To form the partial differential equation (PDE) by eliminating the arbitrary function from $z = f(x^2 + y^2)$, follow these steps:

Step 1: Define a new variable

Let $u = x^2 + y^2$, so that $z = f(u)$, where $f(u)$ is an arbitrary function of $u$.

Step 2: Find the first-order partial derivatives

We need to compute the partial derivatives of $z$ with respect to $x$ and $y$.

Partial derivative of $z$ with respect to $x$:

Using the chain rule: $\frac{\partial z}{\partial x} = \frac{d f}{d u} \cdot \frac{\partial u}{\partial x}$ Since $u = x^2 + y^2$, we have: $\frac{\partial u}{\partial x} = 2x$ Therefore: $\frac{\partial z}{\partial x} = \frac{d f}{d u} \cdot 2x$

Partial derivative of $z$ with respect to $y$:

Similarly, using the chain rule: $\frac{\partial z}{\partial y} = \frac{d f}{d u} \cdot \frac{\partial u}{\partial y}$ Since $u = x^2 + y^2$, we have: $\frac{\partial u}{\partial y} = 2y$ Therefore: $\frac{\partial z}{\partial y} = \frac{d f}{d u} \cdot 2y$

Step 3: Find the second-order partial derivatives

Now, compute the second-order partial derivatives of $z$ with respect to $x$ and $y$.

Second-order partial derivative with respect to $x$:

$\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{d f}{d u} \cdot 2x \right)$ Using the product rule: $\frac{\partial^2 z}{\partial x^2} = \frac{d^2 f}{d u^2} \cdot 4x^2 + \frac{d f}{d u} \cdot 2$

Second-order partial derivative with respect to $y$:

Similarly: $\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y} \left( \frac{d f}{d u} \cdot 2y \right)$ Using the product rule: $\frac{\partial^2 z}{\partial y^2} = \frac{d^2 f}{d u^2} \cdot 4y^2 + \frac{d f}{d u} \cdot 2$

Step 4: Form the PDE

The required PDE involves the sum of the second-order partial derivatives: $\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2}$ Substitute the expressions for $\frac{\partial^2 z}{\partial x^2}$ and $\frac{\partial^2 z}{\partial y^2}$: $\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = \left( \frac{d^2 f}{d u^2} \cdot 4x^2 + \frac{d f}{d u} \cdot 2 \right) + \left( \frac{d^2 f}{d u^2} \cdot 4y^2 + \frac{d f}{d u} \cdot 2 \right)$ Simplify the expression: $\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = \frac{d^2 f}{d u^2} \cdot 4(x^2 + y^2) + 4 \frac{d f}{d u}$ Since $u = x^2 + y^2$, this becomes: $\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 4u \frac{d^2 f}{d u^2} + 4 \frac{d f}{d u}$

Thus, the PDE is: $\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 4 \left( u \frac{d^2 f}{d u^2} + \frac{d f}{d u} \right)$

This is the partial differential equation after eliminating the arbitrary function $f$.

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Do you want more details or have any questions? Here are 5 related questions for further exploration:

1. How can this PDE be solved for a specific form of the function $f(u)$?
2. What is the physical significance of $z = f(x^2 + y^2)$ in various applications like wave equations?
3. How do you derive second-order PDEs from other types of arbitrary functions?
4. Can this method be extended to functions of more variables, like $f(x^2 + y^2 + z^2)$?
5. What are the boundary conditions that could be applied to this PDE?

Tip: When dealing with arbitrary functions, substitution with new variables often simplifies differentiation steps.