PDEMOC16

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Consider the problem u_{x}+u_{y}+u=e^{{x+2y}} with initial conditions: u(x,0)=0


We may write this as (1,1)\cdot \nabla u=e^{{x+2y}}-u, with the initial curve as \Gamma (s_{1},0,0)


So... \partial _{t}x=1 and \partial _{t}y=1

Note:\partial _{t}\equiv {\frac  {\partial }{\partial t}}

This implies x=t+C_{1} and y=t+C_{2}

Applying initial conditions: x(0)=C_{1}=s_{1} and y(0)=C_{2}=0

So, now x=t+s_{1} and y=t

The characteristic curves are given by the following equation: x-y=s_{1}

Now, \partial _{t}u+u=e^{{3t+s_{1}}}, applying the integrating factor e^{t}

e^{t}\partial _{t}u+e^{t}u=\partial _{t}\left(e^{t}u\right)=e^{{4t+s_{1}}}

e^{t}u={\frac  {1}{4}}e^{{4t+s_{1}}}+C

And now applying initial conditions...

u(0)={\frac  {1}{4}}e^{{s_{1}}}+C=0

So finally,

u={\frac  {1}{4}}\left(e^{{3t+s_{1}}}-e^{{s_{1}-t}}\right)

or u(x,y)={\frac  {1}{4}}\left(e^{{x+2y}}-e^{{x-2y}}\right)