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Green's Theorem 📂Partial Differential Equations

Green's Theorem

Theorem

Let’s assume u,v∈C2(Uˉ)u, v \in C^2( \bar{U}). Then, the following expressions hold:

  • (i) ∫UΔudx=∫∂U∂u∂νdS\displaystyle \int_{U} \Delta u dx=\int_{\partial U} \dfrac{\partial u}{\partial \nu}dS

  • (ii) ∫UDv⋅Dudx=−∫UuΔvdx+∫∂U∂v∂νudS\displaystyle \int_{U} Dv \cdot Du dx = -\int_{U} u \Delta v dx+\int_{\partial U}\dfrac{\partial v}{\partial \nu}udS

  • (iii) ∫U(uΔv−vΔu)dx=∫∂U(∂v∂νu−∂u∂νv)dS\displaystyle \int_{U} (u\Delta v - v\Delta u )dx = \int_{\partial U} \left( \dfrac{\partial v}{\partial \nu}u - \dfrac{\partial u}{\partial \nu} v\right)dS

These are collectively referred to as Green’s formula.

Proof

Partial integration formula

Let’s assume u,v∈C1(Uˉ)u, v \in C^1(\bar{U}). Then, the following formula holds:

∫Uuxivdx=−∫Uuvxidx+∫∂UuvνidS(i=1,…,n) \int_{U} u_{x_{i}}vdx = -\int_{U} uv_{x_{i}}dx + \int_{\partial U} uv\nu^{i} dS\quad (i=1,\dots , n)

(i)

By substituting uu with uxiu_{x_{i}}, and vv with 11 in the partial integration formula, we obtain the following formula.

∫Uuxixidx=∫∂UuxiνidS(i=1,⋯ ,n) \int_{U} u_{x_{i} x_{i}}dx = \int_{\partial U} u_{x_{i}}\nu^{i} dS \quad (i=1,\cdots , n)

Summing up for all i=1,⋯ ,ni=1,\cdots, n gives:

∫U(ux1x1+⋯+uxnxn)dx=∫∂U(ux1ν1+⋯uxnνn)dS \int_{U} (u_{x_{1} x_{1}}+\cdots +u_{x_{n} x_{n}} )dx = \int_{\partial U}( u_{x_{1}}\nu^{1} +\cdots u_{x_{n}}\nu^n)dS

By the definition of Laplacian and ∂u∂ν:=ν⋅Du\dfrac{\partial u}{\partial \nu}:=\boldsymbol{\nu}\cdot Du, the following is true.

∫UΔudx=∫∂U∂u∂νdS \int_{U} \Delta u dx=\int_{\partial U} \dfrac{\partial u}{\partial \nu}dS

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(ii)

By substituting vv with vxiv_{x_{i}} in the partial integration formula, we obtain the following formula.

∫Uuxivxidx=−∫Uuvxixidx+∫∂UuvxiνidS(i=1,⋯ ,n) \int_{U} u_{x_{i}}v_{x_{i}}dx = -\int_{U} uv_{x_{i}x_{i}}dx + \int_{\partial U} uv_{x_{i}}\nu^{i} dS \quad (i=1,\cdots , n)

Summing up for all i=1,⋯ ,ni=1,\cdots ,n gives:

∫U(ux1vx1+⋯+uxnvxn)dx=−∫Uu(vx1x1+⋯vxnxn)dx+∫∂U(vx1ν1+⋯vxnνn)udS \int_{U} (u_{x_{1}}v_{x_{1}}+\cdots +u_{x_{n}}v_{x_{n}} )dx = -\int_{U} u(v_{x_{1}x_{1}}+\cdots v_{x_{n} x_{n}})dx + \int_{\partial U} ( v_{x_{1}}\nu^1 +\cdots v_{x_{n}}\nu^n )udS

Which simplifies to:

∫UDu⋅Dvdx=−∫UuΔvdx+∫∂U∂v∂νudS \int_{U} Du\cdot Dvdx = -\int_{U} u\Delta vdx + \int_{\partial U} \dfrac{\partial v}{\partial \nu}u dS

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(iii)

Switching the places of uu and vv in (ii) gives the following formula.

∫UDu⋅Dvdx=−∫UvΔudx+∫∂U∂u∂νvdS \int_{U} Du \cdot Dv dx = -\int_{U} v \Delta u dx+\int_{\partial U}\dfrac{\partial u}{\partial \nu}vdS

Subtracting (ii) from the above formula gives:

0=−∫U(vΔu−uΔv)dx+∫∂U(∂u∂νv−∂v∂νu)dS 0= -\int_{U} ( v \Delta u -u\Delta v) dx+\int_{\partial U} \left( \dfrac{\partial u}{\partial \nu}v -\dfrac{\partial v}{\partial \nu}u \right)dS

Which simplifies to:

−∫U(vΔu−uΔv)dx=∫∂U(∂v∂νu−∂u∂νv)dS -\int_{U} ( v \Delta u -u\Delta v) dx=\int_{\partial U} \left( \dfrac{\partial v}{\partial \nu}u -\dfrac{\partial u}{\partial \nu}v \right)dS

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See also