First transcribe the question then Provide the complete solution in structural manner from start to end, i need complete detailed solution in the format of - 1st - complete solution with all the calcu

Question

First transcribe the question then Provide the complete solution in structural manner from start to end, i need complete detailed solution in the format of - 1st - complete solution with all the calculation, 2nd - explanation of each and every parts, 3rd final conclusion statement of all part at the end. ( Note :- don't use pronoun in the solution)Exercise 4. Lt 1 > 0 bo given aD (0.1) Supens tht 0 x Rom sbven
Bu —dilru+ du) = fx)e™, ED, t>0,
Dott) =3,
LL) +d) = 2.
th tal dot 5,0) = (), wher ae are oth sh oct.
Comptes (fwd)
b) Asst that wale) = 0. Compute ful, (de for all > 0.

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**Transcription of the Question:**

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**Exercise 4.** Let $ L > 0 $ be given and $ D = (0, L) $. Suppose that $ u : D \times \mathbb{R}_{\geq 0} $ solves
$$
\begin{cases}
\partial_t u - \partial_x (x u + \partial_x u) = f(x) e^{-t}, & x \in D, \ t > 0, \
\partial_x u(0, t) = 3, \
L u(L, t) + \partial_x u(L, t) = 2,
\end{cases}
$$
with initial data $ u(x, 0) = u_0(x) $, where $ f $ and $ u_0 $ are both smooth functions.

a) Compute $ \partial_t \left( \int_0^L u(x, t) f(x) dx \right) $.

b) Assume that $ \int_0^L u_0(x) f(x) dx = 0 $. Compute $ \int_0^L u(x, t) f(x) dx $ for all $ t \geq 0 $.

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## 1. Complete Solution with All Calculations

### (a) Compute $ \partial_t \left( \int_0^L u(x, t) f(x) dx \right) $.

Let $ I(t) = \int_0^L u(x, t) f(x) dx $.

Differentiate under the integral:
$$
\partial_t I(t) = \int_0^L \partial_t u(x, t) f(x) dx
$$

From the PDE:
$$
\partial_t u = \partial_x (x u + \partial_x u) + f(x) e^{-t}
$$

Substitute into the integral:
$$
\partial_t I(t) = \int_0^L \left[ \partial_x (x u + \partial_x u) + f(x) e^{-t} \right] f(x) dx
$$
$$
= \int_0^L \partial_x (x u + \partial_x u) f(x) dx + \int_0^L f(x)^2 e^{-t} dx
$$

For the first term, use integration by parts:
$$
\int_0^L \partial_x (x u + \partial_x u) f(x) dx = \left[ (x u + \partial_x u) f(x) \right]_0^L - \int_0^L (x u + \partial_x u) f'(x) dx
$$

Now, compute boundary terms:

- At $ x = 0 $:
  - $ x = 0 $, so $ x u(0, t) = 0 $.
  - $ \partial_x u(0, t) = 3 $.
  - $ f(0) $ remains.

Therefore, boundary at $ x=0 $: $ (0 + 3) f(0) = 3 f(0) $.

- At $ x = L $:
  - $ x u(L, t) = L u(L, t) $.
  - $ \partial_x u(L, t) $.
  - $ f(L) $ remains.

Therefore, boundary at $ x=L $: $ (L u(L, t) + \partial_x u(L, t)) f(L) $.
  - From the boundary condition: $ L u(L, t) + \partial_x u(L, t) = 2 $.
  - So boundary at $ x=L $: $ 2 f(L) $.

Combine the boundary terms:
$$
\left[ (x u + \partial_x u) f(x) \right]_0^L = 2 f(L) - 3 f(0)
$$

Therefore,
$$
\partial_t I(t) = 2 f(L) - 3 f(0) - \int_0^L (x u + \partial_x u) f'(x) dx + \int_0^L f(x)^2 e^{-t} dx
$$

---

### (b) Assume $ \int_0^L u_0(x) f(x) dx = 0 $. Compute $ \int_0^L u(x, t) f(x) dx $ for all $ t \geq 0 $.

From part (a):
$$
\partial_t I(t) = 2 f(L) - 3 f(0) - \int_0^L (x u + \partial_x u) f'(x) dx + \int_0^L f(x)^2 e^{-t} dx
$$

But this is not closed in terms of $ I(t) $. However, if $ f'(x) = 0 $ (that is, if $ f $ is constant), then:

- $ \int_0^L (x u + \partial_x u) f'(x) dx = 0 $
- $ \int_0^L f(x)^2 e^{-t} dx = f^2 L e^{-t} $, if $ f(x) = f $ constant.

So, for general $ f $, the equation is not closed unless $ f $ is a constant. But, assuming $ f $ is constant, the ODE is:
$$
\partial_t I(t) = 2 f - 3 f + f^2 L e^{-t} = -f + f^2 L e^{-t}
$$

This ODE can be solved with integrating factor or direct integration.

Integrate both sides:
$$
\frac{dI}{dt} = -f + f^2 L e^{-t}
$$
$$
I(t) = C - f t - f^2 L e^{-t}
$$

But this is incorrect; the integration should be:

$$
\frac{dI}{dt} = -f + f^2 L e^{-t}
$$
Integrate from $ 0 $ to $ t $:

$$
I(t) - I(0) = \int_0^t \left(-f + f^2 L e^{-s} \right) ds
$$
$$
= -f t + f^2 L \int_0^t e^{-s} ds
$$
$$
= -f t + f^2 L [ -e^{-s} ]_0^t
$$
$$
= -f t + f^2 L (1 - e^{-t})
$$

Given $ I(0) = 0 $ (by assumption):

$$
I(t) = -f t + f^2 L (1 - e^{-t})
$$

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## 2. Explanation of Each and Every Part

### (a) Explanation

- The derivative $ \partial_t \left( \int_0^L u(x, t) f(x) dx \right) $ is computed by differentiating under the integral sign and substituting the given PDE for $ \partial_t u $.
- The PDE is manipulated to separate terms, and integration by parts is used to handle the derivative in $ \partial_x (x u + \partial_x u) $.
- Boundary conditions are applied to evaluate the boundary terms.
- The remaining integral is simplified as much as possible, but if $ f $ is not constant, the answer involves $ u(x, t) $ and its derivatives.

### (b) Explanation

- The result from (a) is used as an ODE for $ I(t) $.
- Under the assumption that $ f $ is constant, the ODE closes and can be integrated directly.
- The initial condition $ I(0) = 0 $ is applied to find the integration constant.
- The explicit formula for $ I(t) $ is obtained.

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## 3. Final Conclusion Statement

- The time derivative $ \partial_t \left( \int_0^L u(x, t) f(x) dx \right) $ is given by
  $$
  2 f(L) - 3 f(0) - \int_0^L (x u + \partial_x u) f'(x) dx + \int_0^L f(x)^2 e^{-t} dx.
  $$
- If $ f $ is constant and $ \int_0^L u_0(x) f(x) dx = 0 $, then
  $$
  \int_0^L u(x, t) f(x) dx = -f t + f^2 L (1 - e^{-t})
  $$
  for all $ t \geq 0 $.

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