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Use separation of variables to solve the following BVP with mixed boundary conditions; $$ \begin{cases} u_t = u_{xx} &x\in (0,l), t>0 \\ u(0,t) = T_0 & t>0 \\ u_x(l,t) = 0 & t>0 \\ u(x, 0) = T_1 &x\in(0,l) \end{cases} $$ where $T_0$ and $T_1$ are constants. In particular solve the above for $T_0 = 1, T_1=5$.

I tried to do it and got stuck. In particular, the condition $T_0 \neq T_1$ threw me of. This problem is in a section on how to solve boundary value problems by separation of variables and using Fourier series.

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    $\begingroup$ In this case, since they are just constants, you should have no problem carrying out SepVar, The only thing that shows up is a slightly different boundary problem when you get to the system of ODE's. Where exactly do you get stuck? $\endgroup$ – DaveNine Dec 14 '17 at 0:52
  • $\begingroup$ In addition to DaveNines question, I would say edit your post to include what you have tried so others can see where you are having difficulties. $\endgroup$ – Mattos Dec 14 '17 at 4:10
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Separation of variables problems such as this require homogeneous endpoint conditions, which is usually achieved by a substitution in a case like this. $$ v(x,t)=u(x,t)-T_0. $$ The new equation for $v$ is $$ v_t = v_{xx} \\ v(0,t)=u(0,t)-T_0=0 \\ v_{x}(l,t)=u_{x}(l,t)=0 \\ v(x,0)=u(x,0)-T_0=T_1-T_0. $$ Now when you separate $v(x,t)=X(x)T(t)$, the ODE in $x$ becomes $$ \frac{X''}{X}=\lambda,\;\; X(0)=0,\; X'(l)=0. $$ The eigenfunctions are $$ \sin(\pi x/2l), \sin(3\pi x/2l),\sin(5\pi x/2l),\cdots. $$ with corresponding eigenvalues $(2n+1)^2\pi^2/4l^2$. These are automatically orthogonal, and you have to expand $v(x,t)=\sum_{n=1}^{\infty}C_nX_n(x)T_n(t)$ in such a way that $v(x,0)=T_1-T_0$, which leads to a Fourier series problem in the orthogonal eigenfunctions $X_n$, and gives the constants $C_n$.

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