Question Number 214678 by issac last updated on 16/Dec/24
$$\mathrm{solve} \\ $$$$\mathrm{partial}\:\mathrm{differantial}\:\mathrm{equation} \\ $$$${x}\frac{\partial{f}\left({x},{y}\right)}{\partial{x}}+{y}\frac{\partial{f}\left({x},{y}\right)}{\partial{y}}={f}\left({x},{y}\right)\mathrm{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$\frac{\partial^{\mathrm{2}} {f}\left({x},{y}\right)}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {f}\left({x},{y}\right)}{\partial{y}^{\mathrm{2}} }=\mathrm{0} \\ $$
Answered by MrGaster last updated on 09/Feb/25
![x=r cos θ ,y=r sin θ f(x,y)=f(r,cos θ,r sin θ)=g(r,θ) ⇒(∂f/∂x)=(∂g/∂r) (∂r/∂x)+(∂g/∂θ) (∂θ/∂x)=cos θ(∂g/∂r)−((sinθ)/r) (∂g/∂θ) ⇒(∂f/∂y)=(∂g/∂r) (∂r/∂y)+(∂g/∂θ) (∂θ/∂y)=sin θ(∂g/∂r)+((cosθ)/r) (∂g/∂θ) ⇒(∂^2 f/∂x^2 )=cos^2 θ(∂^2 g/∂r^2 )−((2 cos θ sin θ)/r) (∂^2 g/(∂r∂θ))+((cos^2 θ)/r) (∂g/∂r)−((2 cos θ sin θ)/r^2 ) (∂g/∂θ) ⇒x(∂f/∂x)+y(∂f/∂y)=r cos θ(cosθ(∂g/∂r)−((sinθ)/r) (∂g/∂θ))+r sin θ(sin θ(∂g/∂r)+((cos θ)/r) (∂g/∂θ)) =r cos^2 θ(∂g/∂r)−cos θ sin θ(∂g/∂θ)+r sin^2 θ(∂g/∂r)+cos θ sin θ(∂g/∂θ) =r(∂g/∂r) =g(r,θ)ln(r^2 ) ⇒r(∂g/∂r)=g(r,θ)ln(r^2 ) r(∂g/∂r)=g(r,θ)ln(r^2 ) (∂g/g)=((ln(r^2 ))/r)dr ln g=∫((2 ln r)/r)dr=ln^2 r+C(θ) g(r,θ)=e^(C(θ)) r^(2 ln r) (∂^2 f/∂x^2 )+(∂^2 f/∂y^2 )=0 ⇒(∂^2 g/∂r^2 )+(1/r^2 ) (∂^2 g/∂θ^2 )+(1/r) (∂g/∂r)=0 Substitute g(r,θ)=e^(C(θ)) r^(2 ln r) : (∂g/∂r)=e^(C(θ)) (2 ln r+2)r^(2 ln r−1) (∂^2 g/∂r^2 )=e^(C(0)) (2 ln r+2)(2 ln r−1)r^(2 ln r−2) (∂^2 g/∂θ^2 )=e^(C(θ)) ((d^2 C(θ))/∂θ^2 )r^2 ln r ⇒e^(C(θ)) (2 ln r+2)(2 ln r−1)r^(2 ln r−2) +(1/r^2 )e^(C(θ)) ((d^2 C(θ))/dθ^2 )r^(2 ln r) +(1/r)e^(C(θ)) (2 ln r+2)r^(2 ln r−1) =0 e^(C(θ)) r^(2 ln r−2) [(2 ln r+2)(2 ln r−1)+(1/r^2 ) ((d^2 C(θ))/dθ^2 )+(1/r)(2 ln r+2)]=0 ∵e^(C(θ)) r^(2 ln r−2) ≠0,have: (2 ln r+2)(2 ln r−1)+(1/r^2 ) ((d^2 C(θ))/dθ^2 )+(1/r)(2 ln r+2)=0 This implies: ((d^2 C(θ))/dθ^2 )=−r^2 [(2 ln r+2)(2 ln r−1)+(1/r)(2 ln r+2) Since the left side is independent of r,the right side must be constant.Thefore,C(θ)is a liear function of θ: C(θ)=C_1 θ+C_2 Thus,the solution is: g(r,θ)=e^(C_1 θ+C_2 ) r^(2 ln r) f(x,y)=e^(C_1 θ+C_2 ) r^(2 ln r) determinant (((f(x,y)=C_1 e^C_2 (x^2 +y^2 )^(ln(x^2 +y^2 )) )))](Q216501.png)
$${x}={r}\:\mathrm{cos}\:\theta\:,{y}={r}\:\mathrm{sin}\:\theta \\ $$$${f}\left({x},{y}\right)={f}\left({r},\mathrm{cos}\:\theta,{r}\:\mathrm{sin}\:\theta\right)={g}\left({r},\theta\right) \\ $$$$\Rightarrow\frac{\partial{f}}{\partial{x}}=\frac{\partial{g}}{\partial{r}}\:\frac{\partial{r}}{\partial{x}}+\frac{\partial{g}}{\partial\theta}\:\frac{\partial\theta}{\partial{x}}=\mathrm{cos}\:\theta\frac{\partial{g}}{\partial{r}}−\frac{\mathrm{sin}\theta}{{r}}\:\frac{\partial{g}}{\partial\theta} \\ $$$$\Rightarrow\frac{\partial{f}}{\partial{y}}=\frac{\partial{g}}{\partial{r}}\:\frac{\partial{r}}{\partial{y}}+\frac{\partial{g}}{\partial\theta}\:\frac{\partial\theta}{\partial{y}}=\mathrm{sin}\:\theta\frac{\partial{g}}{\partial{r}}+\frac{\mathrm{cos}\theta}{{r}}\:\frac{\partial{g}}{\partial\theta} \\ $$$$\Rightarrow\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }=\mathrm{cos}^{\mathrm{2}} \theta\frac{\partial^{\mathrm{2}} {g}}{\partial{r}^{\mathrm{2}} }−\frac{\mathrm{2}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta}{{r}}\:\frac{\partial^{\mathrm{2}} {g}}{\partial{r}\partial\theta}+\frac{\mathrm{cos}^{\mathrm{2}} \theta}{{r}}\:\frac{\partial{g}}{\partial{r}}−\frac{\mathrm{2}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta}{{r}^{\mathrm{2}} }\:\frac{\partial{g}}{\partial\theta} \\ $$$$\Rightarrow{x}\frac{\partial{f}}{\partial{x}}+{y}\frac{\partial{f}}{\partial{y}}={r}\:\mathrm{cos}\:\theta\left(\mathrm{cos}\theta\frac{\partial{g}}{\partial{r}}−\frac{\mathrm{sin}\theta}{{r}}\:\frac{\partial{g}}{\partial\theta}\right)+{r}\:\mathrm{sin}\:\theta\left(\mathrm{sin}\:\theta\frac{\partial{g}}{\partial{r}}+\frac{\mathrm{cos}\:\theta}{{r}}\:\frac{\partial{g}}{\partial\theta}\right) \\ $$$$={r}\:\mathrm{cos}^{\mathrm{2}} \theta\frac{\partial{g}}{\partial{r}}−\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\frac{\partial{g}}{\partial\theta}+{r}\:\mathrm{sin}^{\mathrm{2}} \theta\frac{\partial{g}}{\partial{r}}+\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\frac{\partial{g}}{\partial\theta} \\ $$$$={r}\frac{\partial{g}}{\partial{r}} \\ $$$$={g}\left({r},\theta\right)\mathrm{ln}\left({r}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{r}\frac{\partial{g}}{\partial{r}}={g}\left({r},\theta\right)\mathrm{ln}\left({r}^{\mathrm{2}} \right) \\ $$$${r}\frac{\partial{g}}{\partial{r}}={g}\left({r},\theta\right)\mathrm{ln}\left({r}^{\mathrm{2}} \right) \\ $$$$\frac{\partial{g}}{{g}}=\frac{\mathrm{ln}\left({r}^{\mathrm{2}} \right)}{{r}}{dr} \\ $$$$\mathrm{ln}\:{g}=\int\frac{\mathrm{2}\:\mathrm{ln}\:{r}}{{r}}{dr}=\mathrm{ln}^{\mathrm{2}} {r}+{C}\left(\theta\right) \\ $$$${g}\left({r},\theta\right)={e}^{{C}\left(\theta\right)} {r}^{\mathrm{2}\:\mathrm{ln}\:{r}} \\ $$$$\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {f}}{\partial{y}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\frac{\partial^{\mathrm{2}} {g}}{\partial{r}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:\frac{\partial^{\mathrm{2}} {g}}{\partial\theta^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}}\:\frac{\partial{g}}{\partial{r}}=\mathrm{0} \\ $$$$\mathrm{Substitute}\:{g}\left({r},\theta\right)={e}^{{C}\left(\theta\right)} {r}^{\mathrm{2}\:\mathrm{ln}\:{r}} : \\ $$$$\frac{\partial{g}}{\partial{r}}={e}^{{C}\left(\theta\right)} \left(\mathrm{2}\:\mathrm{ln}\:{r}+\mathrm{2}\right){r}^{\mathrm{2}\:\mathrm{ln}\:{r}−\mathrm{1}} \\ $$$$\frac{\partial^{\mathrm{2}} {g}}{\partial{r}^{\mathrm{2}} }={e}^{{C}\left(\mathrm{0}\right)} \left(\mathrm{2}\:\mathrm{ln}\:{r}+\mathrm{2}\right)\left(\mathrm{2}\:\mathrm{ln}\:{r}−\mathrm{1}\right){r}^{\mathrm{2}\:\mathrm{ln}\:{r}−\mathrm{2}} \\ $$$$\frac{\partial^{\mathrm{2}} {g}}{\partial\theta^{\mathrm{2}} }={e}^{{C}\left(\theta\right)} \frac{{d}^{\mathrm{2}} {C}\left(\theta\right)}{\partial\theta^{\mathrm{2}} }{r}^{\mathrm{2}} \mathrm{ln}\:{r} \\ $$$$\Rightarrow{e}^{{C}\left(\theta\right)} \left(\mathrm{2}\:\mathrm{ln}\:{r}+\mathrm{2}\right)\left(\mathrm{2}\:\mathrm{ln}\:{r}−\mathrm{1}\right){r}^{\mathrm{2}\:\mathrm{ln}\:{r}−\mathrm{2}} +\frac{\mathrm{1}}{{r}^{\mathrm{2}} }{e}^{{C}\left(\theta\right)} \frac{{d}^{\mathrm{2}} {C}\left(\theta\right)}{{d}\theta^{\mathrm{2}} }{r}^{\mathrm{2}\:\mathrm{ln}\:{r}} +\frac{\mathrm{1}}{{r}}{e}^{{C}\left(\theta\right)} \left(\mathrm{2}\:\mathrm{ln}\:{r}+\mathrm{2}\right){r}^{\mathrm{2}\:\mathrm{ln}\:{r}−\mathrm{1}} =\mathrm{0} \\ $$$${e}^{{C}\left(\theta\right)} {r}^{\mathrm{2}\:\mathrm{ln}\:{r}−\mathrm{2}} \left[\left(\mathrm{2}\:\mathrm{ln}\:{r}+\mathrm{2}\right)\left(\mathrm{2}\:\mathrm{ln}\:{r}−\mathrm{1}\right)+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:\frac{{d}^{\mathrm{2}} {C}\left(\theta\right)}{{d}\theta^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}}\left(\mathrm{2}\:\mathrm{ln}\:{r}+\mathrm{2}\right)\right]=\mathrm{0} \\ $$$$\because{e}^{{C}\left(\theta\right)} {r}^{\mathrm{2}\:\mathrm{ln}\:{r}−\mathrm{2}} \neq\mathrm{0},\mathrm{have}: \\ $$$$\left(\mathrm{2}\:\mathrm{ln}\:{r}+\mathrm{2}\right)\left(\mathrm{2}\:\mathrm{ln}\:{r}−\mathrm{1}\right)+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:\frac{{d}^{\mathrm{2}} {C}\left(\theta\right)}{{d}\theta^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}}\left(\mathrm{2}\:\mathrm{ln}\:{r}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{This}\:\mathrm{implies}: \\ $$$$\frac{{d}^{\mathrm{2}} {C}\left(\theta\right)}{{d}\theta^{\mathrm{2}} }=−{r}^{\mathrm{2}} \left[\left(\mathrm{2}\:\mathrm{ln}\:{r}+\mathrm{2}\right)\left(\mathrm{2}\:\mathrm{ln}\:{r}−\mathrm{1}\right)+\frac{\mathrm{1}}{{r}}\left(\mathrm{2}\:\mathrm{ln}\:{r}+\mathrm{2}\right)\right. \\ $$$$\mathrm{Since}\:\mathrm{the}\:\mathrm{left}\:\mathrm{side}\:\mathrm{is}\:\mathrm{independent}\:\mathrm{of}\:{r},\mathrm{the}\:\mathrm{right} \\ $$$$\mathrm{side}\:\mathrm{must}\:\mathrm{be}\:\mathrm{constant}.\mathrm{Thefore},{C}\left(\theta\right)\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{liear}\:\mathrm{function}\:\mathrm{of}\:\theta: \\ $$$${C}\left(\theta\right)={C}_{\mathrm{1}} \theta+{C}_{\mathrm{2}} \\ $$$$\mathrm{Thus},\mathrm{the}\:\mathrm{solution}\:\mathrm{is}: \\ $$$${g}\left({r},\theta\right)={e}^{{C}_{\mathrm{1}} \theta+{C}_{\mathrm{2}} } {r}^{\mathrm{2}\:\mathrm{ln}\:{r}} \\ $$$${f}\left({x},{y}\right)={e}^{{C}_{\mathrm{1}} \theta+{C}_{\mathrm{2}} } {r}^{\mathrm{2}\:\mathrm{ln}\:{r}} \\ $$$$\begin{array}{|c|}{{f}\left({x},{y}\right)={C}_{\mathrm{1}} {e}^{{C}_{\mathrm{2}} } \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} }\\\hline\end{array} \\ $$