I am trying to show that $$u=f(2x+y^2)+g(2x-y^2)$$ satisfies the PDE, $$y^2u_{xx}+\frac{1}{y}u_y-u_{yy}=0,$$ for arbitrary functions $f$ and $g$.

I first start by finding $u_x$. I construct a diagram as follows: enter image description here

From this, I see that $$u_x=\frac{\partial u}{\partial f}\frac{\partial f}{\partial x}+\frac{\partial u}{\partial g}\frac{\partial g}{\partial x}.$$

Is the logic in this first step correct? I am having trouble conceptualising the chain rule in this question.

  • 1
    $\begingroup$ You need to be a little bit careful in computing the partial derivatives. Try setting $s = 2x + y^2$ and $t = 2x - y^2$ (so that $u(x,y) = f(s) + g(t)$). Then $u_x$ can be computed as $f_s s_x + g_tt_x$. $\endgroup$
    – ayberk
    Feb 19, 2019 at 7:16

2 Answers 2


Ultimately, $u$ is just a function of $x$ and $y$ since $f$ and $g$ are both functions of $x$ and $y$ as well. We could introduce new variables $\xi = 2x + y^2$ and $\eta = 2x - y^2$ to make things a little neater so $u(x,y) = f(\xi) + g(\eta)$ such that \begin{align} \frac{\partial u}{\partial x} &= \frac{\partial}{\partial x}\Big(f(\xi) + g(\eta) \Big) \\ &= \frac{\partial f}{\partial x} + \frac{\partial g}{\partial x}\\ &= \frac{\mathrm{d}f}{\mathrm{d}\xi} \frac{\partial \xi}{\partial x} + \frac{\mathrm{d}g}{\mathrm{d}\eta} \frac{\partial \eta}{\partial x}. \end{align} Really all we're doing when using the chain rule is applying partial derivatives, in this case $x$, to each side and accounting for all paths where $x$ is accessible. For instance, if instead we had for appropriate $f$ and $g$, \begin{align} w(x,y) = \sqrt{f(x,y)} + \cos g(\sin y), \end{align} and we wanted the $x$ partial, then we would have \begin{align} w_x = \frac{\partial}{\partial x}\sqrt{f(x,y)} + \frac{\partial}{\partial x}\cos g(\sin y). \end{align} We can immediately take care of $\big(\!\cos g(\sin y)\big)_x$ because $g(\sin y)$ has nothing to do with $x$, so its partial with respect to $x$ is nothing. On the other hand, we already know how to treat $\big(\sqrt{f(x,y)}\big)_x$ from single variable calculus because $y$ is held constant - it's exactly the same procedure. \begin{align} w_x &= \frac{\partial}{\partial x}\sqrt{f(x,y)} \\ &= \frac{\mathrm{d}}{\mathrm{d} \big(f(x,y)\big)}\left(\sqrt{f(x,y)}\right) \frac{\partial f(x,y)}{\partial x} \\ &= \frac{f_x(x,y)}{2\sqrt{f(x,y)}}. \end{align} Similarly, if we wanted $w_y$, we would have \begin{align} w_y &= \frac{\partial}{\partial y}\sqrt{f(x,y)} + \frac{\partial}{\partial y}\cos g(\sin y) \\ &= \frac{\mathrm{d}}{\mathrm{d} \big(f(x,y)\big)}\left(\sqrt{f(x,y)}\right) \frac{\partial f(x,y)}{\partial y} + \frac{\mathrm{d}}{\mathrm{d}\big(g(\sin y)\big)}\Big( \!\cos g(\sin y) \Big) \frac{\mathrm{d}}{\mathrm{d}\big(\!\sin y\big)}\Big(g(\sin y)\Big)\frac{\mathrm{d}\sin y}{\mathrm{d}y} \\ &= \frac{f_y(x,y)}{2\sqrt{f(x,y)}} - \sin g(\sin y) \, g'(\sin y) \, \cos y, \end{align} and so on. That's all the chain rule says.


The multivariable chain rule is not needed !

If $f$ and $g$ are twice differentiable and $u(x,y)=f(2x+y^2)+g(2x-y^2)$ then we have

$u_x(x,y)=2f'(2x+y^2)+2g'(2x-y^2)$ and $u_y(x,y)=2yf'(2x+y^2)-2yg'(2x-y^2).$

Now it is your turn to compute $u_{xx}$ and $u_{yy}$. Then it is easy to see that


is satiesfied.


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