heat equation with inhomogenous BC and IC I'm Zekeriya Özkan from Turkey, I'm a master student in Turkey
Can you solve the heat equation with conditions
$$\frac{\partial^2u}{\partial x^2}=\frac{\partial u}{\partial t}$$
IC: $u(0,t)=1$
BC : $u_x(0,t)=U$,
     $u_x(1,t)=-U$
 A: Let $u(x,t)=v(x,t)-Ux^2+Ux+1$ ,
Then $\dfrac{\partial u(x,t)}{\partial t}=\dfrac{\partial v(x,t)}{\partial t}$
$\dfrac{\partial u(x,t)}{\partial x}=\dfrac{\partial v(x,t)}{\partial x}-2Ux+U$
$\dfrac{\partial^2u(x,t)}{\partial x^2}=\dfrac{\partial^2v(x,t)}{\partial x^2}-2U$
$\therefore\dfrac{\partial v(x,t)}{\partial t}=\dfrac{\partial^2v(x,t)}{\partial x^2}-2U$ with $v(0,t)=0$ , $v_x(0,t)=0$ and $v_x(1,t)=0$
Let $v(x,t)=\sum\limits_{n=0}^\infty C(n,t)\cos n\pi x$ so that it automatically satisfies $v_x(0,t)=0$ and $v_x(1,t)=0$ ,
Then $\sum\limits_{n=0}^\infty C_t(n,t)\cos n\pi x=-\sum\limits_{n=0}^\infty n^2\pi^2C(n,t)\cos n\pi x-2U$
$\sum\limits_{n=0}^\infty C_t(n,t)\cos n\pi x+\sum\limits_{n=0}^\infty n^2\pi^2C(n,t)\cos n\pi x=-2U$
$\sum\limits_{n=0}^\infty(C_t(n,t)+n^2\pi^2C(n,t))\cos n\pi x=-2U$
$\sum\limits_{n=0}^\infty(C_t(n,t)+n^2\pi^2C(n,t))\cos n\pi x=\sum\limits_{n=0}^\infty k\cos n\pi x$ , where $k=\begin{cases}-2U&\text{when}~n=0\\0&\text{when}~n\neq0\end{cases}$
$\therefore\begin{cases}C_t(n,t)=-2U&\text{when}~n=0\\C_t(n,t)+n^2\pi^2C(n,t)=0&\text{when}~n\neq0\end{cases}$
$C(n,t)=\begin{cases}A(0)-2Ut&\text{when}~n=0\\A(n)e^{-n^2\pi^2t}&\text{when}~n\neq0\end{cases}$
$\therefore u(x,t)=\sum\limits_{n=0}^\infty A(n)e^{-n^2\pi^2t}\cos n\pi x-Ux^2+Ux-2Ut+1$
$u(0,t)=1$ :
$\sum\limits_{n=0}^\infty A(n)e^{-n^2\pi^2t}-2Ut+1=1$
$\sum\limits_{n=0}^\infty A(n)e^{-n^2\pi^2t}=2Ut$
$\therefore u(x,t)=\sum\limits_{n=0}^\infty A(n)e^{-n^2\pi^2t}\cos n\pi x-Ux^2+Ux-2Ut+1$ , where $A(n)$ is the solution of $\sum\limits_{n=0}^\infty A(n)e^{-n^2\pi^2t}=2Ut$
