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

1. chihiroasleaf

[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?

2. klimenkov

Now put your boundary conditions into the expression for $$u(x,y)$$.

3. chihiroasleaf

$$u(x,0) = \sin{x} \implies g(x) + F(0) = \sin{x} \implies g(x) = \sin{x} - F(0)$$ $$u(0,y) = y \implies g(0) + F(y) = y \implies F(y) = y- g(0)$$ $$u(x,y) = F(y) + g(x) = y - g(0) + \sin{x} - F(0)$$ like this? what should I do with $$F(0)$$ and $$g(0)$$ ?

4. klimenkov

$$u(x,0)=\sin x=\sin x-g(0)-F(0)$$ $$u(0,y)=y=y-g(0)-F(0)$$ $$g(0)=-F(0)$$ If you put this into the expression for $$u(x,y)$$ you will get $$u(x,y)=\sin x+y$$

5. chihiroasleaf

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..

6. klimenkov

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 $$C=0$$.

7. klimenkov

You have enough conditions to find the definite solution.

8. chihiroasleaf

ah.., I see.. :) my other question, do you get $$g(0) = -F(0)$$ from $$-g(0) - F(0) = 0$$ ?

9. klimenkov

Yes.

10. chihiroasleaf

ok..., thank you... :)