PDEMOC5

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

The characteristics are:

{\frac  {dx}{dt}}=1\,, {\frac  {dy}{dt}}=x\,, {\frac  {dz}{dt}}=z^{2}\,

The intial curve was not given so assume \Gamma =(0,0,z_{0})\,.


The solution to the first DE is:

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

x(s,0)=c_{1}(s)=0\,

x=t\,


The second DE is:

{\frac  {dy}{dt}}=x=t\,

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

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

y={\frac  {1}{2}}t^{2}={\frac  {1}{2}}x^{2}\,


The solution to the third DE is:

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

z(s,0)={\frac  {-1}{c_{3}(s)}}=z_{0}\,

c_{3}(s)=-z_{0}^{{-1}}\,

z={\frac  {z_{0}}{1-z_{0}x}}\,


We now must find two new functions of x,y,z that are constant along gamma.

Let \phi (x,y,z)=y-{\frac  {1}{2}}x^{2}\, and \psi (x,y,z)=z^{{-1}}+x\,.

So the solution is y-{\frac  {1}{2}}x^{2}=f(x+{\frac  {1}{z}})\, where f\, is an arbitrary function as long as f({\frac  {1}{z_{0}}})=0\, because those are the values along \Gamma \,.


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