How to use the initial conditions in that PDE?

Question

I have the following pde:

$xu_{x}-yu_{y}+u=x $, with the initial condition $u=1$, when $y=x^2$

Solving it with the method of characteristics:

$\frac{dx}{ds}=x $, $\hspace{5mm} x(r,0)=r $

$ \frac{dy}{ds}=-y $, $\hspace{5mm} y(r,0)=r^2 $

$\frac{dz}{ds} = x-z $, $\hspace{5mm} z(r,r^2)=1$

And I find

$x(r,s)=re^s$, $\hspace{5mm} y(r,s)=r^2e^{-s}$ and $\hspace{5mm} z(r,s) = \frac{1}{2}re^{2s} +e^{r^2}-\frac{1}{2}r $

But function $z$ doesn't apply the 3rd equation. I think I am not getting right the initial condition $u(x,x^2)=1$

Thank you

Answer 1

You have $z=z(r,s)$, not $z=z(x,y)$ as the third equation for the characteristics seems to assume. So, the correct intial conditions for $z$ are that $z=1$ along the curve with $s=0$ and $r$ free. So is,

$\dfrac{dz}{ds} = x-z$

$z(r,0)=1$

Solvig it and applying the initial conditions:

$z=\dfrac{r}{2}e^s+C e^{-s}\implies 1=\dfrac{r}{2}+C$ and $C=1-\dfrac{r}{2}$

$z=\dfrac{r}{2}e^s+\left(1-\dfrac{r}{2}\right)e^{-s}$

Isolating $r$ and $s$ from the first two equations:

$e^s=\sqrt[3]{\dfrac{x^2}{y}};\;r=\sqrt[3]{xy}$

From here, $u=z(x,y)=\dfrac{x}{2}+\left(1-\dfrac{\sqrt[3]{xy}}{2}\right)\dfrac{\sqrt[3]{xy}}{x}$

that satisfies the inital conditions.

Answer 2

$$xu_{x}-yu_{y}+u=x$$ First DE is: $$\dfrac {dx}{x}=-\dfrac {dy}{y}$$ $$ \ln x= - \ln y +k$$ $$\implies c_1=xy$$ Second DE gives us: $$\dfrac {dx}{x}=\dfrac {dz}{x-z}$$ $$\dfrac {dx-dz}{z}=\dfrac {dz}{x-z}$$ $$(x-z) d(x-z)=z {dz}$$ $$(x-z)^2=z^2+k$$ $$\implies c_2=2xz-x^2$$ Therefore the solution is $$f(c_1)=c_2$$ $$2xz=f(xy)+x^2$$ $$\boxed {z(x,y)=\dfrac x 2 +\dfrac {F(xy)}{2x}}$$ Apply the initial condtion to find $F$ $$z(x,x^2)=1$$ $$ \implies \dfrac x2+\dfrac {F(x^3)}{2x}=1$$ $$ {F(x^3)}=2x-x^2=x(2-x)$$ $$............$$ It's easy to finish.

Answer 3

$$xu_x-yu_y=x-u$$ Charpit-Lagrange system of characteristic ODEs : $$\frac{dx}{x}=\frac{dy}{-y}=\frac{du}{x-u}$$ A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{-y}$ $$xy=c_1 $$ A second characteristic equation comes from solving $\frac{dx}{x}=\frac{du}{x-u}$ $$xu-\frac12x^2=c_2$$ The general solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ is : $$xu-\frac12x^2=F(xy)$$ $$\boxed{u(x,y)=\frac{x}{2}+\frac{1}{x}F(xy)}$$ $F$ is an arbitrary function (to be determined according to the specified condition).

CONDITION : $\quad u(x,y=x^2)=1$ $$\frac{x}{2}+\frac{1}{x}F(x(x^2))=1$$ $$F(x^3)=x-\frac12 x^2$$ Let $X=x^3\quad ; \quad x=X^{1/3}$ $$F(X)=X^{1/3}-\frac12 X^{2/3}$$ Now the function $F$ is determined. We put it into the above general solution where $\quad X=xy\quad.$ Thus $$F(X)=F(xy)=X^{1/3}-\frac12 X^{2/3}=(xy)^{1/3}-\frac12 (xy)^{2/3}$$ $$u(x,y)=\frac{x}{2}+\frac{1}{x}\left((xy)^{1/3}-\frac12 (xy)^{2/3}\right)$$ $$\boxed{u(x,y)=\frac{x}{2}+x^{-2/3}y^{1/3}-\frac12 x^{-1/3}y^{2/3}}$$