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Solve of the following PDE using method of characteristics: $$(1)\qquad xu_x-xyu_y=u \qquad u(x,x)=x^2 e^x$$ $$(2)\qquad xu_x-xyu_y=u, \forall x,y$$

I assume that $(x,y)$ are functions of a parameter, say, $t$, i.e. $(x(t), y(t))$. In that case, the chain rule gives

$$\frac{du}{dt}=\frac{\partial u}{\partial x}.\frac{d x}{dt}+\frac{\partial u}{\partial y}.\frac{dy}{dt}=0$$

Comparing the equations, we need to solve the ODEs to find the characteristic curves.

$$\frac{du}{dt}=u,\quad\frac{d x}{dt}=x \qquad\&\quad\frac{d y}{dt}=-xy$$

For second one, How can get the constant values of those odes without any initial condition? Do I am missing something? Here is an existing solution on M.SE but couldn't manage how did he/she bring initial conditions and write $u(x,y)=f(y_0)e^t$:

$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$ $\dfrac{dy}{dt}=-xy=-e^ty$ , letting $y(0)=y_0$ , we have $y=y_0e^{1-e^t}=y_0e^{1-x}$ $\dfrac{du}{dt}=u$ , letting $u(0)=f(y_0)$ , we have $u(x,y)=f(y_0)e^t=f(e^{x-1}y)x=F(e^xy)x$

Any hint will be appreciated.

Thanks in Advance.

Lutz Lehmann
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falamiw
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  • As you know $x(t)$, the last equation is separable. You get the exponential of an exponential, but it is still fully solvable. – Lutz Lehmann May 18 '22 at 07:05
  • I manage to solve the last equation. I forgot to plugging the $x(t)$ value in that ode. Could you help me with the $(2)$ one? I googled a lot but couldn't manage to understand without initial solution who to construct the solution. I understand there will be one arbitrary function, but how to write that. @LutzLehmann – falamiw May 18 '22 at 17:39

1 Answers1

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If you eliminate $t$ for $x$, you get $u=c_1x$ and $y=c_2e^{-x}$. The idea now is that the family of characteristics in a solution surface only has one parameter, so the constants have a dependency. For almost all solutions one can set $$ c_1=f(c_2)\implies u=f(ye^{x})x. $$

Lutz Lehmann
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  • The family of characteristics in a solution surface only has one parameter, because of our solution contain only one constant parameter $c_1$? @LutzLehmann – falamiw May 18 '22 at 18:07
  • No, you have a 2D surface $z=u(x,y)$. This is generated by a family of curves. So the factor space is one-dimensional. So there can not be 2 independent parameters. The dependency of parameters can, generically and at least locally, be resolved as a functional dependency. – Lutz Lehmann May 18 '22 at 18:23