[proofplan] The proof proceeds in two stages. First, we exploit the [Radial Symmetry of Laplace's Equation](/theorems/1) to reduce the PDE $\Delta u = 0$ to an ODE in the radial variable $r = |x|$: the substitution $u(x) = v(r)$ transforms the Laplacian into $v''(r) + \frac{n-1}{r}v'(r) = 0$. We solve this ODE by separation of variables to obtain the general radial harmonic function, which has the form $b \log r + c$ when $n = 2$ and $b r^{2-n} + c$ when $n \geq 3$. Second, we verify by direct computation that each candidate satisfies the radial ODE for all $r > 0$, confirming that $\Phi$ is harmonic on $\mathbb{R}^n \setminus \{0\}$. [/proofplan] [step:Reduce Laplace's equation to a radial ODE via the substitution $u(x) = v(|x|)$] By the [Radial Symmetry of Laplace's Equation](/theorems/1), if $u$ solves $\Delta u = 0$, then so does $x \mapsto u(Ox)$ for every orthogonal matrix $O$. We therefore seek solutions of the form \begin{align*} u: \mathbb{R}^n \setminus \{0\} &\to \mathbb{R} \\ x &\mapsto v(|x|), \end{align*} where $v: (0,\infty) \to \mathbb{R}$ is a twice-differentiable function of the radial variable $r := |x| = (x_1^2 + \cdots + x_n^2)^{1/2}$. For each $i = 1, \ldots, n$, the chain rule gives \begin{align*} \partial_{x_i} r = \frac{x_i}{r}, \quad r > 0. \end{align*} Applying the chain rule to $u(x) = v(r)$: \begin{align*} \partial_{x_i} u &= v'(r) \cdot \frac{x_i}{r}. \end{align*} Differentiating once more by the product rule and chain rule: \begin{align*} \partial_{x_i}^2 u &= v''(r) \cdot \frac{x_i^2}{r^2} + v'(r) \cdot \frac{1}{r}\left(1 - \frac{x_i^2}{r^2}\right). \end{align*} Summing over $i = 1, \ldots, n$ and using $\sum_{i=1}^n x_i^2 = r^2$: \begin{align*} \Delta u = \sum_{i=1}^n \partial_{x_i}^2 u = v''(r) \cdot \frac{r^2}{r^2} + v'(r) \cdot \frac{1}{r}\left(n - \frac{r^2}{r^2}\right) = v''(r) + \frac{n-1}{r}\,v'(r). \end{align*} The equation $\Delta u = 0$ therefore holds if and only if $v$ satisfies the radial ODE \begin{align*} v''(r) + \frac{n-1}{r}\,v'(r) = 0, \quad r > 0. \end{align*} [guided] We want to find solutions of $\Delta u = 0$ on $\mathbb{R}^n \setminus \{0\}$. By the [Radial Symmetry of Laplace's Equation](/theorems/1), the Laplacian commutes with rotations: if $u$ solves $\Delta u = 0$, then $x \mapsto u(Ox)$ also solves $\Delta u = 0$ for every orthogonal matrix $O \in O(n)$. This suggests looking for radially symmetric solutions — functions that depend only on $r = |x|$, not on direction. We make the ansatz \begin{align*} u: \mathbb{R}^n \setminus \{0\} &\to \mathbb{R} \\ x &\mapsto v(|x|), \end{align*} where $v: (0,\infty) \to \mathbb{R}$ is twice differentiable and $r := |x| = (x_1^2 + \cdots + x_n^2)^{1/2}$ is the radial variable. Our goal is to express $\Delta u$ purely in terms of $v$, $v'$, $v''$, and $r$. **Computing $\partial_{x_i} r$.** For each $i = 1, \ldots, n$, we differentiate $r = (x_1^2 + \cdots + x_n^2)^{1/2}$ by the chain rule: \begin{align*} \partial_{x_i} r = \frac{1}{2}(x_1^2 + \cdots + x_n^2)^{-1/2} \cdot 2x_i = \frac{x_i}{r}, \quad r > 0. \end{align*} This is valid only for $r > 0$, which is why the origin must be excluded. **First partial derivative of $u$.** By the chain rule applied to $u(x) = v(r(x))$: \begin{align*} \partial_{x_i} u = v'(r) \cdot \partial_{x_i} r = v'(r) \cdot \frac{x_i}{r}. \end{align*} **Second partial derivative of $u$.** Differentiating $\partial_{x_i} u = v'(r) \cdot \frac{x_i}{r}$ with respect to $x_i$ using the product rule: \begin{align*} \partial_{x_i}^2 u &= \partial_{x_i}\!\left[v'(r)\right] \cdot \frac{x_i}{r} + v'(r) \cdot \partial_{x_i}\!\left[\frac{x_i}{r}\right]. \end{align*} For the first term, the chain rule gives $\partial_{x_i}[v'(r)] = v''(r) \cdot \frac{x_i}{r}$. For the second term, the quotient rule gives \begin{align*} \partial_{x_i}\!\left[\frac{x_i}{r}\right] = \frac{r - x_i \cdot (x_i/r)}{r^2} = \frac{1}{r} - \frac{x_i^2}{r^3} = \frac{1}{r}\left(1 - \frac{x_i^2}{r^2}\right). \end{align*} Combining: \begin{align*} \partial_{x_i}^2 u = v''(r) \cdot \frac{x_i^2}{r^2} + v'(r) \cdot \frac{1}{r}\left(1 - \frac{x_i^2}{r^2}\right). \end{align*} **Summing over $i$.** Using $\sum_{i=1}^n x_i^2 = r^2$: \begin{align*} \Delta u &= \sum_{i=1}^n \partial_{x_i}^2 u \\ &= v''(r) \cdot \frac{1}{r^2} \sum_{i=1}^n x_i^2 + v'(r) \cdot \frac{1}{r} \sum_{i=1}^n \left(1 - \frac{x_i^2}{r^2}\right) \\ &= v''(r) \cdot \frac{r^2}{r^2} + v'(r) \cdot \frac{1}{r}\left(n - 1\right) \\ &= v''(r) + \frac{n-1}{r}\,v'(r). \end{align*} Why does the sum $\sum_{i=1}^n (1 - x_i^2/r^2)$ simplify to $n - 1$? Because $\sum_{i=1}^n 1 = n$ and $\sum_{i=1}^n x_i^2/r^2 = r^2/r^2 = 1$, so the sum equals $n - 1$. This is the key algebraic simplification that produces the clean ODE. The $n$-dimensional Laplacian of a radial function $u(x) = v(|x|)$ reduces to a second-order ODE in $r$. The PDE $\Delta u = 0$ on $\mathbb{R}^n \setminus \{0\}$ becomes \begin{align*} v''(r) + \frac{n-1}{r}\,v'(r) = 0, \quad r > 0. \end{align*} The term $\frac{n-1}{r}v'(r)$ reflects the geometric fact that in $n$ dimensions, the surface area of a sphere of radius $r$ grows as $r^{n-1}$. The ODE is singular at $r = 0$, which foreshadows the singularity of the fundamental solution at the origin. [/guided] [/step] [step:Solve the radial ODE by separation of variables] Suppose $v'(r) \neq 0$. Rewriting the ODE $v'' + \frac{n-1}{r}v' = 0$ as \begin{align*} \frac{v''(r)}{v'(r)} = -\frac{n-1}{r}, \end{align*} we recognise the left-hand side as $\frac{d}{dr}[\log |v'(r)|]$. Integrating both sides over an interval $[r_0, r] \subset (0,\infty)$ with respect to $\mathcal{L}^1$: \begin{align*} \log |v'(r)| - \log |v'(r_0)| = -(n-1)(\log r - \log r_0). \end{align*} Exponentiating: \begin{align*} v'(r) = \frac{a}{r^{n-1}}, \end{align*} where $a := v'(r_0) \cdot r_0^{n-1}$ is a constant determined by initial conditions. Integrating $v'$ with respect to $\mathcal{L}^1$ on $(0, \infty)$: \begin{align*} v(r) = \begin{cases} a \log r + c & \text{if } n = 2, \\ \dfrac{a}{2-n} \cdot r^{2-n} + c = b\, r^{2-n} + c & \text{if } n \geq 3, \end{cases} \end{align*} where $b := \frac{a}{2-n}$ and $c \in \mathbb{R}$ are constants. The case split arises because $\int r^{-(n-1)} \, d\mathcal{L}^1(r)$ equals $\log r$ when $n - 1 = 1$ and $\frac{r^{2-n}}{2-n}$ when $n - 1 \geq 2$. Selecting the specific constants that define $\Phi$ (namely $a = -\frac{1}{2\pi}$, $c = 0$ for $n = 2$ and $b = \frac{1}{n(n-2)\alpha(n)}$, $c = 0$ for $n \geq 3$), we obtain the fundamental solution. It remains to verify that these candidates satisfy the ODE. [guided] We now solve the ODE $v''(r) + \frac{n-1}{r}v'(r) = 0$ for $r > 0$. The strategy is separation of variables: isolate $v''/v'$ on one side and $r$ on the other. **Separating variables.** Assuming $v'(r) \neq 0$ (the case $v' \equiv 0$ gives only constant solutions, which are not the fundamental solution), we divide both sides by $v'(r)$: \begin{align*} \frac{v''(r)}{v'(r)} = -\frac{n-1}{r}. \end{align*} The left-hand side is the logarithmic derivative $\frac{d}{dr}[\log |v'(r)|]$. This recognition is the key step — it converts a second-order ODE into a first-order equation for $\log |v'|$. **Integrating.** We integrate both sides over the interval $[r_0, r] \subset (0,\infty)$ with respect to $\mathcal{L}^1$: \begin{align*} \int_{r_0}^{r} \frac{v''(s)}{v'(s)} \, d\mathcal{L}^1(s) &= -\int_{r_0}^{r} \frac{n-1}{s} \, d\mathcal{L}^1(s) \\ \log |v'(r)| - \log |v'(r_0)| &= -(n-1)(\log r - \log r_0). \end{align*} Exponentiating both sides yields \begin{align*} |v'(r)| = |v'(r_0)| \cdot \left(\frac{r_0}{r}\right)^{n-1}, \end{align*} so $v'(r) = a \cdot r^{-(n-1)}$ where $a = v'(r_0) \cdot r_0^{n-1}$ absorbs the sign and the reference-point data into a single constant. **Integrating $v'$ to find $v$.** We now integrate $v'(r) = a \cdot r^{-(n-1)}$ with respect to $\mathcal{L}^1$. When $n = 2$: $v'(r) = a/r$, so $\int r^{-1} \, d\mathcal{L}^1(r) = \log r$, giving \begin{align*} v(r) = a \log r + c. \end{align*} When $n \geq 3$: $v'(r) = a \cdot r^{-(n-1)}$ with $-(n-1) \neq -1$, so $\int r^{-(n-1)} \, d\mathcal{L}^1(r) = \frac{r^{2-n}}{2-n}$, giving \begin{align*} v(r) = \frac{a}{2-n} \cdot r^{2-n} + c = b\, r^{2-n} + c, \end{align*} where $b := \frac{a}{2-n}$. Notice that $2 - n < 0$ for $n \geq 3$, so $v(r) \to \infty$ as $r \to 0^+$. This singularity at the origin is not a defect — it is precisely what allows $\Phi$ to act as the kernel of a [Green's function](/pages/Laplace%27s%20Equation). **Why do the two cases split at $n = 2$?** The antiderivative of $r^{-(n-1)}$ is $\frac{r^{2-n}}{2-n}$ when $2 - n \neq 0$, i.e., when $n \neq 2$. When $n = 2$, the exponent $-(n-1) = -1$ and the antiderivative is instead $\log r$. This is the same dichotomy that distinguishes the logarithmic potential in two dimensions from the Newtonian potential in higher dimensions. **Choosing constants.** The fundamental solution $\Phi$ corresponds to the specific choice of constants $c = 0$ and $a = -\frac{1}{2\pi}$ (when $n = 2$) or $b = \frac{1}{n(n-2)\alpha(n)}$ (when $n \geq 3$). These normalisation constants are chosen so that $-\Delta \Phi = \delta_0$ in the [distributional](/pages/Distribution) sense (see the [Fundamental Solution of Laplace's Equation](/theorems/566)), but the harmonicity of $\Phi$ away from the origin holds for any choice of $a$, $b$, and $c$. [/guided] [/step] [step:Verify $\Delta \Phi = 0$ for $n \geq 3$ by direct computation] For $n \geq 3$ and $r > 0$, the candidate is $v(r) = b\, r^{2-n}$ with $b = \frac{1}{n(n-2)\alpha(n)}$. Computing the derivatives: \begin{align*} v'(r) &= b(2-n)\, r^{1-n}, \\ v''(r) &= b(2-n)(1-n)\, r^{-n}. \end{align*} Substituting into the radial ODE: \begin{align*} v''(r) + \frac{n-1}{r}\,v'(r) &= b(2-n)(1-n)\, r^{-n} + \frac{n-1}{r} \cdot b(2-n)\, r^{1-n} \\ &= b(2-n)\, r^{-n} \bigl[(1-n) + (n-1)\bigr] \\ &= b(2-n)\, r^{-n} \cdot 0 \\ &= 0. \end{align*} The cancellation $(1-n) + (n-1) = 0$ confirms that $v(r) = b\, r^{2-n}$ satisfies the radial ODE for every $r > 0$, and therefore $\Delta \Phi(x) = 0$ for all $x \in \mathbb{R}^n \setminus \{0\}$ when $n \geq 3$. [/step] [step:Verify $\Delta \Phi = 0$ for $n = 2$ by direct computation] For $n = 2$ and $r > 0$, the candidate is $v(r) = -\frac{1}{2\pi} \log r$. Computing the derivatives: \begin{align*} v'(r) &= -\frac{1}{2\pi} \cdot \frac{1}{r}, \\ v''(r) &= \frac{1}{2\pi} \cdot \frac{1}{r^2}. \end{align*} Substituting into the radial ODE with $n = 2$: \begin{align*} v''(r) + \frac{2-1}{r}\,v'(r) &= \frac{1}{2\pi r^2} + \frac{1}{r} \cdot \left(-\frac{1}{2\pi r}\right) \\ &= \frac{1}{2\pi r^2} - \frac{1}{2\pi r^2} \\ &= 0. \end{align*} Therefore $\Delta \Phi(x) = 0$ for all $x \in \mathbb{R}^2 \setminus \{0\}$. In both cases, $\Phi$ satisfies $\Delta \Phi = 0$ on the punctured space $\mathbb{R}^n \setminus \{0\}$, completing the proof. [/step]