Heat Equation Fourier 80E41D

1. **Problem statement:** Solve the heat equation $$u_t = u_{xx}$$ for $0 < x < 1$ with boundary conditions $$u(t,0) = 0$$ (left end held at 0 degree) and $$u_x(t,1) = 0$$ (right end insulated), and initial condition $$u(0,x) = f(x).$$ 2. **Method:** Use separation of variables by assuming $$u(t,x) = X(x)T(t).$$ Substitute into the PDE: $$X(x)T'(t) = X''(x)T(t).$$ Divide both sides by $X(x)T(t)$: $$\frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} = -\lambda,$$ where $\lambda$ is a separation constant. 3. **Spatial problem:** Solve $$X''(x) + \lambda X(x) = 0,$$ with boundary conditions $$X(0) = 0, \quad X'(1) = 0.$$ 4. **Solve the spatial ODE:** The general solution is $$X(x) = A \sin(\sqrt{\lambda} x) + B \cos(\sqrt{\lambda} x).$$ Apply $X(0) = 0$: $$X(0) = B = 0,$$ so $$X(x) = A \sin(\sqrt{\lambda} x).$$ Apply $X'(1) = 0$: $$X'(x) = A \sqrt{\lambda} \cos(\sqrt{\lambda} x),$$ so $$X'(1) = A \sqrt{\lambda} \cos(\sqrt{\lambda}) = 0.$$ Since $A \neq 0$, we require $$\cos(\sqrt{\lambda}) = 0.$$ 5. **Eigenvalues:** The solutions to $$\cos(\sqrt{\lambda}) = 0$$ are $$\sqrt{\lambda_n} = \frac{(2n-1)\pi}{2}, \quad n=1,2,3,\dots$$ Thus, $$\lambda_n = \left(\frac{(2n-1)\pi}{2}\right)^2.$$ 6. **Eigenfunctions:** $$X_n(x) = \sin\left(\frac{(2n-1)\pi}{2} x\right).$$ 7. **Temporal problem:** Solve $$T_n'(t) + \lambda_n T_n(t) = 0,$$ which has solution $$T_n(t) = C_n e^{-\lambda_n t}.$$ 8. **General solution:** The solution is a Fourier sine series: $$u(t,x) = \sum_{n=1}^\infty C_n e^{-\lambda_n t} \sin\left(\frac{(2n-1)\pi}{2} x\right).$$ 9. **Initial condition:** Use $$u(0,x) = f(x) = \sum_{n=1}^\infty C_n \sin\left(\frac{(2n-1)\pi}{2} x\right).$$ 10. **Fourier coefficients:** Compute $$C_n = 2 \int_0^1 f(x) \sin\left(\frac{(2n-1)\pi}{2} x\right) dx.$$ **Final answer:** $$\boxed{u(t,x) = \sum_{n=1}^\infty \left[2 \int_0^1 f(s) \sin\left(\frac{(2n-1)\pi}{2} s\right) ds \right] e^{-\left(\frac{(2n-1)\pi}{2}\right)^2 t} \sin\left(\frac{(2n-1)\pi}{2} x\right)}.$$