Finding a potential function · Differential Equations

Intuition

Once the test passes, the potential is recovered in two integrations and one comparison. Integrate M in x, holding y fixed: that gives F up to a function of y alone, because anything that depends only on y vanishes when differentiated in x. Then differentiate the result in y and match it with N, which pins down the missing function. The solutions are the level curves of F.

Rebuilding a hill from two sets of survey notes, one giving the east–west slope everywhere and the other the north–south slope. The east–west notes give the shape of every east–west slice but not how the slices sit relative to one another; the north–south notes fix that last piece.

plane diagram: ellipses-of-one-potential

The exact equation $(2x+y)\,dx+(x+2y)\,dy=0$ has potential $F=x^{2}+xy+y^{2}$ , and its solutions are the level curves $F=C$ : tilted ellipses, drawn for $C=0.8$ , $2.5$ and $5.5$ .

Integrate, differentiate, match

For an exact $M\,dx+N\,dy=0$ : integrate $M$ in $x$ with $y$ fixed, adding an unknown ; differentiate the result in and set it equal to ; the equation that results gives as a function of alone — if it does not, the equation is not exact or an integration went wrong; integrate for and write .

F=\int M\,dx+g(y),\qquad F_{y}=N\ \Longrightarrow\ g'(y)

The steps, and the check

For $(2x+y)\,dx+(x+2y)\,dy=0$ : $F =$ , then gives , , and .

Why the missing function depends on y alone

M_{y}=N_{x},\ G_{x}=M\ \Longrightarrow\ \frac{\partial}{\partial x}\bigl(N-G_{y}\bigr)=0

After integrating M in x, the candidate G has G_x = M. What is still missing is the difference between N and G_y, and the construction needs it to depend on y alone so that it can be g prime. Differentiate it in x: N_x minus the mixed derivative of G, which may be taken in the other order and is then M_y. The test says N_x equals M_y, so the derivative in x is zero and the difference depends on y alone. That is exactly where the test enters the construction.

Proof steps

G=\int M\,dx\ \Longrightarrow\ G_{x}=M

Integrate M in x with y held fixed.

h=N-G_{y}

What is still missing is the part of N that G does not produce.

h_{x}=N_{x}-G_{yx}=N_{x}-G_{xy}=N_{x}-M_{y}

Differentiate in x and swap the order of the mixed derivative.

h_{x}=0

The exactness test makes this zero, so h depends on y alone.

F=G+\int h(y)\,dy

Adding an antiderivative of h gives F with both partial derivatives right: a potential.

Applications

Practice

Integrate M, Keep a Function of y

Integrating in $x$ treats $y$ as a constant, so the constant of integration may be any function of $y$ .

\int(2x+y)\,dx=x^{2}+xy+g(y)

Try it

For $(2xy+1)\,dx+(x^{2}+3y^{2})\,dy=0$ , integrating $M$ in $x$ gives which expression?