Quasilinear PDE using method of characteristics

Question

The equation is: $yu_x+uu_y=-xy$ with initial conditions $u=y$ on $x=0$

I first find that

$\frac{dx}{y}=\frac{dy}{u}=-\frac{du}{xy}$

Solving $\frac{dx}{y}=\frac{dy}{u}$ we get, $ux=\frac{1}{2}y^2+A$

Now solving $\frac{dy}{u}=-\frac{du}{xy}$ we get $\frac{1}{2}u^2=B-\frac{1}{2}xy^2$

Applying the initial conditions yields $A=-\frac{1}{2}y^2$ and $\frac{1}{2}y^2=B$, so $A=-B$, or $A+B=0$

This was obtained on the line $x=0$, but contains only constants, thus holds for all characteristics intersecting $x=0$, so we have,

$A=-B \implies ux-\frac{1}{2}y^2=-\frac{1}{2}u^2-\frac{1}{2}xy^2$

Is this the correct solution?

Thanks!

Answer 1

No, already the first equation you solve is wrong. As the last denominator is not zero, $u$ is not constant along characteristic curves.

You can solve the first and last quotient as $$ x\,dx = -du\implies 2u+x^2=c_1 $$ This you can then insert into the other equations. $$ 0=u\,dx -y\,dy\implies 0=(c_1-x^2)\,dx-2y\,dy\implies c_2=c_1x-\frac{x^3}3-y^2 $$ Because each characteristic curve only has one constant, there is some functional dependence like $c_2=\phi(c_1)$, so that $$ ϕ(2u+x^2)=2xu+\frac23x^3-y^2. $$

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With the initial conditions $u=y$, $x=0$ this reduces to $$ ϕ(2y)=-y^2\implies ϕ(t)=-\frac{t^2}4 $$ so that the surface follows the equation $$ -(u^2+x^2u+\frac{x^4}4)=2xu+\frac23x^3-y^2 $$ which is a quadratic equation for the value of $u$.

Answer 2

$$\frac{dx}{y}=\frac{dy}{u}=-\frac{du}{xy}\qquad \text{OK.}$$ First characteristic curves from $\frac{dx}{y}=-\frac{du}{xy}$ : $$u+\frac{x^2}{2}=c_1$$ Second characteristic curves from $\frac{dx}{y}=\frac{dy}{c_1-\frac{x^2}{2}}$ :

$c_1x-\frac{x^3}{6}=\frac{y^2}{2}+c_2$

$(u+\frac{x^2}{2})x-\frac{x^3}{6}-\frac{y^2}{2}=c_2$

$$xu+\frac{x^3}{3}-\frac{y^2}{2}=c_2$$ General solution on the form of implicite equation : $$xu+\frac{x^3}{3}-\frac{y^2}{2}=F\left(u+\frac{x^2}{2}\right)$$ $F$ is an arbitrary function to be determined according to the boundary condition.

Condition $u(0,y)=y$ :

$0y+\frac{0^3}{3}-\frac{y^2}{2}=F\left(y+\frac{0^2}{2}\right)=-\frac{y^2}{2}=F\left(y\right)$

The function $F$ is determined : $$F(X)=-\frac{X^2}{2}$$ We put it into the general solution where $X=u+\frac{x^2}{2}$ $$xu+\frac{x^3}{3}-\frac{y^2}{2}=-\frac{\left(u+\frac{x^2}{2}\right)^2}{2}$$ Solving for $u$ : $$u(x,y)=-x-\frac{x^2}{2}+y\:\sqrt{1+\frac{x^2}{y^2}+\frac{x^3}{3y^2}}$$

This result has been successfully checked in putting it into the PDE $yu_x+uu_y=-xy$.