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I have the following partial differential equation:

$u_t = u_xu_y$

I know that the solution can be formed via power series. I want to find a solution of degree $2$ that satisfies an initial condition $u(0,x,y) = x+y-2x^2$.

I understand the strategy of the solution would be to take partial derivative combinations of $u$ up to the second degree. So these would involve

$u_t,u_x,u_y,u_{tt},u_{xx},u_{yy},u_{tx},u_{ty},u_{xy},u_{xt},u_{yx},u_{yt}$

For a grand total of 12 terms in the series. Then, the function would need to be evaluated at (0,0,0)...(0,0,1)...(0,1,0)..etc to get the coefficients.

 

After that, I can construct coefficients of partial derivative combinations to form a series and plug in my IVP condition.

 

However...$what$ am I differentiating exactly? Where is $u$? Do I just rearrange the terms $u_t$, $u_x$, and $u_y$ to get what I need?

Rod Smith
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  • Try to make a polynomial where the monomials are the same as the indexes of the partial derivatives you listed – N74 Apr 26 '16 at 20:16

1 Answers1

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Assume a solution of the form $$ u(t, x, y) = c_{000} + c_{100}t + c_{010}x + c_{001}y + c_{200}t^{2} + c_{020}x^{2} + c_{002}y^{2} + c_{110}tx + c_{011}xy + c_{101}ty $$ Since $u(0, x, y) = x + y - 2x^{2}$, we find that $$ u(t, x, y) = c_{100}t + x + y + c_{200}t^{2} - 2x^{2} + c_{110}tx + c_{101}ty $$ Write out partial derivatives: \begin{align} \partial_{t}u &= c_{100} + 2c_{200}t + c_{110}x + c_{101}y \\ \partial_{x}u &= 1 - 4x + c_{110}t \\ \partial_{y}u &= 1 + c_{101}t \end{align} Hence \begin{align} c_{100} + 2c_{200}t + c_{110}x + c_{101}y &= 1 + (c_{101}+c_{110})t -4x -4c_{101}xt + c_{110}c_{101}t^{2} \end{align} and as such $c_{101} = 0$ and $c_{100} = 1$, $c_{110} = -4$, and $c_{200} = -2$. \begin{align} \boxed{u(t, x, y) = t + x + y -2t^2 -2x^2 -4tx} \end{align} I don't know why you need all those other terms.

user14717
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