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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!

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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 $$ ϕ(2u+x^2)=2xu+\frac23x^3-y^2. $$


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 $u$.

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  • $\begingroup$ how do you know which equations to solve? $\endgroup$ – Brad Scott Dec 11 '18 at 17:39
  • $\begingroup$ You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises. $\endgroup$ – LutzL Dec 11 '18 at 17:43
  • $\begingroup$ You could also just multiply all with $y$ so that $ds=\frac{dx}1=\frac{y\,dy}u=-\frac{du}x$, which reduces to the linear system $\frac{d(y^2)}{dx}=2u$ and $\frac{du}{dx}=-x$. $\endgroup$ – LutzL Dec 11 '18 at 17:49
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$$\frac{dx}{y}=\frac{dy}{u}=-\frac{du}{xy}\qquad \text{OK.}$$ 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 $yu_x+uu_y=-xy$.

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  • $\begingroup$ Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used. $\endgroup$ – Brad Scott Dec 11 '18 at 19:03
  • $\begingroup$ In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used". $\endgroup$ – JJacquelin Dec 11 '18 at 21:55

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