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