I was wondering if anyone could help me with this exercise?
(i) Determine a stationary solution $u_S(x)$ of the heat equation with the following boundary conditions: \begin{cases} u_t(x,t)-3u_{xx}(x,t)=0,& (x,t)\in(0,5)\times(0,+\infty);\\ u(0,1)=1,&u_x(5,t)=-2,\quad\text{for every}\ t>0. \end{cases} (ii) Determine as a consequence the solution of the following Cauchy and boundary value problem: \begin{cases} u_t(x,t)-3u_{xx}(x,t)=0,& (x,t)\in(0,5)\times(0,+\infty);\\ u(0,1)=1,&u_x(5,t)=-2,\quad\text{for every}\ t>0;\\ u(x,0)=u_S(x)+2\sin(\frac\pi{10}x)-3\sin(\frac7{10}\pi x),&x\in(0,2). \end{cases}
Since I am not very experienced with PDEs, I would really appreciate if someone could explain me how to solve this step by step.
Do you understand what a "stationary solution" is? It is a solution that does not change with time (the fact the solution is denoted by "us(x)", as a function of x only should have been a hint). If the function depends on x only then the derivatives with respect to t are 0. The heat equation reduces to $-3u_xx= 0$ or simply $u_xx= 0$. Integrating with respect to x, $u_x= A$, some constant. Integrating again, $u= Ax+ B$.
Since $us(0, t)= us(0)= B= 1$ and $us_x(5,t)= us'(5)= A= -2$, we must have $us(x, t)= -2x+ 1$.
Then, since we are told that, for all solutions, including non stationary, $u(x, 0)= us(x)+ 2sin\left(\frac{\pi}{10}x\right)- 3sin\left(\frac{7}{10}\pi x\right)$ the solution to the entire problem is $u(x,t)= \left(2sin\left(\frac{\pi}{10}x\right)- 3sin\left(\frac{7}{10}\pi x\right)\right)e^t- 2x+ 1$