1

If $u= \arcsin\left(\dfrac{x+y}{\sqrt x+\sqrt y}\right)$ then show that :

$$x^2\frac{\partial^2 u}{\partial x^2}+ y^2\frac{\partial^2 u}{\partial y^2}+ 2xy\frac{\partial^2 u}{\partial x \partial y}=-\dfrac{\sin u\cos 2u}{4\cos^3 u}$$

I tried using euler's formula : Where f=sinu $$x^2\frac{\partial^2 f}{\partial x^2}+y^2\frac{\partial^2 f}{\partial x^2}+2xy\frac{\partial^2 f}{\partial x \partial y}=n(n-1) f$$ where n is the degree of homogenous function f which comes out to be 1/2.

Now I get: $\dfrac{\partial^2 f}{\partial x^2}=-\sin u\dfrac{\partial u}{\partial x}+\cos u\dfrac{\partial^2 u}{\partial x^2}$

$\dfrac{\partial^2 f}{\partial y^2}=-\sin u\dfrac{\partial u}{\partial y}+\cos u\dfrac{\partial^2 u}{\partial y^2}$

$\dfrac{\partial^2 f}{\partial x \partial y}=-\sin u\cos u\dfrac{\partial u}{\partial x}$$\dfrac {\partial u}{\partial y}+\cos u\dfrac{\partial u}{\partial x \partial y}$

Substituting this in my euler formula is not giving me answer..... Any help please.

Max Wong
  • 541
Aladdin
  • 282

1 Answers1

2

Since $$ \frac{{\partial f}} {{\partial x}} = \left[ {\cos u} \right]\frac{{\partial u}} {{\partial x}} $$ you have that $$ \frac{{\partial ^2 f}} {{\partial x^2 }} = \left[ { - \sin \left( u \right)\frac{{\partial u}} {{\partial x}}} \right]\frac{{\partial u}} {{\partial x}} + \cos \left( u \right)\frac{{\partial ^2 u}} {{\partial x^2 }} $$ so that $$ \frac{{\partial ^2 f}} {{\partial x^2 }} = - \sin \left( u \right)\left( {\frac{{\partial u}} {{\partial x}}} \right)^2 + \cos \left( u \right)\frac{{\partial ^2 u}} {{\partial x^2 }} $$ and so on...

Luca Goldoni Ph.D.
  • 2,969
  • Ah i am getting solution using your way. But can u tell what's wrong with my solution – Aladdin Sep 18 '19 at 14:12
  • Well, when yo calculate the second derivative you have to calculate the partial derivative of cos(u)= cos(u(x,y). Therefore, according whit the chain rule you have to write not only -sin(u) but you have consider also the partial derivative of u with respect to x. This is what is missing in your calculation. I hope to be clear. – Luca Goldoni Ph.D. Sep 18 '19 at 17:00