PDEMOC7

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

The initial curve will be parameterized by \Gamma (s,0,s^{2})\,.

The characteristics are

{\frac  {dx}{dt}}=x,{\frac  {dy}{dt}}=1,{\frac  {dz}{dt}}=y\,

For x\,,

{\frac  {dx}{dt}}=x\,

x^{{-1}}dx=dt\,

\ln x=t+c_{1}(s)\,

x(s,t)=c_{4}(s)e^{t}\,

x(s,0)=c_{4}(s)=s\,

So x(s,t)=se^{t}\,


For y\,,

{\frac  {dy}{dt}}=1

y(s,t)=t+c_{2}(s)\,

y(s,0)=c_{2}(s)=0\,

So y(s,t)=t\,


For z\,,

{\frac  {dz}{dt}}=y=t\,

z(s,t)={\frac  {1}{2}}t^{2}+c_{s}(3)\,

z(s,0)=c_{s}(0)=s^{2}\,

z(s,t)={\frac  {1}{2}}t^{2}+s^{2}\,


The solution is

z(s,t)=u(x,y)={\frac  {1}{2}}y^{2}+(xe^{{-y}})^{2}\,


Main Page : Partial Differential Equations : Method of Characteristics