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 $x = 0$


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 $$ x_0 = x_1 = x_2 = 0. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.