PDEMOC13

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Solve the initial value problem a(u)u_{x}+u_{y}=0\, with u(x,0)=h(x)\, and show the solution becomes singular for some y>0\, unless a(h(s))\, is a nondecreasing function of s\,.


The solution is u(x,y)=h(x-a(u)y)\,.

Suppose x=s\, at y=0\, so that u(s,0)=h(s)\,.

Let u=u_{x}\, along a characteristic line \Gamma _{\xi }\,.

{\frac  {du}{dy}}={\frac  {\partial u_{x}}{\partial x}}x'(y)+{\frac  {\partial u_{x}}{\partial y}}=u_{{xx}}a(u)+u_{{xy}}\,

From the equation u_{y}+a(u)u_{x}=0\,

we get u_{{xy}}+a(u)u_{{xx}}+a'(u)u_{x}u_{x}=0\,

\implies {\frac  {dw}{dt}}=-a'(u)u_{x}^{2}=-a'(h(s))w^{2}\,

Therefore,

w(y)={\frac  {h'(s)}{1+a'(h(s))h'(s)y}}\,

Thus, u_{x}\, becomes \infty \, on the characteristic line for some y>0\, unless a'(h(s))h'(s)y\geq 0\, which is equivalent to saying that a(h(s))\, is a non-decreasing function of s\,.


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