i want to find a solution $u(x,t)$ to the following $$u_t=u_{xx}$$ in the region $[0,\frac{\pi}{2}] \times [0,\infty)$ with the mixed boundary conditions: $$u(0,t)=u_x(\frac{\pi}{2},t)=0$$ and I.C of:$$u(x,0)=x$$
i thought maybe defining a new function $w(x,t)$ that depends on $u(x,t)$ and something else, and the B.C's of the new function will be of the Dirichlet or Neumann form.
We use separation of variables, i.e., we assume $u(x,t) = X(x)T(t)$. This leads to the equation $$\frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)}.$$ As usual we see that $X$ and $T$ are independent of each other, hence, we assume they are constant $c$: $$\frac{T'(t)}{T(t)} = -c = \frac{X''(x)}{X(x)}.$$ Moreover, the boundary condition $u(0,t)=X(0)T(t) = 0 = u_x(\pi/2,t) = X'(\pi/2)T(t)$ leads to $X(0)=0=X'(\pi/2)$. A general solution of $X''=c X$ is given by $$X(x) = A\sin(\sqrt{c} x)$$, where we used that $X(0)=0$. For the second boundary condition we get $X'(\pi/2)=A \sqrt{c} \cos(\sqrt{c}\pi / 2)=0$. This is true for different $c_n = (2n+1)^2$. So we get $$X_n(x) = A_n \sin((2n+1)x).$$ Solving $T'(t)+(2n+1)T(t)=0$ gives us $$T_n(t) = B_n e^{-(2n+1)t}.$$ Putting everything together we get solutions $$u_n(x,t) = C_n e^{-(2n+1)t} \sin((2n+1)x),$$ where $C_n$ are new arbitrary constants. Since the heat equation is linear we can construct new equations from different $u_n$ via linear combination. So we get that $$u(x,t) = \sum_n C_n e^{-(2n+1)t} \sin((2n+1)x)$$ solves our equation. Now the only thing left is to choose the constants $C_n$ such that $u(x,0)=x$ is satisfied, i.e., $$\sum_n C_n \sin((2n+1)x) = x.$$