# Thread: IBVP with a nonzero BCs

1. ## IBVP with a nonzero BCs

I must solve this initial boundary value problem by transforming it into homogeneous BCs.

$$\\(PDE)\; u_{t}=\alpha ^{2}u_{xx} \: \: \: \: \: \: \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 0 \\ $ $(BCs)\; \left\{\begin{matrix} u(0,t)=1) & 0 u_{x}(1,t)+hu(1,t)=1 & 0 \end{matrix}\right.\\$ $(IC)\;\: \: u(x,0)=sin(\pi x)+x\; \: \: \: \: \: \: \: \: \: \: 0\leq x\leq 1$ ---> My work I set $\[s(x,t)=a(t)x+b(t)(1-x)+U(x,t)$
$
where S(x,t) is the steady state and U(x,t) is the transient

So i substitute s(x,t) into the BCs
$$s(0,t)=1 $ $s_{x}(1,t)+hs(1,t)=1$ These two equations can give us b(t)=1 and a(t)=2/1-h ? and therefore u(x,t)=2x/(1-h) + (1-x) + U(x,t) = 1 - (1+h/1-h)x + U(x,t) MY PROBLEM is, how did they get b(t)=1 and a(t) = 2/1-h. I know i have have to make s(0,t) = to something like a(t) + b(t) = 1 and s_x(1,t) = ?something? so that i can plug it into the $\[ s_{x}(1,t)+hs(1,t)=1$
$
but I am not sure what s and s_x should be.
Can someone please tell me what they are so I can finally get a(t) and b(t).
Then I can finally finish solving this problem.

$u(x,t) = a(t)x + b(t)(1-x) + U(x,t)$
then $u(0,x) = b(t) + U(0,t) = 1$ so choosing $b(t) = 1$ gives $U(0,t) = 0$
Further $u_x(1,t) + hu(1,t) = a(t)-b(t) +U_x(1,t) + h(a(t) + U(1,t)) = 1$
and choosing $a(t) - b(t) + h a(t) = 1$ gives $U_x(1,t) + hU(1,t) = 0.$
With $b(t) = 1$ solving $a(t) - b(t) + h a(t) = 1$ for $a(t)$ gives what you want.