Jacobian Engineering Maths, Btech first year

Jacobian

If $u$ and $v$ are the functions of two independent variables $x$ and $y$ then Jacobian of $u$ and $v$ w.r.t. $x,y$ is: $$J=J\left(\frac{u,v}{x,y}\right)=\frac{\partial (u,v)}{\partial (x,y)}$$ $$=\left|\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right|$$ Similarly, for three variables, $$J=J\left(\frac{u,v,w}{x,y,z}\right)=\frac{\partial (u,v,w)}{\partial (x,y,z)}$$ $$=\left|\begin{array}{ccc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}& \frac{\partial u}{\partial z}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}& \frac{\partial v}{\partial z}\\ \frac{\partial w}{\partial x}& \frac{\partial w}{\partial y}& \frac{\partial w}{\partial z}\end{array}\right|$$ Applications : (1) In change of variables in Multiple Intgration. Example: changing Cartesian coordinates $x,y$ to Polar coordinates $r,\theta$ in Double Integral $$x=r\cos \theta ,y=r\sin \theta$$ $$\text{In }\int xydxdy⟶dxdy=Jdrd\theta$$ (2) To find Functional Relationship between two variables

Properties :
(1) If $u$ and $v$ are functions of $x$ and $y$ then $$J=J\left(\frac{u.v}{x,y}\right)=\frac{\partial (u,v)}{\partial (x,y)}$$ $$J^{\prime }=J\left(\frac{x.y}{u,v}\right)=\frac{\partial (x,y)}{\partial (u,v)}$$ $$J\times J^{\prime }=1$$ (2) Chain Rule : If If $u$ and $v$ are functions of $x$ and $y$ and $x$ and $y$ are functions of $r$ and $s$ then $$J\left(\frac{u.v}{r,s}\right)=J\left(\frac{u.v}{x,y}\right)\times J\left(\frac{x,y}{r,s}\right)$$ $$\frac{\partial (u,v)}{\partial (r,s)}=\frac{\partial (u,v)}{\partial (x,y)}\times \frac{\partial (x,y)}{\partial (r,s)}$$

Sample Problem 1 :

If $x=r\cos \theta ,y=r\sin \theta$, then show that $J\left(\frac{x,y}{r,\theta }\right)=r$

Solution :

$$J\left(\frac{x,y}{r,\theta }\right)=r$$ $$=\left|\begin{array}{cc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }\end{array}\right|$$ Partially differentiating $x,y$ w.r.t. $r,\theta$ $$\frac{\partial x}{\partial r}=\cos \theta ,\frac{\partial y}{\partial r}=\sin \theta$$ $$\frac{\partial x}{\partial \theta }=-r\sin \theta ,\frac{\partial y}{\partial \theta }=r\cos \theta$$ Using Jacobian formula, $$\left|\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right|$$ $$=r{\cos }^2\theta +r{\sin }^2\theta$$ $$=r$$ Hence Proved

Sample Problem 2 :

If $u=\frac{yz}{x},v=\frac{zx}{y},w=\frac{xy}{z}$, then find $\frac{\partial (u,v,w)}{\partial (x,y,z)}$

Solution :

Partially differentiating equations w.r.t. $x,y,z$ $$\frac{\partial u}{\partial x}=\frac{-yz}{x^2},\frac{\partial u}{\partial y}=\frac{z}{x},\frac{\partial u}{\partial z}=\frac{y}{x}$$ $$\frac{\partial v}{\partial x}=\frac{z}{y},\frac{\partial v}{\partial y}=\frac{-zx}{y^2},\frac{\partial v}{\partial z}=\frac{x}{y}$$ $$\frac{\partial w}{\partial x}=\frac{y}{z},\frac{\partial w}{\partial y}=\frac{x}{z},\frac{\partial w}{\partial z}=\frac{-xy}{z^2}$$ $$J=\frac{\partial (u,v,w)}{\partial (x,y,z)}$$ $$=\left|\begin{array}{ccc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}& \frac{\partial u}{\partial z}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}& \frac{\partial v}{\partial z}\\ \frac{\partial w}{\partial x}& \frac{\partial w}{\partial y}& \frac{\partial w}{\partial z}\end{array}\right|$$ $$=\left|\begin{array}{ccc}\frac{-yz}{x^2}& \frac{z}{x}& \frac{y}{x}\\ \frac{z}{y}& \frac{-zx}{y^2}& \frac{x}{y}\\ \frac{y}{z}& \frac{x}{z}& \frac{-xy}{z^2}\end{array}\right|$$ $$=\frac{-yz}{x^2}[\left(\frac{-zx}{y^2}\right)\left(\frac{-xy}{z^2}\right)-\left(\frac{x}{y}\right)\left(\frac{x}{z}\right)]$$ $$-\frac{z}{x}[\left(\frac{z}{y}\right)\left(\frac{-xy}{z^2}\right)-\left(\frac{y}{z}\right)\left(\frac{x}{y}\right)]$$ $$+\frac{y}{x}[\left(\frac{z}{y}\right)\left(\frac{x}{z}\right)+\left(\frac{zx}{y^2}\right)\left(\frac{y}{z}\right)]$$ $$=\frac{-yz}{x^2}[\frac{x^2}{yz}-\frac{x^2}{yz}]-\frac{z}{x}\left[\frac{-zx}{z}\right]+\frac{y}{x}[\frac{x}{y}+\frac{x}{y}]$$ $$=4$$

Sample Problem 3 :

If $x=r\cos \theta ,y=r\sin \theta$ then prove that $\frac{\partial (x,y)}{\partial (r,\theta )}\times \frac{\partial (r,\theta )}{\partial (x,y)}=1$

Solution :

$$\frac{\partial (x,y)}{\partial (r,\theta )}=\left|\begin{array}{cc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }\end{array}\right|$$ $$=\left|\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right|$$ $$=r{\cos }^2\theta +r{\sin }^2\theta$$ $$=r$$ $$\frac{\partial (r,\theta )}{\partial (x,y)}=\left|\begin{array}{cc}\frac{\partial r}{\partial x}& \frac{\partial r}{\partial y}\\ \frac{\partial \theta }{\partial x}& \frac{\partial \theta }{\partial y}\end{array}\right|$$ $$x=r\cos \theta \Rightarrow r=x\sec \theta$$ $$y=r\sin \theta \Rightarrow r=ycosec\theta$$ $$\frac{\partial r}{\partial x}=\sec \theta ,\frac{\partial r}{\partial y}=cosec\theta$$ If you are thinking like that, then you are wrong, because $r$ and $\theta$ must be functions of same group of variables i.e. $x$ and $y$ just like $x$ and $y$ are functions of $r$ and $\theta$. That means we have to remove $\theta$ from $r$ and reduce $r$ into function of $x$ and $y$. Similarly we will remove $r$ from $\theta$ and reduce $\theta$ into function of $x$ and $y$. $$x=r\cos \theta \Rightarrow \frac{x}{r}=\cos \theta$$ $$y=r\sin \theta \Rightarrow \frac{y}{r}=\sin \theta$$ $$\text{As we know, }{\cos }^2\theta +{\sin }^2\theta =1$$ $$\frac{x^2}{r^2}+\frac{y^2}{r^2}=1$$ $$x^2+y^2=r^2$$ $$r=\sqrt{x^2+y^2}$$ $$\text{Now, }\frac{\partial r}{\partial x}=\frac{1}{2}(x^2+y^2{)}^{\frac{1}{2}}\times 2x$$ $$\Rightarrow \frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}$$ $$\text{Similarly, }\frac{\partial r}{\partial y}=\frac{y}{\sqrt{x^2+y^2}}$$ $$\text{Now for }\frac{\partial \theta }{\partial x}\text{ and }\frac{\partial \theta }{\partial y}$$ $$\frac{y}{x}=\frac{r\sin \theta }{r\cos \theta }$$ $$\Rightarrow \frac{y}{x}=\tan \theta$$ $$\Rightarrow \theta ={\tan }^{-1}\frac{y}{x}$$ $$\frac{\partial \theta }{\partial x}=\frac{1}{1+{\left(\frac{y}{x}\right)}^2}\left(\frac{-y}{x^2}\right)$$ $$\frac{\partial \theta }{\partial x}=\frac{-y}{x^2+y^2}$$ $$\frac{\partial \theta }{\partial y}=\frac{1}{1+{\left(\frac{y}{x}\right)}^2}\left(\frac{1}{x}\right)$$ $$\frac{\partial \theta }{\partial y}=\frac{x}{x^2+y^2}$$ Now putting derivatives, $$J=\left|\begin{array}{cc}\frac{x}{\sqrt{x^2+y^2}}& \frac{y}{\sqrt{x^2+y^2}}\\ \frac{-y}{x^2+y^2}& \frac{x}{x^2+y^2}\end{array}\right|$$ $$=\left(\frac{x}{\sqrt{x^2+y^2}}\right)\left(\frac{x}{x^2+y^2}\right)-\left(\frac{-y}{x^2+y^2}\right)\left(\frac{y}{\sqrt{x^2+y^2}}\right)$$ $$=\frac{1}{\sqrt{x^2+y^2}}$$ $$=\frac{1}{r}$$ $$\text{Now, }\frac{\partial (x,y)}{\partial (r,\theta )}\times \frac{\partial (r,\theta )}{\partial (x,y)}$$ $$=r\times \frac{1}{r}$$ $$=1$$ = R.H.S.

Sample Problem 4 - Composite function :

If $u=x^2-y^2,v=2xy\text{ and }x=x\cos \theta ,y=r\sin \theta$, find $J\left(\frac{u,v}{r,\theta }\right)$

Solution :

Here, as you can see, $u$ and $v$ are functions of $x$ and $y$, and $x$ and $y$ are functions of $r$ and $\theta$. So there are two methods to solve this problem:
Method 1 : Substitution
Method 2 : Use Chain Rule $$J\left(\frac{u,v}{r,\theta }\right)=J\left(\frac{u,v}{x,y}\right)\times J\left(\frac{x,y}{r,\theta }\right)$$ $$\text{Now, }J\left(\frac{u,v}{x,y}\right)=\frac{\partial (u,v)}{\partial (x,y)}$$ $$=\left|\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial u}{\partial y}\end{array}\right|$$ $$=\left|\begin{array}{cc}2x& -2y\\ 2y& 2x\end{array}\right|$$ $$=4x^2+4y^2$$ $$J\left(\frac{x,y}{r,\theta }\right)=\left|\begin{array}{cc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }\end{array}\right|$$ $$=\left|\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right|$$ $$=r({\cos }^2\theta +{\sin }^2\theta )$$ $$=r$$ Using Chain Rule $$J\left(\frac{u,v}{r,\theta }\right)=4(x^2+y^2)\times r$$ $$=4r(x^2+y^2)$$ $$=4r(r^2{\cos }^2\theta +r^2{\cos }^2\theta )$$ $$=4r^3$$

Sample Problem 5

If $x=r\sin \theta \cos \varphi ,y=r\sin \theta \sin \varphi$ and $z=r\cos \theta$, then show that $J=r^2\sin \theta$

Solution :

$$x=r\sin \theta \cos \varphi$$ $$y=r\sin \theta \sin \varphi$$ $$z=r\cos \theta$$ Now, $$J=J\left(\frac{x,y,z}{r,\theta ,\varphi }\right)$$ $$=\left|\begin{array}{ccc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }& \frac{\partial x}{\partial \varphi }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }& \frac{\partial y}{\partial \varphi }\\ \frac{\partial z}{\partial r}& \frac{\partial z}{\partial \theta }& \frac{\partial z}{\partial \varphi }\end{array}\right|$$ $$=\left|\begin{array}{ccc}\sin \theta \cos \varphi & r\cos \theta \cos \varphi & -r\sin \theta \sin \varphi \\ \sin \theta \sin \varphi & r\cos \theta \sin \varphi & r\sin \theta \cos \varphi \\ \cos \theta & -r\sin \theta & 0\end{array}\right|$$ Expanding along $R_3$, $$=\cos \theta [(r\sin \theta \cos \varphi )(r\cos \theta \cos \varphi )-(r\cos \theta \sin \varphi )(-r\sin \theta \sin \varphi )]$$ $$-(-r\sin \theta )[(\sin \theta \cos \varphi )(r\sin \theta \cos \varphi )-(-r\sin \theta \sin \varphi )(\sin \theta \sin \varphi )]$$ $$+0$$ $$=\cos \theta (r^2\sin \theta \cos \theta {\cos }^2\varphi +r^2\cos \theta \sin \theta {\sin }^2\varphi )$$ $$+r\sin \theta (r{\sin }^2\theta {\cos }^2\varphi +r{\sin }^2\theta {\sin }^2\varphi )$$ $$=\cos \theta [r^2\cos \theta \sin \theta ({\cos }^2\varphi +{\sin }^2\varphi )]$$ $$+r\sin \theta [r{\sin }^2\theta ({\cos }^2\varphi +{\sin }^2\varphi )]$$ $$=r^2{\cos }^2\theta \sin \theta +r^2{\sin }^3\theta$$ $$=r^2\sin \theta ({\cos }^2\theta +{\sin }^2\theta )$$ $$=r^2\sin \theta$$ Hence Proved

Sample Problem 6 - Implicit Function (reducable to explicit)

If $x+y+z=u;y+z=uv;z=uvw$, find $J\left(\frac{x,y,z}{u,v,w}\right)$

Solution :

$$x+y+z=u⟶(1)$$ $$y+z=uv⟶(2)$$ $$z=uvw⟶(3)$$ Using equation (3) in (2) $$y+uvw=uv$$ $$\boxed{{\displaystyle y=uv-uvw}}$$ Putting value of $y+z$ in equation (1) $$\boxed{{\displaystyle x=u-uv}}$$ $$\text{Now, }J\left(\frac{x,y,z}{u.v.w}\right)=\frac{\partial (x,y,z)}{\partial (u,v,w)}$$ $$=\left|\begin{array}{ccc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}& \frac{\partial x}{\partial w}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}& \frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u}& \frac{\partial z}{\partial v}& \frac{\partial z}{\partial w}\end{array}\right|$$ $$=\left|\begin{array}{ccc}1-v& -u& 0\\ v-vw& u-uw& -uv\\ vw& uw& uv\end{array}\right|$$ $$=(1-v)[(u-uw)uv+u^{2}vw]$$ $$+u[(u-vw)uv+u{v}^{2}w]$$ $$=(1-v)(u^{2}v-\cancel{u^{2}vw}+\cancel{u^{2}vw})+u(u{v}^2-\cancel{u{v}^{2}w}+\cancel{u{v}^{2}w})$$ $$=u^{2}v-\cancel{u^2{v}^2}+\cancel{u^2{v}^2}$$ $$=u^{2}v$$

Sample Problem 7 - Implicit function (not reducable to explicit) :

If $$\begin{array}{rl}u& =xyz\\ v& =x^2+y^2+z^2\\ w& =x+y+z\end{array}$$ find $J\left(\frac{x,y,z}{u,v,w}\right)$

Solution :

There are two ways to solve this:
Method 1 : Using formula $J\times J^{\prime }=1\Rightarrow \boxed{{\displaystyle J^{\prime }=\frac{1}{J}}}$
Method 2 : Using another formula, used especially for such problems
Method 1 is easy, so here we are showing Method 2: $$\begin{array}{rl}u=xyz& \Rightarrow xyz-u=0\\ v=x^2+y^2+z^2& \Rightarrow x^2+y^2+z^2-v=0\\ w=x+y+z& \Rightarrow x+y+z-w=0\end{array}$$ Let $$f_1=xyz-u$$ $$f_2=x^2+y^2+z^2-w$$ $$f_3=x+y+z-w$$ Using formula: $$J\left(\frac{x,y,z}{u,v,w}\right)=(-1{)}^n\frac{J\left(\frac{f_1,f_2,f_3}{u,v,w}\right)}{J\left(\frac{f_1,f_2,f_3}{x,y,z}\right)}$$ $$\text{where }n=\text{ number of variables in one set, here }n=3$$ $$\text{Now, }J\left(\frac{f_1,f_2,f_3}{x,y,z}\right)=\frac{\partial (f_1,f_2,f_3)}{\partial (x,y,z)}$$ $$=\left|\begin{array}{ccc}\frac{\partial f_1}{\partial x}& \frac{\partial f_2}{\partial y}& \frac{\partial f_1}{\partial z}\\ \frac{\partial f_2}{\partial x}& \frac{\partial f_2}{\partial y}& \frac{\partial f_2}{\partial z}\\ \frac{\partial f_3}{\partial x}& \frac{\partial f_3}{\partial y}& \frac{\partial f_3}{\partial z}\end{array}\right|$$ $$=\left|\begin{array}{ccc}yz& xz& xy\\ 2x& 2y& 2z\\ 1& 1& 1\end{array}\right|$$ $$=yz(2y-2z)-xz(2x-2z)+xy(2x-2y)$$ $$=2y^{2}z-2y{z}^2-2x^{2}z+2x{z}^2+2x^{2}y-2x{y}^2$$ $$\text{Now, }J\left(\frac{f_1,f_2,f_3}{u,v,w}\right)$$ $$=\left|\begin{array}{ccc}\frac{\partial f_1}{\partial u}& \frac{\partial f_2}{\partial v}& \frac{\partial f_1}{\partial w}\\ \frac{\partial f_2}{\partial u}& \frac{\partial f_2}{\partial v}& \frac{\partial f_2}{\partial w}\\ \frac{\partial f_3}{\partial u}& \frac{\partial f_3}{\partial v}& \frac{\partial f_3}{\partial w}\end{array}\right|$$ $$=\left|\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right|$$ $$=-1$$ $$\text{Now, }J\left(\frac{x,y,z}{u,v,w}\right)=(-1{)}^3\frac{J\left(\frac{f_1,f_2,f_3}{u,v,w}\right)}{J\left(\frac{f_1,f_2,f_3}{x,y,z}\right)}$$ $$=\frac{(-1)(-1)}{2(x^{2}y-x{y}^2+y^{2}z-y{z}^2+x{z}^2-x^{2}z)}$$ $$=\frac{1}{2(x^{2}y-x{y}^2+y^{2}z-y{z}^2+x{z}^2-x^{2}z)}$$

Functional Dependancy :

If $u,v$ are functons of $x,y$, then the necessary condition for the existance of functional dependancy is $$J\left(\frac{u,v}{x,y}\right)=0$$

Sample Problem 8 :

Show that function $u=\frac{x+y}{1-xy}$, $v={\tan }^{-1}x+{\tan }^{-1}y$ are dependant and find relationship between them.

Solution :

$$u=\frac{x+y}{1-xy}$$ $$\frac{\partial u}{\partial x}=\frac{(1-xy)(1)-(x+y)(-y)}{(1-xy{)}^2}$$ $$=\frac{1-\cancel{xy}+\cancel{xy}-y^2}{(1-xy{)}^2}$$ $$\frac{\partial u}{\partial x}=\frac{1-y^2}{(1-xy{)}^2}$$ $$\frac{\partial u}{\partial y}=\frac{(1-xy)(1)-(x+y)(-x)}{(1-xy{)}^2}$$ $$\frac{\partial u}{\partial y}=\frac{1+x^2}{(1-xy{)}^2}$$ $$\frac{\partial v}{\partial x}=\frac{1}{1+x^2}$$ $$\frac{\partial v}{\partial y}=\frac{1}{1+y^2}$$ $$J\left(\frac{u,v}{x,y}\right)=\left|\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right|$$ $$=\left|\begin{array}{cc}\frac{1+y^2}{(1-xy{)}^2}& \frac{1+x^2}{(1-xy{)}^2}\\ \frac{1}{1+x^2}& \frac{1}{1+y^2}\end{array}\right|$$ $$=\left[\frac{1+y^2}{(1-xy{)}^2}\right]\left[\frac{1}{1+y^2}\right]-\left[\frac{1+x^2}{(1-xy{)}^2}\right]\left[\frac{1}{1+x^2}\right]$$ $$=0$$ Now, to find relationship between $u$ and $v$ $$u=\frac{x+y}{1-xy}$$ $$v={\tan }^{-1}x+{\tan }^{-1}y$$ $$v={\tan }^{-1}\frac{x+y}{1-xy}$$ $$[\text{As, }{\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)]$$ $$\Rightarrow \boxed{{\displaystyle v=ta{n}^{-1}u}}$$

Sample Problem 9 :

Show that functions $$\begin{array}{rl}u& =x+y+z\\ v& =x^2+y^2+z^2-2xy-2yz-2zx\\ w& =x^3+y^3+z^3-3x{y}^2\end{array}$$ are functionally related. Find the relation.

Solution :

$$J\left(\frac{u,v,w}{x,y,z}\right)$$ $$=\left|\begin{array}{ccc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}& \frac{\partial u}{\partial z}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}& \frac{\partial v}{\partial z}\\ \frac{\partial w}{\partial x}& \frac{\partial w}{\partial y}& \frac{\partial w}{\partial z}\end{array}\right|$$ $$=\left|\begin{array}{ccc}1& 1& 1\\ 2x-2y-2z& 2y-2x-2z& 2z-2y-2x\\ 3x^2-3yz& 3y^2-3xz& 3z^2-3xy\end{array}\right|$$ $$=1[(2y-2x-2z)(3z^2-3xy)-(2z-2y-2x)(3y^2-3zx)]$$ $$-1[(2x-2y-2z)(3z^2-3xy)-(2z-2y-2x)(3x^2-3yz)]$$ $$+1[(2x-2y-2z)(3y^2-3zx)-(2y-2x-2z)(3x^2-3yz)]$$ $$=0\Rightarrow \text{ u and v are functionally related }$$ $$\text{Now, }w=x^3+y^3+z^3-3xyz$$ $$=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$ $$=u[(v+2xy+2yz+2zx)-xy-yz-zx]$$ $$=u(v+xy+yz+zx)$$ $$=\frac{4}{4}\times u(v+xy+yz+zx)$$ $$=\frac{u}{4}(4v+4xy+4yz+4zx)$$ $$=\frac{u}{4}[4v+(2xy+2yz+2zx)-(-2xy-2yz-2zx)]$$ $$=\frac{u}{4}[4v+(x^2+y^2+z^2+2xy+2yz+2zx)-(x^2+y^2+z^2-2xy-2yz-2zx)]$$ $$=\frac{u}{4}[4v+(x+y+z{)}^2-v]$$ $$=\frac{u}{4}(3v+u^2)$$ $$\Rightarrow \boxed{{\displaystyle 4w=u^3+3uv}}$$