Heat equation $u_t=u_{xx}$ with $u(x,0)=0, u(1,t)=5t$, and $u_x(0,t)=0$ Heat equation $u_t=u_{xx}$ with $u(x,0)=0, u(1,t)=5t$, and $u_x(0,t)=0$
I saw this in Japanese Q&amp;A website and tried to solve it but I couldn't.
So I am not sure that this can be solved analytically.
What I tried: From Wikipedia, I found a solution for $u(x,0)=0, u(0,t)=f(t)$ as:
$u(x,t)[f] = \int_0^t x/\sqrt{4\pi (t-s)^3}\ \exp(-x^2/4(t-s))\ f(s)\ ds$
In order to satisfy $u_x(0,t)=0$, I considered using symmetry: $u(x,t) = u(x+1,t)[f] + u(x-1,t)[f]$ for some $f$, but I couldn't adjust the result to satisfy $u(1,t)=5t$.
(My calculation)
$u(x,t)[f(t)=t] = \int_0^t x/\sqrt{4\pi (t-s)^3} \exp(-x^2/4(t-s))\ s\ ds \\\
 = [ (x^2/2+t) erf(x/2\sqrt{t-s}) - x\sqrt{t-s}/\sqrt{\pi}\ \exp(-x^2/4(t-s)) ]_0^t \\\
 = (x^2/2+t)(1-erf(x/2\sqrt{t})) - x\sqrt{t}/\sqrt{\pi}\ \exp(-x^2/4t)$
$u(x,t)[f(t)=1] = \int_0^t x/\sqrt{4\pi (t-s)^3}\ \exp(-x^2/4(t-s))\ ds \\\
 = [ erf(x/2\sqrt{t-s})  ]_0^t \\\
 = 1 - erf(x/2\sqrt{t})$
From the comment by @Gregory, let $u=U+V, V=5t+\frac{5}{2}(x^2-1)$.
Then $U(x,t)$ should satisfies
$U(x,0) = \frac{5}{2}(1-x^2), U(1,t) = 0$
Now I can use the stable solutions:
$u(x,t)[n] = \exp(-n^2\pi^2t) \cos(nπx)$
and the Fourier series expansion
$1 - x^2 = 2/3 - \sum_{n=1}^{\infty} (-1)^n  \frac{4}{(nπ)^2} \cos(nπx)\ [-1\leq x\leq 1]$.
I obtained
$U(x,t) = 5/2\ (2/3 - \sum_{n=1}^{\infty} (-1)^n \frac{4}{(nπ)^2} u(x,t)[n])\\\
 = \frac{5}{3} - \frac{10}{\pi^2}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} u(x,t)[n] )$,
and $u(x,t) = 5t + \frac{5}{2}x^2 - \frac{5}{6} - \frac{10}{\pi^2}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \exp(-n^2\pi^2t) \cos(nπx)$.
From the comment by @Gregory, let $u=U+V, V=5t+\frac{5}{2}(x^2-1)$.
Then $U(x,t)$ should satisfies
$U(x,0) = \frac{5}{2}(1-x^2), U(1,t) = 0$
Now I can use the stable solutions:
$u(x,t)[n] = \exp(-n^2\pi^2t) \cos(nπx)$
and the Fourier series expansion
$1 - x^2 = 2/3 - \sum_{n=1}^{\infty} (-1)^n  \frac{4}{(nπ)^2} \cos(nπx)\ [-1\leq x\leq 1]$.
I obtained
$U(x,t) = 5/2\ (2/3 - \sum_{n=1}^{\infty} (-1)^n \frac{4}{(nπ)^2} u(x,t)[n])\\\
 = \frac{5}{3} - \frac{10}{\pi^2}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} u(x,t)[n] )$,
and $u(x,t) = 5t + \frac{5}{2}x^2 - \frac{5}{6} - \frac{10}{\pi^2}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \exp(-n^2\pi^2t) \cos(nπx)$.