Show that the function satisfies the given partial differential equation.

Question

Show that the function satisfies the given partial differential equation.

The function is $z=\sqrt{x^2+y^2}$ and they want me to show $x \cdot \displaystyle\frac{\partial z}{\partial x}+y \cdot \displaystyle\frac{\partial z}{\partial y}=z$

I have that $\displaystyle\frac{\partial z}{\partial x}=\frac{1}{2\sqrt{x^2+y^2}} \cdot 2x = \frac{2x}{2\sqrt{x^2+y^2}}=\frac{x}{\sqrt{x^2+y^2}}$

And I have $\displaystyle\frac{\partial z}{\partial y}=\frac{1}{2\sqrt{x^2+y^2}} \cdot 2y = \frac{2y}{2\sqrt{x^2+y^2}}=\frac{y}{\sqrt{x^2+y^2}}$

Then $x \cdot \displaystyle\frac{x}{\sqrt{x^2+y^2}}+ y \cdot \displaystyle\frac{y}{\sqrt{x^2+y^2}}=z$

$$\displaystyle\frac{x^2}{\sqrt{x^2+y^2}}+\displaystyle\frac{y^2}{\sqrt{x^2+y^2}}=z$$ $$\displaystyle\frac{x^2+y^2}{\sqrt{x^2+y^2}}=z$$

But how can I show that $\displaystyle\frac{x^2+y^2}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}$?

Answer 1

You should have taken a look at Euler's Homogeneous Function Theorem

But you can do this $$\frac{x^2+y^2}{\sqrt{x^2+y^2}}=\frac{\sqrt{x^2+y^2}\cdot\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}$$ $$\text{or}$$

$$\frac{x^2+y^2}{\sqrt{x^2+y^2}}=\frac{x^2+y^2}{\sqrt{x^2+y^2}}\cdot\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}= \frac{x^2+y^2}{x^2+y^2}(\sqrt{x^2+y^2}) =\sqrt{x^2+y^2}$$

Note: I'm assuming that $x\neq0 \text{ and } y\neq0$

Answer 2

This a far longer answer than @Digital Brain's one but I hope it will add some interest on your question. Perform the usual change of variable to planar polar coordinates:

$$ x = r \cos{t}, \quad y = r \sin{t}, \quad (r,t) \in [0,\infty) \times [0,2\pi). $$

We then are given that:

$$ \left( \begin{array}{c} \partial_r\\ \partial_t \end{array} \right) = \left( \begin{array}{cc} \cos{t} & \sin{t} \\ -r \sin{t} & r \cos{t} \end{array} \right) \left( \begin{array}{c} \partial_x\\ \partial_y \end{array} \right),$$ where the matrix stands for the jacobian, $\mathsf{J}$, of the transformation $(x,y) \to (r,t)$, with $\det{(\mathsf{J})}=r$. If we consider the inverse of the transformation, i.e. from $(r,t) \to (x,y)$, we would then have:

$$ \left( \begin{array}{c} \partial_x\\ \partial_y \end{array} \right) = \left( \begin{array}{cc} \cos{t} & \sin{t} \\ -r \sin{t} & r \cos{t} \end{array} \right)^{-1} \left( \begin{array}{c} \partial_r\\ \partial_t \end{array} \right),$$ solving for the inverse, if possible, we obtain:

\begin{align} \partial_x & = \cos{t} \, \partial_r - \frac{\sin{t}}{r} \, \partial_t \\ \partial_y & = \sin{t} \, \partial_r + \frac{\cos{t}}{r} \, \partial_t \end{align} and hence your PDE can be written as:

$$ r \cos{t} \left( \cos{t} \, z_r \color{red}{- \frac{\sin{t}}{r} z_t} \right) + r \sin{t} \left(\sin{t} \, z_r \color{red}{+ \frac{\cos{t}}{r} \, z_t }\right) = z,$$

but remember that $z = \sqrt{x^2+y^2} = r$, so $z_r = 1$. Also, recall that $\sin^2 t + \cos^2 t= 1$ and you are done!

Hope this helps.

Cheers!

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Question

What happens if $x = y = 0$ and hence $z = r = 0$? Do the results above hold? Is the transformation $(x,y) \to (r,t)$ through the jacobian valid? Does it matter?