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[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
 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
 one year ago

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chihiroasleafBest 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?
 one year ago

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

chihiroasleafBest 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 \(g(0) \) ?
 one year ago

klimenkovBest 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 \(u(x,y)=\sin x+y\)
 one year ago

chihiroasleafBest ResponseYou've already chosen the best response.0
the 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..
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
As 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 \(C=0\).
 one year ago

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

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

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