Substitution of variables in differential expressions.

Substitution of variables in differential expressions.

Often in differential expressions, derivatives with respect to certain variables need to be expressed in terms of derivatives with respect to new variables.

Examples.

1. Transform the equation $$x^4\frac{d^{2}y}{d{x}^2}+2x^3\frac{dy}{dx}-y=0,$$ assuming $x=\frac{1}{t}$.

Solution.

Let's express the derivatives of $y$ with respect to $x$ in terms of derivatives of $y$ with respect to $t$:

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{dy}{dt}}{-\frac{1}{t^2}}=-t^2\frac{dy}{dt},$$

$$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}=\frac{-2t\frac{dy}{dt}-t^2\frac{d^{2}y}{d{t}^2}}{-\frac{1}{t^2}}=2t^3\frac{dy}{dt}+t^4\frac{d^{2}y}{d{t}^2}.$$

Let's substitute the found values of derivatives and the expression $x=\frac{1}{t}$ into the given equation.

$$\frac{1}{t^4}(2t^3\frac{dy}{dt}+t^4\frac{d^{2}y}{d{t}^2})+\frac{2}{t^3}\cdot (-t^2)\frac{dy}{dt}-y=\frac{2}{t}\frac{dy}{dt}+\frac{d^{2}y}{d{t}^2}-\frac{2}{t}\frac{dy}{dt}-y=0.$$

Therefore, $$\frac{d^{2}y}{d{t}^2}-y=0.$$

Answer: $\frac{d^{2}y}{d{t}^2}-y=0.$

2. Transform the equation $$3{\left(\frac{d^{2}y}{d{x}^2}\right)}^2-\frac{dy}{dx}\frac{d^{3}y}{d{x}^3}-\frac{d^{2}y}{d{x}^2}{\left(\frac{dy}{dx}\right)}^2=0,$$ taking $y$ as the argument.

Solution.

Let's express the derivatives of $y$ with respect to $x$ in terms of derivatives of $x$ with respect to $y$: $$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}},$$

$$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{1}{\frac{dx}{dy}}\right)=\frac{d}{dy}\left(\frac{1}{\frac{dx}{dy}}\right)\frac{dy}{dx}=-\frac{\frac{d^{2}x}{d{y}^2}}{{\left(\frac{dx}{dy}\right)}^2}\cdot \frac{1}{\frac{dx}{dy}}=-\frac{\frac{d^{2}x}{d{y}^2}}{{\left(\frac{dx}{dy}\right)}^3},$$

$$\frac{d^{3}y}{d{x}^3}=\frac{d}{dx}(-\frac{\frac{d^{2}x}{d{y}^2}}{{\left(\frac{dx}{dy}\right)}^3})=\frac{d}{dy}(-\frac{\frac{d^{2}x}{d{y}^2}}{{\left(\frac{dx}{dy}\right)}^3})\cdot \frac{dy}{dx}=$$

$$=-\frac{\frac{d^{3}x}{d{y}^3}{\left(\frac{dx}{dy}\right)}^3-3{\left(\frac{dx}{dy}\right)}^2\frac{d^{2}x}{d{y}^2}\cdot \frac{d^{2}x}{d{y}^2}}{{\left(\frac{dx}{dy}\right)}^6}\cdot \frac{1}{\frac{dx}{dy}}$$

Let's substitute the obtained expressions of derivatives into the given equation. We get

$$3{(-\frac{\frac{d^{2}x}{d{y}^2}}{{\left(\frac{dx}{dy}\right)}^3})}^2-\frac{1}{\frac{dx}{dy}}(-\frac{\frac{d^{3}x}{d{y}^3}{\left(\frac{dx}{dy}\right)}^3-3{\left(\frac{dx}{dy}\right)}^2\frac{d^{2}x}{d{y}^2}\cdot \frac{d^{2}x}{d{y}^2}}{{\left(\frac{dx}{dy}\right)}^6}\cdot \frac{1}{\frac{dx}{dy}})-$$ $$(-\frac{\frac{d^{2}x}{d{y}^2}}{{\left(\frac{dx}{dy}\right)}^3}){\left(\frac{1}{\frac{dx}{dy}}\right)}^2=0\Rightarrow$$

$$\Rightarrow 3\frac{{\left(\frac{d^{2}x}{d{y}^2}\right)}^2}{{\left(\frac{dx}{dy}\right)}^6}+\frac{\frac{d^{3}x}{d{y}^3}\frac{dx}{dy}-3{\left(\frac{d^{2}x}{d{y}^2}\right)}^2}{{\left(\frac{dx}{dy}\right)}^6}+\frac{\frac{d^{2}x}{d{y}^2}}{{\left(\frac{dx}{dy}\right)}^5}=0\Rightarrow$$

$$\Rightarrow \frac{\frac{d^{3}x}{d{y}^3}+\frac{d^{2}x}{d{y}^2}}{{\left(\frac{dx}{dy}\right)}^5}=0\Rightarrow$$

$$\Rightarrow \frac{d^{3}x}{d{y}^3}+\frac{d^{2}x}{d{y}^2}=0.$$

Thus, we have obtained the solution.

Answer: $$\frac{d^{3}x}{d{y}^3}+\frac{d^{2}x}{d{y}^2}=0.$$

3. Transform the equation $$(x{y}^{\prime }-y{)}^2=2xy(1+y^{\prime 2}),$$ by switching to polar coordinates.

Solution.

We have

$$x=r\cos \phi ,y=r\sin \phi ,$$

$$dx=\cos \phi dr-r\sin \phi d\phi ,dy=\sin \phi dr+r\cos \phi d\phi ,$$

$$y^{\prime }=\frac{dy}{dx}=\frac{\sin \phi dr+r\cos \phi d\phi }{\cos \phi dr-r\sin \phi d\phi }.$$

We substitute the expressions for $x,,,y,,,y^{\prime }$ into the given equation. $${(r\cos \phi \frac{\sin \phi dr+r\cos \phi d\phi }{\cos \phi dr–r\sin \phi d\phi }-r\sin \phi )}^2=$$ $$=2r^2\cos \phi \sin \phi (1+{\left(\frac{\sin \phi dr+r\cos \phi d\phi }{\cos \phi dr-r\sin \phi d\phi }\right)}^2)\Rightarrow$$

$${\left(\frac{r\sin \phi \cos \phi dr+r^2{\cos }^2\phi d\phi -r\sin \phi \cos \phi dr+r^2{\sin }^2\phi d\phi }{\cos \phi dr-r\sin \phi d\phi }\right)}^2=$$

$$=2r^2\cos \phi \sin \phi \frac{{(cos\phi dr-r\sin \phi d\phi )}^2+{(\sin \phi dr+r\cos \phi d\phi )}^2}{{(\cos \phi dr-r\sin \phi d\phi )}^2}\Rightarrow$$

$${\left(\frac{r^{2}d\phi }{\cos \phi dr-r\sin \phi d\phi }\right)}^2=\frac{2r^2\cos \phi \sin \phi (d{r}^2+r^{2}d{\phi }^2)}{{(\cos \phi dr-r\sin \phi d\phi )}^2}\Rightarrow$$

$$r^{4}d{\phi }^2=r^2\sin 2\phi d{r}^2+r^4\sin 2\phi d{\phi }^2\Rightarrow$$

$$\sin 2\phi d{r}^2=(1-\sin 2\phi )r^{2}d{\phi }^2\Rightarrow {\left(\frac{dr}{d\phi }\right)}^2=\frac{1-\sin 2\phi }{\sin 2\phi }{r}^2\Rightarrow$$

$$r^{\prime 2}=\frac{1-\sin 2\phi }{\sin 2\phi }{r}^2$$

Answer: $r^{\prime 2}=\frac{1-\sin 2\phi }{\sin 2\phi }{r}^2$

4. To transform the equation $$(x+y)\frac{\partial z}{\partial x}-(x-y)\frac{\partial z}{\partial y}=0,$$ перейдя к новым независимым переменным $u$ и $v,$ если $u=\ln \sqrt{x^2+y^2},v=arctg\frac{y}{x}.$

using the new independent variables $u$ and $v$, where $u=\ln \sqrt{x^2+y^2},,,v=\arctan \frac{y}{x}.$

Solution.

We express the partial derivatives of $z$ with respect to $x$ and $y$ in terms of the partial derivatives of $z$ with respect to $u$ and $v$.

We have $$\frac{\partial u}{\partial x}=\frac{1}{\sqrt{x^2+y^2}}(\sqrt{x^2+y^2}{)}_x^{\prime }=\frac{1}{\sqrt{x^2+y^2}}\frac{2x}{2\sqrt{x^2+y^2}}=\frac{x}{x^2+y^2};$$

$$\frac{\partial u}{\partial y}=\frac{1}{\sqrt{x^2+y^2}}(\sqrt{x^2+y^2}{)}_y^{\prime }=\frac{1}{\sqrt{x^2+y^2}}\frac{2y}{2\sqrt{x^2+y^2}}=\frac{y}{x^2+y^2};$$

$$\frac{\partial v}{\partial x}=\frac{1}{1+{\left(\frac{y}{x}\right)}^2}{\left(\frac{y}{x}\right)}_x^{\prime }=\frac{1}{1+{\left(\frac{y}{x}\right)}^2}\frac{-y}{x^2}=\frac{-y}{x^2+y^2};$$

$$\frac{\partial v}{\partial y}=\frac{1}{1+{\left(\frac{y}{x}\right)}^2}{\left(\frac{y}{x}\right)}_y^{\prime }=\frac{1}{1+{\left(\frac{y}{x}\right)}^2}\frac{1}{x}=\frac{x}{x^2+y^2}.$$

Next, we find

$$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial z}{\partial u}\frac{x}{x^2+y^2}+\frac{\partial z}{\partial v}\frac{-y}{x^2+y^2};$$

$$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}=\frac{\partial z}{\partial u}\frac{y}{x^2+y^2}+\frac{\partial z}{\partial v}\frac{x}{x^2+y^2}.$$

Let's substitute the found expressions of derivatives into the given equation:

$$(x+y)(\frac{\partial z}{\partial u}\frac{x}{x^2+y^2}+\frac{\partial z}{\partial v}\frac{-y}{x^2+y^2})-(x-y)(\frac{\partial z}{\partial u}\frac{y}{x^2+y^2}+\frac{\partial z}{\partial v}\frac{x}{x^2+y^2})=0\Rightarrow$$

$$\Rightarrow \frac{\partial z}{\partial u}\frac{x^2+xy-xy+y^2}{x^2+y^2}+\frac{\partial z}{\partial v}\frac{-xy-y^2-x^2+xy}{x^2+y^2}=0\Rightarrow$$

$$\Rightarrow \frac{\partial z}{\partial u}-\frac{\partial z}{\partial v}=0.$$

Answer: $\frac{\partial z}{\partial u}=\frac{\partial z}{\partial v}.$

5. To transform the equation $$(xy+z)\frac{\partial z}{\partial x}+(1-y^2)\frac{\partial z}{\partial y}=x+yz,$$ taking $u=yz-x,,,v=xz-y$ as new independent variables and $w=xy-z$ as the new function.

Solution.

We express the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ in terms of $\frac{\partial w}{\partial u}$ and $\frac{\partial w}{\partial v}$.

We have:

$$du=-dx+zdy+ydz;$$

$$dv=zdx+xdz-dy;$$

$$dw=ydx+xdy-dz.$$

Taking into account the formula

$$\frac{du}{dt}=\frac{\partial u}{\partial x_1}\cdot \frac{d{x}_1}{dt}+\frac{\partial u}{\partial x_2}\cdot \frac{d{x}_2}{dt}+\dots +\frac{\partial u}{\partial x_n}\cdot \frac{d{x}_n}{dt},$$ we find

$$dw=\frac{\partial w}{\partial u}\cdot du+\frac{\partial w}{\partial v}\cdot dv\Rightarrow$$

$$ydx+xdy-dz=\frac{\partial w}{\partial u}\cdot (-dx+zdy+ydz)+\frac{\partial w}{\partial v}\cdot (zdx+xdz-dy)\Rightarrow$$

$$ydx+xdy-\frac{\partial w}{\partial u}\cdot (-dx+zdy)-\frac{\partial w}{\partial v}\cdot (zdx-dy)=dz(1+y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v})\Rightarrow$$

$$dz=\frac{y+\frac{\partial w}{\partial u}-z\frac{\partial w}{\partial v}}{1+y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}}dx+\frac{x-z\frac{\partial w}{\partial u}+\frac{\partial w}{\partial v}}{1+y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}}dy\Rightarrow$$

$$\frac{dz}{dx}=\frac{y+\frac{\partial w}{\partial u}-z\frac{\partial w}{\partial v}}{1+y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}};$$

$$\frac{dz}{dy}=\frac{x-z\frac{\partial w}{\partial u}+\frac{\partial w}{\partial v}}{1+y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}}.$$

Let's substitute the found expressions for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ into the given equation. We get

$$(xy+z)\frac{y+\frac{\partial w}{\partial u}-z\frac{\partial w}{\partial v}}{1+y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}}+(1-y^2)\frac{x-z\frac{\partial w}{\partial u}+\frac{\partial w}{\partial v}}{1+y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}}=x+yz\Rightarrow$$

$$\frac{x{y}^2+zy+xy\frac{\partial w}{\partial u}+z\frac{\partial w}{\partial u}-xyz\frac{\partial w}{\partial v}-z^2\frac{\partial w}{\partial v}+x-z\frac{\partial w}{\partial u}+\frac{\partial w}{\partial v}-x{y}^2+y^{2}z\frac{\partial w}{\partial u}-y^2\frac{\partial w}{\partial v}}{1+y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}}=$$ $$=x+yz\Rightarrow$$

$$\frac{zy+xy\frac{\partial w}{\partial u}-xyz\frac{\partial w}{\partial v}-z^2\frac{\partial w}{\partial v}+x+\frac{\partial w}{\partial v}+y^{2}z\frac{\partial w}{\partial u}-y^2\frac{\partial w}{\partial v}}{1+y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}}=x+yz\Rightarrow$$

$$zy+xy\frac{\partial w}{\partial u}-xyz\frac{\partial w}{\partial v}-z^2\frac{\partial w}{\partial v}+x+\frac{\partial w}{\partial v}+y^{2}z\frac{\partial w}{\partial u}-y^2\frac{\partial w}{\partial v}=$$ $$=(x+yz)\left(1+y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}\right)\Rightarrow$$

$$zy+xy\frac{\partial w}{\partial u}-xyz\frac{\partial w}{\partial v}-z^2\frac{\partial w}{\partial v}+x+\frac{\partial w}{\partial v}+y^{2}z\frac{\partial w}{\partial u}-y^2\frac{\partial w}{\partial v}=$$

$$=x+xy\frac{\partial w}{\partial u}+x^2\frac{\partial w}{\partial v}+yz+y^{2}z\frac{\partial w}{\partial u}+xyz\frac{\partial w}{\partial v}\Rightarrow$$

$$-xyz\frac{\partial w}{\partial v}-z^2\frac{\partial w}{\partial v}+\frac{\partial w}{\partial v}-y^2\frac{\partial w}{\partial v}=x^2\frac{\partial w}{\partial v}+xyz\frac{\partial w}{\partial v}.$$

Thus,

$$\frac{\partial w}{\partial v}(x^2+y^2+z^2+2xyz-1)=0\Rightarrow \frac{\partial w}{\partial v}=0.$$

Answer: $\frac{\partial w}{\partial v}=0.$

Tags: Substitution of variables, calculus, derivative, differential expressions, mathematical analysis