Chapter 4: Chapter 4

Problem 105

Show that if a function $\mathrm{F}: \mathrm{R}^{2} \rightarrow \mathrm{R}$ satisfies $\partial \mathrm{F} / \partial \mathrm{x}=\partial \mathrm{F} / \partial \mathrm{y}$, then $F(x, y)=f(x+y)$ where $f$ is some differentiable function of one variable, $\mathrm{s}=\mathrm{x}+\mathrm{y}$

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To show that a function $ F: R^2 \rightarrow R $ satisfying $ \frac{\partial F}{\partial x} = \frac{\partial F}{\partial y} $ is of the form $ F(x, y) = f(x+y) $, we first introduce a new variable $ s = x + y $. Then, we calculate the partial derivative of $ F $ with respect to $ s $, and using the chain rule, we obtain $ \frac{\partial F}{\partial s} = \frac{\partial F}{\partial x}(1+1) = 2\frac{\partial F}{\partial x} $. Integrating the partial derivative with respect to $ s $ yields $ F(s) = \int 2\frac{\partial F}{\partial x} (dx + dy) $. By separating the integral and integrating both parts, we get $ F(s) = f(x) + g(y) $, where $ f(x) $ and $ g(y) $ are the antiderivatives of each integral. Given that $ s = x + y $, we can express $ F $ as a function of $ s $, $ F(x, y) = f(s) - g(y) $, which simplifies to $ F(x, y) = f(x+y) $. This confirms our claim.

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TABLE OF CONTENTS

Step 1: Introduce a new variable s

Let's introduce a new variable s, such that: \[s = x + y\] Our goal is to express F(x, y) as a function of s instead of two variables x and y.

Step 2: Partially differentiate F with respect to s

We start by finding the partial derivative of F with respect to s. To do this, we use the chain rule: \[\frac{\partial F}{\partial s} = \frac{\partial F}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial F}{\partial y}\frac{\partial y}{\partial s}\] Since we know that $\frac{\partial F}{\partial x} = \frac{\partial F}{\partial y}$, the above equation simplifies to: \[\frac{\partial F}{\partial s} = \frac{\partial F}{\partial x}(\frac{\partial x}{\partial s} + \frac{\partial y}{\partial s})\] But $\frac{\partial x}{\partial s} = 1$ and $\frac{\partial y}{\partial s} = 1 $, so: \[\frac{\partial F}{\partial s} = \frac{\partial F}{\partial x}(1+1) = 2\frac{\partial F}{\partial x}\]

Step 3: Integrate the partial derivative with respect to s

Now we need to integrate the partial derivative of F with respect to s to find F(x, y) or rather F(s): \[F(s) = \int \frac{\partial F}{\partial s} ds = \int 2\frac{\partial F}{\partial x} ds\] Recall that $s = x + y$, thus $ds = dx + dy$. We can then rewrite the integral as: \[F(s) = \int 2\frac{\partial F}{\partial x} (dx + dy)\] Separating the integral: \[F(s) = \int 2\frac{\partial F}{\partial x} dx + \int 2\frac{\partial F}{\partial x} dy\] Since $\frac{\partial F}{\partial x} = \frac{\partial F}{\partial y}$, the second integral becomes: \[F(s)=\int 2\frac{\partial F}{\partial x} dx + \int 2\frac{\partial F}{\partial y} dy\] Now, integrating both parts, we get: \[F(s) = f(x) + g(y)\] where f(x) and g(y) are the antiderivatives of each integral.

Step 4: Express F as a function of s

Since $s = x + y$, we can rewrite F(s) as: \[F(x, y) = f(x+y) - g(y)\] which can be simplified to: \[F(x, y) = f(s) - g(y)\] Since f(s) is a function of s and g(y) is a function of y, (and not including x, as partial f(s)/partial x = 0), we can conclude that F(x, y) is indeed a function of the form f(s) for some differentiable function f. In conclusion, we showed that if F satisfies the condition $\frac{\partial F}{\partial x} = \frac{\partial F}{\partial y}$, then F(x, y) can be written as a function of s = x + y, in the form F(x, y) = f(x+y), for some differentiable function f.

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Chapter 4: Chapter 4

Show that if a function \(\mathrm{F}: \mathrm{R}^{2} \rightarrow \mathrm{R}\) satisfies $\partial \mathrm{F} / \partial \mathrm{x}=\partial \mathrm{F} / \partial \mathrm{y}$, then \(F(x, y)=f(x+y)\) where \(f\) is some differentiable function of one variable, \(\mathrm{s}=\mathrm{x}+\mathrm{y}\)

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Step 1: Introduce a new variable s

Step 2: Partially differentiate F with respect to s

Step 3: Integrate the partial derivative with respect to s

Step 4: Express F as a function of s

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