# I Laplacian in integration by parts in Jackson

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1. Feb 1, 2017

### Angelo Cirino

I am reviewing Jackson's "Classical Electromagnetism" and it seems that I need to review vector calculus too. In section 1.11 the equation $W=-\frac{\epsilon_0}{2}\int \Phi\mathbf \nabla^2\Phi d^3x$ through an integration by parts leads to equation 1.54 $W=\frac{\epsilon_0}{2}\int |\mathbf \nabla\Phi|^2 d^3x=\frac{\epsilon_0}{2}\int |\mathbf E|^2 d^3x$. My problem is that I tried to derive the result with back to basics integration by parts with $\int u dv = uv - \int vdu$ with $$u = \Phi, dv = \mathbf \nabla^2\Phi d^3x=\mathbf \nabla \cdot (\mathbf \nabla \Phi) d^3x, du=\mathbf \nabla \Phi d^3x, v = \mathbf \nabla \Phi\\ \int \Phi \mathbf \nabla^2\Phi d^3x=\Phi \mathbf \nabla \Phi - \int \mathbf \nabla \Phi \cdot \mathbf \nabla \Phi d^3x=\Phi \mathbf \nabla \Phi - \int \mathbf |\mathbf \nabla \Phi|^2 d^3x$$ that is obviously wrong, the term $\Phi \mathbf \nabla \Phi=-\Phi \mathbf E$ shouldn't be there and it is a vector quantity summed to a scalar. How I should proceed?

2. Feb 1, 2017

Suggest you try $\nabla \cdot (\Phi \nabla \Phi)=\Phi \nabla^2 \Phi+\nabla \Phi \cdot \nabla \Phi$ to rewrite $\Phi \nabla^2 \Phi$. Then use Gauss' law on the left side of the first equation, and the surface integral may vanish at infinity. $\\$ Editing. Note: The equation $du=\nabla \Phi d^3x$ when $u=\Phi$ is incorrect. Similarly for your $dv$ and $v$.

Last edited: Feb 1, 2017
3. Feb 2, 2017

### Angelo Cirino

I understand that the integration by parts that I posted is a very naive attempt at following the statement by Jackson that the result can be obtained by an integration by parts. Using your suggestion of $\Phi \nabla^2 \Phi = \nabla \cdot (\Phi \nabla \Phi) - |\nabla \Phi|^2$ is very straightforward and economic:
$$\int_V \Phi\nabla^2\Phi d^3x=\int_V \nabla(\Phi\nabla\Phi)d^3-\int_V |\nabla\Phi|^2d^3x$$
By the Divergence Theorem
$$\int_V \nabla\cdot(\Phi \nabla \Phi)d^3x=\oint_S\Phi\nabla\Phi\cdot \mathbf n da=\oint_S\Phi \frac{\partial \Phi}{\partial n}da$$
Now, the integration is over all space and then the surface S goes to infinity and so the surface integral vanishes, leaving
$$\int_V \Phi\nabla^2\Phi d^3x=-\int_V |\nabla\Phi|^2d^3x$$
But I wanted to follow Jackson when he states that the result can be obtained by an integration by parts. This can be done by decomposing the vector in its components
$$\int \Phi\nabla^2\Phi d^3x=\int\Phi\left(\frac{\partial^2 \Phi}{\partial x^2_1}+\frac{\partial^2 \Phi}{\partial x^2_2}+\frac{\partial^2 \Phi}{\partial x^2_3}\right)d^3x=\int\Phi\frac{\partial^2 \Phi}{\partial x^2_1}d^3x+\int\Phi\frac{\partial^2 \Phi}{\partial x^2_2}d^3x+\int\Phi\frac{\partial^2 \Phi}{\partial x^2_3}d^3x$$
Now, integrating by parts
$$u_i=\Phi,~du_i=d_i\Phi=\frac{\partial \Phi}{\partial x_i}dx_i\\ d_iv=\nabla^2_i\Phi dx_i=\frac{\partial^2\Phi}{\partial x^2_i}dx_i=\frac{\partial}{x_i}\left(\frac{\partial \Phi}{\partial x_i}\right)dx_i,~v_i=\frac{\partial\Phi}{\partial x_i}\\ \int\Phi\frac{\partial^2 \Phi}{\partial x^2_i}dx_i=\Phi\frac{\partial \Phi}{\partial x_i}-\int\frac{\partial\Phi}{\partial x_i}\frac{\partial\Phi}{\partial x_i}dx_i= \Phi\frac{\partial \Phi}{\partial x_i}-\int\left(\frac{\partial \Phi}{\partial x_i}\right)^2dx_i$$
Now
$$\int dx_j\int dx_k\left[\phi\frac{\partial\Phi}{\partial x_i}-\int\left(\frac{\partial\Phi}{\partial x_i}\right)^2 dx_i\right]=\int\Phi\frac{\partial\Phi}{\partial x_i}dx_j dx_k-\int\left(\frac{\partial\Phi}{\partial x_i}\right)^2 d^3x\\$$
and
$$\int \Phi\nabla^2\Phi d^3x=\int\Phi\frac{\partial\Phi}{\partial x_1}dx_2 dx_3+\int\Phi\frac{\partial\Phi}{\partial x_2}dx_1 dx_3+\int\Phi\frac{\partial\Phi}{\partial x_3}dx_1 dx_2-\int\left[\left(\frac{\partial\Phi}{\partial x_1}\right)^2+\left(\frac{\partial\Phi}{\partial x_2}\right)^2+\left(\frac{\partial\Phi}{\partial x_3}\right)^2\right]d^3x$$
From Green's First Identity it can be seen that
$$\int\Phi\frac{\partial\Phi}{\partial x_1}dx_2 dx_3+\int\Phi\frac{\partial\Phi}{\partial x_2}dx_1 dx_3+\int\Phi\frac{\partial\Phi}{\partial x_3}dx_1 dx_2=\oint_S\Phi\frac{\partial\Phi}{\partial n}da$$
and we have
$$\int\left[\left(\frac{\partial\Phi}{\partial x_1}\right)^2+\left(\frac{\partial\Phi}{\partial x_2}\right)^2+\left(\frac{\partial\Phi}{\partial x_3}\right)^2\right]d^3x=\int|\nabla\Phi|^2d^3x$$
and then
$$\int \Phi\nabla^2\Phi d^3x=\oint_S\Phi\frac{\partial\Phi}{\partial n}da-\int|\nabla\Phi|^2d^3x$$
Now, again for the same argument, the integration is over all space and then the surface S goes to infinity and so the surface integral vanishes, leaving
$$\int \Phi\nabla^2\Phi d^3x=-\int|\nabla\Phi|^2d^3x$$
the same result as using $\Phi \nabla^2 \Phi = \nabla \cdot (\Phi \nabla \Phi) - |\nabla \Phi|^2$, but way lengthier.

I value a book by the way the author induces us to work out tools and techniques to solve problems, that's why I wanted to see an easy way to perform the integration by parts in the vector integral equation. But vector calculus isn't as straightforward as single variable calculus as we must deal with the components and the scalar and cross products. I really wished that there were an easy way to perform the integration by parts in vector calculus. It would be a technique worthy of keeping in memory.

4. Feb 2, 2017

I think J.D. Jackson's calling it "integration by parts is used loosely to mean doing exactly what I did. Things like this come up quite frequently in J.D. Jackson's book that use a lot of vector calculus. e.g. Given a term such as $B \times \nabla \times B$, the conversion for this typically comes from $\\$ $\nabla (A \cdot B)=(A \cdot \nabla) B +(B \cdot \nabla) A +A \times \nabla \times B + B \times \nabla \times A$. $\\$ Again, when J.D. Jackson calls it "integration by parts", what he means is using the vector identities, (in the previous case operating on $\nabla \cdot (\Phi \nabla \Phi)$ ), in a way similar to $d(uv)=udv + vdu$. For vector calculus, there are about half a dozen or more of these on the cover of his book. Another one,(this one I haven't memorized), is $\nabla \cdot (A \times B )$. They are all very useful and used throughout his book.

5. Feb 2, 2017

### Angelo Cirino

I agree with your assessment of the way Jackson uses the identities, I just wanted to exercise myself and try to see if there really was an easy approach to integrate by parts in the conventional sense. I completed two semesters studying EM from Jackson's book, but it was over fifteen years ago.