Please who can say: why the equation <br clear="all" /><table border="0" width="100%"><tr><td> <table align="center" cellspacing="0" cellpadding="0"><tr><td nowrap="nowrap" align="center"> </td><td nowrap="nowrap" align="center"> *<font face="symbol">¶</font>u <div class="hrcomp"><hr noshade="noshade" size="1"/></div><font face="symbol">¶</font>t
</td><td nowrap="nowrap" align="center"> + a</td><td nowrap="nowrap" align="center"> *<font face="symbol">¶</font>u <div class="hrcomp"><hr noshade="noshade" size="1"/></div><font face="symbol">¶</font>x
</td><td nowrap="nowrap" align="center"> =0</td></tr></table> </td></tr></table> is named hyperbolic?

Please consult any book on partial differential equations. The answer will be on the first ten pages.

Yours,

Philipp

<title>Books consult: an eqn is hyperbolic</title>

<div class="p"> </div> Books consult: an eqn is hyperbolic if a₁₁a₂₂<font face="symbol">-</font>a₁₂² < 0 but here we have a₁₁=a₂₂=a₁₂ <font face="symbol">=</font> 0. Equations with a₁₁a₂₂<font face="symbol">-</font>a₁₂²=0 are called parabolic.

The definition you are refering to applies to second order equations. For first order equations you can use the following definition:

PDE u_t + A u_x = 0

where A could be a matrix, is said to be hyperbolic if A has only real eigenvalues and a complete set of linearly independent eigenvectors. The scalar equation

u_t + au_x=0

is trivially hyperbolic by this definition. The distinct characteristic of hyperbolic pde is that they admit wave-like solutions.

You can write the second order pde as a system of first order pde and try to check the above definition which will lead to the same inequality.

Thank you. I forgot that there is a separate classification for the first-order equations.