Solving an equation with partial derivatives

Question

a) Find all the solutions $g : \mathbb R^2 → \mathbb R$ that have partial derivatives up to the second order that are continuous on $\mathbb R^2$ and such that $\frac{∂^2g}{∂u∂v} = 0$

b) Find all the solutions $f : \mathbb R^2 → \mathbb R$ that have partial derivative up to the second order that are continuous and that satisfy $\frac{∂^2f} {∂x^2} +\frac{∂^2f}{∂x∂y} − 2\frac{∂^2f}{∂y^2} = 0 $

[Hint:introduce $g(u, v) := f(x, y)$ with $(u, v) := (x + y, y − 2x)$ ]

My attempts:

a) First, I integrate $\frac{∂^2g}{∂u∂v}$ wrt $v$, I get: $\frac{∂g}{∂u} = f(u)$. Then I integrate again wrt $u$, and I get $g(u, v) = h(u) + s(v) $. Where $h(u)$ is a function of $u$ and $s(v)$ is a function of $v$.

b) $\frac{∂f(x,y)}{∂x}= \frac{∂g(u,v)}{∂x} = \frac{∂g(u,v)}{∂u}.\frac{∂u}{∂x} + \frac{∂g(u,v)}{∂v}.\frac{∂v}{∂x} = \frac{∂g(u,v)}{∂u}.1 + \frac{∂g(u,v)}{∂v}.(-2)$

And $\frac{∂f(x,y)}{∂y} = \frac{∂g(u,v)}{∂y} = \frac{∂g(u,v)}{∂u}.\frac{∂u}{∂y} + \frac{∂g(u,v)}{∂v}.\frac{∂v}{∂y} = \frac{∂g(u,v)}{∂u}.1 + \frac{∂g(u,v)}{∂v}.1$

How to proceed from here?

Answer 1

• a) First, integrate $\frac{\partial^2 g}{\partial u\partial v }(u,v)$ $ w.r.t.$ $v$, so we get $\frac{\partial g}{\partial u}(u,v) = h(u)$. Then let's integrating in $u$ regarding $v$ is fixed. We get the solution $$g(u,v) = H(u)+G(v), $$where $H' = h$.
• b)Let $$\begin{bmatrix}u\\v\end{bmatrix} = \begin{bmatrix} 1&1\\-2&1\\\end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}$$ By Chain rule $$\begin{split} \frac{\partial f}{\partial x} &=\frac{\partial g}{\partial u}\frac{\partial u}{\partial x}+ \frac{\partial g}{\partial v}\frac{\partial v}{\partial x}\\ &= \frac{\partial g}{\partial u} - 2 \frac{\partial g}{\partial v} \end{split} $$and $$\begin{split} \frac{\partial^2 f}{\partial x^2} &=1\cdot(\frac{\partial^2 g}{\partial u^2} \frac{\partial u}{\partial x}+\frac{\partial^2 g}{\partial u \partial v} \frac{\partial v}{\partial x})-2\cdot(\frac{\partial^2 g}{\partial v^2}\frac{\partial v}{\partial x}+\frac{\partial^2 g}{\partial v \partial u}\frac{\partial u}{\partial x})\\ &=\frac{\partial^2 g}{\partial u^2} -2\frac{\partial^2 g}{\partial u \partial v} -2\cdot(-2 \frac{\partial^2 g}{\partial v^2} +\frac{\partial^2 g}{\partial u \partial v})\\ &=\frac{\partial^2 g}{\partial u^2}-4\frac{\partial^2 g}{\partial u\partial v} +\frac{\partial^2 g}{\partial v^2}.\\ \end{split} $$Similarly,$$\frac{\partial f}{y} = \frac{\partial g}{\partial u}+\frac{\partial g}{\partial v},$$ $$\frac{\partial^2f}{\partial x\partial y} = \frac{\partial^2 g}{\partial u^2}-2\frac{\partial^2g}{\partial v^2}-\frac{\partial ^2g}{\partial u \partial v}, $$ $$\frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 g}{\partial u^2}+2\frac{\partial^2 g}{\partial u \partial v}+\frac{\partial^2 g}{\partial v^2}.$$ Then $$\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2f}{\partial x \partial y}-2\frac{\partial^2 f}{\partial y^2} = -9\frac{\partial^2 g}{\partial u \partial v} = 0.$$ Hence we get the same solution$$g(u,v) = H(u)+G(v), $$where $H' = h$.