IM Gelfand and SV Fomin, Calculus of Variations
Sec. 4.2, Example 2: Partial Walkthrough

Mark Vuletic

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Source: IM Gelfand, SV Fomin. 1963. Calculus of Variations. Mineola, NY: Dover. Tr. RA Silverman. p. 20.

Problem: At one point in Sec. 4.2, Example 2, Gelfand and Fomin move from

$$x + C_1 = C \ln \frac{y + \sqrt{y^2 - C^2}}{C}$$

to

$$y = C \cosh \frac{x + C_1}{C}$$

This move did not seem immediately transparent to me, so I fill in the details here.

Walkthrough:

When I started this, all I remembered about $$\cosh$$ is that it is expressed as a function of some exponential terms. I realized that in isolating $$y$$—something I would have to do anyway—I would end up with exponential terms (see steps 3 and 4), so my strategy was just to isolate $$y$$ and then look up $$\cosh$$ to see whether it was algebraically equivalent to what I had at that point.

$$x + C_1 = C \ln \frac{y + \sqrt{y^2 - C^2}}{C} \tag{1}$$

Divide through by $$C$$:

$$\frac{x + C_1}{C} = \ln \frac{y + \sqrt{y^2 - C^2}}{C} \tag{2}$$

Since for all expressions $$\psi$$ and $$\phi$$, $$\psi = \phi$$ implies $$e^\psi = e^\phi$$, we can change 2 to

$$e^{\frac{x + C_1}{C}} = e^{\ln \frac{y + \sqrt{y^2 - C^2}}{C}} \tag{3}$$

Since for any expression $$\psi$$, $$e^{\ln \psi} = \psi$$, we get

$$e^{\frac{x + C_1}{C}} = \frac{y + \sqrt{y^2 - C^2}}{C}. \tag{4}$$

Multiply through by $$C$$:

$$C e^{\frac{x + C_1}{C}} = y + \sqrt{y^2 - C^2} \tag{5}$$

Subtract $$y$$ from both sides:

$$C e^{\frac{x + C_1}{C}} - y = \sqrt{y^2 - C^2} \tag{6}$$

Square both sides, remembering that for any expression $$\psi$$, $$(e^\psi)^2 = e^{2\psi}$$:

$$C^2e^{2\left(\frac{x+C_1}{C}\right)} -2Ce^{\frac{x+C_1}{C}}y + y^2 = y^2 - C^2 \tag{7}$$

Subtract $$y^2$$ from both sides:

$$C^2e^{2\left(\frac{x+C_1}{C}\right)} -2Ce^{\frac{x+C_1}{C}}y = - C^2 \tag{8}$$

Divide through by $$C$$:

$$Ce^{2\left(\frac{x+C_1}{C}\right)} -2e^{\frac{x+C_1}{C}}y = - C \tag{9}$$

Add $$2e^{\frac{x+C_1}{C}}y + C$$ to both sides:

$$Ce^{2\left(\frac{x+C_1}{C}\right)} + C = 2e^{\frac{x+C_1}{C}}y \tag{10}$$

Divide through by $$2e^{\frac{x+C_1}{C}}$$ and then reverse the order:

$$y = \frac{Ce^{2\left(\frac{x+C_1}{C}\right)} + C}{2e^{\frac{x+C_1}{C}}} \tag{11}$$

At this point, I look up $$\cosh$$ and discover that by one of its definitions, $$\cosh \psi = \frac{e^{2\psi} + 1}{2e^\psi}$$. Comparing this with 11, I see what to do. Pull $$C$$ out of the numerator:

$$y = C \frac{e^{2\left(\frac{x+C_1}{C}\right)} + 1}{2e^{\frac{x+C_1}{C}}} \tag{12}$$

This has the form of $$\cosh \psi$$ described above, with $$\psi = \frac{x+C_1}{C}$$, so we have

$$y = C \cosh \left( \frac{x+C_1}{C}\right) \tag{13}$$

And there you have it.

Last updated: 25 Nov 2016

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