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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, $$\mathrm{s}=\mathrm{x}+\mathrm{y}$$

Expert verified
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.
See the step by step solution

## 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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