ecuación con BC no homogénea del calor y del IC

Question

Zekeriya Özkan de Turquía, yo soy estudiante de master en Turquía

Se puede resolver la ecuación del calor con condiciones $$\frac{\partial^2u}{\partial x^2}=\frac{\partial u}{\partial t}$ $ IC: $u(0,t)=1$

BC: $u_x(0,t)=U$, $u_x(1,t)=-U$

Answer 1

Que $u(x,t)=v(x,t)-Ux^2+Ux+1$,

Entonces $\dfrac{\partial u(x,t)}{\partial t}=\dfrac{\partial v(x,t)}{\partial t}$

$\dfrac{\partial u(x,t)}{\partial x}=\dfrac{\partial v(x,t)}{\partial x}-2Ux+U$

$\dfrac{\partial^2u(x,t)}{\partial x^2}=\dfrac{\partial^2v(x,t)}{\partial x^2}-2U$

$\therefore\dfrac{\partial v(x,t)}{\partial t}=\dfrac{\partial^2v(x,t)}{\partial x^2}-2U$ $v(0,t)=0$, $v_x(0,t)=0$ y $v_x(1,t)=0$

Que $v(x,t)=\sum\limits_{n=0}^\infty C(n,t)\cos n\pi x$ para que automáticamente satisface $v_x(0,t)=0$ y $v_x(1,t)=0$,

Entonces $\sum\limits_{n=0}^\infty Ct(n,t)\cos n\pi x=-\sum\limits{n=0}^\infty n^2\pi^2C(n,t)\cos n\pi x-2U$

$\sum\limits_{n=0}^\infty Ct(n,t)\cos n\pi x+\sum\limits{n=0}^\infty n^2\pi^2C(n,t)\cos n\pi x=-2U$

$\sum\limits_{n=0}^\infty(C_t(n,t)+n^2\pi^2C(n,t))\cos n\pi x=-2U$

$\sum\limits_{n=0}^\infty(Ct(n,t)+n^2\pi^2C(n,t))\cos n\pi x=\sum\limits{n=0}^\infty k\cos n\pi x$, donde $k=\begin{cases}-2U&\text{when}~n=0\0&\text{when}~n\neq0\end{cases}$

$\therefore\begin{cases}C_t(n,t)=-2U&\text{when}~n=0\C_t(n,t)+n^2\pi^2C(n,t)=0&\text{when}~n\neq0\end{cases}$

$C(n,t)=\begin{cases}A(0)-2Ut&\text{when}~n=0\A(n)e^{-n^2\pi^2t}&\text{when}~n\neq0\end{cases}$

$\therefore u(x,t)=\sum\limits_{n=0}^\infty A(n)e^{-n^2\pi^2t}\cos n\pi x-Ux^2+Ux-2Ut+1$

$u(0,t)=1$ :

$\sum\limits_{n=0}^\infty A(n)e^{-n^2\pi^2t}-2Ut+1=1$

$\sum\limits_{n=0}^\infty A(n)e^{-n^2\pi^2t}=2Ut$

$\therefore u(x,t)=\sum\limits{n=0}^\infty A(n)e^{-n^2\pi^2t}\cos n\pi x-Ux^2+Ux-2Ut+1$, $A(n)$ Dónde está la solución de $\sum\limits{n=0}^\infty A(n)e^{-n^2\pi^2t}=2Ut$