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$f(x,y)$ is a differentiable function satisfying the following properties:

  1. $f(x+t, y)= f(x,y) + ty$ and $f(x, y+t)= f(x,y) + tx$, $\forall x, y, t \in\mathbb{R}$ and

  2. $f(z, 0) = k$ for any $z\in\mathbb{R}$ and $k$ is an arbitrary constant. Find $f(x,y).$

Ellya
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rajib
  • 81

2 Answers2

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Use first principles:

$\frac{\partial f}{\partial x}=\lim_{t\to 0}\frac{f(x+t,y)-f(x,y)}{t}=\lim_{t\to 0}\frac{ty}{t}=y$

$\frac{\partial f}{\partial y}=\lim_{t\to 0}\frac{f(x,y+t)-f(x,y)}{t}=\lim_{t\to 0}\frac{tx}{t}=x$

Thus $f(x,y)=\int x dy=xy+g(x)$

and $f(x,y)=\int y dx=xy+h(y)$

$f(z,0)=g(z)=h(0)=k$

So $g(x)=k, h(0)=k$, but also we need $h(y+t)=h(y)$, $\forall y,t\in\mathbb{R}\Rightarrow h(y)=k$

Thus $f(x,y)=xy+k$

Ellya
  • 11,783
1

Differentiate the two given equations with respect to $t$ to get \begin{align} f_1(x + t, y) & = y \\ f_2(x, y + t) & = x \end{align} where $f_i$ is the partial derivative with respect to the $i$-th parameter. Plug in $t = 0$ to get \begin{align} f_1(x, y) & = y \tag{1}\\ f_2(x, y) & = x. \tag{2} \end{align} Integrate (1) with respect to the first parameter to get $$ f(x, y) = xy + c(y) \tag{3} $$ where $c$ is some function of one variable.

Differentiate (3) with respect to $y$ to get $f_2(x, y) = x + c'(y)$. Equate with (2) to get $c'(y) = 0$. Therefore, $c(y)$ is a constant. Let's call it $c$. Now we get $f(x, y) = xy + c$. The condition $f(x, 0) = k$ makes $c = k$. Therefore, $f(x, y) = xy + k$ is a potential solution. Go back and check if this satisfies the two equations we started with.

Tunococ
  • 10,303