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Take the spatial interval $0<x<L$ with $L>0$, and write $v=|c|>0$. A separated nonzero solution satisfiesThe zero-endpoint Dirichlet boundary conditions rule out nonzero solutions for $\lambda \le 0$: linear or hyperbolic spatial factors cannot vanish at both ends. For $\lambda >0$, they give $X_n=\sin (n\pi x/L)$ and ${\lambda }_n=(n\pi /L{)}^2$, $n\ge 1$. The temporal factors solve $T_n^{\prime \prime }+{\omega }_n^2{T}_n=0$, with the fixed-end standing wave frequencies ${\omega }_n=n\pi v/L$. Superposition givesThe constants are arbitrary subject to the convergence and regularity required of the solution. For initial displacement $f$ and initial velocity $g$, their Fourier sine series give $A_n=(2/L){\int }_0^{L}f(x)\sin (n\pi x/L)dx$ and $B_n=(2/(L{\omega }_n)){\int }_0^{L}g(x)\sin (n\pi x/L)dx$. Smooth compatible data give classical solutions; weaker data have the corresponding energy interpretation. A negative printed $L$ describes the interval with endpoints reversed and gives the same modes after using its positive length.

Seek a separated nonzero solution $u=X(x)Y(y)$. Division by $XY$ on intervals where the factors are nonzero gives $X^{\prime \prime }/X+Y^{\prime \prime }/Y=-25$. Write $X^{\prime \prime }+\lambda X=0$ and $Y^{\prime \prime }+(25-\lambda )Y=0$, with both factors zero at their endpoints $0,\pi$.

For $X$, a nonpositive $\lambda$ gives only the zero solution under the two endpoint conditions: linear functions vanish identically when $\lambda =0$, and the hyperbolic solutions do likewise for $\lambda <0$. For $\lambda >0$, the endpoints force $X=A\sin (mx)$ with positive integer $m$ and $\lambda =m^2$. Applying the same argument to $Y$ gives $Y=B\sin (ny)$ and $25-m^2=n^2$, with positive integer $n$.

The only positive integer pairs satisfying $m^2+n^2=25$ are $(3,4)$ and $(4,3)$. By superposition,To establish that this is the most general sufficiently regular solution, expand $u$ in the complete product Fourier sine series on the square. Substitution, or integration by parts against each sine mode, gives $(25-m^2-n^2)c_{mn}=0$ for every coefficient. Hence only the two displayed modes remain; all boundary terms vanish because of the given zero values. This is the eigenvalue-$25$ case of the Dirichlet eigensystem of a rectangle. The zero solution is included when $A=B=0$.

Set $z=e^{i\theta }$, so $2i\sin \theta =z-z^{-1}$. The binomial theorem givesUse $z^k-z^{-k}=2i\sin (k\theta )$ and $(2i{)}^5=32i$ to obtainThis is a finite Fourier sine series, with only odd harmonics.

Take the odd extension $H(x)=x(\pi -|x|)$ on $[-\pi ,\pi ]$ and then extend with period $2\pi$. It agrees with the required arch for $0\le x\le \pi$ and is continuous at its joins. Its Fourier sine series hasFor clarity, let $h=x(\pi -x)$, so $h^{\prime }=\pi -2x$ and $h^{\prime \prime }=-2$. Two integrations by parts, with the first endpoint term zero, giveConsequently $b_n=4[1-(-1{)}^n]/(\pi n^3)$: the even coefficients vanish and the odd ones are $8/(\pi n^3)$. This provesThe Fourier sine series of an odd quadratic arch converges to this continuous extension; its coefficients are also absolutely summable. Finally evaluate at $x=\pi /2$. The left side is ${\pi }^2/4$ and the sine factors are $(-1{)}^n$, soThe evaluation is at a continuity point, so it does not invoke a jump midpoint.

The original PDF contains the product $\sin x(1+4\cos x)$; the transcription drops the $x$ after the first sine. Using $2\sin x\cos x=\sin 2x$ gives the finite Fourier sine seriesThus $b_1=1$, $b_2=2$, and all other sine coefficients vanish. No cosine or constant terms are present.

The required property is odd symmetry, $\boxed{f(-x)=-f(x)}$, up to irrelevant changes on a set of measure zero when discussing Fourier coefficients. An odd function has zero mean, and its products with cosine are odd, so every cosine coefficient vanishes. Its sine coefficients may remain nonzero.

For pointwise representation, the Fourier series takes the average of the one-sided limits at a jump. In particular a pure Fourier sine series has value zero at the origin and at the endpoints; arbitrary assigned values there need not equal the series sum.

Use the double-angle identity $2\sin x\cos x=\sin 2x$. The Fourier series is already a finite trigonometric sum:In the usual Fourier sine series convention, $b_1=b_2=b_3=1$, all other $b_n$ vanish, and all cosine coefficients and the constant coefficient vanish.

For a nonzero separated solution $u=X(x)T(t)$ of the reaction-diffusion equation, substitution and division by $XT$ givesThe two sides depend on different variables, so both equal a separation constant. The homogeneous Dirichlet boundary conditions select a negative spatial constant, written $-k^2$:For $k>0$, $X=A\sin kx+B\cos kx$. The boundary value at zero makes $B=0$; the boundary value at $L$ then requires $kL=n\pi$, with $n=1,2,\dots$. A zero spatial constant gives only a linear function vanishing at both endpoints, hence the zero function; a positive spatial constant similarly gives no nonzero admissible hyperbolic solution. ThereforeBy superposition, the general solution with the stated boundary conditions isThe coefficients are fixed only if initial data are supplied: for $u(x,0)=f(x)$, they are the Fourier sine series coefficients $c_n=(2/L){\int }_0^{L}f(x)\sin (n\pi x/L)dx$. Here they remain arbitrary subject to convergence. Every mode has the ordinary diffusive decay rate plus the additional reaction decay rate one.

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 $m^2{\pi }^2+9{\pi }^2/4+{\beta }^2/4$.