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 one year ago
[Partial Differential Equation]
solve the following boundary value problems
\[ \LARGE \frac{\partial^{2} u}{\partial x \partial y} \left( x,y \right) = 0 , u(x,0) = \sin x , u(0,y) = y \]
 one year ago
[Partial Differential Equation] solve the following boundary value problems \[ \LARGE \frac{\partial^{2} u}{\partial x \partial y} \left( x,y \right) = 0 , u(x,0) = \sin x , u(0,y) = y \]

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chihiroasleaf
 one year ago
Best ResponseYou've already chosen the best response.0[Partial Differential Equation] solve the following boundary value problems \[ \LARGE \frac{\partial^{2} u}{\partial x \partial y} \left( x,y \right) = 0 \] \[ \LARGE \frac{\partial}{\partial x } \left( \frac{\partial u}{\partial y } \left( x,y \right) \right)= 0 \] integrate with respect to \(x\) yields \[ \LARGE \frac{\partial u}{\partial y } \left( x,y \right)= f(y) \] integrate with respect to \(y\) yields \( \LARGE u( x,y )= F(y) + g(x) \) with \[ \LARGE \frac{\partial }{\partial y } F(y) = f(y) \] is this correct? what's next?

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.1Now put your boundary conditions into the expression for \(u(x,y)\).

chihiroasleaf
 one year ago
Best ResponseYou've already chosen the best response.0$$u(x,0) = \sin{x} \implies g(x) + F(0) = \sin{x} \implies g(x) = \sin{x}  F(0) $$ $$u(0,y) = y \implies g(0) + F(y) = y \implies F(y) = y g(0) $$ $$u(x,y) = F(y) + g(x) = y  g(0) + \sin{x}  F(0) $$ like this? what should I do with \(F(0) \) and \(g(0) \) ?

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.1\(u(x,0)=\sin x=\sin xg(0)F(0)\) \(u(0,y)=y=yg(0)F(0)\) \(g(0)=F(0)\) If you put this into the expression for \(u(x,y)\) you will get \(u(x,y)=\sin x+y\)

chihiroasleaf
 one year ago
Best ResponseYou've already chosen the best response.0the third line.., you get \( g(0) = F(0) \) from \( g(0)  F(0) = 0\) right? don't we need to change \( g(0)  F(0) \) become \(C\) (constant) ? There usually the 'C' part in the general solution..

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.1As you can see, \(C\) will not do. If we let \(g(0)F(0)=C\), then \(u(x,y)=\sin x+y+C\). Now check your boundary conditions again: \(u(0,y)=y+C\) It will satisfy only if \(C=0\).

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.1You have enough conditions to find the definite solution.

chihiroasleaf
 one year ago
Best ResponseYou've already chosen the best response.0ah.., I see.. :) my other question, do you get \( g(0) = F(0)\) from \( g(0)  F(0) = 0 \) ?

chihiroasleaf
 one year ago
Best ResponseYou've already chosen the best response.0ok..., thank you... :)
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