We are tasked with solving the following initial-boundary value problem for the function $u(x,t)$:

$\frac{\partial u}{\partial t} = 2 \frac{\partial^2 u}{\partial x^2}, \quad 0 < x < \pi, \quad t > 0$ with boundary conditions: $u(0,t) = u(\pi,t) = 0, \quad t > 0$ and initial condition: $u(x,0) = \sin x + 2 \sin 4x, \quad 0 < x < \pi$ We need to find $u\left(\frac{\pi}{2}, \ln(5)\right)$ rounded to 22 decimal places.

Step 1: Solve the PDE using separation of variables

We start by assuming the solution can be written as a product of functions in $x$ and $t$: $u(x,t) = X(x)T(t)$ Substitute this form into the PDE:

$\frac{\partial}{\partial t} (X(x)T(t)) = 2 \frac{\partial^2}{\partial x^2} (X(x)T(t))$

This gives: $X(x) \frac{dT}{dt} = 2 T(t) \frac{d^2 X}{dx^2}$

Dividing both sides by $X(x)T(t)$, we get the separation of variables equation: $\frac{1}{T(t)} \frac{dT}{dt} = 2 \frac{1}{X(x)} \frac{d^2 X}{dx^2} = -\lambda$ where $\lambda$ is a separation constant. This gives two ODEs:

1. $\frac{d^2 X}{dx^2} + \lambda X = 0$
2. $\frac{dT}{dt} = -2\lambda T(t)$

Step 2: Solve for $X(x)$

We solve the spatial equation $\frac{d^2 X}{dx^2} + \lambda X = 0$ with boundary conditions $X(0) = X(\pi) = 0$. The solution is of the form: $X_n(x) = \sin(nx), \quad \lambda_n = n^2 \quad \text{for} \quad n = 1, 2, 3, \dots$ Thus, the eigenvalues are $\lambda_n = n^2$, and the corresponding eigenfunctions are $X_n(x) = \sin(nx)$.

Step 3: Solve for $T(t)$

From the ODE $\frac{dT}{dt} = -2\lambda T(t)$, we solve for $T(t)$: $T_n(t) = e^{-2n^2 t}$

Step 4: General solution

The general solution is a sum over all modes: $u(x,t) = \sum_{n=1}^{\infty} A_n \sin(nx) e^{-2n^2 t}$

Step 5: Apply initial condition

The initial condition is $u(x,0) = \sin x + 2 \sin 4x$. This implies: $\sum_{n=1}^{\infty} A_n \sin(nx) = \sin x + 2 \sin 4x$ Comparing coefficients, we get: $A_1 = 1, \quad A_4 = 2, \quad A_n = 0 \text{ for } n \neq 1, 4$

Thus, the solution simplifies to: $u(x,t) = \sin x e^{-2t} + 2 \sin 4x e^{-2 \cdot 16 t}$

Step 6: Evaluate $u\left( \frac{\pi}{2}, \ln(5) \right)$

Substitute $x = \frac{\pi}{2}$ and $t = \ln(5)$ into the solution: $u\left( \frac{\pi}{2}, \ln(5) \right) = \sin\left( \frac{\pi}{2} \right) e^{-2 \ln(5)} + 2 \sin(2\pi) e^{-2 \cdot 16 \ln(5)}$

We know: $\sin\left( \frac{\pi}{2} \right) = 1, \quad \sin(2\pi) = 0$ Thus, the second term vanishes, and we are left with: $u\left( \frac{\pi}{2}, \ln(5) \right) = e^{-2 \ln(5)}$ Using properties of logarithms: $e^{-2 \ln(5)} = \frac{1}{5^2} = \frac{1}{25}$

Final Answer

$u\left( \frac{\pi}{2}, \ln(5) \right) = 0.04$

This is the value of $u\left( \frac{\pi}{2}, \ln(5) \right)$ rounded to 22 decimal places.

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Would you like a more detailed explanation of any step or further assistance with this? Here are some related questions you might find interesting:

1. How do the boundary conditions affect the solution of the PDE?
2. What happens to the solution if the initial condition involves a different combination of sine functions?
3. What are the implications of the eigenvalue problem in physical contexts, such as heat conduction or vibration modes?
4. How can we approximate the solution for large $t$ using asymptotic methods?
5. How would you solve a similar problem with a non-homogeneous boundary condition?

Tip: The key to solving such problems is recognizing the role of separation of variables in transforming a PDE into simpler ODEs.