# Find the solution $u$ of $x~u~u_x+y~u~u_y=-x~y,$ that $u(x,1/x)=5,~x>0$.

Solve the following Cauchy problem:

$$x~u~u_x+y~u~u_y=-x~y,$$ under the condition that $u(x,1/x)=5,~x>0$.

Attempt. The characteristic curves for this quasilinear pde satisfy the system of equations: $$\frac{dx}{xz}=\frac{dy}{yz}=\frac{dz}{-xy}.$$ From $\displaystyle \frac{dx}{xz}=\frac{dy}{yz}$, we easily get $g_1(x,y,z)=y/x=c_1$. We are still looking for a second surface $g_2(x,y,z)=c_2$, such that $\nabla g_1\times \nabla g_2\neq (0,0,0)$, so that the general solution can be given as $F(g_1,g_2)=0,$ for $F\in C^1.$ This is where I am stuck: so far I have not figured out a standard method of solving systems, like the above - this is a case where one integration comes easily, while the rest do not seem to me that obvious.

$$x~u~u_x+y~u~u_y=-x~y,$$ under the condition that $u(x,1/x)=5,~x>0$.

You attempt is correct, but mut be completed :

$$\frac{dx}{xu}=\frac{dy}{yu}=\frac{du}{-xy}.$$

The equation of a first characteristic curve comes from $\frac{dx}{xu}=\frac{dy}{yu}$ $$\frac{y}{x}=c_1$$

The equation of a second characteristic curve comes from :

$\frac{dx}{xu}=\frac{dy}{yu}=\frac{ydx+xdy}{y(xu)+x(yu)}=\frac{d(xy)}{2xyu}=\frac{du}{-xy} \quad\to\quad d(xy)=-2udu$ $$u^2+xy=c_2$$ The general solution expressed on the form of implicit equation is : $$\Phi\left(\frac{y}{x}\;,\;u^2+xy\right)=0$$ where $\Phi$ is any differentiable function of two variables.

Solving the implicit equation for the second variable leads to the explicit form of general solution : $$u^2=-xy+F\left(\frac{y}{x}\right)$$ where $F$ is any differentiable function.

The condition $u(x,1/x)=5$ implies $u^2(x,1/x)=25=-x\frac{1}{x}+F\left(\frac{\frac{1}{x}}{x}\right) \quad\to\quad F\left(\frac{1}{x^2}\right)=26$

Thus, $F$ is a constant function equal to 26. $$u^2(x,y)=-xy+26$$

• thank you for the answer! One question: from $\Phi(y/x,u^2+xy)=0$, how do you derive that $u^2+xy=F(y/x)$? – Nikolaos Skout Sep 9 '16 at 9:51
• For an implicit equation of the form $\quad \Phi(X,Y)=0\quad$ where $\Phi$ is any function of two variables an equivalent relationship is that one variable is any fonction of the other variable. That is, it's equivalent to $\quad Y=F(X)\quad$ where $F$ is any function. Also it is equivalent to $\quad X=G(Y)\quad$ where $G$ is any function. But $F$ and $G$ are related (not independent). – JJacquelin Sep 9 '16 at 14:05
• Makes sense. Thank you! – Nikolaos Skout Sep 9 '16 at 14:08

Mentioned equation can be solved using automodel substitution $z=xy$ $$xuu_x+yuu_y+xy\to xuu_zy+yuu_zx+xy=2zuu_z+z\to\\ z(2uu_z+1)=0$$ Case of $z=0$ implies $x=0$ or $y=0$. Which is out of interest in given Cauchy problem. $$2uu_z+1=0\\ udu=-\frac{1}{2}dz\to \frac{u^2}{2}=C^*-\frac{1}{2}z\to u=\pm\sqrt{C-z}$$ Or in terms of initial variables: $$u=\pm\sqrt{C-xy}$$ where $C-const$.

Thus, for Cauchy problem $$u\left(x,\frac{1}{x}\right)=\pm\sqrt{C-1}=5\to \{+\}\text{ and }C=26\\ u(x,y)=\sqrt{26-xy}$$