a^{b}w(x)\overline{f(x)}g(x)dx$ with positive weight $w$. For a self-adjoint operator $L$ and nonzero eigenfunction $Ly=\lambda y$, self-adjointness gives $\lambda ⟨y,y{⟩}_w=⟨y,Ly{⟩}_w=⟨Ly,y{⟩}_w=\overline{\lambda }⟨y,y{⟩}_w$. The positive norm therefore forces $\lambda$ to be real. For two eigenfunctions of distinct eigenvalues, the same calculation gives $({\lambda }_2-{\lambda }_1)⟨y_1,y_2{⟩}_w=0$, proving their orthogonality.

The Chebyshev differential equation takes the Sturm-Liouville formIts operator is $Ly=-(1-x^2)y^{\prime \prime }+x{y}^{\prime }$, so $Lx=x$ as required. For $y=x^3+Bx$, direct substitution gives $Ly=9x^3+(B-6)x$. Comparing coefficients yieldsExplicitly, using $x=\cos \theta$,The endpoints are singular because $p$ vanishes and $w$ diverges there. The polynomial solutions use the natural condition $p{y}^{\prime }\to 0$ at each end; under $x=\cos \theta$ these are the Neumann eigenfunctions of the Chebyshev differential operator, proportional to $\cos (n\theta )$. They differ from the zero-value endpoint realization of the same differential expression.

Use the standard regular self-adjoint assumptions on the Sturm-Liouville problem, including positive weight $w$. The normalized eigenfunctions obey orthogonalityCompleteness means every function in the weighted $L^2$ space has expansion $f(x)={\sum }_n{y}_n(x){\int }_a^{b}w(\xi )y_n(\xi {)}^{\ast }f(\xi )d\xi$, with convergence in that space. Comparing the action of this kernel on test functions with the Dirac delta gives, in the distributional sense,If zero is not an eigenvalue, divide each term by ${\lambda }_n$. Since $L{y}_n={\lambda }_n{y}_n$, application of $L$ recovers the displayed delta expansion, provingThe factor $w(\xi )$ is required because the source equation uses an ordinary, unweighted delta. It distinguishes the normalized Sturm-Liouville spectral Green kernel from a kernel integrated against weighted source measure. If there is a zero eigenvalue, an inverse for arbitrary forcing does not exist; the unqualified division by that eigenvalue is invalid.

For $L=-d^2/d{x}^2+{\omega }^2$ with real $\omega$ and zero endpoints on $[0,1]$, the weight is one, $y_n=\sqrt{2}\sin (n\pi x)$ and ${\lambda }_n=n^2{\pi }^2+{\omega }^2>0$. Substitution yieldsThe series is absolutely and uniformly convergent, although applying the differential operator is understood distributionally at its source point.

Use the real weighted inner product $⟨u,v{⟩}_w={\int }_a^{b}wuvdx$ and the energy bilinear form $B(u,v)={\int }_a^b(p{u}^{\prime }{v}^{\prime }+quv)dx$. Integration by parts, the endpoint conditions, and the Sturm-Liouville problem giveThus orthonormality in the stated expansion givesand hence the Rayleigh quotientThe expansion is understood for nonzero admissible functions with finite norm and energy.