Heat equation with unusual boundary conditions

Question

There seem to be a lot of posts on the subject of heat equations with weird boundary conditions, but after a brief perusal of these I couldn't find quite what I'm looking for. The following 1D problem has arisen in my research, and I'm not sure how best to solve it:

$$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}; \\u(0,t) = f(t); \\ \frac{\partial u}{\partial x}(0,t) = g(t)$$

on the interval $(x,t)\in[0,\infty)\times (-\infty,\infty)$, where $f$ and $g$ are some specified smooth functions.

First of all: is this problem even well-posed? I don't know much existence/uniqueness theory for PDEs so I'm not sure.

Secondly, how can I go about solving this equation analytically? The unusual, nonhomogeneous B.C.'s are really tripping me up.

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EDIT: Here's what I've tried. Letting $\tilde{u}(x,\omega)=\int dt \, e^{-i\omega t} u(x,t),$ we can transform the original equation into a 2nd order ODE in $x$ with Cauchy boundary conditions and solution

$$\tilde{u}(x,\omega) = \tilde{f}(\omega) \cosh\left((1-i)\sqrt{\frac{\omega}{2}}x\right)+\frac{1+i}{\sqrt{2\omega}}\tilde{g}(\omega) \sinh\left((1-i)\sqrt{\frac{\omega}{2}}x\right),$$

where $\tilde{f}$ and $\tilde{g}$ are of course the Fourier transforms of $f$ and $g$ respectively.

Formally, then, \begin{align}u(x,t) &= \int \frac{d\omega}{2\pi} \, e^{i \omega t} \left[ \tilde{f}(\omega) \cosh\left((1-i)\sqrt{\frac{\omega}{2}}x\right)+\frac{1+i}{\sqrt{2\omega}}\tilde{g}(\omega) \sinh\left((1-i)\sqrt{\frac{\omega}{2}}x\right) \right] \\ &=\int \frac{d \omega}{2 \pi} \int d\tau \, e^{i\omega(t-\tau)}\left[f(\tau) \cosh\left((1-i)\sqrt{\frac{\omega}{2}}x\right)+\frac{1+i}{\sqrt{2\omega}}g(\tau) \sinh\left((1-i)\sqrt{\frac{\omega}{2}}x\right)\right] \\ &= \int_{-\infty}^\infty d\tau \, \left(F(x,t-\tau)f(\tau)+G(x,t-\tau)g(\tau)\right), \end{align} where $$F(x,t):= \int_{-\infty}^{\infty} \frac{d\omega}{2\pi} \, e^{i \omega t} \cosh\left((1-i)\sqrt{\frac{\omega}{2}}x\right)$$

and

$$G(x,t):= \int_{-\infty}^{\infty} \frac{d\omega}{2\pi} \, e^{i \omega t} \frac{1+i}{\sqrt{2\omega}} \sinh\left((1-i)\sqrt{\frac{\omega} {2}}x\right). $$

I am getting discouraged, however, attempting to evaluate the kernels $F$ and $G$, which besides simply being challenging (for me anyway) to work with, appear to suffer from divergences.

Have I done something wrong? Are there closed-form, or at least nicer, expressions for $F$ and $G$? Does this approach even work?

Answer 1

Approach $1$: separation of variables

Case $1$: $\text{Re}(t)\geq0$

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=X''(x)T(t)$

$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-s^2\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s^2)e^{-ts^2}\\X(x)=\begin{cases}c_1(s^2)\sin xs+c_2(s^2)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1x+C_2+\int_0^\infty C_3(s^2)e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds$

$\dfrac{\partial u(x,t)}{\partial x}=C_1+\int_0^\infty sC_3(s^2)e^{-ts^2}\cos xs~ds-\int_0^\infty sC_4(s^2)e^{-ts^2}\sin xs~ds$

$\dfrac{\partial u(x,t)}{\partial x}(x=0)=g(t)$ :

$C_1+\int_0^\infty sC_3(s^2)e^{-ts^2}~ds=g(t)$

$\int_0^\infty\dfrac{C_3(s^2)e^{-ts^2}}{2}d(s^2)=g(t)-C_1$

$\int_0^\infty\dfrac{C_3(s)e^{-ts}}{2}ds=g(t)-C_1$

$\mathcal{L}_{s\to t}\biggl\{\dfrac{C_3(s)}{2}\biggr\}=g(t)-C_1$

$C_3(s)=2\mathcal{L}^{-1}_{t\to s}\{g(t)\}-2C_1\delta(s)$

$\therefore u(x,t)=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(t)\}e^{-ts^2}\sin xs~ds-2C_1\int_0^\infty\delta(s^2)e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(t)\}e^{-ts^2}\sin xs~ds-C_1\int_0^\infty\dfrac{\delta(s^2)e^{-ts^2}\sin xs}{s}d(s^2)+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(t)\}e^{-ts^2}\sin xs~ds-C_1\int_0^\infty\dfrac{\delta(s)e^{-ts}\sin x\sqrt{s}}{\sqrt{s}}ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(t)\}e^{-ts^2}\sin xs~ds-C_1\lim\limits_{s\to 0}\dfrac{e^{-ts}\sin x\sqrt{s}}{\sqrt{s}}+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(t)\}e^{-ts^2}\sin xs~ds-C_1\lim\limits_{s\to 0}\dfrac{\dfrac{xe^{-ts}\cos x\sqrt{s}}{2\sqrt{s}}-te^{-ts}\sin x\sqrt{s}}{\dfrac{1}{2\sqrt{s}}}+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(t)\}e^{-ts^2}\sin xs~ds-C_1\lim\limits_{s\to 0}(xe^{-ts}\cos x\sqrt{s}-2t\sqrt{s}e^{-ts}\sin x\sqrt{s})+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(t)\}e^{-ts^2}\sin xs~ds-C_1x+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(t)\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds$

$u(0,t)=f(t)$ :

$C_2+\int_0^\infty C_4(s^2)e^{-ts^2}~ds=f(t)$

$\int_0^\infty\dfrac{C_4(s^2)e^{-ts^2}}{2s}d(s^2)=f(t)-C_2$

$\int_0^\infty\dfrac{C_4(s)e^{-ts}}{2\sqrt{s}}ds=f(t)-C_2$

$\mathcal{L}_{s\to t}\biggl\{\dfrac{C_4(s)}{2\sqrt{s}}\biggr\}=f(t)-C_2$

$C_4(s)=2\sqrt{s}\mathcal{L}^{-1}_{t\to s}\{f(t)\}-C_2\delta(\sqrt{s})$

$\therefore u(x,t)=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(t)\}e^{-ts^2}\sin xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(t)\}e^{-ts^2}\cos xs~ds-C_2\int_0^\infty\delta(s)e^{-ts^2}\cos xs~ds=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(t)\}e^{-ts^2}\sin xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(t)\}e^{-ts^2}\cos xs~ds-C_2=2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(t)\}e^{-ts^2}\sin xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(t)\}e^{-ts^2}\cos xs~ds$

Case $2$: $\text{Re}(t)\leq0$

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=X''(x)T(t)$

$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}=s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=s^2\\X''(x)-s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s^2)e^{ts^2}\\X(x)=\begin{cases}c_1(s^2)\sinh xs+c_2(s^2)\cosh xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1x+C_2+\int_0^\infty C_3(s^2)e^{ts^2}\sinh xs~ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds$

$\dfrac{\partial u(x,t)}{\partial x}=C_1+\int_0^\infty sC_3(s^2)e^{ts^2}\cosh xs~ds+\int_0^\infty sC_4(s^2)e^{ts^2}\sinh xs~ds$

$\dfrac{\partial u(x,t)}{\partial x}(x=0)=g(t)$ :

$C_1+\int_0^\infty sC_3(s^2)e^{ts^2}~ds=g(t)$

$\int_0^\infty\dfrac{C_3(s^2)e^{ts^2}}{2}d(s^2)=g(t)-C_1$

$\int_0^\infty\dfrac{C_3(s)e^{ts}}{2}ds=g(t)-C_1$

$\int_0^\infty\dfrac{C_3(s)e^{-ts}}{2}ds=g(-t)-C_1$

$\mathcal{L}_{s\to t}\biggl\{\dfrac{C_3(s)}{2}\biggr\}=g(-t)-C_1$

$C_3(s)=2\mathcal{L}^{-1}_{t\to s}\{g(-t)\}-2C_1\delta(s)$

$\therefore u(x,t)=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(-t)\}e^{ts^2}\sinh xs~ds-2C_1\int_0^\infty\delta(s^2)e^{ts^2}\sinh xs~ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(-t)\}e^{ts^2}\sinh xs~ds-C_1\int_0^\infty\dfrac{\delta(s^2)e^{ts^2}\sinh xs}{s}d(s^2)+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(-t)\}e^{ts^2}\sinh xs~ds-C_1\int_0^\infty\dfrac{\delta(s)e^{ts}\sinh x\sqrt{s}}{\sqrt{s}}ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(-t)\}e^{ts^2}\sinh xs~ds-C_1\lim\limits_{s\to 0}\dfrac{e^{ts}\sinh x\sqrt{s}}{\sqrt{s}}+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(-t)\}e^{ts^2}\sinh xs~ds-C_1\lim\limits_{s\to 0}\dfrac{\dfrac{xe^{ts}\cosh x\sqrt{s}}{2\sqrt{s}}+te^{ts}\sinh x\sqrt{s}}{\dfrac{1}{2\sqrt{s}}}+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(-t)\}e^{ts^2}\sinh xs~ds-C_1\lim\limits_{s\to 0}(xe^{ts}\cosh x\sqrt{s}+2t\sqrt{s}e^{ts}\sinh x\sqrt{s})+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(-t)\}e^{ts^2}\sinh xs~ds-C_1x+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(-t)\}e^{ts^2}\sinh xs~ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds$

$u(0,t)=f(t)$ :

$C_2+\int_0^\infty C_4(s^2)e^{ts^2}~ds=f(t)$

$\int_0^\infty\dfrac{C_4(s^2)e^{ts^2}}{2s}d(s^2)=f(t)-C_2$

$\int_0^\infty\dfrac{C_4(s)e^{ts}}{2\sqrt{s}}ds=f(t)-C_2$

$\int_0^\infty\dfrac{C_4(s)e^{-ts}}{2\sqrt{s}}ds=f(-t)-C_2$

$\mathcal{L}_{s\to t}\biggl\{\dfrac{C_4(s)}{2\sqrt{s}}\biggr\}=f(-t)-C_2$

$C_4(s)=2\sqrt{s}\mathcal{L}^{-1}_{t\to s}\{f(-t)\}-C_2\delta(\sqrt{s})$

$\therefore u(x,t)=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(-t)\}e^{ts^2}\sinh xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(-t)\}e^{ts^2}\cosh xs~ds-C_2\int_0^\infty\delta(s)e^{ts^2}\cosh xs~ds=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(-t)\}e^{ts^2}\sinh xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(-t)\}e^{ts^2}\cosh xs~ds-C_2=2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(-t)\}e^{ts^2}\sinh xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(-t)\}e^{ts^2}\cosh xs~ds$

Hence $u(x,t)=\begin{cases}2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(t)\}e^{-ts^2}\sin xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(t)\}e^{-ts^2}\cos xs~ds&\text{when Re}(t)\geq0\\2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{g(-t)\}e^{ts^2}\sinh xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(-t)\}e^{ts^2}\cosh xs~ds&\text{when Re}(t)\leq0\end{cases}$

Approach $2$: power series method

Similar to diffusion equation, inhomogenous boundary conditions (the subtraction method):

Let $u(x,t)=\sum\limits_{n=0}^\infty\dfrac{x^n}{n!}\dfrac{\partial^nu(0,t)}{\partial x^n}$ ,

Then $u(x,t)=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\dfrac{\partial^{2n}u(0,t)}{\partial x^{2n}}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(0,t)}{\partial x^{2n+1}}=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\dfrac{\partial^nu(0,t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\dfrac{\partial^nu_x(0,t)}{\partial t^n}=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\dfrac{\partial^nf(t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\dfrac{\partial^ng(t)}{\partial t^n}$

Answer 2

The diffusion equation is invariant wrt. to the gauge of the zero of the field. A boundary condition with $u(0,t))=c$ is equivalent to $(u-c)(t)=0$ that mimics reflection of the solution function around the point $(0,c)$. The underlying proces can be implemented physically by a mirrored solution of Brownian antiparticles that annihilate in pairs when they meet at the boundary, a particle carrying substance u, the antiparticle 2c -u with mean c.

On the other hand, a boundary value of the derivative is an imprinted current. In order not ot blow up, the total current must be zero or positive outward over the boundary.

A mixed boundary condition is a convex combination of absorption and prescribed source gradient at each boundary segment.

Fixing a start distribution at $t=0$ and one of the three types of boundary conditions (Dirchlet, Neumann, mixed - Newton/Robin) - depending on the points of the boundary, determine the unique time forward solution; any additional incompatible boundary condition makes the problem ill posed.

Numerically on a lattice, one cannot describe boundary value funcrtions with its tangent derivative implicitly given, and the normal derivative without destroying ellipticity of the solutiion at the boundary.

Ellipticity means nothing more than the fact, the value at time $t+dt$ at $x$ is a weighted mean over a small neighborhood at time $t$ .