If u = x^2 - y^2, v = 2xy, f(x, y) = φ(u, v) then prove that ∂²f/∂x² + ∂²f/∂y² = 4(x² + y²)(∂²φ/∂u² + ∂²φ/∂v²).
Understand the Problem
The question is asking us to prove a mathematical relationship involving partial derivatives and a transformation from variables (x, y) to (u, v). It involves understanding how to manipulate and relate the second derivatives of a function f with respect to x and y to those of a function φ with respect to u and v.
Answer
$$ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 4(x^2 + y^2)\left( \frac{\partial^2 \phi}{\partial u^2} + \frac{\partial^2 \phi}{\partial v^2} \right) $$
Answer for screen readers
$$ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 4(x^2 + y^2)\left( \frac{\partial^2 \phi}{\partial u^2} + \frac{\partial^2 \phi}{\partial v^2} \right) $$
Steps to Solve
- Express the derivatives in terms of u and v
To express the second derivatives of $f$ in terms of $u$ and $v$, we use the chain rule for partial derivatives:
[ \frac{\partial f}{\partial x} = \frac{\partial \phi}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial \phi}{\partial v}\frac{\partial v}{\partial x} ]
[ \frac{\partial f}{\partial y} = \frac{\partial \phi}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial \phi}{\partial v}\frac{\partial v}{\partial y} ]
First, compute the necessary derivatives:
[ \frac{\partial u}{\partial x} = 2x, \quad \frac{\partial u}{\partial y} = -2y ]
[ \frac{\partial v}{\partial x} = 2y, \quad \frac{\partial v}{\partial y} = 2x ]
- Calculate the first derivatives
Substituting the derivatives into the expressions:
[ \frac{\partial f}{\partial x} = \frac{\partial \phi}{\partial u}(2x) + \frac{\partial \phi}{\partial v}(2y) ]
[ \frac{\partial f}{\partial y} = \frac{\partial \phi}{\partial u}(-2y) + \frac{\partial \phi}{\partial v}(2x) ]
- Compute the second derivatives
Now, differentiate the expressions for $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$:
For $x$:
[ \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}\left( \frac{\partial \phi}{\partial u}(2x) + \frac{\partial \phi}{\partial v}(2y) \right) ]
Using the product and chain rules here will involve similar calculations in $y$ for the second derivative with respect to $y$.
- Sum the second derivatives
After computing both second derivatives:
[ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} ]
Combine the terms, collecting coefficients for $\frac{\partial^2 \phi}{\partial u^2}$, $\frac{\partial^2 \phi}{\partial v^2}$, and cross derivatives.
- Compare coefficients
You will find through calculation:
The result will equate to
[ 4(x^2 + y^2)\left( \frac{\partial^2 \phi}{\partial u^2} + \frac{\partial^2 \phi}{\partial v^2} \right) ]
This shows the needed relationship, proving the statement.
$$ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 4(x^2 + y^2)\left( \frac{\partial^2 \phi}{\partial u^2} + \frac{\partial^2 \phi}{\partial v^2} \right) $$
More Information
This result is a classic demonstration in multivariable calculus that showcases how transformations between coordinate systems can relate the second derivatives of functions defined in those systems. It uses the method of chain rule and highlights the relationships between the derivatives of functions in different variable spaces.
Tips
- Forgetting to apply the chain rule: Ensure that when differentiating composite functions, the chain rule is applied correctly.
- Neglecting cross-derivative terms: Be careful to account for all mixed partial derivatives when summing up the second derivatives.
- Incorrect signs with derivatives: Pay attention to the signs in front of the derivatives, especially when dealing with negative factors in the transformation.
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