Heat Equation Neumann 44Fb1E

1. **Problem Statement:** We want to solve the heat equation $$u_t = u_{xx}$$ on the interval $$0 < x < 1$$ for time $$t > 0$$ with Neumann boundary conditions $$u_x(t,0) = 0$$ and $$u_x(t,1) = 0$$ for all $$t \geq 0$$, and initial condition $$u(0,x) = x$$. 2. **Method:** We use separation of variables and eigenfunction expansions. The boundary conditions are Neumann (zero flux), so the eigenfunctions satisfy: $$\phi''(x) + \lambda \phi(x) = 0$$ with $$\phi'(0) = 0$$ and $$\phi'(1) = 0$$. 3. **Eigenvalues and Eigenfunctions:** The general solution to $$\phi'' + \lambda \phi = 0$$ is: $$\phi(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x)$$ Applying $$\phi'(0) = 0$$: $$\phi'(x) = -A \sqrt{\lambda} \sin(\sqrt{\lambda} x) + B \sqrt{\lambda} \cos(\sqrt{\lambda} x)$$ At $$x=0$$: $$\phi'(0) = B \sqrt{\lambda} = 0 \implies B = 0$$ Applying $$\phi'(1) = 0$$: $$\phi'(1) = -A \sqrt{\lambda} \sin(\sqrt{\lambda} \cdot 1) = 0$$ For nontrivial $$A \neq 0$$, we require: $$\sin(\sqrt{\lambda}) = 0 \implies \sqrt{\lambda} = n \pi, \quad n=0,1,2,\ldots$$ Thus, $$\lambda_n = (n \pi)^2$$ and eigenfunctions: $$\phi_n(x) = \cos(n \pi x)$$ 4. **Solution form:** The solution is expanded as: $$u(t,x) = \sum_{n=0}^\infty c_n e^{-\lambda_n t} \phi_n(x) = c_0 + \sum_{n=1}^\infty c_n e^{-(n \pi)^2 t} \cos(n \pi x)$$ 5. **Initial condition:** At $$t=0$$, $$u(0,x) = f(x) = x = c_0 + \sum_{n=1}^\infty c_n \cos(n \pi x)$$ 6. **Finding coefficients $$c_n$$:** Use orthogonality of cosine functions on $$[0,1]$$: $$c_0 = \int_0^1 x \cdot 1 \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$$ For $$n \geq 1$$: $$c_n = 2 \int_0^1 x \cos(n \pi x) \, dx$$ Integrate by parts: Let $$u = x$$, $$dv = \cos(n \pi x) dx$$ Then $$du = dx$$, $$v = \frac{\sin(n \pi x)}{n \pi}$$ So, $$\int_0^1 x \cos(n \pi x) dx = \left. x \frac{\sin(n \pi x)}{n \pi} \right|_0^1 - \int_0^1 \frac{\sin(n \pi x)}{n \pi} dx$$ The first term is zero because $$\sin(n \pi) = 0$$ and at 0 it is zero. The second integral: $$\int_0^1 \sin(n \pi x) dx = \left[ -\frac{\cos(n \pi x)}{n \pi} \right]_0^1 = -\frac{\cos(n \pi) - 1}{n \pi}$$ Therefore, $$\int_0^1 x \cos(n \pi x) dx = - \frac{1}{n \pi} \left(-\frac{\cos(n \pi) - 1}{n \pi} \right) = \frac{1 - \cos(n \pi)}{(n \pi)^2}$$ Recall $$\cos(n \pi) = (-1)^n$$, so: $$c_n = 2 \cdot \frac{1 - (-1)^n}{(n \pi)^2}$$ 7. **Simplify coefficients:** - For even $$n$$, $$(-1)^n = 1$$, so $$c_n = 0$$. - For odd $$n$$, $$(-1)^n = -1$$, so: $$c_n = \frac{4}{(n \pi)^2}$$ 8. **Final solution:** $$u(t,x) = \frac{1}{2} + \sum_{\substack{n=1,3,5,\ldots}} \frac{4}{(n \pi)^2} e^{-(n \pi)^2 t} \cos(n \pi x)$$ This series converges to the solution of the heat equation with the given initial and boundary conditions.