PDEMOC17

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


The characteristic equations are

{\frac  {dx}{1}}={\frac  {dy}{1}}={\frac  {du}{1-u}}\,

Using the first two equations,

x=y+c_{1}\implies c_{1}=x-y\,

Using the second two equations,

y=-\ln(1-u)+c_{2}\implies c_{2}=y+\ln(1-u)\,

So the general solution is an arbitrary function of these two constants.

f(x-y,y+\ln(1-u))=0\,

This can be equivalently written using another arbitrary function g\,.

y+\ln(1-u)=g(x-y)\,

The general solution is

u(x,y)=1-e^{{g(x-y)-y}}\,


Now using the condition that u=\sin(x)\, on y=x^{2}+x\,,

\sin(x)=1-e^{{g(-x^{2})-x^{2}-x}}\,

g(-x^{2})=\ln(1-\sin(x))+x^{2}+x\,


But we have an arbitrary function of x-y\, in our solution, so set -x^{2}=x-y\,.

x={\sqrt  {y-x}}\,

g(-x^{2})=g(x-y)=\ln(1-\sin {\sqrt  {y-x}})+y-x+{\sqrt  {y-x}}\,

So now

u(x,y)=1-e^{{\ln(1-\sin {\sqrt  {y-x}})+y-x+{\sqrt  {y-x}}-y}}\,


The final solution is

u(x,y)=1-(1-\sin {\sqrt  {y-x}})e^{{-x+{\sqrt  {y-x}}}}\,