Show that $A(x^2-y^2)$ satisfies Laplace's equation

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Verification of the solvability condition:

First one has to calculate $\nabla u \cdot \hat{n}$ along the boundary. Assuming $H,L>0$ and $\hat{n}$ is the exterior normal of the square $[0,L]\times [0,H]$, (e.g. along $x = 0$ this is just $(-1,0)$), we simply get $\nabla u \cdot \hat{n}=-\frac{\partial u}{\partial x} = 0 $. Similarly, along $ x = L$ we have $\nabla u \cdot \hat{n} =\frac{\partial u}{\partial x} = g $ while for $y=0$ we have $\nabla u \cdot \hat{n} =-\frac{\partial u}{\partial y}=0$ and $\nabla u \cdot \hat{n} =\frac{\partial u}{\partial y}=f$ along the last part of the boundary $y=H$.

So we only have to calculate the integral of the constant $g$ along $x=L$ and the integral of the constant $f$ along $y = H$. Since these are constants the integral is just length of the side of the square times the constant, which means $= H\cdot g$ for $x=L$ and $=L\cdot f$ along $y=H$. So the integral is $H\cdot g + L \cdot f$. For this to vanish you need $ H\cdot g = -f \cdot L $

You can now verify this holds, since $u_x = 2Ax , u_y = -2 A y$ so if $u_x(L,y) = 2 A L = g$ and $ u_y(H,y) = -2AH = f$ then $H\cdot g = 2ALH =-f\cdot L $.

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Updated on November 06, 2020

Comments

  • continental
    continental about 3 years

    Consider $u(x,y)$ that satisfies Laplace's equation given by $\nabla^2 u = 0$. Subject to the boundary conditions $\frac{\partial u}{\partial x}(0,y) = 0$, $\frac{\partial u}{\partial x}(L,y) = g(y)$, $\frac{\partial u}{\partial y}(x,0) = 0$, $\frac{\partial u}{\partial y}(x,H) = f(x)$.

    The solvability condition is that $$\oint \nabla u \cdot\hat{n}dS = 0$$

    Show that $u(x,y) = A(x^2-y^2)$ is a solution if $g(y)$ and $f(x)$ are constants [under the condition stated above].

    However I am not sure how to show this? I know $\nabla^2 u = 0$ but I don't think this what the question is asking for.

    • Thomas
      Thomas about 8 years
      The question ask you to calculate $\nabla^2 u$, verify it vanishes, and to check the given function satisfies the boundary values (by inspection/calculation). If you already calculated $\nabla^2 u$ you are about 90% through.
    • continental
      continental about 8 years
      Must I not show that the line integral also vanishes? If so how do I do this?
    • Thomas
      Thomas about 8 years
      The solvability condition shows that a solution exists (if it is fulfilled). The point is that you can check this without knowing the exact behaviour in the interior of the domain. If you already happen to know a solution then there is no need to verify the solvability condition (unless of course this is homework and you are asked to do just that). But the question you cited just asks you to show that $u$ solves the equation for certain $g$, $f$.
    • continental
      continental about 8 years
      Could you perhaps demonstrate how I would verify the solvability condition?