Problem 5 - Advanced Calculus

Question

Recall that a function \(f\) is said to be homogeneous of degree \(k\) if $$ f(x t, y t, z t)=t^{k} f(x, y, z) $$ for all \(t \geq 0\) and all \((x, y, z)\) in a sphere about the origin. Let $$ \omega=A d x+B d y+C d z $$ be an exact 1 -form whose coefficuents are all homogeneous of degree \(k, k \geq 0 .\) Show that \(\omega=d f\), where $$ f(x, y, z)=\frac{x A(x, y, z)+y B(x, y, z)+z C(x, y, z)}{k+1} $$

Step-by-Step Solution

Answer

If the computation shows that the partial derivatives of f with respect to x, y, and z are indeed equal to A, B, and C, respectively, this will confirm that the given 1-form \(\omega\) is exact and equals the differential of the function f.

Step 1: Recall the definition

First, let's recall the definition of an exact 1-form. A 1-form \(\omega=A dx + B dy + C dz\) is exact if there exists a function f such that \(\omega=df\). This is the case if and only if the following partial differential equations hold:\[\begin{align*}\frac{\partial f}{\partial x} &= A, \\frac{\partial f}{\partial y} &= B, \\frac{\partial f}{\partial z} &= C. \end{align*}\]Since this is what needs to be shown, it's important to understand how to compute partial derivatives. The function f has been given, so the derivatives can be directly calculated.

Step 2: Compute the partial derivatives

Now, calculate the partial derivatives of f. Firstly, calculate the x-derivative of f:\[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x A(x, y, z)+y B(x, y, z)+z C(x, y, z)}{k+1}\right).\]By applying the product rule and the chain rule, if necessary, compute this derivative.Then, calculate the y-derivative of f:\[\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\left(\frac{x A(x, y, z)+y B(x, y, z)+z C(x, y, z)}{k+1}\right).\]Use the same rules as before.Lastly, compute the z-derivative:\[\frac{\partial f}{\partial z} = \frac{\partial}{\partial z}\left(\frac{x A(x, y, z)+y B(x, y, z)+z C(x, y, z)}{k+1}\right).\]Again, apply the product and chain rules, if necessary.

Step 3: Apply the homogeneity conditions

The exercise stated that the coefficients A, B, C of the 1-form are homogeneous of degree k. Therefore, these coefficients satisfy the equation\[f(x t, y t, z t)=t^{k} f(x, y, z)\]To prove the derivation, use this equation in the calculations. Also, remember that homogeneous functions have a property that relates their derivative to the original function: \(\frac{\partial f}{\partial x} = k f(x, y, z)\). Use this property in the calculations.

Step 4: Comparing results

Now, compare the results of the computations (from Step 2 and Step 3) with the coefficients of \(\omega\). This will now tell whether the calculated partial derivatives of f are indeed equal to the coefficients of \(\omega\). By `equal', it is meant that these functions are the same as functions of \(x, y, z\).

Step 5: Summarise the results

Summarise the findings. If the results from Step 4 match as expected, it can be concluded that \(\omega\) is exact and that it is the differential of the given function f.

Key Concepts

Exact Differential Forms

Understanding exact differential forms begins with the concept of a differential form. A differential form is essentially an expression like \(\omega = A dx + B dy + C dz\), where \(A, B,\) and \(C\) are functions of variables like \(x, y, \) and \(z\). It is called exact if there exists a differentiable function \(f\) such that \(df = \omega\).
Exact differential forms are crucial for converting differential expressions into integrable ones, greatly simplifying the process of solving differential equations.

• In practical terms, checking whether a differential form is exact involves the concept of potential functions \(f\).
• This means that if you can find \(f\) such that its partial derivatives match the coefficients of the differential form, your form is exact.
• Exact differential forms often appear when dealing with conservative vector fields in physics, particularly in energy-related problems.

Partial Derivatives

Partial derivatives are a fundamental concept in calculus used to understand how a function changes as one specific variable changes, assuming the others remain constant. In the context of your exercise, partial derivatives are used to demonstrate that \(\omega\) can indeed be represented as \(df\).
Each partial derivative \(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), and \(\frac{\partial f}{\partial z}\) needs to equal the coefficients \(A, B,\) and \(C\) respectively for \(\omega\) to be exact.

• To calculate a partial derivative, keep all other variables constant while differentiating with respect to your variable of interest.
• For the function \(f\), you compute each variable's partial derivative separately, resulting in a system of equations that help verify the exactness of \(\omega\).

Understanding partial derivatives simplifies complex relations between variables and is a cornerstone for multivariable calculus and most sciences.

Homogeneity Conditions

Homogeneity conditions are important when dealing with functions that scale uniformly. A function \(f\) is said to be homogeneous of degree \(k\) if scaling all inputs by a factor \(t\) results in the output being scaled by \(t^k\). Specifically, \(f(xt, yt, zt) = t^k f(x, y, z)\).
In your specific exercise, the coefficients \(A, B,\) and \(C\) are each homogeneous functions of degree \(k\). This condition plays a pivotal role in simplifying the expression of \(f\).

• Homogeneity helps in transforming a function into manageable parts by leveraging properties like Euler's theorem, which links the differentiation of homogeneous functions to multiplying its degree by the function itself.
• It allows the expression for \(f\) to be rewritten easily, ensuring that all partial derivatives align perfectly with the coefficients \(A, B,\) and \(C\).

Considering homogeneity greatly eases solving problems involving symmetry and scaling.

Differential Equations

Differential equations are equations that involve functions and their derivatives, capturing how a change in one variable affects another. In this situation, you're dealing with differential equations to solve for \(f\) given the differential form \(\omega\).
The task boils down to proving that a specific differential form can be associated with a function through its derivatives.

• Differential equations are classified based on being ordinary or partial, depending on the number of variables involved.
• Solving them often includes finding functions \(f\) that satisfy the conditions imposed by these equations.
• In more complex cases, techniques such as separation of variables or integrating factors may be used, but in your exercise, leveraging the exactness simplification is sufficient.

Differential equations are powerful tools in modeling real-world phenomena, from physics to biology.