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For positive thermal diffusivity $\nu$, thermal conductivity $k$ and length $L$, subtract the stationary solution $w(x)=Hx/k$. The remainder satisfies the heat equation with a homogeneous Dirichlet boundary condition at zero and a homogeneous Neumann boundary condition at $L$. The mixed Dirichlet-Neumann modes on an interval are $\sin (k_{n}x)$, where $k_n=(n+1/2)\pi /L$. Thus for an initial temperature profile $u_0$,For zero initial temperature, $b_n=-2H(-1{)}^n/(kL{k}_n^2)$. The series gives the initial profile in the usual Fourier series sense and a smooth solution for $t>0$. If $H\ne 0$, the zero initial profile does not satisfy the prescribed endpoint derivative at $t=0$: the temperature can be continuous there without its spatial derivative extending continuously to that corner. The sign $k{u}_x(L,t)=H>0$ describes incoming heat flux, since the outward conductive heat flux is $-k{u}_x$.

Use the mixed Dirichlet-Neumann modes on an interval. To justify their Fourier series, extend $y$ evenly across $x=1$ to $[0,2]$, then oddly across $x=0$, and repeat with period four. The condition $y(0)=0$ makes the odd extension continuous, and $y^{\prime }(1)=0$ makes the even reflection continuously differentiable. The resulting periodic function has a convergent Fourier sine series. Reflection about $x=1$ eliminates the even sine harmonics, leavingAlternatively these values follow directly from $\sin {\lambda }_{n}0=0$ and ${\lambda }_n\cos {\lambda }_n=0$. Fourier orthogonality givesThe extension argument also supplies completeness, not merely orthogonality, of this Fourier sine basis.

Expand the forcing as $x\cos \pi x=\sum b_n\sin {\lambda }_{n}x$. Product-to-sum and integration by parts giveEach mode of the differential operator has eigenvalue $-({\lambda }_n^2+{\alpha }^2)$. Hence, for real $\alpha$, the solution isThe coefficients are $O(n^{-4})$, so the series and its first two differentiated series converge uniformly. Therefore it solves the equation and the boundary conditions term by term. Uniqueness follows by multiplying the homogeneous equation by its real solution and integrating: the boundary terms vanish and ${\int }_0^1[(y^{\prime }{)}^2+{\alpha }^2{y}^2],dx=0$, forcing $y=0$.

Put $a=f^{\prime }(0)>0$. Linearizing the morphogen reaction-diffusion equation at $C=0$ givesSeek separated modes $C(x,t)=X(x)e^{\lambda t}$. ThenThe mixed Dirichlet-Neumann modes on an interval areIndeed, the Dirichlet condition selects sine functions, while the Neumann condition requires $\cos (k_{n}L)=0$. Their growth rates areThe largest growth rate is the lowest mode,Linear stability requires ${\lambda }_0<0$, after which all higher modes also decay. Therefore the critical length for a linearly growing morphogen isEquality gives a neutral lowest mode, while a larger domain is linearly unstable.

Subtract the constant steady state profile: set $v=T-T_0$. ThenThe last equation follows from Fourier's law and zero heat flux at the insulated end. For a nonzero separated solution $v=X(x)H(t)$, separation of variables givesTo see that $\mu >0$ for a nonzero eigenfunction, multiply the spatial equation by $X$ and use integration by parts:Both boundary terms vanish. If the right side were zero, $X$ would be constant and $X(0)=0$ would make it zero. Thus write $\mu ={\lambda }^2$ with $\lambda >0$. Solving gives $X=A\sin (\lambda x)$, and $X^{\prime }(L)=0$ requires $\cos (\lambda L)=0$. The mixed Dirichlet-Neumann modes on an interval therefore haveThe eigenvalues of $-d^2/d{x}^2$ are ${\mu }_n={\lambda }_n^2$; ${\lambda }_n$ are their positive square roots. By superposition, with coefficients for which the resulting series and its required derivatives converge at positive times,The eigenfunctions are orthogonal on $[0,L]$ and have squared norm $L/2$. Expanding the initial excess temperature $v(x,0)=T_{0}x/L$ givesThus the requested temperature isAt the insulated endpoint $\sin ({\lambda }_{n}L)=(-1{)}^{n-1}$ cancels the alternating coefficient sign, leavingEvery summand is positive. The endpoint temperature never drops below $T_0$; it decreases towards $T_0$ as $t\to \infty$ and starts at $T(L,0)=2T_0$.

For completeness, the Sturm-Liouville eigenfunction expansion gives the initial profile in the $L^2$ sense. Here the coefficients are absolutely summable, so the Weierstrass M-test also gives uniform convergence on $[0,L]\times [0,\infty )$, and the initial values are recovered everywhere. For positive times the exponential factors justify all required termwise derivatives, giving a classical solution. There is, however, an initial incompatibility with a Neumann boundary condition at $(L,0)$: the initial slope is $T_0/L$, while the insulated Neumann boundary condition requires zero slope for $t>0$. The temperature is continuous to the initial edge, but $T_x$ cannot extend continuously to that corner. One must not differentiate the initial series term by term at $x=L$.

For a nonzero separated solution $v=X(x)T(t)$, separation of variables gives $X^{\prime \prime }+{\kappa }^{2}X=0$ and $T^{\prime }=-\nu {\kappa }^{2}T$. The endpoint conditions require $X=\sin \kappa x$ and $\cos \kappa L=0$. Thus the mixed Dirichlet-Neumann modes on an interval have ${\kappa }_n=(n+1/2)\pi /L$, $n\ge 0$. Their squared L2 norms are $L/2$, so the initial Fourier coefficients areThe integration uses $\cos {\kappa }_{n}L=0$ and $\sin {\kappa }_{n}L=(-1{)}^n$. ThereforeAt positive time the series can be differentiated termwise and satisfies both homogeneous endpoint conditions for $v$. At time zero it recovers $-Hx/k$, so $u$ recovers the initial zero temperature. If $H\ne 0$, the prescribed derivative at $(L,0)$ is incompatible with differentiating the initial zero profile; the temperature is still continuous there, while its spatial derivative has a corner layer. At long times the decaying modes leave $u\to Hx/k$.