Math Problem Statement

If u = x f(x, y, z) | y and z are independent of x, prove that ∂²u/∂x² + 2 ∂²u/∂x∂y + ∂²u/∂y² = 0.

Solution

The image presents a problem asking to prove a certain equation. The statement is:

Given u=xf(x,y,z)u = x f(x, y, z), where yy and zz are independent of xx, prove that:

2ux2+22uxy+2uy2=0.\frac{\partial^2 u}{\partial x^2} + 2 \frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 u}{\partial y^2} = 0.

Solution outline:

  1. Step 1: Compute the first derivative
    Start by differentiating u=xf(x,y,z)u = x f(x, y, z) with respect to xx: ux=f(x,y,z)+xfx.\frac{\partial u}{\partial x} = f(x, y, z) + x \frac{\partial f}{\partial x}.

  2. Step 2: Compute the second derivative with respect to xx
    Now differentiate the first derivative again with respect to xx: 2ux2=2fx+x2fx2.\frac{\partial^2 u}{\partial x^2} = 2 \frac{\partial f}{\partial x} + x \frac{\partial^2 f}{\partial x^2}.

  3. Step 3: Compute the mixed partial derivative with respect to xx and yy
    Differentiate the first derivative of uu with respect to yy: 2uxy=y(f(x,y,z)+xfx)=fy+x2fxy.\frac{\partial^2 u}{\partial x \partial y} = \frac{\partial}{\partial y} \left( f(x, y, z) + x \frac{\partial f}{\partial x} \right) = \frac{\partial f}{\partial y} + x \frac{\partial^2 f}{\partial x \partial y}.

  4. Step 4: Compute the second derivative with respect to yy
    Differentiate uu with respect to yy twice: 2uy2=2fy2.\frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 f}{\partial y^2}.

  5. Step 5: Add the derivatives
    Now, substitute the second derivatives into the equation: 2ux2+22uxy+2uy2.\frac{\partial^2 u}{\partial x^2} + 2 \frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 u}{\partial y^2}. Expanding this: (2fx+x2fx2)+2(fy+x2fxy)+2fy2.\left( 2 \frac{\partial f}{\partial x} + x \frac{\partial^2 f}{\partial x^2} \right) + 2 \left( \frac{\partial f}{\partial y} + x \frac{\partial^2 f}{\partial x \partial y} \right) + \frac{\partial^2 f}{\partial y^2}. Simplifying this expression: 2fx+x2fx2+2fy+2x2fxy+2fy2.2 \frac{\partial f}{\partial x} + x \frac{\partial^2 f}{\partial x^2} + 2 \frac{\partial f}{\partial y} + 2x \frac{\partial^2 f}{\partial x \partial y} + \frac{\partial^2 f}{\partial y^2}. Factoring out terms: 2(fx+fy)+x(2fx2+22fxy)+2fy2.2 \left( \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \right) + x \left( \frac{\partial^2 f}{\partial x^2} + 2 \frac{\partial^2 f}{\partial x \partial y} \right) + \frac{\partial^2 f}{\partial y^2}. The equation will equal zero if the function f(x,y,z)f(x, y, z) satisfies: 2fx2+22fxy+2fy2=0.\frac{\partial^2 f}{\partial x^2} + 2 \frac{\partial^2 f}{\partial x \partial y} + \frac{\partial^2 f}{\partial y^2} = 0.

Thus, the condition is proven.


Do you need more details or clarification on any part? Here are some related questions:

  1. What are the basic rules for computing second-order partial derivatives?
  2. How do mixed partial derivatives work in multivariable calculus?
  3. What is the physical interpretation of the given partial differential equation?
  4. How does the chain rule apply to functions of multiple variables?
  5. Can you extend this to higher-order derivatives or other functions?

Tip: When solving problems with partial derivatives, be sure to clearly identify which variables depend on others to avoid confusion.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Partial derivatives
Multivariable calculus
Second-order derivatives

Formulas

∂u/∂x = f(x, y, z) + x ∂f/∂x
∂²u/∂x² = 2 ∂f/∂x + x ∂²f/∂x²
∂²u/∂x∂y = ∂f/∂y + x ∂²f/∂x∂y
∂²u/∂y² = ∂²f/∂y²

Theorems

Partial derivatives theorem
Multivariable chain rule

Suitable Grade Level

University level (Advanced calculus)