Initial-value problem for non-linear partial differential equation y x 2 = k / y t 2 − 1For this problem, y is a function of two variables: one space variable x and one time variable t.k&gt;0 is some constant.And x takes is value in the interval [0,1] and t ≥ 0At the initial time, y follows a parabolic profile, like y ( x , 0 ) = 1 − ( x − 1 2 ) 2 Finally, y satisfies this PDE: ( ∂ y ∂ x ) 2 = k ( ∂ y ∂ t ) 2 − 1.Does anyone have an idea how to solve this problem (and find the expression of y(x,t)) ?About: The problem arise in physics, when studying the temporal shift of a front of iron particles in a magnetic field.

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Initial-value problem for non-linear partial differential equation        y    x    2    =  k      /        y    t    2    −  1For this problem, y is a function of two variables: one space variable x and one time variable t.k&amp;gt;0 is some constant.And x takes is value in the interval [0,1] and   t  ≥  0At the initial time, y follows a parabolic profile, like   y  (  x  ,  0  )  =  1  −  (  x  −      1    2        )    2  Finally, y satisfies this PDE:            (                        ∂          y                          ∂          x                    )        2    =      k                  (                              ∂            y                                ∂            t                          )            2        −  1.Does anyone have an idea how to solve this problem (and find the expression of y(x,t)) ?About: The problem arise in physics, when studying the temporal shift of a front of iron particles in a magnetic field.

Marshall Flowers · Accepted Answer

Your equation can be written as follows:  F  (  p  ,  q  ,  x  ,  t  ,  y  )  =  (      p    2    +  1  )      q    2    −  k  =  0  ,    p  =      y    x    ,    q  =      y    t    ,and hence:                              F          p                                    =        2        p                  q          2                ,                                      F          q                                    =        2        (        1        +                  p          2                )        q        ,                                      F          t                =                  F          x                =                  F          y                                    =        0        ,            so the Lagrange-Charpit equations read:                    d            x              2      p              q        2              =                    d            t              2      (      1      +              p        2            )      q        =                    d            y              2              p        2                    q        2            +      2      (      1      +              p        2            )              q        2              =  −                    d            p        0    =  −                    d            q        0    ,which tells you that dp=dq=0 and, thus, p=A (or q=B) constant. Since p=yx we have that   y  (  x  ,  t  )  =  A  x  +  f  (  t  ). Plug this into the original PDE to find:  f  (  t  )  =  ±                    k            t              1      +              A        2              +  C  ,    C  ∈      R    ,so the solution is finally given by:      y    (    x    ,    t    )    =    A    x    ±                            k                t                    1        +                  A          2                      +    C    ,this is known as a complete integral of your PDE. It remains to set the initial condition y(x,0). Can you take it from here?

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