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From ODE-Constrained Optimization to Bochner Spaces

University of Pisa

Overview

The previous lectures developed the continuous and discrete theory for elliptic optimal control, including:

At this stage, the next genuine change of perspective is the introduction of time-dependent dynamics. The simplest rigorous entry point is not yet a parabolic PDE, but a linear control problem governed by an ordinary differential equation. This lecture has two goals:

  1. derive the optimality system for a linear-quadratic control problem governed by an ODE;

  2. introduce the vector-valued function spaces needed to pass from ODEs to parabolic PDEs.

The logical path of the lecture is the following:

  1. formulate a linear-quadratic control problem on (0,T)(0,T);

  2. prove well-posedness of the state equation and continuity of the control-to-state map;

  3. derive the reduced derivative through a linearized state equation;

  4. introduce the adjoint ODE and obtain the gradient formula;

  5. state the first-order optimality conditions, both unconstrained and box-constrained;

  6. isolate the structural difference with the elliptic case: a forward-backward optimality system;

  7. introduce strongly measurable vector-valued functions and Bochner spaces;

  8. define weak time derivatives in dual spaces and the space

    W(0,T):={y∈L2(0,T;V):yt∈L2(0,T;V′)};W(0,T):=\{y\in L^2(0,T;V): y_t\in L^2(0,T;V')\};
  9. state the basic embedding and integration-by-parts results that make parabolic optimal control possible.


Model Problem

Let T>0T>0 be fixed, and let

A∈Rn×n,B∈Rn×m,y0∈Rn.A\in \mathbb R^{n\times n}, \qquad B\in \mathbb R^{n\times m}, \qquad y_0\in \mathbb R^n.

Let also

f∈L2(0,T;Rn),yd∈L2(0,T;Rn),yT∈Rn,f\in L^2(0,T;\mathbb R^n), \qquad y_d\in L^2(0,T;\mathbb R^n), \qquad y_T\in \mathbb R^n,

and let α>0\alpha>0, β≥0\beta\ge 0. The control space is

U:=L2(0,T;Rm).U:=L^2(0,T;\mathbb R^m).

The admissible set is a nonempty, closed, convex subset

Uad⊂U.U_{\mathrm{ad}}\subset U.

For u∈Uu\in U, the state y=y(u)y=y(u) is defined by the linear ODE

y˙(t)=Ay(t)+Bu(t)+f(t)for a.e. t∈(0,T),y(0)=y0.\dot y(t)=Ay(t)+Bu(t)+f(t) \qquad \text{for a.e. }t\in(0,T), \qquad y(0)=y_0.

We consider the optimal control problem

min⁡u∈Uadj(u):=J(y(u),u),\min_{u\in U_{\mathrm{ad}}} j(u):=J(y(u),u),

where

J(y,u):=12∫0T∣y(t)−yd(t)∣2 dt+α2∫0T∣u(t)∣2 dt+β2∣y(T)−yT∣2.J(y,u) := \frac12\int_0^T |y(t)-y_d(t)|^2\,dt +\frac\alpha2\int_0^T |u(t)|^2\,dt +\frac\beta2 |y(T)-y_T|^2.

The three terms are:

This model already contains all structural ingredients of parabolic optimal control:


State Equation and Control-to-State Map

We begin with the well-posedness of the state equation. Since the right-hand side belongs to L2(0,T;Rn)L^2(0,T;\mathbb R^n), one expects the solution to belong to H1(0,T;Rn)H^1(0,T;\mathbb R^n).

Define

Y:=H1(0,T;Rn).Y:=H^1(0,T;\mathbb R^n).

Recall that in finite dimensions

H1(0,T;Rn)↪C([0,T];Rn),H^1(0,T;\mathbb R^n)\hookrightarrow C([0,T];\mathbb R^n),

hence the terminal value y(T)y(T) is meaningful.

Proposition. For every u∈Uu\in U, the state equation admits a unique solution

y∈Y=H1(0,T;Rn).y\in Y=H^1(0,T;\mathbb R^n).

Moreover, the solution is given by the variation-of-constants formula

y(t)=etAy0+∫0te(t−s)A(Bu(s)+f(s)) ds,y(t)=e^{tA}y_0+\int_0^t e^{(t-s)A}(Bu(s)+f(s))\,ds,

and there exists a constant C>0C>0, depending only on AA, BB, and TT, such that

∥y∥H1(0,T;Rn)+∥y∥C([0,T];Rn)≤C(∣y0∣+∥u∥L2(0,T;Rm)+∥f∥L2(0,T;Rn)).\|y\|_{H^1(0,T;\mathbb R^n)} + \|y\|_{C([0,T];\mathbb R^n)} \le C\bigl(|y_0|+\|u\|_{L^2(0,T;\mathbb R^m)}+\|f\|_{L^2(0,T;\mathbb R^n)}\bigr).

Proof. Fix u∈Uu\in U and define

g(t):=Bu(t)+f(t)∈L2(0,T;Rn).g(t):=Bu(t)+f(t)\in L^2(0,T;\mathbb R^n).

Since AA is a constant matrix, the matrix exponential etAe^{tA} is well defined and continuous. The formula

y(t):=etAy0+∫0te(t−s)Ag(s) dsy(t):=e^{tA}y_0+\int_0^t e^{(t-s)A}g(s)\,ds

defines a continuous function on [0,T][0,T]. Differentiating under the integral sign yields

y˙(t)=AetAy0+g(t)+∫0tAe(t−s)Ag(s) ds=Ay(t)+g(t)\dot y(t)=Ae^{tA}y_0+g(t)+\int_0^t A e^{(t-s)A} g(s)\,ds =Ay(t)+g(t)

for a.e. t∈(0,T)t\in(0,T), and clearly y(0)=y0y(0)=y_0. Thus y∈H1(0,T;Rn)y\in H^1(0,T;\mathbb R^n) and solves the ODE.

Uniqueness follows from linearity. Indeed, if y1y_1 and y2y_2 solve the same problem, then w:=y1−y2w:=y_1-y_2 satisfies

w˙=Aw,w(0)=0.\dot w = A w, \qquad w(0)=0.

Hence w(t)=etAw(0)=0w(t)=e^{tA}w(0)=0 for all tt.

To estimate yy, let

MA:=max⁡0≤t≤T∥etA∥.M_A:=\max_{0\le t\le T} \|e^{tA}\|.

Then

∣y(t)∣≤MA∣y0∣+MA∫0T∣g(s)∣ ds≤MA∣y0∣+MAT1/2∥g∥L2(0,T).|y(t)| \le M_A |y_0| + M_A \int_0^T |g(s)|\,ds \le M_A |y_0| + M_A T^{1/2}\|g\|_{L^2(0,T)}.

Taking the supremum in tt gives

∥y∥C([0,T])≤C(∣y0∣+∥g∥L2(0,T))≤C(∣y0∣+∥u∥L2(0,T)+∥f∥L2(0,T)).\|y\|_{C([0,T])} \le C\bigl(|y_0|+\|g\|_{L^2(0,T)}\bigr) \le C\bigl(|y_0|+\|u\|_{L^2(0,T)}+\|f\|_{L^2(0,T)}\bigr).

Since y˙=Ay+g\dot y=Ay+g,

∥y˙∥L2(0,T)≤∥A∥ ∥y∥L2(0,T)+∥g∥L2(0,T)≤C(∣y0∣+∥u∥L2(0,T)+∥f∥L2(0,T)).\|\dot y\|_{L^2(0,T)} \le \|A\|\,\|y\|_{L^2(0,T)} + \|g\|_{L^2(0,T)} \le C\bigl(|y_0|+\|u\|_{L^2(0,T)}+\|f\|_{L^2(0,T)}\bigr).

Combining the bounds for yy and y˙\dot y yields the claim. □\square


The proposition allows us to define the control-to-state map

S:U→Y,u↦y=S(u).S:U\to Y, \qquad u\mapsto y=S(u).

Because the state equation is linear, SS is linear and continuous. The reduced functional is therefore

j(u)=J(S(u),u).j(u)=J(S(u),u).

Existence and Uniqueness of an Optimal Control

The previous lectures already established an abstract existence theorem in Hilbert spaces. In the present setting the argument becomes particularly transparent.

Theorem. Assume α>0\alpha>0 and let Uad⊂UU_{\mathrm{ad}}\subset U be nonempty, closed, and convex. Then the reduced problem

min⁡u∈Uadj(u)\min_{u\in U_{\mathrm{ad}}} j(u)

admits a unique solution uˉ∈Uad\bar u\in U_{\mathrm{ad}}.

Proof. Since S:U→YS:U\to Y is continuous and JJ is continuous on Y×UY\times U, the map j:U→Rj:U\to \mathbb R is continuous. Moreover,

j(u)=12∥S(u)−yd∥L2(0,T;Rn)2+α2∥u∥L2(0,T;Rm)2+β2∣S(u)(T)−yT∣2≥α2∥u∥U2.j(u) = \frac12\|S(u)-y_d\|_{L^2(0,T;\mathbb R^n)}^2 +\frac\alpha2\|u\|_{L^2(0,T;\mathbb R^m)}^2 +\frac\beta2 |S(u)(T)-y_T|^2 \ge \frac\alpha2\|u\|_U^2.

Hence jj is coercive on UU. Let (uk)(u_k) be a minimizing sequence in UadU_{\mathrm{ad}}. Coercivity implies that (uk)(u_k) is bounded in UU. Since UU is a Hilbert space, there exists a subsequence, not relabeled, and an element uˉ∈U\bar u\in U such that

uk⇀uˉweakly in U.u_k \rightharpoonup \bar u \qquad \text{weakly in }U.

Because UadU_{\mathrm{ad}} is closed and convex, it is weakly closed, hence uˉ∈Uad\bar u\in U_{\mathrm{ad}}. By continuity and linearity of SS,

S(uk)⇀S(uˉ)weakly in Y.S(u_k) \rightharpoonup S(\bar u) \qquad \text{weakly in }Y.

Since the trace map

y↦y(T)y\mapsto y(T)

is a continuous linear functional on H1(0,T;Rn)H^1(0,T;\mathbb R^n), we also have

S(uk)(T)⇀S(uˉ)(T)in Rn.S(u_k)(T) \rightharpoonup S(\bar u)(T) \qquad \text{in } \mathbb R^n.

Since the norm is weakly lower semicontinuous in Hilbert spaces, each term in jj is weakly lower semicontinuous, hence

j(uˉ)≤lim inf⁡k→∞j(uk),j(\bar u) \le \liminf_{k\to\infty} j(u_k),

so uˉ\bar u is a minimizer.

To prove uniqueness, let u1≠u2u_1\neq u_2 and θ∈(0,1)\theta\in(0,1). Since SS is linear,

S(θu1+(1−θ)u2)=θS(u1)+(1−θ)S(u2).S(\theta u_1+(1-\theta)u_2)=\theta S(u_1)+(1-\theta)S(u_2).

The first and third terms of jj are convex, while the control term is strictly convex because α>0\alpha>0:

∥θu1+(1−θ)u2∥U2<θ∥u1∥U2+(1−θ)∥u2∥U2.\|\theta u_1+(1-\theta)u_2\|_U^2 < \theta \|u_1\|_U^2 + (1-\theta)\|u_2\|_U^2.

Therefore jj is strictly convex and the minimizer is unique. □\square


Linearized State Equation

To differentiate the reduced cost, we perturb the control by h∈Uh\in U. Since the state equation is linear, the state increment is described by a linear ODE independent of the base point.

Let u∈Uu\in U and h∈Uh\in U. Define

zh:=S′(u)h.z_h := S'(u)h.

Then zhz_h solves

z˙h(t)=Azh(t)+Bh(t)for a.e. t∈(0,T),zh(0)=0.\dot z_h(t)=A z_h(t)+B h(t) \qquad \text{for a.e. } t\in(0,T), \qquad z_h(0)=0.

Because the control-to-state map is linear, in fact

S′(u)h=S(h)−S(0)S'(u)h = S(h)-S(0)

for every u∈Uu\in U. Equivalently, one may derive the linearized equation directly from

y˙=Ay+Bu+f,\dot y = Ay+Bu+f,

by replacing uu with u+εhu+\varepsilon h, subtracting the equation for uu, dividing by ε\varepsilon, and passing to the limit.

The directional derivative of the reduced functional therefore reads

j′(u)h=∫0T(y(t)−yd(t))⋅zh(t) dt+β(y(T)−yT)⋅zh(T)+α∫0Tu(t)⋅h(t) dt.j'(u)h = \int_0^T (y(t)-y_d(t))\cdot z_h(t)\,dt +\beta (y(T)-y_T)\cdot z_h(T) +\alpha \int_0^T u(t)\cdot h(t)\,dt.

The difficulty is that this formula still contains the linearized state zhz_h. As in the elliptic case, the adjoint variable removes this dependence.


Adjoint Equation

The adjoint equation is the backward-in-time equation dual to the linearized state equation. Its role is exactly the same as in the elliptic case, but the time direction is reversed.

Let u∈Uu\in U be fixed, and let y=S(u)y=S(u). We define the adjoint state p∈H1(0,T;Rn)p\in H^1(0,T;\mathbb R^n) by

−p˙(t)=ATp(t)+y(t)−yd(t)for a.e. t∈(0,T),p(T)=β (y(T)−yT).-\dot p(t)=A^T p(t)+y(t)-y_d(t) \qquad \text{for a.e. } t\in(0,T), \qquad p(T)=\beta\,(y(T)-y_T).

This is again a linear ODE, now with terminal condition prescribed at t=Tt=T. Its unique solution is obtained by solving backward in time, or equivalently by the change of variable s=T−ts=T-t.

Proposition. Let u∈Uu\in U, y=S(u)y=S(u), and let pp solve the adjoint equation above. Then for every h∈Uh\in U, with zhz_h solving the linearized state equation,

∫0T(y(t)−yd(t))⋅zh(t) dt+β(y(T)−yT)⋅zh(T)=∫0TBTp(t)⋅h(t) dt.\int_0^T (y(t)-y_d(t))\cdot z_h(t)\,dt +\beta (y(T)-y_T)\cdot z_h(T) = \int_0^T B^T p(t)\cdot h(t)\,dt.

Proof. From the linearized state equation,

z˙h−Azh−Bh=0.\dot z_h - A z_h - B h = 0.

Multiply by pp and integrate over (0,T)(0,T):

∫0Tp(t)⋅(z˙h(t)−Azh(t)−Bh(t)) dt=0.\int_0^T p(t)\cdot \bigl(\dot z_h(t)-A z_h(t)-B h(t)\bigr)\,dt = 0.

Using (p,Azh)=(ATp,zh)(p,Az_h)=(A^Tp,z_h) and integrating by parts in time,

∫0Tp⋅z˙h dt=p(T)⋅zh(T)−p(0)⋅zh(0)−∫0Tp˙⋅zh dt.\int_0^T p\cdot \dot z_h\,dt = p(T)\cdot z_h(T)-p(0)\cdot z_h(0)-\int_0^T \dot p\cdot z_h\,dt.

Since zh(0)=0z_h(0)=0,

p(T)⋅zh(T)−∫0T(p˙+ATp)⋅zh dt−∫0TBTp⋅h dt=0.p(T)\cdot z_h(T)-\int_0^T \bigl(\dot p + A^T p\bigr)\cdot z_h\,dt -\int_0^T B^T p\cdot h\,dt = 0.

By the adjoint equation,

−(p˙+ATp)=y−yd,-(\dot p + A^T p)=y-y_d,

hence

p(T)⋅zh(T)+∫0T(y−yd)⋅zh dt−∫0TBTp⋅h dt=0.p(T)\cdot z_h(T)+\int_0^T (y-y_d)\cdot z_h\,dt -\int_0^T B^T p\cdot h\,dt = 0.

Finally, the terminal condition gives

p(T)⋅zh(T)=β(y(T)−yT)⋅zh(T),p(T)\cdot z_h(T)=\beta (y(T)-y_T)\cdot z_h(T),

which yields the claim. □\square


Reduced Gradient Formula

We can now eliminate the linearized state from the derivative.

Theorem. The reduced functional j:U→Rj:U\to \mathbb R is Fréchet differentiable and, for every u∈Uu\in U,

j′(u)h=∫0T(αu(t)+BTp(t))⋅h(t) dt∀h∈U,j'(u)h = \int_0^T \bigl(\alpha u(t)+B^T p(t)\bigr)\cdot h(t)\,dt \qquad \forall h\in U,

where pp is the adjoint associated with uu. Hence the gradient of jj in the Hilbert space U=L2(0,T;Rm)U=L^2(0,T;\mathbb R^m) is

∇j(u)=αu+BTp.\nabla j(u)=\alpha u + B^T p.

Proof. We already computed

j′(u)h=∫0T(y−yd)⋅zh dt+β(y(T)−yT)⋅zh(T)+α∫0Tu⋅h dt.j'(u)h = \int_0^T (y-y_d)\cdot z_h\,dt +\beta (y(T)-y_T)\cdot z_h(T) +\alpha\int_0^T u\cdot h\,dt.

The adjoint identity from the previous proposition gives

∫0T(y−yd)⋅zh dt+β(y(T)−yT)⋅zh(T)=∫0TBTp⋅h dt.\int_0^T (y-y_d)\cdot z_h\,dt +\beta (y(T)-y_T)\cdot z_h(T) = \int_0^T B^T p\cdot h\,dt.

Therefore

j′(u)h=∫0T(αu+BTp)⋅h dt.j'(u)h = \int_0^T (\alpha u + B^T p)\cdot h\,dt.

Since this is a bounded linear functional of hh, the Fréchet derivative is represented in UU by the function αu+BTp\alpha u + B^T p. □\square


First-Order Optimality System

We can now write the necessary and sufficient optimality conditions. Because the problem is convex and the reduced functional is strictly convex, first-order conditions characterize the unique minimizer.

Unconstrained case

If

Uad=U=L2(0,T;Rm),U_{\mathrm{ad}}=U=L^2(0,T;\mathbb R^m),

then the stationarity condition is simply

∇j(uˉ)=0,\nabla j(\bar u)=0,

i.e.

αuˉ+BTpˉ=0in L2(0,T;Rm).\alpha \bar u + B^T \bar p = 0 \qquad \text{in }L^2(0,T;\mathbb R^m).

Thus the optimal control satisfies the explicit relation

uˉ(t)=−1αBTpˉ(t)for a.e. t∈(0,T).\bar u(t) = -\frac1\alpha B^T \bar p(t) \qquad \text{for a.e. } t\in(0,T).

The optimality system becomes

{yˉ˙(t)=Ayˉ(t)+Buˉ(t)+f(t),a.e. t∈(0,T),yˉ(0)=y0,−pˉ˙(t)=ATpˉ(t)+yˉ(t)−yd(t),a.e. t∈(0,T),pˉ(T)=β(yˉ(T)−yT),αuˉ(t)+BTpˉ(t)=0,a.e. t∈(0,T).\begin{cases} \dot{\bar y}(t)=A\bar y(t)+B\bar u(t)+f(t), & \text{a.e. } t\in(0,T),\\ \bar y(0)=y_0,\\ -\dot{\bar p}(t)=A^T\bar p(t)+\bar y(t)-y_d(t), & \text{a.e. } t\in(0,T),\\ \bar p(T)=\beta(\bar y(T)-y_T),\\ \alpha \bar u(t)+B^T\bar p(t)=0, & \text{a.e. } t\in(0,T). \end{cases}

This is a forward-backward system:

This two-sided time structure is the first major structural difference from elliptic control.

Constrained case

Assume now that Uad⊂UU_{\mathrm{ad}}\subset U is closed and convex. Then uˉ∈Uad\bar u\in U_{\mathrm{ad}} is optimal if and only if

j′(uˉ)(u−uˉ)≥0∀u∈Uad.j'(\bar u)(u-\bar u)\ge 0 \qquad \forall u\in U_{\mathrm{ad}}.

Using the adjoint representation of the derivative, this becomes

∫0T(αuˉ(t)+BTpˉ(t))⋅(u(t)−uˉ(t)) dt≥0∀u∈Uad.\int_0^T \bigl(\alpha \bar u(t)+B^T\bar p(t)\bigr)\cdot \bigl(u(t)-\bar u(t)\bigr)\,dt \ge 0 \qquad \forall u\in U_{\mathrm{ad}}.

Equivalently, in normal-cone form,

0∈αuˉ+BTpˉ+NUad(uˉ)in U.0\in \alpha \bar u + B^T \bar p + N_{U_{\mathrm{ad}}}(\bar u) \qquad \text{in }U.

Hence the full optimality system is

{yˉ˙=Ayˉ+Buˉ+f,yˉ(0)=y0,−pˉ˙=ATpˉ+yˉ−yd,pˉ(T)=β(yˉ(T)−yT),0∈αuˉ+BTpˉ+NUad(uˉ).\begin{cases} \dot{\bar y}=A\bar y+B\bar u+f, \qquad \bar y(0)=y_0,\\[0.3em] -\dot{\bar p}=A^T\bar p+\bar y-y_d, \qquad \bar p(T)=\beta(\bar y(T)-y_T),\\[0.3em] 0\in \alpha \bar u + B^T \bar p + N_{U_{\mathrm{ad}}}(\bar u). \end{cases}

Box Constraints and Projection Formula

A particularly important case is the box-constrained set

Uad:={u∈L2(0,T;Rm):ua(t)≤u(t)≤ub(t) for a.e. t∈(0,T)},U_{\mathrm{ad}} := \left\{ u\in L^2(0,T;\mathbb R^m): u_a(t)\le u(t)\le u_b(t) \text{ for a.e. } t\in(0,T) \right\},

where ua,ub∈L∞(0,T;Rm)u_a,u_b\in L^\infty(0,T;\mathbb R^m) and the inequalities are understood componentwise.

In this case the optimality condition is equivalent to the pointwise projection formula

uˉ(t)=P[ua(t),ub(t)] ⁣(−1αBTpˉ(t))for a.e. t∈(0,T),\bar u(t)=P_{[u_a(t),u_b(t)]}\!\left(-\frac1\alpha B^T\bar p(t)\right) \qquad \text{for a.e. } t\in(0,T),

where for ξ∈Rm\xi\in \mathbb R^m

P[ua(t),ub(t)](ξ)=min⁡(max⁡(ξ,ua(t)),ub(t))P_{[u_a(t),u_b(t)]}(\xi) = \min\bigl(\max(\xi,u_a(t)),u_b(t)\bigr)

componentwise.

The proof is identical in structure to the elliptic case:

Thus the only genuinely new analytical ingredient introduced by time dependence is not the control condition, but the forward-backward evolution structure of state and adjoint.


Forward-Backward Interpretation

It is worth isolating the conceptual meaning of the adjoint in the dynamical case.

For a fixed control uu:

The terminal condition

p(T)=β(y(T)−yT)p(T)=\beta(y(T)-y_T)

encodes the derivative of the terminal observation. If β=0\beta=0, then p(T)=0p(T)=0 and only the distributed tracking term drives the adjoint. If instead the cost is purely terminal,

J(y,u)=α2∫0T∣u(t)∣2 dt+β2∣y(T)−yT∣2,J(y,u)=\frac\alpha2\int_0^T |u(t)|^2\,dt + \frac\beta2 |y(T)-y_T|^2,

then the adjoint satisfies

−p˙=ATp,p(T)=β(y(T)−yT).-\dot p = A^T p, \qquad p(T)=\beta(y(T)-y_T).

This is exactly the same phenomenon that one encounters later for parabolic PDEs:


From ODEs to Evolution Equations

The ODE model can be rewritten abstractly as

yt+Ay=Bu+f,y_t + \mathcal A y = \mathcal B u + f,

where now the state y(t)y(t) is an element of the finite-dimensional Hilbert space H=RnH=\mathbb R^n and

A:=−A,B:=B.\mathcal A := -A, \qquad \mathcal B := B.

For parabolic PDEs, the same formula remains formally correct, but the state at each time is no longer a vector in Rn\mathbb R^n. Instead:

This is the reason why the usual scalar-valued Lebesgue and Sobolev spaces are not sufficient. We need function spaces of the form

Lp(0,T;X),L^p(0,T;X),

where XX is itself a Banach or Hilbert space. These are the Bochner spaces.


Strongly Measurable Vector-Valued Functions

Let XX be a Banach space. A function

y:(0,T)→Xy:(0,T)\to X

is called simple if it has the form

y(t)=∑k=1Nxk χEk(t),y(t)=\sum_{k=1}^N x_k\,\chi_{E_k}(t),

where xk∈Xx_k\in X and Ek⊂(0,T)E_k\subset(0,T) are measurable sets.

A function y:(0,T)→Xy:(0,T)\to X is called strongly measurable if there exists a sequence of simple functions (yn)(y_n) such that

yn(t)→y(t)for a.e. t∈(0,T).y_n(t)\to y(t) \qquad \text{for a.e. } t\in(0,T).

This is the natural notion of measurability for vector-valued functions. In the Hilbert spaces used in parabolic PDEs, separability holds, so this definition behaves well.

If yy is strongly measurable and

∫0T∥y(t)∥X dt<∞,\int_0^T \|y(t)\|_X\,dt < \infty,

then yy is Bochner integrable and one may define

∫0Ty(t) dt∈X\int_0^T y(t)\,dt \in X

as the limit of the integrals of simple approximations. This is the vector-valued analogue of the usual Lebesgue integral.


Bochner Spaces Lp(0,T;X)L^p(0,T;X)

Let 1≤p<∞1\le p<\infty. We define

Lp(0,T;X):={y:(0,T)→X:y strongly measurable and ∫0T∥y(t)∥Xp dt<∞}.L^p(0,T;X) := \left\{ y:(0,T)\to X: y \text{ strongly measurable and } \int_0^T \|y(t)\|_X^p\,dt < \infty \right\}.

The norm is

∥y∥Lp(0,T;X):=(∫0T∥y(t)∥Xp dt)1/p.\|y\|_{L^p(0,T;X)} := \left(\int_0^T \|y(t)\|_X^p\,dt\right)^{1/p}.

Similarly,

L∞(0,T;X):={y:(0,T)→X:y strongly measurable and ess sup⁡t∈(0,T)∥y(t)∥X<∞}.L^\infty(0,T;X) := \left\{ y:(0,T)\to X: y \text{ strongly measurable and } \operatorname*{ess\,sup}_{t\in(0,T)} \|y(t)\|_X < \infty \right\}.

Standard facts:

Fundamental examples

Let Ω⊂Rd\Omega\subset \mathbb R^d be a bounded Lipschitz domain and set

QT:=Ω×(0,T).Q_T:=\Omega\times(0,T).

Then:

  1. if X=L2(Ω)X=L^2(\Omega),

    L2(0,T;L2(Ω))≅L2(QT);L^2(0,T;L^2(\Omega)) \cong L^2(Q_T);
  2. if X=H01(Ω)X=H_0^1(\Omega),

    L2(0,T;H01(Ω))L^2(0,T;H_0^1(\Omega))

    consists of functions square integrable in time with values in H01(Ω)H_0^1(\Omega);

  3. if X=H−1(Ω)X=H^{-1}(\Omega),

    L2(0,T;H−1(Ω))L^2(0,T;H^{-1}(\Omega))

    is the natural space for weak time derivatives of parabolic states.

Thus, in parabolic theory, the same function y(x,t)y(x,t) is viewed as a map

t↦y(t):=y(⋅,t)t\mapsto y(t):=y(\cdot,t)

with values in a spatial function space.


Weak Time Derivatives

For parabolic equations, the time derivative does not usually belong to the same space as the state. This forces a dual-space formulation.

Let XX be a Banach space and let

y∈L1(0,T;X).y\in L^1(0,T;X).

A function

z∈L1(0,T;X)z\in L^1(0,T;X)

is called the weak time derivative of yy if for every scalar test function φ∈Cc∞(0,T)\varphi\in C_c^\infty(0,T) and every ℓ∈X′\ell\in X',

∫0T⟨ℓ,y(t)⟩X′,X φ′(t) dt=−∫0T⟨ℓ,z(t)⟩X′,X φ(t) dt.\int_0^T \langle \ell,y(t)\rangle_{X',X}\, \varphi'(t)\,dt = -\int_0^T \langle \ell,z(t)\rangle_{X',X}\, \varphi(t)\,dt.

In this case we write

z=yt.z = y_t.

If XX is Hilbert, one can identify X≅X′X\cong X' by the Riesz map and recover the familiar scalar definition.

The Sobolev space of XX-valued functions is then

H1(0,T;X):={y∈L2(0,T;X):yt∈L2(0,T;X)}.H^1(0,T;X) := \{y\in L^2(0,T;X): y_t\in L^2(0,T;X)\}.

For ODEs, where X=RnX=\mathbb R^n, this is the space used for the state and adjoint variables.

For parabolic PDEs, however, the correct setting is generally not H1(0,T;X)H^1(0,T;X) with a single space XX, but a mixed space involving a Hilbert triple.


Gelfand Triples and the Space W(0,T)W(0,T)

Let VV and HH be Hilbert spaces such that

V↪HV \hookrightarrow H

continuously and densely. By identifying HH with its dual H′H' through the Riesz isomorphism, one obtains the Gelfand triple

V↪H≅H′↪V′.V \hookrightarrow H \cong H' \hookrightarrow V'.

The last embedding is defined by

⟨h,v⟩V′,V:=(h,v)H∀h∈H, ∀v∈V.\langle h,v\rangle_{V',V} := (h,v)_H \qquad \forall h\in H,\ \forall v\in V.

The canonical parabolic example is

V=H01(Ω),H=L2(Ω),V′=H−1(Ω).V=H_0^1(\Omega), \qquad H=L^2(\Omega), \qquad V'=H^{-1}(\Omega).

The natural energy space for parabolic problems is

W(0,T):={y∈L2(0,T;V):yt∈L2(0,T;V′)}.W(0,T) := \{y\in L^2(0,T;V): y_t\in L^2(0,T;V')\}.

It is a Hilbert space with norm

∥y∥W(0,T)2:=∥y∥L2(0,T;V)2+∥yt∥L2(0,T;V′)2.\|y\|_{W(0,T)}^2 := \|y\|_{L^2(0,T;V)}^2 + \|y_t\|_{L^2(0,T;V')}^2.

This is the parabolic analogue of H1(0,T;Rn)H^1(0,T;\mathbb R^n) for ODEs.

The key point is the asymmetry:

This is forced by the weak formulation of the PDE. For the heat equation,

yt−Δy=F,y_t - \Delta y = F,

one expects

y(t)∈H01(Ω),yt(t)∈H−1(Ω).y(t)\in H_0^1(\Omega), \qquad y_t(t)\in H^{-1}(\Omega).

Fundamental Theorem for W(0,T)W(0,T)

The space W(0,T)W(0,T) has a decisive property: its elements possess a continuous representative with values in HH. This is what makes initial and terminal conditions meaningful.

Theorem (Lions-Magenes). Let

V↪H↪V′V \hookrightarrow H \hookrightarrow V'

be a Gelfand triple. Then:

  1. every y∈W(0,T)y\in W(0,T) admits a representative, still denoted by yy, such that

    y∈C([0,T];H);y\in C([0,T];H);
  2. the embedding

    W(0,T)↪C([0,T];H)W(0,T) \hookrightarrow C([0,T];H)

    is continuous;

  3. if y,v∈W(0,T)y,v\in W(0,T), then the scalar map

    t↦(y(t),v(t))Ht\mapsto (y(t),v(t))_H

    is absolutely continuous and satisfies

    ddt(y(t),v(t))H=⟨yt(t),v(t)⟩V′,V+⟨vt(t),y(t)⟩V′,V\frac{d}{dt}(y(t),v(t))_H = \langle y_t(t),v(t)\rangle_{V',V} + \langle v_t(t),y(t)\rangle_{V',V}

    for a.e. t∈(0,T)t\in(0,T).

In particular, taking v=yv=y gives

12ddt∥y(t)∥H2=⟨yt(t),y(t)⟩V′,Vfor a.e. t∈(0,T).\frac12\frac{d}{dt}\|y(t)\|_H^2 = \langle y_t(t),y(t)\rangle_{V',V} \qquad \text{for a.e. } t\in(0,T).

Integrating between ss and tt yields the energy identity

12∥y(t)∥H2−12∥y(s)∥H2=∫st⟨yt(τ),y(τ)⟩V′,V dτ.\frac12\|y(t)\|_H^2 - \frac12\|y(s)\|_H^2 = \int_s^t \langle y_t(\tau),y(\tau)\rangle_{V',V}\,d\tau.

More generally, for y,v∈W(0,T)y,v\in W(0,T),

(y(t),v(t))H−(y(s),v(s))H=∫st⟨yt(τ),v(τ)⟩V′,V dτ+∫st⟨vt(τ),y(τ)⟩V′,V dτ.(y(t),v(t))_H - (y(s),v(s))_H = \int_s^t \langle y_t(\tau),v(\tau)\rangle_{V',V}\,d\tau + \int_s^t \langle v_t(\tau),y(\tau)\rangle_{V',V}\,d\tau.

This is the time-integration-by-parts formula needed for parabolic adjoints.


Abstract Parabolic Problem

With the previous tools in place, one can formulate the prototype parabolic state equation.

Let V↪H↪V′V\hookrightarrow H\hookrightarrow V' be a Gelfand triple. Let

a:V×V→Ra:V\times V\to \mathbb R

be a bilinear form satisfying:

  1. continuity:

    ∣a(w,v)∣≤M∥w∥V∥v∥V∀w,v∈V;|a(w,v)|\le M\|w\|_V\|v\|_V \qquad \forall w,v\in V;
  2. coercivity:

    a(v,v)≥ca∥v∥V2∀v∈V,a(v,v)\ge c_a \|v\|_V^2 \qquad \forall v\in V,

    for some ca>0c_a>0.

Define the operator

A:V→V′,⟨Ay,v⟩V′,V=a(y,v).A:V\to V', \qquad \langle Ay,v\rangle_{V',V}=a(y,v).

Given

F∈L2(0,T;V′),y0∈H,F\in L^2(0,T;V'), \qquad y_0\in H,

the weak parabolic problem is: find y∈W(0,T)y\in W(0,T) such that

⟨yt(t),v⟩V′,V+a(y(t),v)=⟨F(t),v⟩V′,V∀v∈V,for a.e. t∈(0,T),\langle y_t(t),v\rangle_{V',V} + a(y(t),v) = \langle F(t),v\rangle_{V',V} \qquad \forall v\in V, \quad \text{for a.e. } t\in(0,T),

with

y(0)=y0in H.y(0)=y_0 \quad \text{in } H.

Theorem. Under the assumptions above, the parabolic problem admits a unique solution

y∈W(0,T).y\in W(0,T).

Moreover,

∥y∥L2(0,T;V)+∥y∥L∞(0,T;H)+∥yt∥L2(0,T;V′)≤C(∥F∥L2(0,T;V′)+∥y0∥H),\|y\|_{L^2(0,T;V)} + \|y\|_{L^\infty(0,T;H)} + \|y_t\|_{L^2(0,T;V')} \le C\bigl(\|F\|_{L^2(0,T;V')}+\|y_0\|_H\bigr),

for a constant CC depending only on the continuity and coercivity constants and on TT.

This theorem is the infinite-dimensional analogue of the well-posedness proposition for the ODE state equation. The analogies are exact:


Heat Equation as Canonical Example

Take

V=H01(Ω),H=L2(Ω),V′=H−1(Ω),V=H_0^1(\Omega), \qquad H=L^2(\Omega), \qquad V'=H^{-1}(\Omega),

and define

a(y,v)=∫Ω∇y⋅∇v dx.a(y,v)=\int_\Omega \nabla y\cdot \nabla v\,dx.

Then

⟨Ay,v⟩=∫Ω∇y⋅∇v dx\langle Ay,v\rangle = \int_\Omega \nabla y\cdot \nabla v\,dx

corresponds to the operator A=−ΔA=-\Delta in weak form. Given a control u∈L2(0,T;L2(Ω))u\in L^2(0,T;L^2(\Omega)), one may write

F(t)=u(t)+f(t)∈H↪V′.F(t)=u(t)+f(t) \in H \hookrightarrow V'.

The parabolic state equation becomes

⟨yt(t),v⟩H−1,H01+∫Ω∇y(t)⋅∇v dx=∫Ω(u(t)+f(t))v dx\langle y_t(t),v\rangle_{H^{-1},H_0^1} + \int_\Omega \nabla y(t)\cdot \nabla v\,dx = \int_\Omega (u(t)+f(t))v\,dx

for all v∈H01(Ω)v\in H_0^1(\Omega) and a.e. t∈(0,T)t\in(0,T), with y(0)=y0y(0)=y_0.

This is the standard weak formulation of the heat equation

yt−Δy=u+fin Ω×(0,T),y_t - \Delta y = u+f \qquad \text{in } \Omega\times(0,T),

with homogeneous Dirichlet boundary condition.


What Changes in the Optimality System for Parabolic PDEs?

At the formal level, almost nothing changes. The parabolic optimal control problem has the structure

min⁡u∈Uad12∫0T∥y(t)−yd(t)∥H2 dt+α2∥u∥L2(0,T;U)2+β2∥y(T)−yT∥H2\min_{u\in U_{\mathrm{ad}}} \frac12\int_0^T \|y(t)-y_d(t)\|_H^2\,dt + \frac\alpha2 \|u\|_{L^2(0,T;U)}^2 + \frac\beta2 \|y(T)-y_T\|_H^2

subject to

yt+Ay=Bu+f,y(0)=y0.y_t + Ay = Bu + f, \qquad y(0)=y_0.

The optimality system has the same forward-backward pattern as in the ODE case:

The new difficulty is not conceptual but functional-analytic:


Summary

This lecture introduced the time-dependent side of optimal control in two layers.

ODE layer

For the linear-quadratic control problem governed by

y˙=Ay+Bu+f,y(0)=y0,\dot y = Ay+Bu+f, \qquad y(0)=y_0,

we proved:

Hence time-dependent optimality already appears as a forward-backward system.

Functional-analytic layer

To pass from ODEs to parabolic PDEs we introduced:

These are exactly the tools needed for the next lecture, where the same adjoint-based optimality machinery will be applied to parabolic PDEs.


References