Partial differential equations and Green functions

Read a field with two inputs

A scalar field assigns one scalar value to each allowed combination of inputs. Here u(x, t) assigns a length-valued number to a spatial coordinate x and a time label t. Fix a straight coordinate line, origin x = 0 and right as positive; x uses metres and t uses seconds from a declared start. The field value u also uses metres, but it is not the coordinate of a moving body.

The partial derivative uₜ = ∂u/∂t differentiates with respect to t while holding x fixed; uₓ = ∂u/∂x holds t fixed. The notation uₓₓ = ∂²u/∂x² means differentiate twice with respect to x. For our length-valued u, their units are respectively m/s, m/m and 1/m. A dimensionless spatial gradient is not therefore a time velocity.

An ordinary differential equation, or ODE, concerns an unknown function of one independent variable and its ordinary derivatives. A partial differential equation, or PDE, concerns an unknown function of several independent variables and its partial derivatives. A PDE constrains a field; a formula supplying that field is a candidate solution to check. Familiar algebra, derivatives with other inputs held fixed, and definite integrals are enough for the examples below.

Specify an equation, domain and units

Supply the exact coefficient κ = 1 m²/s and the equation uₜ = κuₓₓ. This has diffusion form. Here it is a stipulated mathematical exercise for a length-valued field, without a claim that a particular material or measured process obeys it. Every coefficient and number in this article is supplied exactly for calculation.

The field is supplied on 0 ≤ x ≤ 2 m and 0 ≤ t ≤ 1 s. We impose the PDE on the interior 0 < x < 2 m, 0 < t < 1 s; the edges will carry data. Both PDE sides have m/s units, because (m²/s)(1/m) = m/s. Units are a necessary consistency check, not experimental evidence for the equation.

The order is the highest derivative order appearing in the equation. The selected PDE is second order because of uₓₓ, although it is first order in t. Order does not count how many input variables there are. Neither order alone nor the presence of a time label tells us which data will make an arbitrary problem solvable.

Separate initial and boundary data

An initial condition gives a starting profile over space. A spatial boundary condition constrains the field at the edges over the time interval. A PDE problem includes its domain and appropriate data, not only an equation. For our selected problem prescribe u(x, 0) = 1 m + x²/(1 m) for 0 ≤ x ≤ 2 m.

Prescribe the endpoint values u(0, t) = 1 m + (2 m/s)t and u(2 m, t) = 5 m + (2 m/s)t for 0 ≤ t ≤ 1 s. These specify values rather than slopes; such fixed-value conditions are called Dirichlet conditions. The values vary with t even though the endpoints in x stay fixed.

At the two corners with t = 0, the initial profile gives 1 m and 5 m, matching the boundary values. This simple compatibility check matters for a candidate continuous at the corners. It is not a complete test of whether every PDE and every choice of data has a solution. Different equations and boundary choices require their own conditions.

Check a supplied exact candidate

Supply u(x, t) = 1 m + x²/(1 m) + (2 m/s)t on the declared rectangle. Holding x fixed gives uₜ = 2 m/s. Holding t fixed gives uₓ = 2x/(1 m) and uₓₓ = 2/(1 m). Thus κuₓₓ = 2 m/s, exactly matching uₜ throughout the interior.

Setting t = 0 gives the prescribed initial profile. Setting x = 0 and x = 2 m gives the two prescribed boundary functions. The polynomial is continuous on the closed rectangle and has the needed derivatives in its interior. These checks establish that this candidate satisfies this equation and these data.

For example, at x = 1 m and t = 0.5 s it gives 1 m + 1 m + 1 m = 3 m. Adding a constant 1 m to the whole candidate would leave both derivatives in the PDE unchanged, but would change the initial and boundary values. Solving the equation alone is therefore a different requirement from solving this stated problem. Existence, uniqueness and physical adequacy are separate questions; this substitution does not settle them for other problems.

Use linearity with the data

Define the differential operator D[f] = fₜ − κfₓₓ: it is a rule that takes a sufficiently differentiable field f and returns another field. It is linear because D[af + bg] = aD[f] + bD[g] for dimensionless constant real numbers a and b and compatible fields. Distributing the partial derivatives verifies that property here. Our first candidate satisfies D[u] = 0 m/s.

An equation D[f] = r(x, t) has a supplied forcing or source term r. This word does not by itself mean a mechanical force; here r has m/s units. A zero right side makes the equation homogeneous; a nonzero right side makes it inhomogeneous. Supply v(x, t) = (3 m/s)t, uniform in x. Then D[v] = 3 m/s and D[u + v] = 3 m/s.

Adding fields also adds their initial and boundary values. The field u + v has the same initial profile as u because v(x, 0) = 0 m, but its endpoint values now contain (5 m/s)t. Likewise, 2u still solves D[2u] = 0 m/s but doubles the nonzero prescribed data. For our homogeneous linear equation, linear combinations of solutions with zero initial profiles and zero Dirichlet endpoint values retain those zero data. Always check the data as well as the operator.

Understand the role of a Green function

For a suitable linear problem, a Green function describes the response to a unit localized source. Integrating those responses, weighted by a supplied source, can build its response. The kernel must belong to the specified operator, domain and boundary conditions. A kernel is the two-input weighting function inside such an integral.

This construction is conditional: the appropriate Green function and representation must exist for the problem and class of sources being considered. Nonzero boundary or initial data may require additional response terms. A single kernel is not a universal method for every PDE, and linearity alone does not prove existence or uniqueness. We do not require the more advanced mathematics of ideal point sources here.

For an explicit check, now select a separate equation P[w] = −wₓₓ = h(x, t) on 0 < x < L, with L = 2 m, 0 ≤ t ≤ 1 s and w(0, t) = w(L, t) = 0 m. Here w is length-valued and h has 1/m units. There is no time derivative: for each fixed t this is a one-dimensional spatial boundary problem. The label t selects a profile, rather than an initial-value evolution rule.

Check the interval kernel

Let x be the position where we want the response, and let z be the source position; both use metres in 0 ≤ x, z ≤ L. For this zero-endpoint problem use G(x, z) = x(L − z)/L when x ≤ z, and G(x, z) = z(L − x)/L when x ≥ z. Each expression has metre units. At x = z they give the same value z(L − z)/L, so the definition is unambiguous.

For a fixed interior source position z, the kernel vanishes at x = 0 and x = L. Each piece is linear in x, so Gₓₓ = 0 away from x = z. The x slopes are (L − z)/L to the left and −z/L to the right. The right slope minus the left slope is −1. This jump encodes the unit-source normalization for our negative-second-derivative operator. The kernel has a corner there; its ordinary x derivative, and hence its ordinary second derivative, is not defined at that corner.

The endpoints and length L occur explicitly in G. Replacing a boundary condition or changing the interval requires checking an appropriate kernel again. For example, at x = 0.5 m and z = 1.5 m, the x ≤ z branch gives G = (0.5 m)(0.5 m)/(2 m) = 0.125 m. This kernel calculation supplies no measured spatial response or moving-object path.

Integrate the specified source

For this spatial problem the response representation is w(x, t) = ∫₀ᴸ G(x, z)h(z, t) dz. Integrate in the source coordinate z while holding x and t fixed. The units are m × (1/m) × m = m. Supply the original source h(z, t) = (2/(1 m))(1 + t/τ), uniform in z, with τ = 1 s. The ratio t/τ is dimensionless.

At a fixed t, write this constant source value as h₀(t). Splitting the integral at z = x gives two contributions: h₀(t)(L − x)x²/(2L) and h₀(t)x(L − x)²/(2L). Their sum is h₀(t)x(L − x)/2. Thus w(x, t) = (1/(1 m))(1 + t/τ)x(L − x). Directly, wₓₓ = −h₀(t), and both endpoint values are zero, checking the response against the selected equation and data.

At x = 1 m the response is 1 m for t = 0 and 2 m for t = 1 s. Doubling this source while retaining the zero boundary values doubles this response. These are exact consequences of the selected spatial model, not observed motion between those profiles. A PDE containing time derivatives would need its own initial data and appropriate response construction. Keep equation, domain, data, units and claimed physical meaning together.

Worked examples

Try it yourself

1. An unknown field f(x, t) satisfies fₜ = κfₓₓ with the specified nonzero κ. Is this a PDE or ODE, and what is its overall order?

   Show the answer and reasoning

   It is a second-order PDE, first order in t and second order in x.

   The unknown has two independent inputs and the equation uses partial derivatives. The largest derivative order present is two; the number of variables does not set the order.

2. For the supplied u = 1 m + x²/(1 m) + (2 m/s)t, find uₓ at x = 0.5 m. Does that derivative give a time velocity?

   Show the answer and reasoning

   uₓ = 2x/(1 m) = 1, dimensionless. It is a spatial gradient, not a time velocity.

   t is held fixed while differentiating in x. Both the field and x use length units, so the ratio is m/m. A time derivative here has m/s units and a different meaning.

3. Let g = u + 1 m, using the verified u and the original prescribed data. Does g satisfy the PDE? Does it satisfy the complete original problem?

   Show the answer and reasoning

   It satisfies the PDE but fails the original initial and boundary data.

   The constant addition has zero t derivative and zero second x derivative, so D[g] = 0 m/s. Its initial and endpoint values are all 1 m above the original requirements.

4. The original u solves D[u] = 0 m/s and has u(0, 0) = 1 m. Does 2u solve the same differential equation? Does it retain that original corner value?

   Show the answer and reasoning

   The equation is retained, but the corner value becomes 2 m.

   Linearity gives D[2u] = 2D[u] = 0 m/s. It also doubles every nonzero initial and boundary value, so the complete prescribed problem is changed.

5. For L = 2 m, evaluate G at x = 0.5 m, z = 1.5 m. For this fixed z, what are the x slopes on either side of x = z, and their right-minus-left jump?

   Show the answer and reasoning

   G = 0.125 m; the left slope is 0.25, the right slope is −0.75, and the jump is −1.

   x < z selects x(L − z)/L = (0.5 m)(0.5 m)/(2 m). The slopes are (L − z)/L = 0.25 and −z/L = −0.75. They differ, so no ordinary x derivative exists at the interior corner.

6. At t = 1 s the selected source gives w(1 m, 1 s) = 2 m. If that source is doubled everywhere in x while the zero endpoint conditions and L stay the same, what is the new midpoint response? Does this establish motion from t = 0?

   Show the answer and reasoning

   The new midpoint response is 4 m. It does not establish time evolution or a moving object's path.

   The integral with this unchanged kernel is linear in the source, so doubling the source doubles the selected response and retains zero endpoints. The operator −∂²/∂x² contains no time derivative; t labels the family of spatial problems.

What to carry forward

• A PDE constrains an unknown field and its partial derivatives; order counts the highest derivative order.
• State the domain, compatible units and appropriate initial and boundary data.
• Substitution can verify a supplied candidate without proving a general existence or uniqueness theorem or a physical law.
• Linear combinations combine forcing and data as well as field values.
• A Green function belongs to a specified linear problem, including its domain and boundary conditions.
• The checked interval kernel builds spatial responses; a time label alone does not supply time evolution.