Initial Curve Problems and Integral Surfaces

We now discuss the initial curve problem in PDE. This is the geometrical form of prescribing data for a first-order quasi-linear equation. The main question is simple. Can we find an integral surface which passes through a given curve? Consider the first-order quasi-linear equation

\begin{equation*} a(x,y,u)u_x+b(x,y,u)u_y-c(x,y,u)=0. \quad \text{(1)} \end{equation*}

A solution

u=u(x,y)

gives an integral surface in

(x,y,u)

-space. This surface may be written as

\begin{equation*} f(x,y,u)=u(x,y)-u=0. \quad \text{(2)} \end{equation*}

Suppose the value of

u

is prescribed on a plane curve

C

in the

(x,y)

-plane. Then every point of

C

, together with its assigned value of

u

, gives a point in

(x,y,u)

-space. These points form a space curve, usually denoted by

\Gamma

An initial curve problem for a first-order PDE asks us to find an integral surface which satisfies the PDE and passes through a given space curve

\Gamma

determined by prescribed data on a plane curve

C

\begin{equation*} a(x,y,u)u_x+b(x,y,u)u_y-c(x,y,u)=0. \quad \text{(1)} \end{equation*}

A solution

u=u(x,y)

gives an integral surface in

(x,y,u)

-space. This surface may be written as

\begin{equation*} f(x,y,u)=u(x,y)-u=0. \quad \text{(2)} \end{equation*}

Suppose the value of

u

is prescribed on a plane curve

C

in the

(x,y)

-plane. Then every point of

C

, together with its assigned value of

u

, gives a point in

(x,y,u)

-space. These points form a space curve, usually denoted by

\Gamma

An initial curve problem for a first-order PDE asks us to find an integral surface which satisfies the PDE and passes through a given space curve

\Gamma

determined by prescribed data on a plane curve

C

The initial curve

C

lies in the

(x,y)

-plane, but the actual initial data must also include the corresponding value of

u

. Choose a plane curve and a rule for assigning

u

, then click Lift Curve to see how

C

becomes the space curve

\Gamma

. The curve on the plane shows only

(x,y)

, while the lifted curve shows

(x,y,u)

. This distinction matters because the integral surface must pass through

\Gamma

, not merely through

C

The curve

C

lies in the

(x,y)

-plane. The curve

\Gamma

lies in

(x,y,u)

-space. The curve

\Gamma

carries the actual initial data because it includes the corresponding values of

u

An integral surface through a given curve

\Gamma

is a solution surface of the PDE which contains every point of

\Gamma

An integral surface through

\Gamma

must contain every lifted initial-data point. Choose a sample surface and click Show Surface to see a curve

\Gamma

lying on that surface. The surface represents a possible solution

u=u(x,y)

, and the curve marks the prescribed data. If the curve were not on the surface, the surface would not satisfy the initial condition.

Geometrically, the initial data gives a starting curve. From this curve, we try to grow a surface. The characteristic direction then tells us how the surface should extend from the given curve. For equation (1), the characteristic direction is given by the vector field

(a,b,c)

. Hence the characteristic equations are

\begin{equation*} \frac{dx}{dt}=a(x,y,u), \quad \frac{dy}{dt}=b(x,y,u), \quad \frac{du}{dt}=c(x,y,u). \quad \text{(3)} \end{equation*}

The initial curve supplies starting points, and the characteristic equations tell how to move away from those points. Click Build Surface to send several characteristic curves out from a sample curve

\Gamma

. Together, these curves sweep out a candidate integral surface. This is the geometric idea behind solving an initial curve problem by characteristics.

Starting from points of

\Gamma

, these equations generate characteristic curves. The required integral surface is obtained by placing these characteristic curves together, provided no obstruction occurs.

The initial curve

\Gamma

supplies the starting points. The characteristic equations supply the directions of motion. The union of the characteristic curves starting from

\Gamma

gives the candidate integral surface.

A point on

\Gamma

becomes a starting point for a characteristic curve. Enter a point, choose the characteristic direction values, and click Compute Step to move once along that direction. The calculator uses the rule

\Delta x=h a

\Delta y=h b

, and

\Delta u=h c

. This gives a numerical version of how the surface begins to grow from the initial curve.

This interpretation is very useful. It changes the problem of constructing a surface into the problem of solving ordinary differential equations with initial points on a curve.

Suppose the initial data is given on a plane curve

C

in the

(x,y)

-plane. If a point on

C

(x_0,y_0)

and the prescribed value is

u_0

, then the corresponding point in space is

(x_0,y_0,u_0)

. The set of all such points forms the curve

\Gamma

(x,y,u)

-space.

The lifting process sends each point

(x_0,y_0)

C

to a point

(x_0,y_0,u_0)

\Gamma

. Choose a plane curve and an initial data rule, then click Show Lift to display both curves. The plane curve stays at height

u=0

for reference, while the lifted curve uses the prescribed values of

u

. This shows exactly how initial data becomes a space curve.

This topic is also described by phrases such as geometrical Cauchy problem and integral surface through a curve. We use Initial Curve Problems and Integral Surfaces because it directly connects the given data curve with the solution surface.

BMLabs Mathematics

BMLabs Mathematics

Initial Curve Problems and Integral Surfaces

Initial Curve Problems and Integral Surfaces

Plane Curve to Space Curve Viewer

Integral Surface Through Curve Viewer

Characteristic Surface Builder

Initial Point Characteristic Step Calculator

Initial Data Lifting Explorer

Coordinates and Locus

Invariants under Orthogonal Transformation