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Within this polynomial family, $u(0,0)=0$ already holds and $u(1,0)=1+\beta =0$ gives $\beta =-1$. Thus one solution isTwo specified point values alone do not give uniqueness among all solutions of the heat equation.

The proposed function has partial derivatives $u_t=1$ and $u_{xx}=2\alpha$. Substitution in the heat equation requires $1=\alpha$, henceThe coefficient $\beta$ remains arbitrary.

Write the kernel as $K(s,t)=(4\pi t{)}^{-1/2}{e}^{-s^2/(4t)}$, with $t>0$. Its partial derivatives areThus the heat kernel itself satisfies $K_t=K_{ss}$. The proposed convolution consequently givesThis verifies the heat equation. The parameter is positive time; the real integral formula is not defined at negative time.

The passage of derivatives through the integral needs justification. A sufficient condition is the Gaussian-weighted absolute integrability in heat-kernel convergence for Gaussian-integrable initial data; it holds, in particular, for data of polynomial growth. On compact positive-time intervals strictly below an integrable later time, the differentiated kernels have an integrable Gaussian bound, so differentiation under the integral sign applies.

Even if the PDF's assumed convergence is read as ordinary conditional convergence, one can justify the same calculation. Choose a later time $T>t$ and let $H(\xi )$ be a primitive of $\phi (\xi )e^{-{\xi }^2/(4T)}$. Convergence at spatial point zero and time $T$ makes $H$ bounded on both tails. Put $q(x,\xi ,t)=K(x-\xi ,t)e^{{\xi }^2/(4T)}$. This function, its spatial and time derivatives, and their $\xi$-derivatives decay as Gaussians when $t<T$. Integration by parts rewrites the convolution as $-\int H{q}_{\xi }d\xi$, with vanishing endpoint terms. Boundedness of $H$ now permits differentiation under this absolutely convergent integral; since $q_t=q_{xx}$, the same equation follows. Thus one need not silently equate conditional and absolute convergence.

Set $T=t+a>0$. The similarity solution uses $y=(x+b)T^{-1/2}$, whose partial derivatives at fixed $x$ or $t$ areApply the multivariable chain rule to $u=T^{-1/2}v(y)$:Substituting in the heat equation and cancelling the positive common factor $T^{-3/2}$ gives

The heat equation is linear and homogeneous in $u$: the operator $L={\partial }_t-\lambda {\partial }_x^2$ satisfies $L(A{u}_1+B{u}_2)=AL{u}_1+BL{u}_2$. Thus the superposition principle applies to its solutions. Spatial translations are also allowed because the coefficients are constant.

Choose $a=1/(4\lambda )$ in part (b) and multiply by $\sqrt{a}$ to make the initial amplitude one. Taking the two shifts $b=1$ and $b=-1$ then givesAt $t=0$ this reproduces both initial peaks. Each term is an instance of Gaussian initial data for the heat equation, so their sum satisfies the equation. Its extension to negative times is a real smooth solution precisely on the connected intervalAt the lower endpoint the peaks narrow without bound and their heights diverge; there is no finite classical continuation there. For positive times each peak spreads and decreases in height while preserving its integral.

The steady profile has no time derivative. Substituting $u=\tilde{u}+u_s$ and using its steady equation cancels the forcing, givingThe initial condition isThe resulting heat equation is independent of $G$, as are its boundary conditions. Its initial amplitude still depends on $G$; removing a source from the evolution equation does not remove it from the whole problem.

For the heat equation ${\Theta }_t=\kappa {\Theta }_{xx}$, separation of variables with homogeneous Neumann boundary condition at both ends gives spatial modes $\cos (n\pi x/L)$ and decay factors $e^{-\kappa (n\pi /L{)}^{2}t}$. The $n=0$ mode is constant and does not decay.

The initial profile already contains only two such modes:Evolving each mode therefore yieldsThe endpoint derivatives vanish, and substitution checks both the equation and the initial profile. The spatial integral remains $L{T}_0/2$: no heat escapes through insulated ends, so the limiting uniform temperature is $T_0/2$.

With homogeneous Dirichlet boundary condition, the eigenfunctions are $\sin (n\pi x/L)$ and all modes decay. WriteReflection about the midpoint makes every even-$n$ coefficient zero. For odd $n$, putting $u=\pi x/L$ and using the product-to-sum identity givesIn particular $b_1=8T_0/(3\pi )$ and $b_3=-8T_0/(15\pi )$. ThusThis uses the Fourier sine series of a squared sine. The coefficients are $O(n^{-3})$, so the initial series converges uniformly; for $t>0$ the exponential factors also justify termwise differentiation to check the heat equation. Every mode vanishes at both endpoints, and the initial series gives the specified profile. Unlike part (a), there is no constant mode: the temperature tends to zero, with leading long-time term

Fourth-order spatial formula. For a sufficiently smooth function, pair the Taylor expansions at opposite shifts:Multiply the first sum by sixteen, subtract the second, and subtract $30u(x)$. The constant and fourth-derivative terms cancel, leavingHence the fourth-order centered second derivative satisfiesRearranging proves the requested $u^{\prime \prime }=D_{4,h}u+O(h^4)$, including its leading error coefficient.

Explicit diffusion scheme. Take $x_j=jh$, $h=1/N$, $t_n=n\tau$, and let $U_j^n$ approximate $\psi (x_j,t_n)$. Combining the forward Euler method in time with $D_{4,h}$ in space givesTo check the error, insert the exact solution. The heat equation gives ${\psi }_t={\psi }_{xx}$, while the time Taylor expansion and the spatial formula give the one-step defectThus its unnormalized local truncation error isas requested. Dividing this defect by $\tau$ gives the normalized residual $O(\tau )+O(h^4)$; these are distinct error conventions.

Boundary implementation. The centered stencil requires two ghost values at each end. An even-reflection closure for Neumann diffusion imposesThe reflected first centered difference at either endpoint is zero. Apply the same evolution formula at the endpoints after substituting these ghost values; for example,The ghost values also close the stencil at $j=1$ and $j=N-1$. They are determined auxiliary values, not new physical data. For a smooth solution compatible with the Neumann boundary conditions, differentiating those conditions in time and using the heat equation gives ${\psi }_{xxx}={\psi }_{xxxxx}=\cdots =0$ at the endpoints. Thus even reflection preserves the smoothness needed for fourth-order spatial accuracy. If the initial data fail the required higher compatibility conditions, a uniform smooth error expansion through $t=0$ cannot be assumed.

For completeness, the reflected spatial operator has cosine eigenvectors $\cos (\ell \pi j/N)$ and eigenvaluesThe most negative one is $-16/(3h^2)$. Requiring $|1+\tau {\lambda }_{\ell }|\le 1$ gives the forward Euler method stability restriction $\boxed{\tau /h^2\le 3/8}$ in the natural trapezoid-weighted discrete norm. A fourth-order spatial stencil does not remove the explicit diffusion time-step restriction.

Implicit alternative. The Crank-Nicolson diffusion scheme using the same spatial operator and boundary closure isIt has second-order time accuracy and fourth-order spatial accuracy for smooth compatible solutions; its one-step defect is $O({\tau }^3)+O(\tau h^4)$. Each spatial eigenvalue $\lambda \le 0$ has amplification factor $(1+\tau \lambda /2)/(1-\tau \lambda /2)$ of magnitude at most one, so it removes the explicit restriction for this boundary closure.

The price is solving a banded linear system at every step; the same matrix factorization can be reused for fixed $h$ and $\tau$. Large time steps still reduce accuracy. Moreover, the amplification factor tends to $-1$ as $\tau |\lambda |\to \infty$, so short spatial wavelengths can alternate sign with little numerical damping. Thus unconditional linear stability does not guarantee a monotone or well-resolved transient. Boundary accuracy and initial compatibility remain necessary with either time discretization.

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.

With temperature measured relative to the fixed-temperature end, the heat equation and its data areHere $\nu >0$ is thermal diffusivity and $k>0$ is thermal conductivity. The sign of the derivative condition describes incoming heat flux at $L$: the outward conductive heat flux is $-k{u}_x=-H$.

By Fourier's law, insulation means zero outward heat flux. The Neumann boundary conditions are ${\theta }_x(0,t)={\theta }_x(L,t)=0$. The constant mode equals the initial integral average $L/4$. For $n\ge 1$, the cosine coefficients areOdd modes and modes divisible by four vanish; write the remaining indices as $n=2(2j+1)$. The heat flow from a triangular initial temperature with insulated endpoints givesThe nonconstant modes decay, so $\boxed{{\theta }_N\to L/4}$ uniformly as $t\to \infty$. Integrating the heat equation gives $d({\int }_0^L\theta dx)/dt=\nu [{\theta }_x{]}_0^L=0$, explaining the conserved mean. The cooled case tends instead to zero because heat leaves through its ends.

The initial triangle has nonzero endpoint slopes, so it is not compatible with the insulated condition at the corner $t=0$. The series nevertheless recovers the continuous initial temperature and satisfies the insulated condition for every $t>0$; one should not assert continuity of its spatial derivative at that initial corner.

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.

Use the conventionThese are the Fourier transform and inverse Fourier transform for suitably integrable data. Assuming the required ordinary transforms exist and the lower derivatives vanish at both ends, integration by parts gives $\widehat{g^{\prime }}=[g{e}^{-ikx}{]}_{-\infty }^{\infty }+ik\hat{g}=ik\hat{g}$. Repeating this argument yieldsThe decay assumptions are used to discard the boundary terms at each step.

For positive diffusivity $\alpha$, transforming the heat equation gives ${\partial }_t\hat{f}=-\alpha k^2\hat{f}$. Thus $\hat{f}(k,t)=e^{-\alpha k^{2}t}{\hat{f}}_0(k)$, where $f_0(x)=f(x,0)$. The convolution theorem and the Gaussian heat kernel giveFor $f_0(x)=e^{-x^2}$, the Fourier transform of a Gaussian is $\sqrt{\pi }{e}^{-k^2/4}$. Multiplying by the heat factor and inverting gives the Gaussian initial data for the heat equation solutionIt recovers the initial profile as $t\downarrow 0$ and preserves its spatial integral $\sqrt{\pi }$.

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.

Differentiate the spatial Fourier transform under the integral, and integrate the second spatial derivative by parts twice. For sufficiently decaying temperature and first derivative, the boundary terms vanish:Hence the transformed heat equation is the ordinary differential equation

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.

Complete the square in the Fourier transform exponent:The supplied shifted Gaussian integral therefore givesThis is the Fourier transform of a Gaussian, with angular-wavenumber convention $e^{-ikx}$.

Take the spatial Fourier transform of the heat equation. Two integrations by parts, with decaying fields and first derivatives at infinity, give $\mathcal{F}[{\psi }_{xx}]=-k^2\tilde{\psi }$. Thus each mode satisfies the ordinary differential equationInserting the transformed initial condition and integrating the time equation givesFinally set $A=t+1/(4\lambda )>0$. The inverse Fourier transform, evaluated by the same Gaussian identity, givesEquivalently the Gaussian initial data for the heat equation evolve asThe profile broadens and its peak decreases, while its total integral remains $\sqrt{\pi /\lambda }$, as required by diffusion without loss on the whole line. Direct differentiation verifies the equation and the initial limit.

Substitute $T=e^{-\nu t}R(r)$ into the axisymmetric heat equation and cancel the exponential:Thus the radial ordinary differential equation and its Sturm-Liouville form areThe weight in this Sturm-Liouville problem is therefore $\boxed{w=r}$. The outer Dirichlet boundary condition is $R(1)=0$, and the physical solution is regular at the centre. The factor $r$ in the weighted inner product is also the radial part of the polar area element.