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Boundary Value Problems ( Partial Differential Equations)

Posted November 5th, 2010 by vams42
in

hello....i am unable to understand what kind of problem this is in boundary value problems, whether its a dirichlet problem for a circle or a problem for circular annulus....could anyone of you help me out what kind of problem is this, and if possible, which formula to use to solve it, so that i can try solving it.....

$ \nabla^{2}= 0 , 1 < r < 2 , 0 < \theta < \pi , </p> <p>u(1,\theta) = sin \theta</p> <p>u(2,\theta) = 0</p> <p>u(r,0) = 0</p> <p>u(r,\pi) = 0 $

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Comments

Laplace equation in two

On November 6th, 2010 Structure says:

Laplace equation in two dimension
$ \Delta h=\frac{\partial ^2h}{\partial x^2}+\frac{\partial ^2h}{\partial y^2}=0 $ can be written in polar coordinates $ x=\rho \cos \phi\:y=\rho \sin \phi $ with inverse restriction expressed by $ \rho(x,y)=\sqrt{x^2+y^2}\: and \:\phi(x,y)=arctan \frac {y}{x} $
let $ h(x,y)=g(\rho (x,y),\phi (x,y)) $

Then

$$\frac{\partial h}{\partial x}=\frac{\partial g}{\partial \rho}(\rho,\phi)\frac{\partial \rho}{\partial x}+\frac{\partial g}{\partial \phi}(\rho,\phi)\frac{\partial \phi}{\partial x}$$
$$\frac{\partial h}{\partial y}=\frac{\partial g}{\partial \rho}(\rho,\phi)\frac{\partial \rho}{\partial y}+\frac{\partial g}{\partial \phi}(\rho,\phi)\frac{\partial \phi}{\partial y}$$
$$\frac{\partial ^2h}{\partial x^2}= \frac{\partial ^2g}{\partial \rho^2}(\rho,\phi){\frac{\partial \rho}{\partial x}}{\frac{\partial \rho}{\partial x}}+\frac{\partial ^2g}{\partial \phi\partial \rho}(\rho,\phi)\frac{\partial \phi}{\partial x}\frac{\partial \rho}{\partial x}+ \frac{\partial ^2g}{\partial \rho\partial \phi}(\rho,\phi)\frac{\partial \rho}{\partial x}\frac{\partial \phi}{\partial x}+\frac{\partial ^2g}{\partial \phi^2}(\rho,\phi)\frac{\partial \phi}{\partial x}\frac{\partial \phi}{\partial x}+$$
$$+\frac{\partial g}{\partial \rho}(\rho,\phi)\frac{\partial ^2 \rho}{\partial ^2x}+\frac{\partial g}{\partial \phi}(\rho,\phi)\frac{\partial ^2\phi}{\partial x^2}$$
$$\frac{\partial ^2h}{\partial y^2}= \frac{\partial ^2g}{\partial \rho^2}(\rho,\phi){\frac{\partial \rho}{\partial y}}{\frac{\partial \rho}{\partial y}}+ \frac{\partial ^2g}{\partial \phi\partial \rho}(\rho,\phi)\frac{\partial \phi}{\partial y}\frac{\partial \rho}{\partial y}+ \frac{\partial ^2g}{\partial \rho\partial \phi}(\rho,\phi)\frac{\partial \rho}{\partial y}\frac{\partial \phi}{\partial y}+ \frac{\partial ^2g}{\partial \phi^2}(\rho,\phi)\frac{\partial \phi}{\partial y}\frac{\partial \phi}{\partial y}+$$
$$+\frac{\partial g}{\partial \rho}(\rho,\phi)\frac{\partial ^2 \rho}{\partial ^2y}+\frac{\partial g}{\partial \phi}(\rho,\phi)\frac{\partial ^2\phi}{\partial y^2}$$

We have

$$ grad \rho =(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}})\: grad \phi =(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2})$$

If we suppose that our function is of class $ C^2 $ than mixed partial derivative are equal by Young or Schwartz theorem so we can write
(*)

$$\frac{\partial ^2h}{\partial x^2}= \frac{\partial ^2g}{\partial \rho^2}(\rho,\phi){\frac{\partial \rho}{\partial x}}{\frac{\partial \rho}{\partial x}}+ 2\frac{\partial ^2g}{\partial \phi\partial \rho}(\rho,\phi)\frac{\partial \phi}{\partial x}\frac{\partial \rho}{\partial x}+ \frac{\partial ^2g}{\partial \phi^2}(\rho,\phi)\frac{\partial \phi}{\partial x}\frac{\partial \phi}{\partial x}+ \frac{\partial g}{\partial \rho}(\rho,\phi)\frac{\partial ^2 \rho}{\partial ^2x}+\frac{\partial g}{\partial \phi}(\rho,\phi)\frac{\partial ^2\phi}{\partial x^2}$$

(**)$ \frac{\partial ^2h}{\partial y^2}= \frac{\partial ^2g}{\partial \rho^2}(\rho,\phi){\frac{\partial \rho}{\partial y}}{\frac{\partial \rho}{\partial y}}+ 2\frac{\partial ^2g}{\partial \phi\partial \rho}(\rho,\phi)\frac{\partial \phi}{\partial y}\frac{\partial \rho}{\partial y}+ \frac{\partial ^2g}{\partial \phi^2}(\rho,\phi)\frac{\partial \phi}{\partial y}\frac{\partial \phi}{\partial y}+ \frac{\partial g}{\partial \rho}(\rho,\phi)\frac{\partial ^2 \rho}{\partial ^2y}+\frac{\partial g}{\partial \phi}(\rho,\phi)\frac{\partial ^2\phi}{\partial y^2} $

For a more condensed expression it is useful to introduce

$ grad f=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}) $ and inner product $ <grad f,grad g>=\frac{\partial f}{\partial x}\frac{\partial g}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial g}{\partial y} $

Then summing up (*) and (**) we have
$ \Delta h=<grad\rho ,grad\rho>\frac{\partial ^2g}{\partial \rho^2}(\rho,\phi)+2<grad\rho,grad\phi>\frac{\partial ^2g}{\partial \phi\partial \rho}(\rho,\phi)+<grad \phi,grad \phi>\frac{\partial ^2g}{\partial \phi^2}(\rho,\phi)+ $
$ +\Delta \rho\frac{\partial g}{\partial \rho}(\rho,\phi)+\Delta \phi\frac{\partial g}{\partial \phi}(\rho,\phi) $
We have $ grad \rho =(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}})\: grad \phi =(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}) $ so
$ <grad\rho ,grad\rho>=1\:<grad\rho,grad\phi>=0\:<grad \phi,grad \phi>=\frac{1}{x^2+y^2}=\frac{1}{\rho ^2} $
$ \Delta \rho =\frac{1}{\rho}\:\Delta \phi=0 $ and finally

$ \Delta h=\frac{\partial ^2g}{\partial \rho^2}(\rho,\phi)+\frac{1}{\rho ^2}\frac{\partial ^2g}{\partial \phi^2}(\rho,\phi)+\frac{1}{\rho}\frac{\partial g}{\partial \rho}(\rho,\phi) $

Your problem look for a function u which satisfies

$$ \Delta h=\frac{\partial ^2u}{\partial \rho^2}(\rho,\theta)+\frac{1}{\rho ^2}\frac{\partial ^2u}{\partial \theta^2}(\rho,\theta)+\frac{1}{\rho}\frac{\partial u}{\partial \rho}(\rho,\theta) =0$$

Separating variables you get

$$u(r,\theta)=(\frac{4}{3r}-\frac{r}{3})\sin\theta$$

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