I've been given a PDE of the form $$xu_x+(x^2+y)u_y=1-\left(\frac{y}{x}-x\right)u~~; ~~u(1,y)=0$$ Attempt : Firstly it's a first order linear PDE which has the general form $$a(x,y)u_x+b(x,y)u_y=c(x,y)u+d(x,y)$$Where $a,b,c,d \in \mathcal{C}^1(\Omega)$, where $\Omega \subseteq \mathbb{R}^2$ (open and connected) and we also have $a^2+b^2 \neq 0$ on $\Omega_2$. Then first of all the largest $\Omega_2$ on which $F:=a^2+b^2=x^2+(x^2+y)\neq 0$ is : $(x,y) \in \Omega_2\setminus \{(0,0)\}$ right?
We have $s \mapsto \Gamma_s:=(f(s_0):=1,g(s_0):=s,h(s_0):=0))$. Now, from transversality conditions we get :$$J:=\mathrm{det}\, \begin{pmatrix}a&f'\\b&g'\end{pmatrix}\Bigg{|}_{f(s_0),g(s_0),h(s_0)}=\mathrm{det}\begin{pmatrix}1&0\\1+s&1\end{pmatrix}=1$$ So we atleast expect solutions to exist $\textit{locally}$ for all $s \in (-\delta,\delta),\delta>0$. Now to find the solutions explicitly i'll employ the method of characteristics: \begin{align}\frac{\mathrm{d}x}{\mathrm{d}t}&=x~~;~x(0,s):=1 \implies x(t,s):=e^t\\\frac{\mathrm{d}y}{\mathrm{d}t}&=x^2+y~~;~ y(0,s):=s \\&=y+e^{2t} \end{align} Solving this we get \begin{align} y(t,s)&=e^{2t}+c_2e^{t}~~;~y(0,s):=s \implies y(t,s)=e^{t}\left(e^t+s-1\right)\end{align} Lastly \begin{align}\frac{\mathrm{d}z}{\mathrm{d}t}&=1-\left(\frac{y}{x}-x\right)z~~;~z(0,s):=0 \\&=1-\left(\frac{e^t(e^t+s-1)}{e^t}-e^t\right)z=1+(1-s)z\end{align} Which gives me $$z(t,s)=\frac{1}{s-1}\left(1-e^{t(1-s)}\right)$$ As the Jacobian was non-zero by inverse function theorem i can expect that i can invert the map $$(t,s)\mapsto \left(x(t,s),y(t,s)\right)$$ and i can solve $x=x(t,s),y=y(t,s)$ uniquely for $t$ and $s$ and the map $$(x,y)\mapsto \left(T(x,y);S(x,y)\right)$$ should be of class $\mathcal{C}^1$.
So i have $x(t,s)=e^t \implies t=\ln\,x=:T(x,y)$ and $y(t,s)=e^{\ln\,x}\left(e^{\ln \,x}+s-1\right)=x(x+s-1)$ i.e $s=(y/x)-x+1=:S(x,y)$. From here i can write $$u=z(T(x,y),S(x,y))=\frac{x}{y-x^2}\left(1-x^{\frac{x^2-y}{x}}\right)$$ Which is valid when $y>x^2$. I'm not sure about my workings here, appreciate any hints and help. Thanks.
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Siddhartha
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