Euler’s theorem and harmonic functions

Homogeneity is a scaling rule. A function $f(x,y)$ is homogeneous of degree $n$ if replacing $x\to tx,\ y\to ty$ pulls out an overall factor $t^n$:

where: $n$ is the degree of homogeneity, found by direct substitution rather than guessed from the highest total-degree term — this matters for quotients and roots, where the answer is not always obvious from inspection. For example $f(x,y)=x^3+3x^2y-y^3$ is degree $3$ (every monomial has total degree $3$), while $f(x,y)=\sqrt{x^2+y^2}$ is degree $1$, since $f(tx,ty)=t\sqrt{x^2+y^2}$. A quotient like $f(x,y)=y^2/x$ is also degree $1$: substituting gives $f(tx,ty)=(ty)^2/(tx)=t\,(y^2/x)$, so the $t$'s from numerator and denominator do not simply cancel to $t^0$ — you must carry the algebra through rather than eyeball the exponents. It helps to notice why the definition is stated for polynomials in the first place: a sum of monomials is homogeneous of degree $n$ exactly when every monomial individually has total degree $n$, because substituting $x\to tx,\ y\to ty$ multiplies a monomial $x^py^q$ by $t^{p+q}$, and a sum of terms all carrying the same power of $t$ factors that common power straight out. Mixed-degree sums such as $x^2+y$ fail the test for precisely this reason — no single $t^n$ can be factored out of $t^2x^2+ty$.

Euler's theorem converts scaling into a derivative identity. For a differentiable homogeneous function of degree $n$,

and differentiating this relation once more (applying it to $f_x$, itself homogeneous of degree $n-1$, and to $f_y$) gives the second-order extension

where: $f_x,f_y$ are the first partials and $f_{xx},f_{xy},f_{yy}$ the second partials, with $f_{xy}=f_{yx}$ guaranteed equal by Clairaut's theorem, which is exactly why the mixed term above carries a coefficient of $2$. Both identities let you check a partial-derivative computation independently, or shortcut a problem that only asks for the combination $xf_x+yf_y$: instead of computing $f_x$ and $f_y$ separately and then forming $xf_x+yf_y$ by hand, you can read off the degree $n$ by inspection and immediately write down $nf$, saving a line of algebra on exam problems built around this identity.

Harmonicity is an unrelated second-order condition. A function $f(x,y)$ with continuous second partials is harmonic if it satisfies Laplace's equation:

Verifying this is a direct two-step computation — find $f_{xx}$, find $f_{yy}$, check the sum vanishes. Harmonic functions model steady-state temperature fields, electrostatic potentials, and incompressible irrotational flow; here they are treated purely as real-variable objects, though their deeper identity as the real and imaginary parts of an analytic complex function $f(z)=u+iv$ is the subject of the Cauchy-Riemann equations later in the course. Unlike homogeneity, harmonicity has no simple substitution test — there is no shortcut analogous to plugging in $tx,ty$. Every check genuinely requires computing both second-order partial derivatives from scratch and adding them, which is why it is worth being fast and careful with second partials before attempting this kind of problem under time pressure.

Verify both the first- and second-order Euler relations for $f(x,y)=x^3+2x^2y-xy^2+y^3$, and check whether the same function is harmonic.

Step 1. Confirm the degree. Every term has total degree $3$ ($x^3$, $2x^2y$, $xy^2$, $y^3$), and directly $f(tx,ty)=t^3f(x,y)$, so $n=3$.

Step 2. Compute the first partials: $f_x=3x^2+4xy-y^2$ and $f_y=2x^2-2xy+3y^2$. Forming $xf_x+yf_y$ gives $3x^3+4x^2y-xy^2+2x^2y-2xy^2+3y^3 = 3x^3+6x^2y-3xy^2+3y^3$, which equals $3f=3(x^3+2x^2y-xy^2+y^3)$ exactly — the first-order Euler relation holds.

Step 3. Compute the second partials: $f_{xx}=6x+4y$, $f_{xy}=4x-2y$, $f_{yy}=-2x+6y$. Forming $x^2f_{xx}+2xyf_{xy}+y^2f_{yy}$: $x^2(6x+4y)+2xy(4x-2y)+y^2(-2x+6y) = 6x^3+4x^2y+8x^2y-4xy^2-2xy^2+6y^3 = 6x^3+12x^2y-6xy^2+6y^3$. Compare with $n(n-1)f=6f=6x^3+12x^2y-6xy^2+6y^3$ — the two match, confirming the second-order relation.

Step 4. Check harmonicity separately: $f_{xx}+f_{yy}=(6x+4y)+(-2x+6y)=4x+10y$, which is not identically zero. This $f$ is homogeneous of degree $3$ and satisfies both Euler relations, but it is not harmonic — confirming the two properties are independent even for the same function.

Competitive stretch

Test $u(x,y)=e^x\cos y$: $u_{xx}=e^x\cos y$ and $u_{yy}=-e^x\cos y$, so $u_{xx}+u_{yy}=0$ and $u$ is harmonic everywhere. But $u(tx,ty)=e^{tx}\cos(ty)$ cannot be written as $t^nu(x,y)$ for any fixed $n$, since $t$ sits inside the exponential and cosine rather than as an overall scale factor — so $u$ is harmonic but not homogeneous of any degree, the mirror case to Step 4 above. For a function that is genuinely both, try $g(x,y)=x^2-y^2$: it is homogeneous of degree $2$ (every term has total degree $2$, and $g(tx,ty)=t^2x^2-t^2y^2=t^2g(x,y)$), and $g_{xx}=2,\ g_{yy}=-2$ gives $g_{xx}+g_{yy}=0$, so $g$ is also harmonic. Together, $f$, $u$ and $g$ populate all three non-trivial cells of the two-by-two grid — homogeneous only, harmonic only, and both — leaving only "neither" to find on your own, for example $h(x,y)=x^2+y$.

• Confusing homogeneity with harmonicity. They are independent tests on different derivatives — a function can pass one, both, or neither, as the worked example and its stretch both show.
• Guessing the degree $n$ instead of substituting. For quotients or roots such as $f=y^2/x$ or $f=\sqrt{x^2+y^2}$, always compute $f(tx,ty)$ directly rather than reading $n$ off the highest total-degree term.
• Dropping the factor of $2$ in the second-order relation. The identity $x^2f_{xx}+2xyf_{xy}+y^2f_{yy}=n(n-1)f$ needs the coefficient $2$ on the mixed term, since $f_{xy}$ and $f_{yx}$ both contribute and are equal by Clairaut's theorem.
• Declaring harmonicity from only one second partial. Both $f_{xx}$ and $f_{yy}$ must be computed, and it is specifically their sum that must vanish.

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