Here is the transcription of the handwritten problem, followed by a detailed, step-by-step mathematical breakdown and the Python code required to visualize the result.Problem TranscriptionEquation: The 1D Heat Equation$$\frac{\partial u}{\partial t} = 2 \frac{\partial^2 u}{\partial x^2}$$Domain: A rod of length $L = \pi$Boundary Conditions (BCs): * $u(0,t) = 0$$u(\pi,t) = 0$Initial Condition (IC): * $u(x,0) = \sin(x)$Step-by-Step SolutionWhile you mentioned advanced techniques like Picard iteration and Charpit's method, they are not the appropriate tools for this specific problem. Picard iteration is primarily used for proving the existence and uniqueness of solutions to initial value problems in Ordinary Differential Equations (ODEs). Charpit's method is designed for solving non-linear first-order Partial Differential Equations (PDEs).Because we are dealing with a linear, second-order PDE with homogeneous boundary conditions, the most robust and elegant approach is the Method of Separation of Variables.Step 1: Assume a Separable SolutionWe begin by assuming the solution $u(x,t)$ can be written as the product of two functions, one depending solely on $x$ and the other solely on $t$:$$u(x,t) = X(x)T(t)$$Step 2: Substitute into the PDETaking the partial derivatives of our assumed solution:$\frac{\partial u}{\partial t} = X(x)T'(t)$$\frac{\partial^2 u}{\partial x^2} = X''(x)T(t)$Substitute these back into the original heat equation:$$X(x)T'(t) = 2 X''(x)T(t)$$Step 3: Separate the VariablesDivide both sides by $2X(x)T(t)$ to isolate the variables to opposite sides of the equation:$$\frac{T'(t)}{2T(t)} = \frac{X''(x)}{X(x)}$$Since the left side depends only on $t$ and the right side depends only on $x$, both sides must equal a constant. We call this the separation constant, $-\lambda$:$$\frac{T'(t)}{2T(t)} = \frac{X''(x)}{X(x)} = -\lambda$$This yields two ordinary differential equations:Time ODE: $T'(t) + 2\lambda T(t) = 0$Spatial ODE: $X''(x) + \lambda X(x) = 0$Step 4: Solve the Spatial ODE with Boundary ConditionsWe apply the boundary conditions to the spatial component $X(x)$:$u(0,t) = 0 \implies X(0)T(t) = 0 \implies X(0) = 0$$u(\pi,t) = 0 \implies X(\pi)T(t) = 0 \implies X(\pi) = 0$We now solve $X''(x) + \lambda X(x) = 0$ for different cases of $\lambda$. The only non-trivial solutions occur when $\lambda > 0$. Let $\lambda = n^2$ where $n$ is a positive constant.The general solution to the spatial ODE is:$$X(x) = c_1 \cos(nx) + c_2 \sin(nx)$$Applying the first boundary condition ($X(0) = 0$):$$X(0) = c_1(1) + c_2(0) = 0 \implies c_1 = 0$$Applying the second boundary condition ($X(\pi) = 0$):$$X(\pi) = c_2 \sin(n\pi) = 0$$For a non-trivial solution ($c_2 \neq 0$), we must have $\sin(n\pi) = 0$. This occurs when $n$ is an integer ($n = 1, 2, 3, \dots$).Therefore, our eigenvalues are $\lambda_n = n^2$, and our corresponding eigenfunctions are:$$X_n(x) = \sin(nx)$$Step 5: Solve the Time ODESubstitute $\lambda = n^2$ into the time ODE:$$T_n'(t) + 2n^2 T_n(t) = 0$$This is a standard first-order linear ODE. Its solution represents exponential decay:$$T_n(t) = A_n e^{-2n^2 t}$$Step 6: Construct the General SolutionBy the Principle of Superposition, the general solution is the infinite sum of all valid solutions:$$u(x,t) = \sum_{n=1}^{\infty} A_n \sin(nx) e^{-2n^2 t}$$Step 7: Apply the Initial ConditionFinally, we use the initial condition $u(x,0) = \sin(x)$ to find the specific coefficients $A_n$. Set $t = 0$:$$u(x,0) = \sum_{n=1}^{\infty} A_n \sin(nx) = \sin(x)$$By matching the coefficients on both sides of the equation (a simple case of a Fourier sine series), it is immediately clear that:$A_1 = 1$ (for $n = 1$)$A_n = 0$ (for all $n > 1$)Substituting these back into the general solution yields the final, exact analytical solution:$$u(x,t) = \sin(x) e^{-2t}$$Python VisualizationHere is the data science component. This script uses numpy for the matrix calculations and matplotlib to plot the 3D surface of the temperature distribution across the rod over time.Pythonimport numpy as np import matplotlib.pyplot as plt # 1. Define the parameters and create the meshgrid # We simulate the rod from x = 0 to pi, and time from t = 0 to 2 seconds x = np.linspace(0, np.pi, 100) t = np.linspace(0, 2, 100) # Create 2D coordinate matrices from 1D coordinate vectors X, T = np.meshgrid(x, t) # 2. Compute the exact mathematical solution derived above # u(x,t) = sin(x) * e^(-2t) U = np.sin(X) * np.exp(-2 * T) # 3. Initialize the figure and 3D axis fig = plt.figure(figsize=(10, 8)) ax = fig.add_subplot(111, projection='3d') # 4. Generate the 3D Surface Plot # 'plasma' is an excellent colormap for heat/temperature visualization # as it scales from dark purple (cool) to bright yellow (hot) surface = ax.plot_surface(X, T, U, cmap='plasma', edgecolor='none', alpha=0.9) # 5. Label axes and format the plot for readability ax.set_title('3D Surface Plot of the Heat Equation\n$u(x,t) = \sin(x) e^{-2t}$', fontsize=14, pad=20) ax.set_xlabel('Position along the rod (x)', fontsize=12, labelpad=10) ax.set_ylabel('Time (t)', fontsize=12, labelpad=10) ax.set_zlabel('Temperature (u)', fontsize=12, labelpad=10) # Adjust the viewing angle for optimal perspective ax.view_init(elev=30, azim=45) # Add a color bar map to represent the temperature scale fig.colorbar(surface, ax=ax, shrink=0.5, aspect=10, label='Temperature Magnitude') # Render the plot plt.tight_layout() plt.show()