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Differentiate the second ordinary differential equation and use the first to eliminate $x^{\prime }$:Thus the reduced second-order linear differential equation isNo solutions are lost: given a solution $y$, define $x=y^{\prime }-\cos t$; then both original ordinary differential equations hold.

The differential operator on the left is $(D-3{)}^2$, where $D=d/dt$. Its repeated characteristic root suggests writing $y=e^{3t}v$. By the product rule,so the ordinary differential equation reduces to $v^{\prime \prime }=3(t+1)$. Two antiderivatives giveThe boundary conditions imply $v(0)=1$ and $v(1)=0$, hence $B=1$ and $A=-3$. ThereforeThe calculation of $v^{\prime \prime }$ verifies the ordinary differential equation, and the bracket takes the values one and zero at the two endpoints. These boundary conditions determine both constants uniquely.

Solving the transformed ordinary differential equation with its initial condition gives $\tilde{\theta }(k,t)=e^{-k^{2}t}\tilde{\theta }(k,0)$. The Fourier inversion theorem therefore yieldsFor $t>0$ the Gaussian multiplier suppresses large spatial frequencies, which explains the smoothing produced by heat conduction.

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

For a spherically symmetric number density, the radial Laplacian is $r^{-2}{\partial }_r(r^2{\partial }_{r}n)$. Applying separation of variables and choosing the separation constant $-k^2$ givesso the ordinary differential equations areThe radial problem is an eigenvalue problem for $-\Delta$ with Dirichlet boundary conditions, hence $k^2>0$: multiplying the radial equation by $r^2\overline{R}$ and integrating gives $k^2{\int }_0^a{r}^2|R{|}^{2}dr={\int }_0^a{r}^2|R^{\prime }{|}^{2}dr>0$ for a nonzero regular mode vanishing at $a$.

To solve it explicitly, set $R=f/r$, corresponding to the suggested power $p=-1$. The equation reduces to $f^{\prime \prime }+k^{2}f=0$, soFiniteness at the origin forces $B=0$, and the Dirichlet boundary condition at $r=a$ requires $k=m\pi /a$ for an integer $m\ge 1$. The absence of interior zeros excludes $m\ge 2$, selecting the fundamental normal mode. If $n_c>0$ denotes the initial central density, the result iswhere the sine ratio is defined as $1$ at the origin. It is strictly positive in the interior for every finite time. The arbitrary positive scale $n_c$ cannot be fixed without additional initial data.

The spatial factor is time independent, so the density and the total neutron population have the same exponential growth rate. Thus the spherical reaction-diffusion growth threshold isAt equality the fundamental mode is stationary; below it the mode decays. For fixed $\lambda >0$, the equivalent critical-radius condition is $a>\pi /\sqrt{\lambda }$. This is the competition between multiplication in the bulk and diffusion loss at the surface in the stated dimensionless model.

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.

For separation of variables, take $u(x,t)=X(x)T(t)$ in the damped wave equation. Dividing by $c^{2}XT$, with $c>0$, gives $X^{\prime \prime }/X=(T^{\prime \prime }+\alpha T^{\prime })/(c^{2}T)=-k^2$. The separated ordinary differential equations are thereforeThe negative separation constant anticipates the positive spatial eigenvalues selected by the fixed endpoints.

The Fourier transform of $u_{xx}$ is $-k^2\tilde{u}$: two integrations by parts have no boundary terms under the stated decay assumption. Time differentiation commutes with the spatial transform. ThereforeFor every fixed spatial wavenumber $k$, this is a constant-coefficient ordinary differential equation in time.

For the Gaussian function, direct differentiation gives $f^{\prime }=-tf$. Applying the Fourier transform identities from part (a),Solving this first-order ordinary differential equation givesIn fact, the normalization can also be fixed: the Gaussian integral gives $F(0)={\int }_{\mathbb{R}}{e}^{-t^2/2}dt=\sqrt{2\pi }$, so $C=\sqrt{2\pi }$ with the convention used here. The transform has the same Gaussian shape.

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.

Divide by $2x^2$ to put the ordinary differential equation into normalized form $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$:The coefficients are singular at zero, butare both analytic there. This is precisely the regular singular point criterion for a second-order equation, so $x=0$ is a regular singular point.

Take $\lambda >0$ and $t>0$, the positive diffusivity requirement for the forward heat kernel. The PDF does not explicitly specify the sign of $\lambda$; the stated real Gaussian function formula requires this assumption. For a direct calculation, initially assume sufficiently smooth and rapidly decreasing data, so that the Fourier transform, its inverse and the following integrations are justified. The resulting heat-kernel convolution extends to wider classes of initial data.

Using the stated Fourier transform convention and integration by parts,The transformed ordinary differential equation has solution $\tilde{u}(k,t)=e^{-\lambda k^{2}t}{\tilde{u}}_0(k)$. Fourier inversion then givesComplete the square:Part (a), after scaling by $\sqrt{\lambda t}$, evaluates the remaining Gaussian integral even though its shift is imaginary. ThereforeThe heat kernel has integral one and is an approximate identity as $t\downarrow 0$, which recovers the initial condition. Mere decay at spatial infinity does not by itself imply every Fourier integral is absolutely convergent; the regularity assumption above supplies a rigorous setting for the derivation.

Seek a nonzero separated solution $\psi =T(t)R(r)\Theta (\theta )$. Substitution into the heat equation and division by the product givesThe left side depends only on time, so set it to $-k^2$. Multiplication by $r^2$ then separates the angular dependence, with separation constant $\ell (\ell +1)$. The three ordinary differential equations areThe temporal solution is $T=T_0{e}^{-\kappa k^{2}t}$. With $q=\cos \theta$, the angular equation is the Legendre differential equation; regularity at both angular endpoints selects $\ell =0,1,2,\dots$ and $\Theta \propto P_{\ell }(q)$, the Legendre polynomial. The radial equation has the Spherical Bessel functions $j_{\ell }(kr)$ and $y_{\ell }(kr)$ for $k>0$. Regularity at $r=0$, if imposed, excludes the singular second solution. For $k=0$ the radial powers are $r^{\ell }$ and $r^{-\ell -1}$. Without specified radial boundary conditions, no discrete set of $k$ is selected; $-k^2$ is a convention for the temporal separation constant.

For separation of variables matching the boundary Fourier mode, take ${\psi }_0=X(x)e^{iky}$ with $X(0)=1$. Because ${\partial }_y^3{e}^{iky}=-i{k}^3{e}^{iky}$, the reduced ordinary differential equation isThereforeThe decay of a Fourier mode in a third-order spatial evolution equation requires a sign check: this satisfies the requested limit at positive infinity precisely when $\boxed{k>0}$. For $k=0$ it remains identically one, while for $k<0$ its amplitude grows exponentially. Hence for those nonpositive values there is no separated solution meeting both prescribed boundary conditions; the printed assumption that $k$ is merely real needs this additional restriction for decay.

The Fermat principle makes the optical path length stationary. For a graph $y=\xi (x)$ the variational integrand isThe Euler-Lagrange equation isUse the chain rule $dn(x,\xi (x))/dx=n_x+n_y{\xi }^{\prime }$ and differentiate ${\xi }^{\prime }/\sqrt{1+{\xi }^{\prime 2}}$ to obtainMultiplying through and combining the $n_y$ terms gives the required second-order ordinary differential equation with no derivatives in denominators:All partial derivatives of the refractive index are evaluated along the ray.

Use separation of variables, $u=X(x)Y(y)$, and divide by $XY$ where nonzero. A real separation constant $\kappa$ givesso $X^{\prime \prime }=\kappa X$ and $Y^{\prime }=(\kappa -1-\sin y)Y$. Normalizing the second factor to one at zero givesThe spatial ordinary differential equation has three types of real solution:Products $X{Y}_{\kappa }$ are the separated eigenfunctions. The negative, zero and positive separation constants give trigonometric, linear and hyperbolic function forms respectively. There is no discrete quantization until boundary conditions on $x$ are supplied. Since $\cos y-1$ is bounded, the envelope of $Y_{\kappa }$ decays for $\kappa <1$, grows for $\kappa >1$, and is bounded and periodic for $\kappa =1$. The scalar prefactor can also be derived by removing a variable scalar reaction rate by an integrating factor.

Expand the solution and the forcing in the orthonormal Sturm-Liouville eigenfunction expansion. The forcing coefficients areEvery ${\lambda }_n$ is nonzero, so the solution coefficient is $b_n/{\lambda }_n$. ThusAs an independent check, solving $Y^{\prime \prime }+Y=-1$ directly with the two endpoint conditions givesIt satisfies both Dirichlet boundary conditions and the forced ordinary differential equation.

Use the same Fourier transform convention, now in the sense of tempered distributions, since the homogeneous oscillations are not integrable. Transforming givesSince $q>0$, the two roots are distinct. A distribution annihilated by this polynomial is a linear combination of Dirac delta distributions at its roots:Fourier inversion therefore givesThe constants may be complex. Differentiating the factor $e^{-irt}$ also verifies that the original ordinary differential equation reduces to $\ddot{\psi }+q^2\psi =0$ for $\varphi =e^{-irt}\psi$.

For separation of variables, take $n=T(t)X(x)Y(y)Z(z)$ and divide the reaction-diffusion equation by this product wherever it is nonzero:Each spatial term depends on a different variable and must be constant. Write $X^{\prime \prime }=-k_x^{2}X$ and similarly for the other two factors. The Dirichlet boundary conditions give $X(0)=X(L)=0$. A zero separation constant permits only a linear function and hence the zero solution; a positive value of $X^{\prime \prime }/X$ likewise permits no nonzero function vanishing at both endpoints. The nontrivial choices are thereforeThe temporal ordinary differential equation isso, absorbing its constant multiplier into $a_{p,q,r}$,The superposition principle and completeness of the Fourier sine series give the stated threefold expansion with these factors. For an initial density $n_0$, orthogonality also determines the constants:

No. A necessary condition for consistency of a numerical method is reproduction of the constant solution $y(x)=1$ of the ordinary differential equation $y^{\prime }=0$. In a linear multistep method this givesBut the reverse triangle inequality gives, under the strict coefficient inequality,Thus the necessary order conditions for a linear multistep method already fail at order zero, regardless of the values of the ${\beta }_j$.