Substitution on a Temperature Problem

Question

Given real parameters $A,B,C$ consider the temperature problem with non-homogeneous boundary conditions:

$$u_t=ku_{xx}, \;\;\; u=u(x,t), \;\;\;0\leq x\leq \pi,\;\;\; t,k> 0$$ $$u_x(0,t)=u(0,t)+A$$ $$u_x(\pi,t)=u(\pi,t)+B$$ $$u(x,0)=Cx$$

I am tasked to use a substitution $u(x,t)=U(x,t)+\Phi(x)$ to reduce it to the temperature problem:

$$u_t=ku_{xx}, \;\;\; u=u(x,t), \;\;\;0\leq x\leq \pi,\;\;\; t,k> 0$$ $$u_x(0,t)=u(0,t)$$ $$u_x(\pi,t)=u(\pi,t)$$ $$u(x,0)=f(x)\in C^0([0,\pi])$$

Furthermore I am to indicate new initial temperatures F(x) in the homogeneous problem about U(x,t).

What is this substitution? I know that the second temperature problem has its solution, in the form of an infinite series, as $$u(x,t)=\sum_{n=0}^\infty c_ny_n(x)T_n(t)$$ but I do not know what the constants $c_n$ are, nor do I know the functions $y, T$ (I am currently attempting to solve this problem, I posted another question on it). Will I have to use this infinite series solution to determine the substitution? Will I need it to determine F(x) in the homogenous problem about U(x,t)? I'm not even 100% sure what "the homogeneous problem about U(x,t)" is. I would love clarification on any of these issues, thanks.

Answer 1

You can use a linear function $\Phi(x)=Dx+E$. Here is what you need from this function: $$ \Phi_x(0)=u_x(0,t)-U_x(0,t)=u(0,t)+A-U(0,t)=u(0,t)+A-u(0,t)+\Phi(0)=\Phi(0)+A $$ and $$ \Phi_x(\pi)=u_x(\pi,t)-U_x(\pi,t)=u(\pi,t)+B-U(\pi,t)=u(\pi,t)+B-u(\pi,t)+\Phi(\pi)=\Phi(\pi)+B $$ So you get a system $$ D=A+E;\qquad D=D\pi+E+B $$ where you can easily find $D$ and $E$ in terms of $A$ and $B$.

The initial condition becomes $U(x,0)=(C+D)x+E$.

Answer 2

If $u$ is a solution of $u_t=u_{xx}$, then so is $v(x,t)=u(x,t)+R+Sx$, for any constants $R,S$. This is because $(R+Sx)_t=0=(R+Sx)_{xx}$. This gives you a little flexibility to change your equation in $u$ to an equation in $v$ that has homogeneous endpoint conditions, which can then be solved using separation of variables. The new problem becomes $$ v_t = v_{xx},\\ v_x(0,t)=u_{x}(0,t)+S=(u(0,t)+A)+S=v(0,t)-R+S \\ v_{x}(\pi,t)=u_x(\pi,t)+S=u(\pi,t)+B+S=v(\pi,t)-R+S+B \\ v(x,0)=u(x,0)+R+Sx=Cx+R+Sx=(C+S)x+R. $$