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Let $\phi(x,t)$ be smooth function.

Let $\zeta= x-t$ and $\eta= x+t$. $u=\phi_\zeta$, $v= \phi_\eta$.

Let $u$, $v$ satisfies following equations:

1- $$u_\eta- v_\zeta= 0$$ $$v^2u_\zeta-(1+2uv)u_\eta+ u^2v_\eta=0$$

The roles of dependent and independent variables are then interchanged to give

2-$$\zeta_v-\eta_u=0$$ $$v^2\eta_v+(1+2uv)\zeta_v+u^2\zeta_u=0$$

Can someone please help me to find the required transformation for getting set 2 from set-1. Is this an example of hodographic transformation. Can someone be precise in explaining this transformation.

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2 Answers 2

up vote 7 down vote accepted

Another way to understand this 'transformation' is to think in terms of differential forms. Let $(u,v,\eta,\zeta)$ be coordinates on $\mathbb{R}^4$ and consider the pair of $2$-forms $$ \Upsilon_1 = du\wedge d\zeta + dv\wedge d\eta \quad\text{and}\quad \Upsilon_2 = v^2 du\wedge d\eta + (1{+}2uv)\ du\wedge d\zeta - u^2 dv\wedge d\zeta. $$ A surface $S\subset\mathbb{R}^4$ that can be written as a graph in the form $\bigl(u,v,\eta(u,v),\zeta(u,v)\bigr)$ is an integral surface of these two $2$-forms (i.e., they pull back to $S$ to become $0$) if and only if the functions $\bigl(\eta(u,v),\zeta(u,v)\bigr)$ satisfy your second set of equations. If the surface can also be written as a graph in the form $\bigl(u(\eta,\zeta),v(\eta,\zeta),\eta,\zeta\bigr)$, then it is an integral surface of these two $2$-forms if and only if $\bigl(u(\eta,\zeta),v(\eta,\zeta)\bigr)$ satisfy your first set of equations. In that sense, the two equation systems are equivalent. (Note that the forms $\Upsilon_i$ are linear in $\eta$ and $\zeta$, so one can 'add' solution surfaces to obtain new solution surfaces when they satisfy $du\wedge dv\not=0$.

Thus, each set of equations defines the 'same' set of integral surfaces of the pair of $2$-forms. In some sense, you should regard an integral surface of these $2$-forms as a 'generalized' solution to either set of equations.

Also, looked at in this way, you see that there is no actual `transformation' going on, in the sense that there is no mapping. You are merely changing which variables you want to regard as 'independent', so it's really a change in the way you are describing the same class of surfaces.

Remark: While you didn't ask, I should point out that the type analysis of this system shows that it's elliptic in the region where $1+4uv<0$ and hyperbolic in the region where $1+4uv>0$.

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@Robert Bryant sir, Thanks for the giving a nice interpretation. –  zapkm Apr 5 '12 at 2:14

Forget about $x,t$. Consider a $C^1$ mapping $\phi:(\zeta,\eta)\mapsto (u,v)$. Locally if $|d\phi|\neq 0$ we can invert it. Let the inverse be $\psi: (u,v)\mapsto (\zeta,\eta)$, so $\psi\circ\phi(\zeta,\eta) = (\zeta,\eta)$ and $\phi\circ\psi(u,v) = (u,v)$.

An elementary computation shows that the Jacobian matrices of $\phi$ and $\psi$ are inverses. That is, evaluated at a fixed $u,v,\zeta,\eta$ where $(u,v) = \phi(\zeta,\eta)$,

$$ \begin{pmatrix} \partial_1\psi^1 & \partial_1\psi^2 \newline \partial_2\psi^1 & \partial_2\psi^2\end{pmatrix}^{-1} = \begin{pmatrix} \partial_1\phi^1 & \partial_1\phi^2 \newline \partial_2\phi^1 & \partial_2 \phi^2\end{pmatrix} $$

For disambiguation: write the function $\phi = (U,V)(\zeta,\eta)$ and the function $\psi = (Z,N)(u,v)$. The above implies

$$ \begin{pmatrix} U_\zeta & V_\zeta \newline U_\eta & V_\eta\end{pmatrix} = \frac{1}{Z_u N_v - Z_v N_u} \begin{pmatrix} N_v & - N_u \newline -Z_v & Z_u\end{pmatrix} $$

Your system (1) gives $U_\eta = V_\zeta$ which from the matrix inequality immediately implies that $Z_v = N_u$, which is the first equation of system (2).

Using that $Z_uN_v - Z_vN_u = |d\psi| \neq 0$, the equality of the matrix components also gives you that the second equation of (2) is obtained from the second equation of (1) by the replacement $U_\zeta \to N_v$, $V_\zeta \to -N_u$, $U_\eta \to -Z_v$ and $V_\eta \to Z_u$.

Notice that this works because (a) we have a 2-by-2 system and (b) the system (1) is quasilinear. If it were genuinely nonlinear, we cannot "factor" out from the equation the common factor of $|d\psi|$ to end up with a purely linear equation.

And yes, this is an example of the hodographic transformation. In general the same procedure works for any quasilinear system of first order partial differential equations with 2 dependent and 2 independent variables.

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1  
For a scalar first order quasilinear PDE in arbitrary dimensions, the method of characteristics bears some philosophical similarity to the hodograph methods. –  Willie Wong Apr 4 '12 at 12:23
    
@Wilie wong, Thanks a lot for explaining. –  zapkm Apr 5 '12 at 2:12

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