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Let $\;f:\mathbb R^n\rightarrow \mathbb R^m\;$ and $\;G:\mathbb R^m \rightarrow \mathbb R_{+}\;$ (NOTE: $\;n\;$ is not necessary equal to $\;m\;$). Assume the functional:

$\;I_{\mathbb R^n} (f) = \int_{\mathbb R^n} \frac{1}{2} {\vert \nabla f \vert}^2 + G(f) \;dx\;$ where $\;\nabla f=(\frac{\partial f_i}{\partial x_j})_{1\le i \le m,1\le j \le n}\;$ and $\;\vert \cdot \vert\;$is the Euclidean norm of the matrix.

Prove the Euler-Lagrange equation of the above functional is given by the system : $\;\Delta f -G_f(f)=0\;$ where $\;G_f(f)=(\frac{\partial G}{\partial f_1}, \dots, \frac{\partial G}{\partial f_m})^{T}\;$

My attempt:

I searched on my Evans PDE's book , where I found this:

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The formula $\;(16)\;$, I believe is the solution to my problem. However I have trouble applying it here because I don't know what exactly is the $\;\frac{1}{2} {\vert \nabla f \vert}^2\;$.

I understand that $\;p_i^{k}\;$ from the book stands for $\;\frac{\partial f_k}{\partial x_i}\;$ but I'm a bit unsure how this appears in $\;\frac{1}{2} {\vert \nabla f \vert}^2\;$. How can I compute $\;L_{p_i^{k}}\;$ if I don't know what $\;\frac{1}{2} {\vert \nabla f \vert}^2\;$ looks like?

I would really appreciate any help because I've been stuck here!

Thanks in advance

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Try with this definition (and some index renaming):

$$\left|\nabla f\right|^2=\sum_{1\le i \le m,1\le j \le n}\left(\frac{\partial f_i}{\partial x_j}\right)^2$$

Then, with $L=\dfrac{1}{2}\left|\nabla f\right|^2+G(f)$ and $p_j^i=\dfrac{\partial f_i}{\partial x_j}$, we can apply the given Euler-Lagrange equations, getting for the first term:

$$L_{p_j^i}=\dfrac{\partial L}{\partial {p_j^i}}=\dfrac{\partial L}{\partial\left(\partial f_i/\partial x_j\right)}=2\frac{\partial f_i}{\partial x_j}\implies\sum_{1\le j \le n}L_{p_j^i}=2\sum_{1\le j \le n}\frac{\partial f_i}{\partial x_j}\implies$$

$$\implies \sum_{1\le j \le n}\left(L_{p_j^i}\right)_{x_j}=\sum_{1\le j \le n}\frac{\partial}{\partial x_j}L_{p_j^i}=2\sum_{1\le j \le n}\frac{\partial^2 f_i}{\partial x_j^2}=2\Delta f_i$$

with $1\leq i\leq m$. And for the second term, with $z^i=f_i$:

$$L_{z^i}=\frac{\partial G}{\partial f_i}$$

So, the Euler-Lagrange equations

$$-\sum_{1\le j \le n}L_{p_j^i}+L_{z^i}=0\;,1\leq i\leq m$$

are for this lagrangian:

$$\Delta f_i-\frac{\partial G}{\partial f_i}=0\;,1\leq i\leq m$$

Or, writing $\Delta f=(\Delta f_1,\dots\Delta f_m)^T$ and $G_f(f)=(\dfrac{\partial G}{\partial f_1}, \dots, \dfrac{\partial G}{\partial f_m})^{T}$

$$\Delta f -G_f(f)=0$$

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  • $\begingroup$ Budria Thanks a lot! I found the answer by myself later that day but I was a bit lazy to post it here. Glad to see ,it's the same as yours :) $\endgroup$ – kaithkolesidou Aug 23 '17 at 15:44
  • $\begingroup$ You are welcome. The $1/2$ factor into the lagrangian is a big clue, isn't it? :) $\endgroup$ – Rafa Budría Aug 24 '17 at 14:15

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