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I was trying to solve the following PDE: $$\ln(x_2)\partial_1U+x_2U\partial_2U-U=0$$ with the initial condition $$U(t+1,e)=1$$ I managed to get to the following system: $$\begin{cases} \dot{x}_1(s)=ln(x_2)\\\dot{x}_2(s)=Ux_2 \\\dot{U}(s)=p_1\ln(x_2)+p_2x_2U \\ \dot{p_1}(s)=p_1-p_2p_1x_2 \\ \dot{p_2}(s)=p_2-\frac{p_1}{x_2}-Up_2-p^2_2x_2 \end{cases} $$ where $x_1>0,\quad x_2>0$ and $p_1=\partial_{x_1}U$ and $p_2=\partial_{x_2}U$.

But, I am not able to solve this system of ODEs. Please help!

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  • $\begingroup$ Hi, from the first equation, please precise what are initially in your problem (how they are defined) $x_2$, $U$, and what are the other function that appear after namely $x_1$, $p_1$, p_2$, where do those equations come from ? $\endgroup$
    – EDX
    Commented Jun 23 at 19:27
  • $\begingroup$ $x_2>0, x_1>0$ and $p_1=\partial_{x_1}U$ and $p_2=\partial_{x_2}U$ $\endgroup$ Commented Jun 23 at 20:22
  • $\begingroup$ Ok put it all in the question so it will be perectly clear with all parameter, functions, initial conditions to help you through this problem. $\endgroup$
    – EDX
    Commented Jun 23 at 20:23
  • $\begingroup$ See my answer in : math.stackexchange.com/questions/4936865/… . $\endgroup$
    – JJacquelin
    Commented Jun 24 at 8:36

2 Answers 2

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The characteristic ODEs are

$$\frac{dx_{1}}{\ln x_{2}} = \frac{dx_{2}}{x_{2} u} = \frac{du}{u}$$

The last equality implies ln u $$u - \ln x_{2} = C_{1}$$

The first and third ratios imply

\begin{align} d x_{1} &= \frac{(u - C_{1}) du}{u} \\ \implies u - C_{1} \ln u - x_{1} &= C_{2} \\ \implies u - C_{1} + C_{1} - C_{1} \ln u - x_{1} &= C_{2} \\ \implies \ln x_{2} + (u - \ln x_{2}) (1 - \ln u) - x_{1} &= C_{2} \\ \end{align}

and hence

$$\ln x_{2} + (u - \ln x_{2})(1 - \ln u) - x_{1} = f(u - \ln x_{2})$$

for some arbitrary differentiable function $f$. You can check this satisfies the PDE by differentiation

\begin{align} u_{x_{1}}(1 - \ln u) - (u - \ln x_{2}) u_{x_{1}} u^{-1} - 1 &= u_{x_{1}} f' \\ \implies u_{x_{1}} \left(1 - \ln u - (u - \ln x_{2}) u^{-1} - f' \right) &= 1 \\ \implies u_{x_{1}} &= \frac{1}{\left(1 - \ln u - (u - \ln x_{2}) u^{-1} - f' \right)} \\ x_{2}^{-1} + (u_{x_{2}} - x_{2}^{-1})(1 - \ln u) - (u - \ln x_{2}) u_{x_{2}} u^{-1} &= (u_{x_{2}} - x_{2}^{-1}) f' \\ \implies u_{x_{2}} \left(1 - \ln u - (u - \ln x_{2}) u^{-1} - f' \right) &= - x_{2}^{-1} (\ln u + f') \\ \implies u_{x_{2}} &= \frac{- x_{2}^{-1} (\ln u + f')}{\left(1 - \ln u - (u - \ln x_{2}) u^{-1} - f' \right)} \end{align}

so that

\begin{align} \implies \ln x_{2} u_{x_{1}} + x_{2} u u_{x_{2}} &= \frac{\ln x_{2} - u (\ln u + f')}{\left(1 - \ln u - (u - \ln x_{2}) u^{-1} - f' \right)} \\ &= \frac{u (\ln x_{2} - u (\ln u + f'))}{\left(u - u \ln u - (u - \ln x_{2}) - u f' \right)} \\ &= \frac{u (- u \ln u + \ln x_{2} - u f')}{\left(- u \ln u + \ln x_{2} - u f' \right)} \\ &= u \end{align}

I leave the domain of convergence and initial condition to you (hint for initial condition; the solution to the PDE can also be written as $$u - \ln x_{2} = g(\ln x_{2} + (u - \ln x_{2})(1 - \ln u) - x_{1})$$ for some arbitrary differentiable function $g$, then note that evaluating both sides at $(x_{1}, x_{2}, u) = (t + 1, e, 1)$ makes the LHS vanish identically).

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The fact that $\partial_tU(1+t,e)=0$ seems to imply that $U(x_1,x_2)$ does not depend on $x_1$, which leads us to try the ansatz $U(x_1,x_2)=f(x_2)$. Plugging it into the PDE $$ \ln(x_2)\partial_1U+x_2U\partial_2U-U=0, \tag{1} $$ we get $$ (x_2f'(x_2)-1)f(x_2)=0 \implies f(x_2)=0\quad\text{or}\quad x_2f'(x_2)-1=0. \tag{2} $$ The first alternative must be discarded, as it does not satisfy the condition $$ U(1+t,e)=f(e)=1. \tag{3} $$ The second alternative, on the other hand, yields $$ f'(x_2)=\frac{1}{x_2} \implies f(x_2)=\ln(x_2)+C, \tag{4} $$ which satisfies $(3)$ if $C=0$. Therefore, a solution to $(1)$ (I make no claim that the solution is unique) is $$ U(x_1,x_2)=\ln(x_2). \tag{5} $$

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