How to check if 2 PDEs have the same solution?

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I have a formula that generates PDEs:
$$
partial_t^k_0g(t)partial_x partial_t^n-k_0F(x,t)frac(n-1)!(n-k_0)!(k_0+1)!+partial_t^nF(x,t)=0
$$
I don't think any other information is necessary except that $g(t):mathbbRrightarrowmathbbR$, $F(x,t):mathbbR^2rightarrowmathbbR$ and $n, k_0inmathbbN^*$, with $n>k_0$, $k_0$ being fixed. (I'm giving the equation only to describe the problem)



I'm interested in methods of showing if some function $F$ satisfies the PDE for $n=k_0+1$, it will also satisfy the PDEs $forall n>k_0+1$.



I am not sure if this really is true or not, I just tried to compute for simple cases and prove this using brute force and I got stuck. Getting rid of $partial_t^k_0g(t)$ for 2 values of $n$ did not help me too much so I thought there must be other methods that I am not aware of for proving that 2 or more PDEs have the same solutions. And here comes the question:



Are they?




pde




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    up vote
    2
    down vote

    favorite
    1












    I have a formula that generates PDEs:
    $$
    partial_t^k_0g(t)partial_x partial_t^n-k_0F(x,t)frac(n-1)!(n-k_0)!(k_0+1)!+partial_t^nF(x,t)=0
    $$
    I don't think any other information is necessary except that $g(t):mathbbRrightarrowmathbbR$, $F(x,t):mathbbR^2rightarrowmathbbR$ and $n, k_0inmathbbN^*$, with $n>k_0$, $k_0$ being fixed. (I'm giving the equation only to describe the problem)



    I'm interested in methods of showing if some function $F$ satisfies the PDE for $n=k_0+1$, it will also satisfy the PDEs $forall n>k_0+1$.



    I am not sure if this really is true or not, I just tried to compute for simple cases and prove this using brute force and I got stuck. Getting rid of $partial_t^k_0g(t)$ for 2 values of $n$ did not help me too much so I thought there must be other methods that I am not aware of for proving that 2 or more PDEs have the same solutions. And here comes the question:



    Are they?




    pde




    share|cite|improve this question





















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      I have a formula that generates PDEs:
      $$
      partial_t^k_0g(t)partial_x partial_t^n-k_0F(x,t)frac(n-1)!(n-k_0)!(k_0+1)!+partial_t^nF(x,t)=0
      $$
      I don't think any other information is necessary except that $g(t):mathbbRrightarrowmathbbR$, $F(x,t):mathbbR^2rightarrowmathbbR$ and $n, k_0inmathbbN^*$, with $n>k_0$, $k_0$ being fixed. (I'm giving the equation only to describe the problem)



      I'm interested in methods of showing if some function $F$ satisfies the PDE for $n=k_0+1$, it will also satisfy the PDEs $forall n>k_0+1$.



      I am not sure if this really is true or not, I just tried to compute for simple cases and prove this using brute force and I got stuck. Getting rid of $partial_t^k_0g(t)$ for 2 values of $n$ did not help me too much so I thought there must be other methods that I am not aware of for proving that 2 or more PDEs have the same solutions. And here comes the question:



      Are they?




      pde




      share|cite|improve this question











      I have a formula that generates PDEs:
      $$
      partial_t^k_0g(t)partial_x partial_t^n-k_0F(x,t)frac(n-1)!(n-k_0)!(k_0+1)!+partial_t^nF(x,t)=0
      $$
      I don't think any other information is necessary except that $g(t):mathbbRrightarrowmathbbR$, $F(x,t):mathbbR^2rightarrowmathbbR$ and $n, k_0inmathbbN^*$, with $n>k_0$, $k_0$ being fixed. (I'm giving the equation only to describe the problem)



      I'm interested in methods of showing if some function $F$ satisfies the PDE for $n=k_0+1$, it will also satisfy the PDEs $forall n>k_0+1$.



      I am not sure if this really is true or not, I just tried to compute for simple cases and prove this using brute force and I got stuck. Getting rid of $partial_t^k_0g(t)$ for 2 values of $n$ did not help me too much so I thought there must be other methods that I am not aware of for proving that 2 or more PDEs have the same solutions. And here comes the question:



      Are they?





      pde





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Aug 3 at 17:08









      Victor Palea

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