Seen Questions

Exact differential equation

Approximate a function linearly

Gradient normal to a level set

Exact equations and level-set geometry

Answer the following questions about exact differential equations.

(a)

First classify the differential equation

Select all options which correctly describe this equation:

• non-linear;
• first order;
• second order;
• initial value problem;
• boundary value problem;
• linear;
• constant coefficient;
• fifth order.

Now consider a differential equation of the form

Here $M,N:\mathbb R^2\to\mathbb R$ are functions which are continuous everywhere, differentiable everywhere and whose first-order partial derivatives are also continuous everywhere.

(b)

Suppose equation (1) is not exact. To use the integrating-factor method, determine when multiplying by a differentiable, nonvanishing function $u(x,y)$ produces an equivalent exact equation. State a necessary condition.

(c)

Now suppose equation (1) is exact. For a constant $c$, its solution may be written in the form

where $F(x,y)$ is to be constructed from $M$ and $N$.

(d)

Suppose (1) is an exact differential equation with solution $F(x,y)=c$, where $c$ is a constant. This solution can be interpreted as a level curve of the surface

To understand the behaviour of the solution near a given point $(x_0,y_0)$, Liu wishes to use Taylor polynomials to find an approximation. Give the linear approximation to the surface near $(x_0,y_0)$.

(e)

For the geometry of solutions, consider the following example.

Let $F:\mathbb R^2\to\mathbb R$ be a potential for the exact differential form below, with continuous second-order derivatives. Consider its level curve

which passes through the point $(5,4)$ and for which locally the function $y(x)$ is implicitly defined by the exact differential equation

Compute a normal vector to the surface $z=F(x,y)$ at the point $(5,4,20)$.

Solution

The first equation is in differential form, so it contains $dx$ and $dy$ but no derivative of order greater than one. It is first order. It is non-linear because, after writing it as an equation for $dy/dx$, the terms $y^2$ and $e^y$ occur non-linearly. It is neither an initial-value nor a boundary-value problem because no data select a particular solution and it is not a constant-coefficient linear equation. The correct selections in part (a) are

For equation (1), multiplying by $u$ gives

Exactness requires the mixed partials of the new coefficients to agree. Therefore, a necessary condition is

Expanding both sides gives the equivalent first-order equation for the integrating factor:

The unknown function $u$ must satisfy this equation: multiplying by an arbitrary non-zero function is not enough, because the new coefficients must pass the exactness test. On a simply connected domain, with the required continuous first partial derivatives, this equality is also sufficient for a potential; nonvanishing $u$ is what preserves the original solution curves.

If the original equation is exact, there is a potential function $F$ such that

Integrating $F_x=M$ with respect to $x$, while treating $y$ as constant, gives

where $h$ is an arbitrary function of $y$. Differentiating this expression with respect to $y$ gives

Matching this with $N(x,y)$ determines $h'(y)$ and hence $h(y)$. Therefore the requested construction is

The function $F$ is determined only up to an additive constant, which is absorbed into the constant $c$ in $F(x,y)=c$.

For small changes $\Delta x$ and $\Delta y$, the partial derivatives give the first-order change in height:

Replacing $\Delta x$ and $\Delta y$ by $x-x_0$ and $y-y_0$ gives the tangent plane to $z=F(x,y)$ at $(x_0,y_0,F(x_0,y_0))$:

Since $F_x=M$ and $F_y=N$, the same plane can be written as

For part (e), the differential equation identifies the first partial derivatives of a potential:

Integrating the first expression with respect to $x$ gives

Differentiating with respect to $y$ gives

Comparison with the coefficient of $dy$ gives $h'(y)=0$, so $h$ is a constant. That constant can be chosen so that $F(5,4)=20$.

For the graph $z=F(x,y)$, at $(x,y,F(x,y))$ changing $x$ while holding $y$ fixed gives the tangent direction $(1,0,F_x(x,y))$; changing $y$ while holding $x$ fixed gives $(0,1,F_y(x,y))$. A normal vector must be perpendicular to both. The vector $(-F_x(x,y),-F_y(x,y),1)$ has zero dot product with each direction, since $-F_x+F_x=0$ and $-F_y+F_y=0$. Thus one normal vector is

At $(5,4)$,

Consequently, one normal vector at $(5,4,20)$ is

An exact differential form gives a level curve of a potential. The graph of the same potential is a surface in three dimensions and its partial derivatives supply the tangent-plane and normal information. The linear approximation is the tangent plane, while the original solution curve is obtained by intersecting that surface with the horizontal plane $z=c$.

Integrating factor

Characteristic equation

Undetermined coefficients

First- and second-order price models

Suppose a price $B(t)$ is modelled by

where $G(t)$ is a target price, $a(t)$ is a relaxation rate and $P(t)$ is an additional price-change rate. These are prescribed functions of $t$, not of $B$. Their particular values are not specified in part (a).

(a)

Which classes must Equation (1) belong to? Select all that apply:

• linear;
• non-linear;
• homogeneous;
• inhomogeneous;
• an exact differential equation;
• a separable differential equation.

(b)

At the launch of the coin, measurements gave

Rewrite the equation as

so that

Find $\phi(t)$.

(c)

A second illustrative price $N(t)$ satisfies

(i)

To find the homogeneous equation, solve the associated characteristic equation, written in terms of a variable $\nu$:

(ii)

Find a particular solution $N_p$ that contains no undetermined coefficients.

Solution

Expand the right-hand side of (1) and collect the terms involving $B$:

This is linear in the unknown function $B$. It is not necessarily inhomogeneous: $aG+P$ could vanish identically (for example, $a=G=1$ and $P=-1$). Nor must it be separable, although some choices of the prescribed functions make it so. In differential form,

the partial derivative of the $dt$ coefficient with respect to $B$ is $a(t)$, whereas the partial derivative of the $dB$ coefficient with respect to $t$ is zero. Exactness is therefore not guaranteed either. The only classification that must hold in part (a) is

For part (b), the coefficient of $B$ in standard linear form is $p(t)=a(t)=t^9$. The integrating factor is

Since $\phi'(t)=t^9\phi(t)$, multiplying the standard equation by $\phi$ gives

Thus,

The general solution includes an integration constant:

For part (c), first move all terms involving $N$ to the left:

The associated homogeneous equation is

Using the trial solution $N=e^{\nu t}$ gives

Since $e^{\nu t}\ne0$, the characteristic equation is

The repeated root is $\nu=-5$, so the homogeneous solution is

The forcing term $8e^{-5t}$ has the same exponential as the repeated characteristic root. Both the usual trial $Ce^{-5t}$ and the once-corrected trial $Cte^{-5t}$ belong to the homogeneous solution space, so use

Write the differential operator as

For $N_p=At^2e^{-5t}$,

and hence

Substitution into the left-hand side gives

Matching this with $8e^{-5t}$ gives

Therefore

The complete general solution is

The first model is a first-order linear equation: the integrating factor creates the product rule. The second uses the constant-coefficient method for a second-order equation. Its repeated characteristic root explains the factors $1,t$ in the homogeneous solution and the extra factor $t^2$ in the particular trial.