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Q. Given that $u=\arctan\left(\frac{x^3+y^3}{x-y}\right)$, prove the following :

$$x^2\frac{\partial^2u}{\partial x^2}+2xy\frac{\partial^2u}{\partial x\partial y}+y^2\frac{\partial^2u}{\partial y^2}=(1-4\sin^2 u)\sin(2u)$$

(The relevant partial derivatives are assumed to be continuous)

Attempted incomplete solution:

$$\tan(u)=\frac{x^3+y^3}{x-y}=f~\textrm{(say)}$$

We note that $f$ is a homogeneous function in $x,y$ of degree $2$ and hence, by a general result of Euler's Theorem, we have,

$$x^2\frac{\partial^2f}{\partial x^2}+2xy\frac{\partial^2f}{\partial x\partial y}+y^2\frac{\partial^2f}{\partial y^2}=2(2-1)f=2\tan(u)$$

I'm having trouble expressing the second order partial derivatives of $f$ in terms of that of $u$ (I'm relatively new at this). Can someone help me out? I don't want the complete solution, just how to apply the chain rule to get the partial derivatives of $f$ in terms of $u$. Thanks.

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Here is an outline of one way forward.

Designate the argument of the arctangent by the new variable $t$ so that

$$t(x,y)=\frac{x^3+y^3}{x-y}$$

Therefore, we can write $u(t(x,y))=\arctan (t(x,y))$. Then, we have

$$\begin{align} \sin u&=\frac{t}{\sqrt{1+t^2}} \tag 1\\\\ \sin(2u)&=\frac{2t}{1+t^2}\tag 2\\\\ u'(t)&=\frac{1}{1+t^2}\tag 3\\\\ u''(t)&=\frac{-2t}{(1+t^2)^2}\tag 4 \end{align}$$

And from the Chain Rule, we have

$$\begin{align} \frac{\partial u}{\partial x}&=u'(t)\frac{\partial t}{\partial x}\\\\ \frac{\partial u}{\partial y}&=u'(t)\frac{\partial t}{\partial y}\\\\ \frac{\partial^2 u}{\partial x^2}&=u''(t)\left(\frac{\partial t}{\partial x}\right)^2+u'(t)\frac{\partial^2 t}{\partial x^2} \tag 5\\\\ \frac{\partial^2 u}{\partial y^2}&=u''(t)\left(\frac{\partial t}{\partial y}\right)^2+u'(t)\frac{\partial^2 t}{\partial y^2} \tag 6\\\\ \frac{\partial^2 u}{\partial x \partial y}&=u''(t)\left(\frac{\partial t}{\partial x}\frac{\partial t}{\partial y}\right)+u'(t)\frac{\partial^2 t}{\partial x \partial y} \tag 7\\\\ \end{align}$$

Using $(5)-(7)$ and the general result of Euler's Theorem, we have

$$u''(t)\left(x\frac{\partial t}{\partial x}+y\frac{\partial t}{\partial y}\right)^2+2tu'(t)=(1-4\sin^2(u))\sin(2u)$$

Now, finish by calculating the partial derivatives of $t$ with respect to $x$ and $y$ and using $(1)-(4)$.

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  • $\begingroup$ This one is "messy." Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Nov 29 '15 at 21:23
  • $\begingroup$ This solution does the work but I'd appreciate a neater way that continues from my attempted solution. Is there any such way? Thanks in advance! $\endgroup$ – learner Nov 30 '15 at 12:32
  • $\begingroup$ I used the result from your attempt to arrive at the last equation. And I have the partial derivatives of $u$ in terms of $t$, which is the same as your $f$. $\endgroup$ – Mark Viola Nov 30 '15 at 15:22
  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. - Mark $\endgroup$ – Mark Viola Dec 2 '15 at 0:13
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It's not a bad idea to keep track of your variable dependencies, especially when first learning chain rule problems. Specifically, your statement $\tan u = f$ is clearer when written $$\tan u(x,y) = f(x,y).$$ Now you can take the derivative of both sides with respect to $x$. $$\sec^2 (u(x,y)) \frac{\partial u}{\partial x}(x,y) = \frac{\partial f}{\partial x}(x,y).$$ Differentiating again gives $$2\sec^2 (u(x,y)) \tan (u(x,y)) \left(\frac{\partial u}{\partial x} (x,y) \right)^2 + \sec^2(u(x,y)) \frac{\partial^2 u}{\partial x^2}(x,y) = \frac{\partial ^2 f}{\partial x^2}(x,y).$$

It gets a little messy with all the dependencies, especially on the derivatives, so you can drop them when you feel comfortable doing so. But writing them out can help with chain rule issues.

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  • $\begingroup$ I'm not sure if that works. This is exactly what I tried later on (i.e., using chain rule in a method analogous to the one I used when working with total derivatives, but I don't see how one would get rid of the $\left(\frac{\partial u}{\partial x}\right)^2$ terms at last to get to the statement in question and conclude the proof. $\endgroup$ – learner Nov 27 '15 at 20:22
  • $\begingroup$ I should add that I don't have notational problems, I'm just not seeing how one would prove the statement in question using chain rule like this. Since I wasn't getting anywhere close, I thought that maybe I was using chain rule wrong. From your answer, that doesn't seem to be the case anymore. $\endgroup$ – learner Nov 27 '15 at 20:24

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