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chihiroasleaf
[Partial Differential Equation] solve the following boundary value problems \[ \LARGE \frac{\partial^{2} u}{\partial x \partial y} \left( x,y \right) = 0 , u(x,0) = \sin x , u(0,y) = y \]
[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?
Now put your boundary conditions into the expression for \(u(x,y)\).
$$u(x,0) = \sin{x} \implies g(x) + F(0) = \sin{x} \implies g(x) = \sin{x} - F(0) $$ $$u(0,y) = y \implies g(0) + F(y) = y \implies F(y) = y- g(0) $$ $$u(x,y) = F(y) + g(x) = y - g(0) + \sin{x} - F(0) $$ like this? what should I do with \(F(0) \) and \(g(0) \) ?
\(u(x,0)=\sin x=\sin x-g(0)-F(0)\) \(u(0,y)=y=y-g(0)-F(0)\) \(g(0)=-F(0)\) If you put this into the expression for \(u(x,y)\) you will get \(u(x,y)=\sin x+y\)
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..
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 \(C=0\).
You have enough conditions to find the definite solution.
ah.., I see.. :) my other question, do you get \( g(0) = -F(0)\) from \( -g(0) - F(0) = 0 \) ?
ok..., thank you... :)