PDE:Method of characteristics

From Example Problems
Revision as of 15:40, 11 March 2009 by Todd (Talk | contribs)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

solution u_{t}+au_{x}=0,u(x,0)=f(x)\,

solution u_{t}+uu_{x}=0,u(x,0)=x\,

solution y^{{-1}}u_{x}+u_{y}=0,u(x,1)=x^{2}\,

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

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

solution u_{x}+xu_{y}-u_{z}=u\,, u(x,y,1)=x+y\,

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

solution xu_{x}+yu_{y}+u_{z}=u,u(x,y,0)=h(x,y)\,

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

solution Show that if z=u(x,y)\, is an integral surface of V=<a,b,c>\, containing a point P\,, then the surface contains the characteristic curve \chi \, passing through P\,. (Assume the vector field V\, is C^{1}\,).

solution If S_{1}\, and S_{2}\, are two graphs \left[S_{i}=u_{i}(x,y),i=1,2\right]\, that are integral surfaces of V=<a,b,c>\, and intersect in a curve \chi \,, show that \chi \, is a characteristic curve.

solution (x+u)u_{x}+(y+u)u_{y}=0\,

solution u_{t}+uu_{x}=0,u(x,0)={\begin{cases}1,&x\leq 0\\1-x,&0<x\leq 1\\0,&x>1\end{cases}}\,

solution 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\,.

solution Consider uu_{x}+u_{y}=0\, with the IC u(x,0)=h(x)={\begin{cases}u_{0}>0,&x\leq 0\\u_{0}(1-x),&0<x<1\\0,&x\geq 1\end{cases}}\,

Show that a shock develops at a finite time and describe the weak solution.

solution Consider uu_{x}+u_{t}=0\, with the IC u(x,0)=h(x)={\begin{cases}0,&x<0\\u_{0}(x-1),&x>0\end{cases}}\,

Find the weak solution.

solution Consider the problem u_{x}+u_{y}+u=1\, with condition: u=\sin(x)\, on y=x^{2}+x\,

Main Page : Partial Differential Equations