Could you please check my analytical solution for accuracy and correctness?
Consider the heat equation: $$ u_t(x, t)=u_{x x}(x, t) $$ where $$ 0<x<1, \quad t>0 $$ Neumann Boundary condition: $$ \begin{gathered} u_x(0, t)=u, \quad t>0 \\ u_x(1, t)=-u, \quad t>0 \end{gathered} $$ Initial condition: $$ u(x, 0)=1, \quad 0<x<1 $$
To solve the heat equation with Neumann boundary conditions using the method of separation of variables, we assume that the solution can be expressed as a product of functions of $x$ and $t$ $$ u(x, t)=X(x) T(t) $$ Substituting this into the heat equation, we get $$ X T^{\prime}=X^n T $$ Dividing both sides by XT, we obtain: $$ \frac{T^{\prime}}{T}=\frac{X^n}{X}=-\lambda $$ where $\lambda$ is the separation constant. Thus, we have two ODEs to solve: $$ T^{\prime}+\lambda T=0 \text { and } X^{\prime \prime}+\lambda X=0 $$ The Neumann boundary conditions can be expressed in terms of the solution of $X(x)$, follows: $$ \begin{aligned} & X^{\prime}(0)=u=k_1 \\ & X^{\prime}(1)=-u=-k_2 \end{aligned} $$ where $k_1$ and $k_2$ are constants. Solving the ODE for $T(t)$, we get $$ T(t)=c_1 e^{-\lambda t} $$ where $c_1$ is a constant. Solving the $O D E$ for $X(x)$, we get $$ X(x)=c_2 \cos (\sqrt{\lambda} x)+c_3 \sin (\sqrt{\lambda} x) $$ Using the boundary conditions for $X(x)$. we get. $$ \begin{aligned} & X^{\prime}(0)=-c_2 \sqrt{\lambda}=u \\ & X^{\prime}(1)=-c_2 \sqrt{\lambda} \sin (\sqrt{\lambda})+c_2 \sqrt{\lambda} \cos (\sqrt{\lambda})=-u \end{aligned} $$ Solving for $c_2$ and $c_3$, we get $$ c_2=-\frac{u}{\sqrt{\lambda}} \text { and } c_3=\frac{u}{\sqrt{\lambda}} \sin (\sqrt{\lambda}) $$ Substituting the expressions for $T(t)$ and $X(x)$ into the solution for $u(x, t)$, we get $$ u(x, t)=\sum_{n=1}^{\infty} b_n \cos \left(\sqrt{\lambda_n} x\right) e^{-\lambda_n t} $$ where $\lambda_n = n^2\pi^2$ and $b_n$ is given by:$$ b_n=\frac{2}{u \sqrt{\lambda_n} \sin \left(\sqrt{\lambda_n}\right)} \int_0^1 u(x, 0) \cos \left(\sqrt{\lambda_n} x\right) d x $$ Using the initial condition, we get $$ \begin{aligned} b_n & =\frac{2}{u \sqrt{\lambda_n} \sin \left(\sqrt{\lambda_n}\right)} \int_0^1 \cos \left(\sqrt{\lambda_n} x\right) d x \\ & =\frac{2}{u \sqrt{\lambda_n} \sin \left(\sqrt{\lambda_n}\right)}\left[\frac{\sin \left(\sqrt{\lambda_n} x\right)}{\sqrt{\lambda_n}}\right]_0^1 \\ & =\frac{2}{u \pi n}(-1)^{n+1} \end{aligned} $$ Therefore, the analytic solution for the heat equation with Neumann boundary conditions and initial condition is:
$$ u(x, t)=\frac{4}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cos (n \pi x) e^{-n^2 n^2 t} $$ where $\lambda_n = n^2\pi^2$ is the separation constant, and $n$ is a positive integer. The solution is obtained using the method of separation of variables, where we assume that the solution can be expressed as a product of two functions of x and t: $u(x,t) = X(x)T(t)$. We then solve the resulting ODEs for $X(x)$ and $T(t)$, subject to the given Neumann boundary conditions and initial condition. The solution involves an infinite series of terms, each representing a different mode of heat transfer. The constant $\lambda_n$ is the eigenvalue associated with the $n$th mode, and determines the rate of decay of the corresponding mode as time progresses.