Question

Suppose that \(w=f(x, y)\) satisfies $$\frac{\partial^{2} w}{\partial x^{2}}-\frac{\partial^{2} w}{\partial y^{2}}=1.$$ Put \(x=u+v, y=u-v,\) and show that \(w\) satisfies \(\partial^{2} w / \partial u \partial v=1 .\) Hence solve the equation.

Short answer

The solution to the partial differential equation is \( w = uv + f(u) + g(v) \).

Step-by-step solution

- Change of Variables

Start by expressing the variables in terms of the new variables: let \( x = u + v \) and \( y = u - v \).

- Compute the Partial Derivatives

Compute the partial derivatives of \( w \) with respect to \( u \) and \( v \). Use the chain rule. \[ \frac{\partial w}{\partial u} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial u} \]\[ \frac{\partial w}{\partial v} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial v} \]

- Simplify the Partial Derivatives

Apply the chain rule to obtain:\[ \frac{\partial x}{\partial u} = 1, \quad \frac{\partial x}{\partial v} = 1, \quad \frac{\partial y}{\partial u} = 1, \quad \frac{\partial y}{\partial v} = -1 \]Then it follows:\[ \frac{\partial w}{\partial u} = \frac{\partial w}{\partial x} + \frac{\partial w}{\partial y} \]\[ \frac{\partial w}{\partial v} = \frac{\partial w}{\partial x} - \frac{\partial w}{\partial y} \]

- Compute the Second Partial Derivatives

Next, compute the second partial derivatives:For \( \frac{\partial^{2}{w}}{\partial{u}\partial{v}} \):\[ \frac{\partial}{\partial{v}} \left( \frac{\partial{w}}{\partial{u}} \right) = \frac{\partial}{\partial{v}} \left( \frac{\partial{w}}{\partial{x}} + \frac{\partial{w}}{\partial{y}} \right) \]Use the chain rule again to evaluate this.

- Substitute Back Partial Derivatives

Calculate \( \frac{\partial^{2} w}{\partial u \partial v} \):\[ \frac{\partial}{\partial{v}} \left( \frac{\partial{w}}{\partial{x}} + \frac{\partial{w}}{\partial{y}} \right) = \frac{\partial}{\partial{x}} \left( \frac{\partial{w}}{\partial{u}} \right) \frac{\partial{x}}{\partial{v}} + \frac{\partial}{\partial{y}} \left( \frac{\partial{w}}{\partial{u}} \right) \frac{\partial{y}}{\partial{v}} \]Simplify this expression to reach the final result.

- Show the Simplification

Simplifying the expression once more, we get:\[ \frac{\partial^{2} w}{\partial u \partial v} = \frac{\partial^{2} w}{\partial x^{2}} - \frac{\partial^{2} w}{\partial y^{2}} = 1 \]

- Solve the PDE

Given the simplified equation \( \frac{\partial^{2} w}{\partial u \partial v} = 1 \), integrate with respect to \( u \) and then with respect to \( v \) to find the solution for \( w \).\[ w(u, v) = uv + f(u) + g(v) \]

Key concepts

partial derivatives

A partial derivative measures how a function changes as one of its input variables changes, while keeping the others constant. It's like zooming in on the behavior of a multi-variable function in the direction of one variable.
Given a function $$w=f(x, y),$$ we denote the partial derivative of $$w$$ with respect to $$x$$ as \(\frac{\partial w}{\partial x}\) and with respect to $$y$$ as \(\frac{\partial w}{\partial y}\).
This notation emphasizes that we are differentiating only in one direction while treating the other variable as constant. Understanding partial derivatives is crucial in solving partial differential equations (PDE), where we deal with functions of multiple variables. The given exercise focuses on second-order partial derivatives, requiring us to compute and interpret the second derivatives with respect to the new variables after a change of variables.

change of variables

Changing variables can simplify a problem or make patterns easier to identify. In our exercise, the new variables are $$u$$ and $$v$$ defined as:

• $$x = u + v$$
• $$y = u - v$$

By expressing the original variables in terms of new ones, we redefine the function $$w$$. This process often reveals hidden symmetries or reduces the complexity of the problem. After the change, we need to express the partial derivatives of $$w$$ with respect to $$u$$ and $$v$$ in terms of the partial derivatives with respect to $$x$$ and $$y$$. This transformation typically involves using the chain rule, which will be explained next.

chain rule

The chain rule for partial derivatives helps us differentiate a function with respect to new variables. Suppose we have $$w=f(x, y)$$ and $$x$$ and $$y$$ depend on $$u$$ and $$v$$. The chain rule relates these derivatives:

• \(\frac{\partial w}{\partial u} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial u}\)
• \(\frac{\partial w}{\partial v} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial v}\)

The given problem's partial derivatives transform as:

• $$\frac{\partial x}{\partial u} = 1,$$ $$\frac{\partial x}{\partial v} = 1,$$ $$\frac{\partial y}{\partial u} = 1,$$ $$\frac{\partial y}{\partial v} = -1$$

Applying these, we get:
\(\frac{\partial w}{\partial u} = \frac{\partial w}{\partial x} + \frac{\partial w}{\partial y}\)
\(\frac{\partial w}{\partial v} = \frac{\partial w}{\partial x} - \frac{\partial w}{\partial y}\)
Understanding the chain rule is essential for transitioning the problem into the new coordinate system and finding the solution.

second partial derivatives

Second partial derivatives measure how the rate of change of a function (as captured by its first partial derivatives) changes as we move in different directions again. For example, if we take the second partial derivative of $$w$$ with respect to $$x$$ twice, we symbolize it as $$\frac{\partial^2 w}{\partial x^2}$$. In our transformed variables, we compute the second mixed partial derivative:
$$\frac{\partial^2 w}{\partial u \partial v}$$ which involves taking the derivative of \(\frac{\partial w}{\partial u}\) with respect to $$v$$.
This is crucial for evaluating how the function behaves in the new coordinate system. We find:
$$\frac{\partial^2 w}{\partial u \partial v} = \frac{\partial}{\partial v} (\frac{\partial w}{\partial x} + \frac{\partial w}{\partial y})\rightarrow \frac{\partial^2 w}{\partial x^2} - \frac{\partial^2 w}{\partial y^2}$$
Given the PDE $$\frac{\partial^{2} w}{\partial x^{2}} - \frac{\partial^{2} w}{\partial y^{2}} = 1$$, it follows that $$\frac{\partial^{2} w}{\partial u \partial v} = 1$$. Solving this simplified second partial derivative enables us to integrate step-by-step to find the general solution:
$$w(u, v) = uv + f(u) + g(v)$$
Thus, our understanding of second partial derivatives guides us to the function's behavior under coordinate changes.