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For a separated nonzero solution $T=f(x)g(y)$, substitution and division by $fg$ givesThe homogeneous Dirichlet boundary conditions impose $g(0)=g(1)=0$. For a nonzero spatial factor, integration by parts gives ${\lambda }^2{\int }_0^1{g}^{2}dy={\int }_0^1(g^{\prime }{)}^{2}dy$, so the separation value is nonnegative. The zero value gives only a trivial linear $g$ under these boundary conditions. The nontrivial separated factors, for $a\ne 0$, are thereforeFor the intended forward diffusion equation, assume $a>0$. The superposition principle gives a Fourier sine series, whose coefficients must expand the initial ramp in the interior:Consequently the standard heat-flow solution isIt satisfies both lateral boundary conditions for $x>0$, and tends to $y$ as $x\downarrow 0$ for $0<y<1$, also converging to the ramp in the L2 norm. The exponential coefficient decay justifies differentiating the series for positive $x$.

There is a genuine incompatibility in the printed data: at $(0,1)$ the lateral condition demands zero while the initial condition demands one. Thus no function satisfies every stated value at that corner. The displayed series solves the usual consistent interpretation with initial data on $0<y<1$ and lateral data for $x>0$; it cannot be continuous at the upper initial corner. This discrepancy is present in the original PDF.

The sign of $a$ is also unstated. If $a=0$, the spatial equation with two zero endpoints forces $T=0$, incompatible with the ramp. If $a<0$, the separated Fourier coefficients grow like $e^{n^2{\pi }^{2}x/|a|}/n$ and fail to be square summable for every $x>0$; these data therefore have no forward solution in the standard L2 space class. The positive-$a$ assumption is essential to the stated heat-flow interpretation.

Termwise differentiation of a Fourier series, applied to the continuous periodic extension in part (c), gives for $-1<x<1$Hence the required Fourier sine series isThe endpoint deserves care: at $x=1$ every term is zero, so the Fourier series sums to $0$, the average of the periodic one-sided limits $1$ and $-1$. It therefore does not reproduce the specified endpoint value $g(1)=1$. Changing a value at one point does not change any Fourier coefficient, so this is still the Fourier series of the stated function on $(-1,1]$.

Evaluate the Fourier sine series of part (d) at the interior point $x=1/2$. Only odd indices remain, with alternating signs:Thus $\boxed{A=4}$.

For the second identity, apply Parseval's identity to $h(x)=x^2$. From part (c), $a_0=2/3$, $a_n=4(-1{)}^n/(n^2{\pi }^2)$ and $b_n=0$, soThe left side is $2/5$, giving $\frac{16}{{\pi }^4}\sum n^{-4}=2/5-2/9=8/45$. ConsequentlyThe first series is conditionally convergent, while the second converges absolutely by the P-series criterion.

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.

Multiplication by $x^2$ puts the eigenvalue problem into Sturm-Liouville form, $-(x^2{y}^{\prime }{)}^{\prime }=\lambda x^{2}y$, with weight $x^2$. If $u,v$ correspond to distinct eigenvalues $\lambda ,\mu$, subtract their multiplied equations and integrate to getThe upper boundary term vanishes by the Dirichlet boundary condition. For a bounded solution near zero, $x^2{y}^{\prime }$ has a limit because $(x^2{y}^{\prime }{)}^{\prime }=-\lambda x^{2}y$ is integrable. A nonzero limit would force a $1/x$ divergence in $y$, so that limit is zero; integration then gives $y^{\prime }=O(x)$. The lower boundary term therefore also vanishes, proving the weighted orthogonality.

Set $z=xy$. For $x>0$, $y^{\prime \prime }+2y^{\prime }/x=z^{\prime \prime }/x$, and finiteness of $y(0)$ gives $z(0)=0$. Thus $z^{\prime \prime }+\lambda z=0$ with zero values at zero and $\pi$. The radial Dirichlet spectrum in three dimensions givesThe zero and negative eigenvalues yield only the zero solution under these boundary conditions. Also ${\int }_0^{\pi }{x}^2{y}_n^{2}dx=\pi /2$.

For the forced equation, the same substitution gives $z^{\prime \prime }-\beta z=x$. The Fourier sine series of $x$ on $(0,\pi )$ has coefficientsExpanding $z=\sum c_n\sin nx$ gives $-(n^2+\beta )c_n=b_n$, henceThe value at zero is defined by a limit. In fact $|\sin nx/x|\le n$, so uniform convergence holds even there, by comparison with $2\sum (n^2+\beta {)}^{-1}$. An independent check comes from solving the constant-coefficient equation directly:It satisfies both endpoint conditions and the forcing for $x>0$, and its finite origin limit agrees with the series. For $\beta >0$ the homogeneous boundary value problem has no nonzero solution, so the solution is unique.

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.

Set $T(x,t)=X(x)G(t)$ in the heat equation. Separation of variables gives $G^{\prime }/(c^{2}G)=X^{\prime \prime }/X=-\lambda$. Zero endpoint temperatures impose Dirichlet boundary conditions $X(0)=X(\pi )=0$, selecting $\lambda =n^2$ for $n\ge 1$. The separated factors can therefore be chosen asZero and negative separation eigenvalues give only the zero spatial solution under these boundary conditions.

The initial profile is continuous and triangular, with midpoint height $\pi /2$. Its Fourier sine series coefficients areConsequently the heat flow from a triangular initial temperature isFor $t\ge 0$, each summand has magnitude at most $4/[\pi (2m+1{)}^2]$. The Weierstrass M-test proves absolute and uniform convergence. The initial Fourier sine series equals the continuous piecewise smooth profile, and for $t>0$ the exponential factors justify termwise spatial and time differentiation on every time interval bounded away from zero.

Separation of variables in the wave equation with fixed endpoints selects $H_n(x)=\sin (n\pi x/L)$ and angular frequencies ${\omega }_n=cn\pi /L$, $n\ge 1$. The time factors may be chosen as $F_n(t)=\cos {\omega }_{n}t$, $G_n(t)=\sin {\omega }_{n}t$. By Fourier sine series orthogonality, the initial data giveHence $u={\sum }_{n\ge 1}[a_n\cos {\omega }_{n}t+b_n\sin {\omega }_{n}t]\sin (n\pi x/L)$. The factor $1/{\omega }_n$ in $b_n$ is required because differentiating the time sine supplies ${\omega }_n$.

The given displacement is a symmetric triangular tent: it is zero at both endpoints and has height $k$ at $L/2$. Integrating each linear half gives $a_n=8k\sin (n\pi /2)/(n^2{\pi }^2)$. Release from rest gives $b_n=0$, so the midpoint-plucked fixed-end string isOnly odd modes occur, by symmetry about the midpoint. The initial slope corner makes this a finite-energy weak solution; it is a classical solution away from the propagating corner lines.

Use separation of variables $u=X(x)T(t)$. The fixed-end Dirichlet boundary conditions select $X_n=\sin (nx)$, $n\ge 1$, and each mode satisfies $T_n^{\prime \prime }+2h{T}_n^{\prime }+n^2{T}_n=0$. In the physical resistance range $0\le h<1$, all normal modes are underdamped, with ${\omega }_n=\sqrt{n^2-h^2}>0$. The modal factor normalized by $T_n(0)=1$, $T_n^{\prime }(0)=0$ isConsequently the linearly damped string has solutionThe coefficients are those of the initial Fourier sine series. Differentiating each normalized modal factor at zero confirms that the string starts from rest. Suitable initial data give convergence in the string's energy space; the piecewise linear profile here need not give a twice differentiable solution along every characteristic from its corner.

For the specified tent profile, integrating its two linear pieces by parts givesFor example the first piece gives $-\pi \cos (n\pi /2)/(2n)+\sin (n\pi /2)/n^2$, and the second gives the opposite cosine term and the same sine term. HenceThe zero even coefficients also follow from reflection symmetry about the midpoint. This is the Fourier sine series of a symmetric tent. If $h<1$ were read without the physical assumption $h\ge 0$, the same modal solution extends by replacing imaginary-frequency sine and cosine by their hyperbolic counterparts, and taking $\sin (\omega t)/\omega =t$ when $\omega =0$.

The substitution separates the uniform equilibrium from the decaying perturbation. Direct differentiation givesConsequently the reaction-diffusion equation reduces to the heat equation $v_t=v_{xx}$. Its data become

When the three constants coincide, the Dirichlet boundary conditions for $v$ are homogeneous. Separation of variables gives $X^{\prime \prime }+n^{2}X=0$ and $T^{\prime }=-n^{2}T$, selecting the normal modes $\sin (nx)e^{-n^{2}t}$ for $n\ge 1$. ThusTwo integrations by parts give ${\int }_0^{\pi }x(\pi -x)\sin (nx)dx=2[1-(-1{)}^n]/n^3$, while the constant term uses ${\int }_0^{\pi }\sin (nx)dx=[1-(-1{)}^n]/n$. Therefore the Fourier sine series isFor $u_0\ne 2$, its first coefficient is nonzero. Multiplying by $e^{-\beta t}$ and retaining the slowest-decaying mode givesThe remainder estimate holds uniformly in $x$ for sufficiently large positive time. If $u_0=2$, the fundamental vanishes and the next possible mode has decay rate $\beta +9$, explaining the exclusion.

There is an initial endpoint compatibility issue unless $u_0=0$: the given polynomial profile vanishes at the endpoints, whereas the boundary values equal $u_0$. The series realizes the initial data in L2 space and pointwise in the open interval, imposing the prescribed boundary values for $t>0$. For nonzero $u_0$ it cannot also be continuous at both initial endpoint corners.

In logarithmic coordinates the eigenvalue problem becomes $-Y^{\prime \prime }=(\lambda +1)Y$. A nonzero function with zero values at both ends requires $\lambda +1>0$: multiplying by $Y^{\ast }$ and integrating gives $(\lambda +1){\int }_0^1|Y{|}^{2}dt={\int }_0^1|Y^{\prime }{|}^{2}dt>0$. Write $\kappa =\sqrt{\lambda +1}$. The first endpoint gives $Y=A\sin \kappa t$, and the second requires $\kappa =n\pi$, $n=1,2,\dots$.

For vector normalization, ${\int }_1^e|y{|}^{2}dx/x=|A{|}^2{\int }_0^1{\sin }^2(n\pi t)dt=|A{|}^2/2$. A positive real normalization constant therefore givesThe sine orthogonality identity also gives ${\int }_1^e{y}_m{y}_{n}dx/x={\delta }_{mn}$. Completeness is the usual completeness of the Fourier sine series on $(0,1)$ under the logarithmic coordinate transformation.

After multiplying the initial field by $e^{\beta x/2}$, the $y$ dependence is exactly the $n=2$ cosine mode. Its orthogonality therefore gives $A_{mn}=0$ when $n\ne 2$. For $n=2$, the remaining Fourier coefficient in the Fourier sine series isThusThe coefficients are independent of $\beta$, because the initial exponential exactly cancels the drift-removing weight. Substitution into the preceding expansion gives the full subsequent solution, with decay rates $m^2{\pi }^2+9{\pi }^2/4+{\beta }^2/4$.

Assume the struck interval lies within the string, so $0\le \ell -a/2<\ell +a/2\le L$. The initially stationary straight string has $A_n=0$. Write its initial velocity as a Fourier sine series $g(x)={\sum }_n{g}_n\sin (k_{n}x)$. ThenThus $y(x,t)={\sum }_n(g_n/{\omega }_n)\sin ({\omega }_{n}t)\sin (k_{n}x)$. Immediately after the strike all the energy is kinetic energy:The orthogonality integral ${\int }_0^L\sin (k_{n}x)\sin (k_{m}x)dx=(L/2){\delta }_{nm}$ assigns energy $E_n=\rho L{g}_n^2/4$ to mode $n$. Since the ideal modes evolve independently, their total kinetic-plus-elastic energies remain constant. Therefore the modal energy of a localized velocity impulse on a string givesThe centre factor selects modes having significant motion at the impact location; the width factor accounts for cancellation across the finite hammer footprint. Parseval's identity ensures ${\sum }_{n\ge 1}{E}_n/E=1$. The velocity discontinuities are understood in the mean-square sense and do not alter the energy integrals.

For each sine normal mode, the temporal characteristic equation is $s^2+2\lambda s+{\alpha }_n^2=0$, with roots $s=-\lambda \pm \sqrt{{\lambda }^2-{\alpha }_n^2}$. Since $\lambda L/\pi =M+1/2$, the first $M$ modes are overdamped and all higher modes are underdamped. No integer $n$ gives critical damping.

The zero initial displacement excludes the hyperbolic cosine and cosine terms. HencewhereTo impose the initial velocity, let $f_n=\frac{2}{L}{\int }_0^{L}f(x)\sin ({\alpha }_{n}x)dx$ be its Fourier sine series coefficients. Differentiating the expansion at $t=0$ gives $A_n{\Omega }_n=f_n$ for the overdamped modes and $A_n{\omega }_n=f_n$ for the underdamped modes. These determine all coefficients. Despite the hyperbolic sine, the first modes decay: ${\Omega }_n<\lambda$, so both temporal exponential rates are negative.

Here $\lambda =7/2$ and ${\alpha }_n=n$. LetTwo applications of integration by parts giveSolving these equations,Consequently the required amplitudes areThe stated initial velocity is nonzero at the fixed endpoints, whereas a classical solution differentiable there would require zero endpoint velocity. Thus it is understood on the open interval, or as an L2 space initial datum in the usual finite-energy solution. Its Fourier sine series takes value zero at the endpoints; this changes no initial energy or interior initial data.

First decompose the initial $y$ dependence using ${\sin }^2(2\pi y/b)=[1-\cos (4\pi y/b)]/2$. Only the cosine indices $m=0$ and $m=4$ occur. Expand the remaining factor in a Fourier sine series:For $k_n=n\pi /a$, two integrations by parts giveThus even $n$ vanish, while $b_n=8a^2/(n^3{\pi }^3)$ for odd $n$. The nonzero double-series coefficients are $A_{n0}=b_n/2$ and $A_{n4}=-b_n/2$. Their separate heat-mode decay factors giveAt $t=0$ the bracket is $2{\sin }^2(2\pi y/b)$, which recovers the initial profile with the displayed sine coefficients. Each term satisfies the mixed Dirichlet boundary conditions and Neumann boundary conditions.

Set $z=x^3/3$ and write $y(x)=Y(z)$. Thenso the two first-derivative contributions cancel exactly, leavingThe Dirichlet spectrum of $-d^2/d{z}^2$ on this interval is $(3n\pi {)}^2$, $n\ge 1$. If $\lambda +a\le 0$, the linear or hyperbolic solutions cannot satisfy both zero endpoints nontrivially, so no additional modes have been omitted. ThusTo verify normalization, use $dz=x^{2}dx$:This proves weighted orthonormality and also transfers completeness from the ordinary Fourier sine series. The construction is the cubic-coordinate reduction of a singular Sturm-Liouville operator. The positive constant $a$ shifts the entire spectrum down; a zero eigenvalue occurs if $a=9n^2{\pi }^2$, so the no-zero-eigenvalue proviso in part (a)(iii) must be imposed separately if that inverse formula is used here.

Expand in the complete Fourier sine series $\tilde{y}(\omega ,x)={\sum }_{n\ge 1}{b}_n(\omega )\sin (k_{n}x)$ and apply Fourier inversion. Substitution of part (ii) givesTo check the source normalization, apply ${\partial }_t^2-{\partial }_x^2$ mode by mode. It cancels the denominator; the remaining inverse transform gives $2\pi F(t)$, and completeness givesThe endpoints vanish termwise. These identities establish the equation in the distributional sense, with classical validity wherever the series and derivatives converge suitably.

The displayed real-axis integrals in the paper are not ordinary convergent integrals at their poles. One definite choice is the Cauchy principal value at each $\omega =\pm k_n$. If a causal solution is wanted, use $[k_n^2-(\omega -i0{)}^2{]}^{-1}$ instead. Its inverse transform is $H(t)\sin (k_{n}t)/k_n$, so the causal solution iswhen the past integral exists. The principal-value choice instead uses the time-symmetric kernel $\sin (k_n|t|)/(2k_n)$. No temporal initial data are given, so homogeneous sine and cosine modes remain permissible additions. This is the causal Fourier resolvent of a point-driven fixed string.

Apply separation of variables with $\rho (x,t)=X(x)T(t)$ and temporal factor $T=e^{-\lambda t}$. The spatial eigenvalue equation isThe nontrivial solutions of these Dirichlet boundary conditions are sine eigenfunctions. Normalizing in the usual inner product on $(0,L)$ giveswith ${\int }_0^L{\psi }_n{\psi }_{m}dx={\delta }_{nm}$. The superposition principle and the Fourier sine series then giveThe constants are the initial-data coefficients $a_n={\int }_0^L\rho (x,0){\psi }_n(x)dx$. Here physical diffusion means $D>0$.

Let $f(x)=\rho (x,0)$. The first normalized Fourier sine series coefficient isStrict positivity follows because $f\ge 0$, has positive total mass, and $\sin (\pi x/L)>0$ everywhere inside the interval. In particular, there is no cancellation that could remove the slowest term. This is the positivity of the first Dirichlet reaction-diffusion coefficient.

For finite initial mass $M={\int }_0^{L}f(x)dx$, all coefficients satisfy $|a_n|\le \sqrt{2/L}M$. The eigenfunction expansion therefore gives the uniform boundThe positive first profile thus controls the long-time behavior. We recover exactly the threshold $\boxed{{\gamma }_c=D{\pi }^2/L^2}$, with decay below it and unbounded positive growth at every interior point above it. The finite-mass condition is the usual meaning of an initial density profile here; standard positive-time smoothing gives the same conclusion in the usual broader admissible data classes.