Consider the heat equation: $$\partial_t u-div(A\nabla u)=f$$

with $u(0)=0, u=0$ on the boundary of the domain of definition, call it $U$. Consider a test function $v=v(x,t)$, and perform the following operations:

  1. integrate the strong formulation from $0$ to $t$: $$u(t)-\int_0^t div(A\nabla u)(\tau)d\tau=...$$
  2. integrate in space and take a convolution product with $v$: $$\int_0^T\int_U u(t,x)dx v(T-t,x)dt-\int_0^T\int_U\int_0^t div(A\nabla u)(\tau,x)d\tau v(T-t,x)dx dt=...$$
  3. apply, in space, the divergence theorem: $$\int_U (u(\cdot,x)*v(\cdot,x))(T)dx +\int_U\left(\left (\int_0^\cdot (A\nabla u)(\tau,x)d\tau\right ) * \nabla v(\cdot,x) \right)(T)dx = ...$$

Now, if $A$ is constant in time, we get a symmetric formulation. In fact, doing some simple change of variables yields:

$$\left(\left (\int_0^\cdot \nabla u(\tau,x)d\tau\right ) * \nabla v(\cdot,x) \right)(T) = \left(\left (\int_0^\cdot \nabla v(\tau,x)d\tau\right ) * \nabla u(\cdot,x) \right)(T)$$

This also shows that as soon as $A$ is not constant in time, symmetry cannot be expected. Yet, the authors of this very short paper seem to claim this is the case, in the first page.

Can anyone confirm that the this convolution-based formulation in non-symmetric in the general case, even if $u,v,A$ are taken as continuous piecewise linear functions in time? It seems strange that a peer reviewed paper contains such an error, so that probably I am making a mistake somewhere.


  • I have decided to use a linear heat equation, which one would obtain after linearization of the formulation in the linked paper
  • my point I think could be made with ODEs directly, the space variable does not play a role here. I decided to keep it to mantain a certain similarity with the above paper


Solving the heat equation with a space-time method, naturally yields non-symmetric linear systems to be solved, because of the presence of the time derivative. However, applying the convolution operator, one could hope to make those systems symmetric (albeit dense). This might yield increased computational efficiency.


1 Answer 1


Everything you wrote seems fine (didn't check all the details though) but it isn't what the paper is doing as far as I can tell. In the paper they are convolving the heat equation with the constant 1 function which is just a complicated way of saying they are integrating in time, leading to the integral equation $$ u(t) - \int_0^t div(A\nabla u)(\tau)\ d\tau = \cdots$$ like you said. This weak form of the heat equation is symmetric in the sense that if we multiply through by a test function $v$ and integrate in spacetime we get an expression symmetric in the two variables. This is a little tricky so first a little lemma: for any functions $f,g$, $$ \int_0^T \int_0^t f(t)g(\tau)\ d\tau \ dt = \left(\int_0^Tf(t)\ dt\right)\left(\int_0^Tg(t)\ dt\right) - \int_0^Tf(t)g(t)\ dt.$$

Proof: define $F(t) := \int_0^t f(\tau)\ d\tau$ and similarly for $G$. Then $$\int_0^T \int_0^t f(t)g(\tau)\ d\tau \ dt = \int_0^T f(t)\left(\int_0^t g(\tau)\ d\tau\right) \ dt = \int_0^Tf(t)G(t)\ dt = F(T)G(T) - \int_0^Tf(t)g(t)\ dt$$ where we used the facts that $G' = g$ and $F(0) = G(0) = 0$.

Now taking $$ u(t) - \int_0^t div(A\nabla u)(\tau)\ d\tau = \cdots,$$ multiplying through by $v(t)$, and integrating in space + integration by parts we get $$ \int_{\Omega}u(t)v(t)\ dx - \int_\Omega v(t)\left(\int_0^t div(A\nabla u)(\tau)\ d\tau\right)\ dx = \cdots$$ $$ \implies \int_{\Omega}u(t)v(t)\ dx - \int_0^t\int_\Omega \nabla v(t)\cdot A\nabla u(\tau) \ dx \ d\tau = \cdots$$ so when integrating this whole expression in time from 0 to $T$, by the lemma one gets an expression symmetric in the two variables $u$ and $v$ even if $A$ evolves in time.

The reason this works is that the derivative operator is not self-adjoint in $L^2$ but the integral operator is as the lemma shows; $\int_0^T f(t) G(t)\ dt = \int_0^T F(t) g(t)\ dt$.

  • $\begingroup$ How can you apply the lemma to $\int_\Omega \nabla v(t)\cdot A\nabla u(\tau)$? This is an $h(t,\tau)\neq f(t)g(\tau)$ $\endgroup$
    – Lilla
    May 16, 2023 at 8:34
  • $\begingroup$ $\nabla v(t)\cdot A\nabla u(\tau)$ is a product of the two functions $f(t) =\nabla v(t)$ and $g(\tau) = A\nabla u(\tau)$ in $t$ and $\tau$, respectively $\endgroup$
    – Andrew
    May 16, 2023 at 16:04
  • 1
    $\begingroup$ Ah I get it, I suppose that you are swapping the integrals: $\int_\Omega$ is the outermost one, and you apply the lemma to the integrand of $\int_\Omega$. Anyhow: I still don't follow why your proof works, sorry. Your lemma applied to $\int_0^T\int_0^t\ \nabla v(t)\cdot A(\tau)\nabla u(\tau) \ dx \ d\tau$ yields equality with $\int_0^T\int_0^t\ A(t)\nabla u(t)\cdot \nabla v(\tau) \ dx \ d\tau$. We want equality with $\int_0^T\int_0^t\ \nabla u(t)\cdot A(\tau)\nabla v(\tau) \ dx \ d\tau$ to have a symmetric formulation. $\endgroup$
    – Lilla
    May 16, 2023 at 16:14
  • $\begingroup$ Hi, any thoguhts on the last comment :) ? $\endgroup$
    – Lilla
    Jun 14, 2023 at 11:51

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