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Mathematics LibreTexts

1.1: Examples

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Example 1.1.1:

uy=0, where u=u(x,y). All functions u=w(x) are solutions.

Example 1.1.2:

ux=uy, where u=u(x,y). A change of coordinates transforms this equation into an equation of the first example. Set ξ=x+y, η=xy, then
u(x,y)=u(ξ+η2,ξη2)=:v(ξ,η).
Assume uC1, then
vη=12(uxuy).
If ux=uy, then vη=0 and vice versa, thus v=w(ξ) are solutions for arbitrary C1-functions w(ξ). Consequently, we have a large class of solutions of the original partial differential equation: u=w(x+y) with an arbitrary C1-function w.

Example 1.1.3:

A necessary and sufficient condition such that for given C1-functions M, N the integral
P1P0 M(x,y)dx+N(x,y)dy
is independent of the curve which connects the points P0 with P1 in a simply connected domain ΩR2 is that the partial differential equation (condition of integrability)
My=Nx
in Ω.

Independence of the path

Figure 1.1.1: Independence of the path

This is one equation for two functions. A large class of solutions is given by M=Φx, N=Φy, where Φ(x,y) is an arbitrary C2-function. It follows from Gauss theorem that these are all C1-solutions of the above differential equation.

Example 1.1.4: Method of an integrating multiplier for an ordinary

differential

equation

Consider the ordinary differential equation
M(x,y)dx+N(x,y)dy=0
for given C1-functions M, N. Then we seek a C1-function μ(x,y) such that μMdx+μNdy is a total differential, i. e., that (μM)y=(μN)x is satisfied. This is a linear partial differential equation of first order for μ:
$$
M\mu_y-N\mu_x=\mu (N_x-M_y).
\]

Example 1.1.5:

Two C1-functions u(x,y) and v(x,y) are said to be functionally dependent if
det(uxuyvxvy)=0,
which is a linear partial differential equation of first order for u if v is a given C1-function. A large class of solutions is given by
u=H(v(x,y)),
where H is an arbitrary C1-function.

Example 1.1.6: Cauchy-Riemann equations

Set f(z)=u(x,y)+iv(x,y), where z=x+iy and u, v are given C1(Ω)-functions. Here Ω is a domain in R2. If the function f(z) is differentiable with respect to the complex variable z then u, v satisfy the Cauchy-Riemann equations
ux=vy,  uy=vx.
It is known from the theory of functions of one complex variable that the real part u and the imaginary part v of a differentiable function f(z) are solutions of the Laplace equation
u=0,  v=0,
where u=uxx+uyy.

Example 1.1.7: Newton Potential

The Newton potential
u=1x2+y2+z2
is a solution of the Laplace equation in R3(0,0,0), that is, of
$$
u_{xx}+u_{yy}+u_{zz}=0.
\]

Example 1.1.8: Heat equation

Let u(x,t) be the temperature of a point xΩ at time t, where ΩR3 is a domain. Then u(x,t) satisfies in Ω×[0,) the heat equation
ut=ku,
where u=ux1x1+ux2x2+ux3x3 and k is a positive constant. The condition
u(x,0)=u0(x),  xΩ,
where u0(x) is given, is an initial condition associated to the above heat equation. The condition
u(x,t)=h(x,t),  xΩ, t0,
where h(x,t) is given, is a boundary condition for the heat equation.

If h(x,t)=g(x), that is, h is independent of t, then one expects that the solution u(x,t) tends to a function v(x) if t. Moreover, it turns out that v is the solution of the boundary value problem for the Laplace equation
v=0  in Ωv=g(x)  on Ω.

Example 1.1.9: Wave equation

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Figure 1.1.2: Oscillating string

The wave equation
utt=c2u,
where u=u(x,t), c is a positive constant, describes oscillations of membranes or of three dimensional domains, for example. In the one-dimensional case
utt=c2uxx
describes oscillations of a string.

Associated initial conditions are
u(x,0)=u0(x),  ut(x,0)=u1(x),
where u0, u1 are given functions. Thus the initial position and the initial velocity are prescribed.

If the string is finite one describes additionally boundary conditions, for example
$$
u(0,t)=0,\ \ u(l,t)=0\ \ \mbox{for all}\ t\ge 0.
\]

Contributors and Attributions


This page titled 1.1: Examples is shared under a not declared license and was authored, remixed, and/or curated by Erich Miersemann.

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