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The non-linear PDE that I would like to solve is:

\begin{align} u^2_{x_1} - u^2_{x_2} &= 2u \label{DGL1} \\ u(0,x_2) &= (1+x_2)^2 \label{DGL2}. \end{align}

We define the notion

\begin{align*} z(s) &= u(x(s))\\ p(s) &= (p_1(s),p_2(s)) = (u_{x_1}(x(s)), u_{x_2}(x(s)))). \end{align*}

With this notion we have

\begin{align*} z(0) = u(x(0)) = u(0,x_0) = (1+x_0)^2. \end{align*}

So we can transfrom the PDE into this parameterized PDE: \begin{equation*} F(p,z,x) = p_1^2 - p_2^2 - 2z. \end{equation*}

We can solve this with the method of characteristics: \begin{align*} \dot{p}(s) &= - \begin{pmatrix} 0\\ 0 \end{pmatrix} -(-2)\begin{pmatrix} p_1\\ p_2 \end{pmatrix} = \begin{pmatrix} 2p_1, \, 2p_2 \end{pmatrix} \\ \dot{z}(s) &= \begin{pmatrix} 2p_1, \, -2p_2 \end{pmatrix} \begin{pmatrix} p_1\\ p_2 \end{pmatrix} = 2p_1^2 - 2p_2^2 = 2\left(p_1^2 - p_2^2\right). \end{align*}

Since $\dot{x}_1 = 2p_1 = \dot{p}_1$ is valid with $x_1(s=0)=0$, we have \begin{align*} &x_1(s) = p_{10}(e^{2s}-1)\\ \Rightarrow \quad &\dot{x}_1 = 2p_{10}e^{2s} = 2p_1 = \dot{p}_1\\ \Rightarrow \quad &p_1 = p_{10}e^{2s}. \end{align*}

Since $\dot{x}_2 = -2p_2 = - \dot{p}_2$ is valid, we have $x_2(s=0)=x_0$, we have \begin{align*} &x_2(s) = -p_{20}\left(e^{2s}-1\right) + x_0\\ \Rightarrow \quad &\dot{x}_2 = -2p_{20}e^{2s} = -2p_2 = -\dot{p}_2\\ \Rightarrow \quad &p_2 = p_{20}e^{2s}. \end{align*}

We can use this now for $\dot{z}(s)$ and $z(s)$ respectively: \begin{align*} \dot{z}(s) &= 2\left(p_1^2 - p_2^2\right)\\ &= 2\left((p_{10}e^{2s})^2 - (p_{20}e^{2s})^2\right)\\ &= 2p_{10}^2e^{4s} - 2p_{20}^2e^{4s}\\ \Rightarrow \quad z(s) &= \tfrac{1}{2}p_{10}^2e^{4s} - \tfrac{1}{2}p_{20}^2e^{4s} + z_0. \end{align*}

We have $z_0 = z(0) = u(x(0)) = u(0,x_0) = (1+x_0)^2$.

Now let us calculate $p(0)=(p_{10},p_{20})$: \begin{align*} &p_{20} = p_2(0) = u_{x_2}(x(0)) = u_{x_2}((0,x_0)) = 2(1+x_0) \quad \text{and}\\ &p_{10}^2 - p_{20}^2 = u_{x_1}^2((0,x_0)) - u_{x_2}^2((0,x_0)) \underbrace{=}_{PDE} 2u(0,x_0) = 2z(x(0)) = 2z_0 = 2(1+x_0)^2\\ \Rightarrow \quad &p_{10}^2 - p_{20}^2 = p_{10}^2 - 4(1+x_0)^2 = 2(1+x_0)^2\\ \Leftrightarrow \quad &p_{10}^2 = 6(1+x_0)^2\\ \Leftrightarrow \quad &p_{10} = \sqrt{6}(1+x_0). \end{align*}

With $p_{10}$ and $p_{20}$, we can calculate $x(s)$, $z(s)$ und $p(s)$:

\begin{align*} x_1(s) &= \sqrt{6}(1+x_0)(e^{2s}-1)\\ x_2(s) &= -2(1+x_0)(e^{2s}-1) + x_0\\ z(s) &= \tfrac{1}{2}6(1+x_0)^2e^{4s} - \tfrac{1}{2}4(1+x_0)^2e^{4s} + (1+x_0)^2\\ &=(1+x_0)^2(e^{4s}+1)\\ p_1(s) &= \sqrt{6}(1+x_0)e^{2s}\\ p_2(s) &= 2(1+x_0)e^{2s}. \end{align*}

Let $(x,y) \in \Omega := \mathbb{R}^2$ be arbitrary and select $s$ u and $x_0$: \begin{align*} (x,y) = (x_1(s), \, x_2(s)) = (\sqrt{6}(1+x_0)(e^{2s}-1), \, -2(1+x_0)(e^{2s}-1) + x_0). \end{align*} This is a linear equation system and can be solved by $x_0$ and $e^{2s}$: \begin{align*} x_0 = y+\frac{2x}{\sqrt{6}}, \qquad e^{2s} = \frac{x}{\left(y+1+\frac{2x}{\sqrt{6}}\right)\sqrt{6}}+1. \end{align*}

Finally, we can solve the desired function $u(x,y)$: \begin{align*} u(x,y) &= u(x_1(s), x_2(s)) = z(s)\\ &= (1+x_0)^2\left((e^{2s}\right)^2+1)\\ &= \left(1+y+\frac{2x}{\sqrt{6}}\right)^2 \left(\left(\frac{x}{ (y+1+ \frac{2x}{\sqrt{6}} ) \sqrt{6} }+1\right)^2+1\right). \end{align*}

However, if I want to verify this $u(x,y)$ with the PDEs at the beginning, it doesn't add up! So I must have done something wrong. Does anyone have an idea?

Please also comment on anything thats strange.

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    $\begingroup$ The difficulty is due to an ambiguity in the wording of the question. What is the correst condition ? Is it $u(0,x_2) = (1+x_2^2)$ or is it $u(0,x_0) = (1+x_0)^2$ ? Both are contradictory at the point $(x_1=0\:,\:x_2=x_0)$. $\endgroup$
    – JJacquelin
    May 4, 2020 at 9:03
  • $\begingroup$ The correction condition is $u(0,x_2) = (1+x_2)^2$. If we use the notion $z(s) = u(x(s))$, then $z(0) = u(x(0)) = u(0,x_0) = (1+x_2)^2$. Thanks, I will be more clear about this. $\endgroup$
    – MJimitater
    May 4, 2020 at 11:51

1 Answer 1

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I found my mistake: $z(0)$ isn't equal to $z_0$.

It must be \begin{align*} z(s) &= \tfrac{1}{2}p_{10}^2\left(e^{4s}-1\right) - \tfrac{1}{2}p_{20}^2\left(e^{4s}-1\right) + z_0, \end{align*}

so that $z_0 = z(0) = u(x(0)) = u(0,x_0) = (1+x_0)^2$ is right.

This leads to the right solution

\begin{align*} u(x,y) &= u(x_1(s), x_2(s)) = z(s)\\ &= (1+x_0)^2\left(e^{2s}\right)^2\\ &= \left(1+y+\frac{2x}{\sqrt{6}}\right)^2 \left(\frac{x}{ (y+1+ \frac{2x}{\sqrt{6}} ) \sqrt{6} }+1\right)^2\\ &\underbrace{=}_{WolframAlpha} \left( \frac{\sqrt{6}x}{2} + y +1 \right)^2. \end{align*}

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