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### Problem 7
**Cauchy Problem Solution Evaluation**

Given the partial differential equation (PDE):
\[ u_x + u_y = y \]
with the initial condition:
\[ u(x, 2 - x) = \frac{x + 3}{2} \]

We need to find the value of \( u(2, 2) \).

**Solution:**

1. **Transform Variables:**
   Let \( u(x, y) = f(x, y) \). Then the PDE becomes:
   \[ f_x + f_y = y \]

2. **Method of Characteristics:**
   - The characteristic equations are:
     \[ \frac{dx}{1} = \frac{dy}{1} = \frac{du}{y} \]

   - Solve the first pair:
     \[ \frac{dx}{1} = \frac{dy}{1} \implies x - y = C_1 \]

   - Solve the second pair:
     \[ \frac{dx}{1} = \frac{du}{y} \implies du = y \, dx \]

   Since \( x - y = C_1 \), let \( x - y = C \). So:
   \[ dy = du \implies u = \frac{y^2}{2} + h(x - y) \]

3. **Initial Condition:**
   \[ u(x, 2 - x) = \frac{x + 3}{2} \]

   Substituting \( y = 2 - x \) into the general solution:
   \[ u(x, 2 - x) = \frac{(2 - x)^2}{2} + h(x - (2 - x)) \]
   \[ \frac{x + 3}{2} = \frac{(2 - x)^2}{2} + h(2x - 2) \]

4. **Simplify and Solve for \( h \):**
   \[ \frac{x + 3}{2} = \frac{4 - 4x + x^2}{2} + h(2x - 2) \]
   \[ \frac{x + 3}{2} = 2 - 2x + \frac{x^2}{2} + h(2x - 2) \]

   Assume \( h(z) = k \). Thus, find the constants and check consistency.

5. **Substitute \( x = 2 \):**
   \[ u(2, 2) = \frac{(2)^2}{2} + h(2 - 2) \]
   \[ u(2, 2) = \frac{4}{2} = 2 \]

Thus, the value of \( u(2, 2) \) is \( \boxed{2} \).

### Problem 8
**Cauchy Problem Solution Count**

Given the PDE:
\[ (x + 1)u_x + y u_y = u \]
with the initial condition:
\[ u(x, 1) = x + e^x \]

We need to determine how many solutions exist in a neighborhood of the point \( (2, 1) \).

**Solution:**

1. **Check Characteristics:**
   The PDE can be written as:
   \[ (x + 1)\frac{du}{dx} + y \frac{du}{dy} = u \]

   The characteristic equations are:
   \[ \frac{dx}{x+1} = \frac{dy}{y} = \frac{du}{u} \]

2. **Solving Characteristics:**
   \[ \frac{dx}{x+1} = \frac{dy}{y} \]
   Solving these:
   \[ \ln|y| = \ln|x+1| + C_1 \implies y = C_1(x+1) \]

   \[ \frac{dx}{x+1} = \frac{du}{u} \]
   Solving these:
   \[ \ln|u| = \ln|x+1| + C_2 \implies u = C_2(x+1) \]

3. **Initial Condition:**
   \[ u(x, 1) = x + e^x \]

   Substituting \( y = 1 \):
   \[ u = C_2(x+1) \]

4. **Verification:**
   Check if this fits the initial condition:
   \[ x + e^x = C_2(x + 1) \]
   Find appropriate \( C_2 \) for \( x \).

Therefore, the PDE and initial conditions provide a unique solution:
   \[ \boxed{(B)} \]

Please let me know if you need more details or have any questions.

Here are 8 related questions for further practice:

1. How would you find the solution of a Cauchy problem if the initial condition is a function of two variables?
2. What methods can be used to solve first-order PDEs other than the method of characteristics?
3. How does the presence of non-linear terms affect the solutions of PDEs?
4. What are the implications of having multiple solutions to a PDE in a given region?
5. Can you describe the physical interpretation of the given PDEs in real-world scenarios?
6. How can numerical methods be applied to solve the PDEs given in the problems?
7. What is the general form of the Cauchy problem for second-order PDEs?
8. How does the initial condition affect the nature of solutions in hyperbolic PDEs?

**Tip:** When solving PDEs, always verify the consistency of initial and boundary conditions with the general solution to ensure correctness.

Understand how to evaluate solutions for Cauchy problems involving partial differential equations (PDEs) with specific initial conditions. Learn the method of characteristics and apply it to find unique solutions. Includes step-by-step examples and practice questions.

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