Wikipedia:Reference desk/Archives/Mathematics/2017 October 13

= October 13 =

Differential equations
I had a few problems that I was stuck on:
 * 1. Suppose a cell is suspended in a solution containing a solute of constant concentration $C_{s}$. Suppose further that the cell has constant volume $V$ and that the area of its permeable membrane is the constant $A$. By Fick's law the rate of change of its mass $m$ is directly proportional to the area $A$ and the difference $C_{s} − C(t)$, where $C(t)$ is the concentration of the solute inside the cell at time $t$. Find $C(t)$ if $m = V · C(t)$ and $C(0) = C_{0}$. Use $k > 0$ as the proportionality constant.
 * 2a. According to Stefan's law of radiation the absolute temperature T of a body cooling in a medium at constant absolute temperature $C(t) =$ is given by $T_{m}$, where $dT⁄dt = k(T^{4} - T_{m}^{4})$ is a constant. Stefan's law can be used over a greater temperature range than Newton's law of cooling. Solve the differential equation.
 * 2b. Using the binomial series, expand the right side of the following equation. (Only write the first three terms of the expansion.)
 * 3. A classical problem in the calculus of variations is to find the shape of a curve $k$ such that a bead, under the influence of gravity, will slide from point $T(t) =$ to point $dT⁄dt = k(T^{4} - T_{m}^{4})$ in the least time. It can be shown that a nonlinear differential equation for the shape $dT⁄dt = k[T_{m} + (T - T_{m})]^{4} - T_{m}^{4}$ of the path is $dT⁄dt = kT_{m}^{4}[1 + (T⁄T_{m}))^{4} - 1]$, where $dT⁄dt = kT_{m}^{4}[(____________________) - 1]$ is a constant. First solve for $C$ in terms of $A(0, 0)$ and $B(x_{1}, y_{1})$. Then use the substitution $y(x)$ to obtain a parametric form of the solution. The curve $y(1 + (dy⁄dx)^{2}) = k$ turns out to be a cycloid.
 * 2b. Using the binomial series, expand the right side of the following equation. (Only write the first three terms of the expansion.)
 * 3. A classical problem in the calculus of variations is to find the shape of a curve $k$ such that a bead, under the influence of gravity, will slide from point $dx$ to point $y$ in the least time. It can be shown that a nonlinear differential equation for the shape $dy$ of the path is $y = k sin^{2}(θ)$, where $C$ is a constant. First solve for $x(θ) =$ in terms of $C(t) = m⁄V$ and $dT⁄dt = k(T^{4} - T_{m}^{4})$. Then use the substitution $dT⁄T^{4} - T_{m}^{4} = k dt$ to obtain a parametric form of the solution. The curve $dT⁄(T^{2} + T_{m}^{2})(T + T_{m})(T - T_{m}) = k dt$ turns out to be a cycloid.
 * 3. A classical problem in the calculus of variations is to find the shape of a curve $A⁄(T^{2} + T_{m}^{2}) + B⁄(T + T_{m}) + C⁄(T - T_{m}) = 1$ such that a bead, under the influence of gravity, will slide from point $A, B, C ∈ ℝ$ to point $A(T + T_{m})(T - T_{m})⁄(T^{2} + T_{m}^{2})(T + T_{m})(T - T_{m}) + B(T^{2} + T_{m}^{2})(T - T_{m})⁄(T^{2} + T_{m}^{2})(T + T_{m})(T - T_{m}) + C(T^{2} + T_{m}^{2})(T + T_{m})⁄(T^{2} + T_{m}^{2})(T + T_{m})(T - T_{m}) = 1$ in the least time. It can be shown that a nonlinear differential equation for the shape $A(T + T_{m})(T - T_{m}) + B(T^{2} + T_{m}^{2})(T - T_{m}) + C(T^{2} + T_{m}^{2})(T + T_{m})⁄T^{4} - T_{m}^{4} = 1$ of the path is $B(T^{2} + T_{m}^{2})(T - T_{m})⁄(T^{2} + T_{m}^{2})(T + T_{m})(T - T_{m}) = 1$, where $T = -T_{m}$ is a constant. First solve for $B(2T_{m}^{2})(-2T_{m}) = 1$ in terms of $B = -1⁄4T_{m}^{3}$ and $C(T^{2} + T_{m}^{2})(T + T_{m})⁄(T^{2} + T_{m}^{2})(T + T_{m})(T - T_{m}) = 1$. Then use the substitution $T = T_{m}$ to obtain a parametric form of the solution. The curve $C(2T_{m}^{2})(2T_{m}) = 1$ turns out to be a cycloid.
 * 3. A classical problem in the calculus of variations is to find the shape of a curve $C = 1⁄4T_{m}^{3}$ such that a bead, under the influence of gravity, will slide from point $A(T + T_{m})(T - T_{m}) + B(T^{2} + T_{m}^{2})(T - T_{m}) + C(T^{2} + T_{m}^{2})(T + T_{m})⁄T^{4} - T_{m}^{4} = 1$ to point $A(T + T_{m})(T - T_{m}) + (-1⁄4T_{m}^{3})(T^{2} + T_{m}^{2})(T - T_{m}) + (1⁄4T_{m}^{3})(T^{2} + T_{m}^{2})(T + T_{m}) = T^{4} - T_{m}^{4}$ in the least time. It can be shown that a nonlinear differential equation for the shape $A(T^{2} - T_{m}^{2}) + -1⁄4T_{m}^{3}(T^{3} - T^{2}T_{m} + TT_{m}^{2} - T_{m}^{3}) + 1⁄4T_{m}^{3}(T^{3} + T^{2}T_{m} + TT_{m}^{2} + T_{m}^{3}) = T^{4} - T_{m}^{4}$ of the path is $A(T^{2} - T_{m}^{2}) + -1⁄4T_{m}^{3}(- T^{2}T_{m} - T_{m}^{3}) + 1⁄4T_{m}^{3}(T^{2}T_{m} + T_{m}^{3}) = T^{4} - T_{m}^{4}$, where $A(T^{2} - T_{m}^{2}) + 1⁄2T_{m}^{3}(T^{2}T_{m} + T_{m}^{3}) = T^{4} - T_{m}^{4}$ is a constant. First solve for $A(T^{2} - T_{m}^{2}) = T^{4} - T_{m}^{4} - 1⁄2T_{m}^{3}(T^{2}T_{m} + T_{m}^{3})$ in terms of $A = T^{4} - T_{m}^{4} - 1⁄2T^{2}T_{m}^{4} - 1⁄2T_{m}^{6}⁄T^{2} - T_{m}^{2}$ and $A = - T_{m}^{4} - 1⁄2T_{m}^{6}⁄- T_{m}^{2}$. Then use the substitution $T = 0$ to obtain a parametric form of the solution. The curve $A = T_{m}^{2} + 1⁄2T_{m}^{4}$ turns out to be a cycloid.
 * 3. A classical problem in the calculus of variations is to find the shape of a curve $A = T_{m}^{2} + 1⁄2T_{m}^{4}$ such that a bead, under the influence of gravity, will slide from point $B = -1⁄4T_{m}^{3}$ to point $C = 1⁄4T_{m}^{3}$ in the least time. It can be shown that a nonlinear differential equation for the shape $T_{m}^{2} + 1⁄2T_{m}^{4}⁄T^{2} + T_{m}^{2} - 1⁄4T_{m}^{3}⁄T + T_{m} + 1⁄4T_{m}^{3}⁄T - T_{m} = k dt$ of the path is $T_{m}^{2} + 1⁄2T_{m}^{4}⁄T^{2} + T_{m}^{2} dT -  1⁄4T_{m}^{3}⁄T + T_{m} dT +    1⁄4T_{m}^{3}⁄T - T_{m} dT = k dt$, where $(T_{m}^{2} + 1⁄2T_{m}^{4}) 1⁄T^{2} + T_{m}^{2} dT - 1⁄4T_{m}^{3}   1⁄T + T_{m} dT + 1⁄4T_{m}^{3}   1⁄T - T_{m} dT = k dt$ is a constant. First solve for $(T_{m}^{2} + 1⁄2T_{m}^{4})arctan(T⁄T_{m})⁄T_{m} - 1⁄4T_{m}^{3}ln(T + T_{m}) + 1⁄4T_{m}^{3}ln(T - T_{m}) = kt + D$ in terms of $D ∈ ℝ$ and $T_{m}^{2} + 1⁄2T_{m}^{4}⁄T_{m}arctan(T⁄T_{m}) + 1⁄4T_{m}^{3}(ln(T - T_{m}) - ln(T + T_{m})) = kt + D$. Then use the substitution $T_{m}^{2} + 1⁄2T_{m}^{4}⁄T_{m}arctan(T⁄T_{m}) + 1⁄4T_{m}^{3}ln(T - T_{m}⁄T + T_{m}) = kt + D$ to obtain a parametric form of the solution. The curve $T_{m}^{2} + 1⁄2T_{m}^{4}⁄T_{m}arctan(T⁄T_{m}) + ln(T - T_{m}⁄T + T_{m})^{{{sfrac|1|4}}T_{m}^{3}} = kt + D|undefined$ turns out to be a cycloid.
 * 3. A classical problem in the calculus of variations is to find the shape of a curve $A$ such that a bead, under the influence of gravity, will slide from point $B$ to point $C$ in the least time. It can be shown that a nonlinear differential equation for the shape $T^{4}⁄T_{m}^{4} + 4T^{3}⁄T_{m}^{3} + 6T^{2}⁄T_{m}^{2}$ of the path is $T^{4}⁄T_{m}^{4} - 4T^{3}⁄T_{m}^{3} + 6T^{2}⁄T_{m}^{2}$, where ᙭᙭᙭ is a constant. First solve for ᙭᙭᙭ in terms of ᙭᙭᙭ and ᙭᙭᙭. Then use the substitution ᙭᙭᙭ to obtain a parametric form of the solution. The curve ᙭᙭᙭ turns out to be a cycloid.

For 1, I wrote ᙭᙭᙭, but didn't know how to proceed from there.

For 2a, I tried separation of variables then factoring then partial fractions:

however, this answer was verified to be incorrect. I think I might have the wrong values for ᙭᙭᙭, ᙭᙭᙭, and ᙭᙭᙭.

For 2b, I tried ᙭᙭᙭ and ᙭᙭᙭, both of which were verified to be incorrect.

Any help would be appreciated. 147.126.10.148 (talk) 10:23, 13 October 2017 (UTC)

Differential equation 2a.

 * $$-\frac{dT}{dt} = k(T^4 - T_m^4)$$

Note the minus sign. The body is cooling when it is warmer than the surroundings. Choose your units of time and temperature such that the equation takes the form
 * $$-\frac{dT}{dt} = T^4 - 1$$.

Use perturbation.
 * $$-\frac{dT}{dt} = T^4 - \epsilon $$
 * $$T(t,\epsilon)\approx \sum _{n=0}^\infty T_n(t) \epsilon^n$$ (as $$\epsilon \rarr 0 $$)

The first term in the perturbation series
 * $$T_0(t)=\lim_{\epsilon \rarr 0 }T(t,\epsilon)$$

satisfies the differential equation
 * $$-\frac{dT_0}{dt} = T_0^4 $$
 * $$dt=-T_0^{-4}dT_0$$

Integrate
 * $$t=3^{-1}T_0^{-3}+t_0$$

Let the beginning of time be $$t_0=0$$.
 * $$t=3^{-1}T_0^{-3}$$
 * $$T_0 = (3t)^{-3^{-1}}$$

This is the cooling formula when the surrounding is at absolute zero.

Differentiate
 * $$dT_0 = d((3t)^{-3^{-1}})

= 3^{-3^{-1}}(-3^{-1}) t^{-3^{-1}-1} dt = -3^{-3^{-1}4} t^{-3^{-1}4} dt$$

= -(3^{-3^{-1}} t^{-3^{-1}})^4 dt = -T_0^4 dt$$ checking that the differential equation $$dT_0 = -T_0^4 dt$$ is satisfied.

Insert the next term in the perturbation series
 * $$T\approx T_0+T_1\epsilon$$ (as $$\epsilon\rarr 0$$)

into the differential equation
 * $$-\frac{d(T_0+T_1\epsilon)}{dt} = (T_0+T_1\epsilon)^4 - \epsilon$$

and expand to the first power in $$\epsilon$$
 * $$-\frac{d T_0}{dt} -\epsilon\frac{dT_1}{dt} = T_0^4+4T_0^3 T_1\epsilon - \epsilon$$

The constant terms vanish and the rest is divided by $$\epsilon$$
 * $$-\frac{dT_1}{dt} = 4T_0^3 T_1-1$$

Inserting $$T_0^3 = ((3t)^{-3^{-1}})^3=(3t)^{-1}=3^{-1}t^{-1}$$
 * $$-\frac{dT_1}{dt} = \frac{4 T_1}{3 t}-1$$

or
 * $$(\frac{d}{dt}+ \frac{4 }{3 t})T_1=1$$

This is an inhomogenous Linear differential equation.

The integrating factor is
 * $$e^{\int\frac{4 dt }{3 t}}=e^{\frac{4 }{3}\ln t}=t^{\frac {4}{3}}$$

and the differential equation becomes
 * $$d(T_1 t^{\frac 4 3})=t^{\frac 4 3}dt$$

integrate
 * $$T_1 t^{\frac 4 3}=\frac 3 7 t^{\frac 7 3}+C$$

and divide
 * $$T_1 =\frac 3 7 t+Ct^{-\frac 4 3}$$

Bo Jacoby (talk) 12:24, 13 October 2017 (UTC).


 * (Bo, I don't know what you're doing here, but it's probably well beyond the scope of what's expected.) The original poster's attempt at separating variables, and then integrating via partial fraction decomposition is most likely what's intended.  I'll point out that in general, when you have a term like $$\frac{1}{T^2 + T_m^2},$$ the numerator needs to be $$AT + B,$$ not just $$B.$$  In this case, it happens to work out because $$A = 0,$$ but it won't always be like that.  Otherwise, it's just a matter of being more careful with your algebra, because I think you've got the basic idea right (and I'm not going to try to wade through all your work, sorry ).  --Deacon Vorbis (talk) 15:35, 14 October 2017 (UTC)
 * Thank you! I expect that the differential equation is not solvable in terms of elementary functions. Most differential equations are not. So I tried to find an asymptotic expression. I think I failed. As $$t\rarr \infty$$ you should get $$T\rarr 1$$, and that is not what I got. I must have made mistakes. Bo Jacoby (talk) 18:51, 14 October 2017 (UTC).

Summary
The equation is $$-\frac{dT}{dt} = k(T^4 - T_m^4)$$. Substituting $$T = T_m y$$ gives $$3 dy = -3 k T_m^3 (y^4- 1)dt$$. (The factor 3 is for future convenience). Substituting $$x =3 kT_m^3 t$$ gives the dimensionless differential equation $$3 dy+y^4 dx = dx$$. This is attacked by perturbation theory
 * $$y\approx y_0+y_1\epsilon+y_2\epsilon^2+\cdots$$

satisfies
 * $$3 dy+y^4 dx=\epsilon dx$$

for $$\epsilon\rarr 0$$. Equating coefficients:
 * $$3 dy_0+y_0^4 dx=0$$
 * $$3 dy_1 +4 y_0^3 y_1 dx = dx$$
 * $$3 dy_2 +4 y_0^3 y_2 dx+ 6 y_0^2 y_1^2 dx = 0$$

The solutions are
 * $$y_0=x^{-3^{-1}}$$
 * $$y_1=7^{-1} x+C_1 x^{-3^{-1}4}$$

where $$C_1$$ is a constant of integration.

The next equation is
 * $$3 dy_2 +(4x^{-1} y_2 + 7^{-2}6 x^{3^{-1}4}+6 C^2

x^{-3^{-1}10}+7^{-1}12 C x^{-1}) dx = 0$$

Bo Jacoby (talk) 06:42, 15 October 2017 (UTC).

Forget all that nonsense. The integral is elementary:  Bo Jacoby (talk) 18:26, 19 October 2017 (UTC).