PDEMOC4

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u_{x}+2u_{y}=u^{2},u(x,0)=h(x)\,

The parameterized curve at t=0\, is \Gamma (s,0,h(s))\,. This comes from the symbols x,0,h(x)\, in the IC.

The characteristics are {\frac  {dx}{dt}}=1\,, {\frac  {dy}{dt}}=2\,, {\frac  {dz}{dt}}=z^{2}\,.


These differential equations give x(s,t)=t+c_{1}(s)\,, y(s,t)=2t+c_{2}(s)\,, -z(s,t)^{{-1}}=t-c_{3}(s)^{{-1}}\,.

At t=0\,, x=c_{1}(s)=s\,, y=c_{2}(s)=0\,, and z=-c_{3}(s)^{{-1}}=h(s)\,.

So the solutions are x=t+s\,, y=2t\,, and z=-1/(t-h(s)^{{-1}})\,.


Therefore s=x-t\,, t=y/2\,, and z=u(x,y)={\frac  {h(x-y/2)}{1-{\frac  {1}{2}}yh(x-y/2)}}\,.


Main Page : Partial Differential Equations : Method of Characteristics