If $u=e^x \cos y \text{ and } v=e^x \sin y$ transform the following: $w_{xx}+w_{yy}=0.$

Question

If $u=e^x \cos y \text{ and } v=e^x \sin y$ transform the following: $w_{xx}+w_{yy}=0.$

I was hoping that someone would maybe be familiar to this $w$ function that is stated, because this is the only given information in the question:$w_{xx}+w_{yy}=0.$ This is the sum of the partial derivatives twice with respect to $x$ and $y$. These types of question usually concern doing the partial derivative type of "decomposition". Is there a function $w$ in say, complex analysis? or math in general that im unaware of?

Answer 1

It looks to be of the form of a cauchy reimann equation where $w_x$ is the general first derivative of either $u$ or $v$ with respect to $x$ and with 2 x's it is differentiated twice. if the sum of these 2 $w_{xx}+w_{yy}=0$ is true for some $x$ and $y$ then it means it is continuous and harmonic and is therefore a cauchy reimann equation. This is otherwise known as laplace's equation

EDIT: If you look here under Laplace equations in 2 dimensions it gives a pretty good explanation of its link to cauchy reimann

Furthermore, in this instance

$$ u_{xx}=e^x\cos{y} $$

and

$$ u_{yy}=-e^x\cos{y} $$

Which when summated is equal to zero hence the laplace is satisfied in this instance. This also holds for v(x,y)

Answer 2

Here is an algebraic approach that doesn't rely on complex analysis. From the chain rule, $$w_x=w_uu_x+w_vv_x \\ w_y=w_uu_y+w_vv_y$$ Apply chain rule again... $$w_{xx}=w_uu_{xx}+u_x(w_{uu}u_x+w_{uv}v_x)+w_vv_{xx}+v_x(w_{vu}u_x+w_{vv}v_x) \\ w_{yy}=w_uu_{yy}+u_y(w_{uu}u_y+w_{uv}v_y)+w_vv_{yy}+v_y(w_{vu}u_y+w_{vv}v_y)$$ If we assume the mixed partials are equal this turns into $$w_{xx}=e^{2x}\cos^2(y)w_{uu}+e^{2x}\sin^2(y)w_{vv}+e^x\cos(y)w_u+e^x\sin(y)w_v+2e^{2x}\sin(y)\cos(y)w_{uv} \\ w_{yy}=e^{2x}\sin^2(y)w_{uu}+e^{2x}\cos^2(y)w_{vv}-e^x\cos(y)w_u-e^x\sin(y)w_v-2e^{2x}\sin(y)\cos(y)w_{uv}$$ Therefore $$w_{xx}+w_{yy}=0 \iff e^{2x}w_{uu}+e^{2x}w_{vv}=0 \iff w_{uu}+w_{vv}=0$$