Solve $u_t(x,t)=u_{xx}(x,t)$

Question

$$\begin{cases}u_t(x,y)=u_{xx}(x,t) \\ u(0,t)=0 \\ u(\pi,t)=1 \\ u(x,0)=0 \end{cases}$$

$$u(x,t)=v(x) w(t)$$

$$\frac{v_{xx}(x)}{v(x)}=\frac{w_t(t)}{w(t)}=-\lambda$$
$$-v''(x)=\lambda v(x)$$ $$\lambda=N^2$$ $$v(x)=C_1 \sin(Nx)+C_2 \cos(Nx)$$
$$w'(t)=-\lambda w(t)=-N^2 w(t)$$ $$w(t)=B_1 e^{-N^2 \ t}$$
$$v(0)=0$$ $$v(0)=C_2=0 $$

How can I apply the condition $u(\pi,t)=1$ ?

Thanks

Answer 1

Let $w(x,t)=v(x,t)-x/\pi$. This function will satisfy the equations $$ w_t(x,t) = w_{xx}(x,t) \\ w(0,t)=v(0,t)-0 = 0 \\ w(\pi,t)=v(1,t)-1=0 \\ w(x,0)=v(x,0)-x/\pi=-x/\pi. $$ Now you can solve for $w$. After finding $w$, then $v(x,t)=x/\pi+w(x,t)$ is the solution you want.