Let $x_i=x_i(t,s),i=0,1,2$ be functions of $t,s$ and consider the system of differential equations \begin{cases} \frac{\partial x_0}{\partial t}=0,\\ \frac{\partial x_1}{\partial t}=x_0,\\ \frac{\partial x_2}{\partial t}=2x_1,\\ \frac{\partial x_0}{\partial s}=2x_1,\\ \frac{\partial x_1}{\partial s}=x_2,\\ \frac{\partial x_2}{\partial s}=0, \end{cases} How to solve it? My attemt. First of all I have solved the first 3 equations and get \begin{align} &x_{{0}}=C_1 \left( s \right) ,\\ &x_{{1}} =C_1 \left( s \right) t+C_2(s),\\ &x_{{2}} =C_1 \left( s \right) {t}^{2}+2\,C_2 \left( s \right) t+C_3 \left( s \right), \end{align}

and substitute into the last 3 equation

$$ \begin{cases} \displaystyle {\frac { d}{{ d}s}}C_1 \left( s \right) =2\,C_1 \left( s \right) t+2\,C_2 \left( s \right),\\ \\ \displaystyle t {\frac { d}{{ d}s}}C_1 \left( s \right)+{\frac { d}{{ d}s}}C_2 \left( s \right) =C_1 \left( s \right) {t}^{2}+2\,C_2 \left( s \right) t+C_3 \left( s \right) ,\\ \\ \displaystyle {t}^{2}\, {\frac { d}{{ d}s}}C_1 \left( s \right) +2\,t {\frac { d}{{ d}s} }C_2 \left( s \right) +{\frac { d}{{ d}s}}C_3 \left( s \right) =0, \end{cases} $$ or normalize it: $$ \begin{cases} \displaystyle{\frac { d}{{ d}s}}C_1 \left( s \right)=2\,C_1 \left( s \right) t+2\,C_2 \left( s \right) ,\\ \\ \displaystyle{\frac { d}{{ d}s}}C_2 \left( s \right)=-C_1 \left( s \right) {t}^{2}+C_2 \left( s \right) ,\\ \\ \displaystyle{\frac { d}{{ d}s}}C_3 \left( s \right)=-2\,C_2 \left( s \right) {t}^{ 2}-2\,C_2 \left( s \right) t \end{cases} $$

Then I solve the system for $C_1(s),C_2(s),C_3(s)$ and get

\begin{align*} &C_1 \left( s \right) =-{\frac {2\,C_1'\,{s}^{2}{t}^{2}+4\,C_2'\,s{t}^{2}+2\,C_1'\,st+4\,C_3'\,{ t}^{2}+2\,C_2'\,t+C_1'}{4{t}^{3}}},\\ &C_2 \left( s \right) =\frac{1}{2}\,C_1'\,{s}^{2}+C_2'\,s+C_3', \\ &C_3 \left( s \right) =-\,{\frac {2\,C_1'\,{s}^{2}{t}^{2}+4\,C_2\,s{t}^{2}-2\,C_1'\,st+4\,C_3'\,{t}^{2}-2\,C_2'\,t+C_1'}{4t}}. \end{align*} Here $C_1',C_2',C_3'$ are constants.

Then I substitute it into the above expression for $x_0,x_1,x_2$ and get \begin{align*} &x_{{0}} \left( t,s \right) ={\frac { \left( -2\,C_1'\,{s}^{2}-4\,C_2'\,s-4\,C_3' \right) {t}^{2}+ \left( -2\,C_1'\,s-2\,C_2' \right) t-C_1'}{4{t}^{3}}}, \\&x_{{1} } \left( t,s \right) ={\frac { \left( -2\,C_1'\,s-2\,{C_2'} \right) t-C_1'}{4{t}^{2}}},\\ &x_{{2}} \left( t,s \right) =- \,{\frac {C_1'}{2t}} \end{align*}

But obviously it is not solutions of the initial system!

Where is my mistake?

Note, the system have solution, for example $x_1^2-x_0 x_2$ is its first integral.

  • $\begingroup$ I guess you missed something, since $x_0$ can be expressed only in terms of $s$ $x_0=2C_1(s)t+2C_2(s)$ would only be true if $C_1(s)$ is nil and this deadlocks the system, or maybe i did. $\endgroup$ – Abr001am Aug 23 '19 at 16:54
  • $\begingroup$ no, I have cheked all my steps $\endgroup$ – Leox Aug 23 '19 at 17:14
  • $\begingroup$ wel tbh this is a mystery!! $\endgroup$ – Abr001am Aug 23 '19 at 19:18
  • $\begingroup$ leox, notice that the derivate of $x_1$ in terms of $t$ gives $x_0$ then in term of $s$ gives itself $2x_1$, which kind of function you derivate it twice it yields to itself ? we are not dealing with polynomials here, its either sine or exponential. $\endgroup$ – Abr001am Aug 24 '19 at 10:28

$$ \frac{\partial x}{\partial t}=M_1 x\Rightarrow x = C_1e^{M_1 t}+\phi(s)\\ \frac{\partial x}{\partial s}=M_2 x\Rightarrow x = C_2e^{M_2 s}+\psi(t)\\ $$


$$ x = C_1e^{M_1 t}+C_2e^{M_2 s} $$

but for this solution we have

$$ \frac{\partial x}{\partial t}-M_1 x = -M_1C_2e^{M_2 s}=0\Rightarrow C_2 = 0\\ \frac{\partial x}{\partial s}-M_2 x = -M_1C_1e^{M_1 t}=0\Rightarrow C_1 = 0\\ $$

hence the solution is $x = 0$

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The first and last equations give $x_0(t,s) = x_0(s)$ and $x_2(t,s) = x_2(t)$. Therefore, the system becomes \begin{aligned} \partial_{t} x_1 &= x_0 \\ \partial_{t} x_2 &= 2x_1 \\ \tfrac{\text d}{\text d s} x_0 &= 2x_1 \\ \partial_{s} x_1 &= x_2 \end{aligned} Applying $\partial_{t}$ to the third line yields $0 = 2\partial_{t} x_1$ and the first line gives $\partial_{t} x_1 = x_0$. Hence we know $x_0=0$ and $x_1(t,s) = x_1(s)$. The system becomes \begin{aligned} \partial_{t} x_2 &= 2x_1 \\ 0 &= 2x_1 \\ \tfrac{\text d}{\text d s} x_1 &= x_2 \end{aligned} Since $x_1=0$ as well, we end up with $x_2=0$. The solution must be $$ x_0 = x_1 = x_2 = 0. $$

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