[proofplan] We give two proofs. The first is a direct classical verification: define the auxiliary function $v^s(x,t) := \int_{\mathbb{R}^n}\Phi(x-y,t-s)\,f(y,s)\,d\mathcal{L}^n(y)$, which solves the homogeneous [heat equation](/page/Heat%20Equation) for $t > s$ by the [solution of the Cauchy problem](/theorems/54). The Duhamel integral is $u(x,t) = \int_0^t v^s(x,t)\,d\mathcal{L}^1(s)$. Differentiating in $t$ via the [Leibniz integral rule](/theorems/831) produces a boundary term $v^t(x,t) = f(x,t)$ (from the delta-[limit](/page/Limit) property) plus an integral term $\int_0^t \partial_t v^s\,d\mathcal{L}^1(s) = \Delta u$ (since each $v^s$ satisfies the heat equation). The second proof uses the distributional framework: extend $\Phi$ to all of $\mathbb{R}^{n+1}$ by setting it to zero for $t \leq 0$, verify that this causal extension $E$ satisfies $(\partial_t - \Delta)E = \delta$ in $\mathcal{D}'(\mathbb{R}^{n+1})$, identify $u = E * \tilde{f}$, and conclude $(\partial_t - \Delta)u = \delta * \tilde{f} = f$. [/proofplan] ## Proof 1: Classical verification [step:Define the auxiliary functions $v^s$ and identify $u$ as their integral] For each $s \geq 0$, define the auxiliary function \begin{align*} v^s: \mathbb{R}^n \times (s, \infty) &\to \mathbb{R} \\ (x,t) &\mapsto \int_{\mathbb{R}^n}\Phi(x - y, t - s)\,f(y,s)\,d\mathcal{L}^n(y). \end{align*} Since $f$ has compact support, $f(\cdot, s) \in C^2(\mathbb{R}^n) \cap L^\infty(\mathbb{R}^n)$ for each $s \geq 0$. By the [solution of the Cauchy problem](/theorems/54) applied with initial data $g := f(\cdot, s)$ and time origin shifted to $s$: $v^s \in C^\infty(\mathbb{R}^n \times (s,\infty))$, the function $v^s$ satisfies $\partial_t v^s - \Delta_x v^s = 0$ for $t > s$, and \begin{align*} \lim_{\substack{(x,t) \to (x_0, s) \\ t > s}} v^s(x,t) = f(x_0, s) \quad \text{for every } x_0 \in \mathbb{R}^n. \end{align*} The hypotheses of the [solution of the Cauchy problem](/theorems/54) are satisfied: $f(\cdot, s)$ is continuous (since $f \in C^{2,1}$) and bounded (since $f$ has compact support). The candidate solution is \begin{align*} u(x,t) = \int_0^t v^s(x,t)\,d\mathcal{L}^1(s). \end{align*} [/step] [step:Differentiate in $t$ using the Leibniz integral rule] [claim:Leibniz rule for the Duhamel integral] For all $(x,t) \in \mathbb{R}^n \times (0,\infty)$, \begin{align*} \partial_t u(x,t) = v^t(x,t) + \int_0^t \partial_t v^s(x,t)\,d\mathcal{L}^1(s). \end{align*} [/claim] [proof] The function $u(x,t) = \int_0^t v^s(x,t)\,d\mathcal{L}^1(s)$ has both the integrand and the upper limit depending on $t$. The [Leibniz integral rule](/theorems/831) states: if $F(t) = \int_0^{a(t)} h(s,t)\,d\mathcal{L}^1(s)$ with $a(t) = t$, and if $h$ and $\partial_t h$ are continuous on $\{(s,t) : 0 \leq s \leq t\}$, then \begin{align*} F'(t) = h(t,t) + \int_0^t \partial_t h(s,t)\,d\mathcal{L}^1(s). \end{align*} We verify the hypotheses with $h(s,t) := v^s(x,t)$ for fixed $x$. For $0 \leq s < t$, the function $(s,t) \mapsto v^s(x,t)$ is continuous: continuity in $t$ follows from $v^s \in C^\infty$ for $t > s$, and continuity in $s$ follows from the [dominated convergence theorem](/theorems/4) applied to the integral $\int_{\mathbb{R}^n}\Phi(x-y,t-s)\,f(y,s)\,d\mathcal{L}^n(y)$, since $f$ has compact spatial support and varies continuously in $s$. Similarly, $\partial_t v^s(x,t) = \int_{\mathbb{R}^n}\partial_t\Phi(x-y,t-s)\,f(y,s)\,d\mathcal{L}^n(y)$ is continuous in $(s,t)$ for $0 \leq s < t$ (the differentiation under the integral sign is justified by the Gaussian decay argument from the proof of the [Cauchy problem](/theorems/54)). For the behaviour as $s \to t$: the kernel $\Phi(x-y,t-s)$ concentrates near $y = x$ as $t - s \to 0$, and the delta-limit property (property 4 of the [fundamental solution](/theorems/53)) gives $v^s(x,t) \to f(x,s)$. Since $f$ is continuous in $s$, the function $s \mapsto v^s(x,t)$ extends continuously to $s = t$ with value $f(x,t)$. This gives $h(t,t) = f(x,t)$, and the Leibniz rule applies. [/proof] [guided] Why does the Leibniz rule produce two terms? The Duhamel integral $u(x,t) = \int_0^t v^s(x,t)\,d\mathcal{L}^1(s)$ depends on $t$ in two ways: the upper limit of integration is $t$, and the integrand $v^s(x,t)$ depends on $t$ through its second argument. The [Leibniz integral rule](/theorems/831) for $F(t) = \int_0^{a(t)} h(s,t)\,d\mathcal{L}^1(s)$ with $a(t) = t$ states \begin{align*} F'(t) = h(t,t)\cdot a'(t) + \int_0^{a(t)} \partial_t h(s,t)\,d\mathcal{L}^1(s) = h(t,t) + \int_0^t \partial_t h(s,t)\,d\mathcal{L}^1(s). \end{align*} With $h(s,t) = v^s(x,t) = \int_{\mathbb{R}^n}\Phi(x-y,t-s)\,f(y,s)\,d\mathcal{L}^n(y)$, the formula becomes \begin{align*} \partial_t u(x,t) = v^t(x,t) + \int_0^t \partial_t v^s(x,t)\,d\mathcal{L}^1(s). \end{align*} The boundary term $v^t(x,t)$ is where the source function $f$ enters. Physically, at each instant $s$, the source $f(\cdot, s)$ injects an "impulse" of heat that is then propagated by the heat kernel $\Phi(\cdot, t - s)$ for the remaining time $t - s$. The boundary term captures the impulse injected at $s = t$, which has had zero propagation time, so it contributes exactly $f(x,t)$ by the delta-limit property of $\Phi$. The integral term $\int_0^t \partial_t v^s(x,t)\,d\mathcal{L}^1(s)$ captures the accumulated effect of all past impulses, each evolved forward by the homogeneous heat equation. The most delicate verification is the continuity of $h(s,t) = v^s(x,t)$ as $s \to t^-$. At this limit, $\tau := t - s \to 0^+$ and the heat kernel $\Phi(x-y,\tau)$ becomes singular -- it concentrates as a Dirac delta at $y = x$. By the delta-limit property (property 4 of the [fundamental solution](/theorems/53)): \begin{align*} \lim_{\tau \to 0^+} v^s(x,t) = \lim_{\tau \to 0^+} \int_{\mathbb{R}^n}\Phi(x-y,\tau)\,f(y,t - \tau)\,d\mathcal{L}^n(y) = f(x,t). \end{align*} This uses two inputs: $\Phi(\cdot,\tau) \to \delta_x$ in $\mathcal{D}'(\mathbb{R}^n)$ as $\tau \to 0^+$, and the uniform convergence $f(y, t-\tau) \to f(y,t)$ as $\tau \to 0$ (since $f$ is continuous in $t$ with compact spatial support). The continuity in $s$ on $[0,t)$ follows from the [dominated convergence theorem](/theorems/4) applied to the integral $\int_{\mathbb{R}^n}\Phi(x-y,t-s)\,f(y,s)\,d\mathcal{L}^n(y)$: the integrand varies continuously in $s$ and is dominated by $C\,\exp(-|x-y|^2/(8T))$ (a fixed Gaussian integrable in $y$) for $s$ bounded away from $t$. Together, these ensure $h(s,t)$ extends continuously to $s = t$ with value $f(x,t)$, validating the hypotheses of the [Leibniz integral rule](/theorems/831). [/guided] [/step] [step:Evaluate the boundary term $v^t(x,t) = f(x,t)$ via the delta-limit property] [claim:Boundary term evaluation] For all $(x,t) \in \mathbb{R}^n \times (0,\infty)$, the boundary term satisfies $v^t(x,t) = f(x,t)$. [/claim] [proof] The value $v^t(x,t)$ is defined as the limit \begin{align*} v^t(x,t) = \lim_{s \to t^-} v^s(x,t) = \lim_{s \to t^-} \int_{\mathbb{R}^n}\Phi(x - y, t - s)\,f(y,s)\,d\mathcal{L}^n(y). \end{align*} Set $\tau := t - s$, so $\tau \to 0^+$ as $s \to t^-$, and $f(y,s) = f(y, t - \tau)$. Write the integrand as a sum of two terms: \begin{align*} \Phi(x-y,\tau)\,f(y,t-\tau) = \Phi(x-y,\tau)\,f(y,t) + \Phi(x-y,\tau)\,\bigl(f(y,t-\tau) - f(y,t)\bigr). \end{align*} For the first term: the function $y \mapsto f(y,t)$ belongs to $C(\mathbb{R}^n) \cap L^\infty(\mathbb{R}^n)$ (continuous and bounded since $f$ has compact support). By the delta-limit property (property 4 of the [fundamental solution](/theorems/53)): \begin{align*} \lim_{\tau \to 0^+}\int_{\mathbb{R}^n}\Phi(x-y,\tau)\,f(y,t)\,d\mathcal{L}^n(y) = f(x,t). \end{align*} For the second term: since $f$ has compact spatial support and is continuous in $t$, the supremum $\sup_{y \in \mathbb{R}^n}|f(y,t-\tau) - f(y,t)|$ tends to $0$ as $\tau \to 0$. Using the unit mass property $\int_{\mathbb{R}^n}\Phi(x-y,\tau)\,d\mathcal{L}^n(y) = 1$: \begin{align*} \left|\int_{\mathbb{R}^n}\Phi(x-y,\tau)\,\bigl(f(y,t-\tau) - f(y,t)\bigr)\,d\mathcal{L}^n(y)\right| \leq \sup_{y}|f(y,t-\tau) - f(y,t)| \to 0. \end{align*} Combining: $v^t(x,t) = f(x,t)$. [/proof] [/step] [step:Show the integral term equals $\Delta u$ using the heat equation for $v^s$] [claim:Interchange of Laplacian and time integral] For all $(x,t) \in \mathbb{R}^n \times (0,\infty)$, \begin{align*} \int_0^t \partial_t v^s(x,t)\,d\mathcal{L}^1(s) = \Delta u(x,t). \end{align*} [/claim] [proof] Since $v^s$ solves the homogeneous heat equation for $t > s$ (from the first step), $\partial_t v^s(x,t) = \Delta_x v^s(x,t)$ for each fixed $s \in [0,t)$. Therefore \begin{align*} \int_0^t \partial_t v^s(x,t)\,d\mathcal{L}^1(s) = \int_0^t \Delta_x v^s(x,t)\,d\mathcal{L}^1(s). \end{align*} It remains to show that $\Delta_x$ passes through the $s$-integral. For each $s \in [0,t)$ and each coordinate index $i$, integrate by parts twice in the spatial variable (boundary terms vanish since $f$ has compact support): \begin{align*} \partial_{x_i x_i} v^s(x,t) = \int_{\mathbb{R}^n}\partial_{x_i x_i}\Phi(x-y,t-s)\,f(y,s)\,d\mathcal{L}^n(y) = \int_{\mathbb{R}^n}\Phi(x-y,t-s)\,\partial_{y_i y_i}f(y,s)\,d\mathcal{L}^n(y). \end{align*} The second expression shows that $|\partial_{x_i x_i} v^s(x,t)| \leq \|\partial_{x_i x_i}f\|_{L^\infty}$ uniformly in $s \in [0,t]$ (using the unit mass property of $\Phi$). This uniform bound provides an integrable dominator over $[0,t]$, so the [dominated convergence theorem](/theorems/4) (or Fubini's theorem, since the dominator is constant) justifies the interchange: \begin{align*} \int_0^t \Delta_x v^s(x,t)\,d\mathcal{L}^1(s) = \Delta_x \int_0^t v^s(x,t)\,d\mathcal{L}^1(s) = \Delta u(x,t). \end{align*} [/proof] [guided] The crucial point is the interchange of $\Delta_x$ and $\int_0^t \cdot\,d\mathcal{L}^1(s)$. A direct estimate of $\partial_{x_ix_i}\Phi(x-y,t-s)$ gives a bound proportional to $(t-s)^{-1}$, which is not integrable at $s = t$. This would block the interchange. The resolution uses integration by parts in the spatial variable. Since $f$ has compact support, we transfer the spatial derivatives from $\Phi$ to $f$: \begin{align*} \partial_{x_ix_i}v^s(x,t) = \int_{\mathbb{R}^n}\Phi(x-y,t-s)\,\partial_{y_iy_i}f(y,s)\,d\mathcal{L}^n(y). \end{align*} This uses $\partial_{x_i}\Phi(x-y,t-s) = -\partial_{y_i}\Phi(x-y,t-s)$ (the sign flip from the chain rule applied to $x - y$), together with the vanishing of boundary terms at infinity (since $f$ has compact support). The resulting integrand involves $\Phi$ (with no derivatives) times $\partial_{y_iy_i}f$, and the unit mass property gives \begin{align*} |\partial_{x_ix_i}v^s(x,t)| \leq \|\partial_{x_ix_i}f\|_{L^\infty} \int_{\mathbb{R}^n}\Phi(x-y,t-s)\,d\mathcal{L}^n(y) = \|\partial_{x_ix_i}f\|_{L^\infty}. \end{align*} This bound is uniform in $s$ and integrable over $[0,t]$ (it is a constant). By the [dominated convergence theorem](/theorems/4), the Laplacian and the $s$-integral commute, completing the argument. [/guided] [/step] [step:Combine terms and verify the initial condition $u(\cdot, 0) = 0$] From the Leibniz rule (step 2), the boundary term evaluation (step 3), and the Laplacian interchange (step 4): \begin{align*} \partial_t u(x,t) &= v^t(x,t) + \int_0^t \partial_t v^s(x,t)\,d\mathcal{L}^1(s) = f(x,t) + \Delta u(x,t). \end{align*} Rearranging gives $\partial_t u - \Delta u = f$ in $\mathbb{R}^n \times (0,\infty)$. For the initial condition: at $t = 0$, the domain of integration $[0,0]$ has $\mathcal{L}^1$-measure zero, so \begin{align*} u(x,0) = \int_0^0 v^s(x,0)\,d\mathcal{L}^1(s) = 0. \end{align*} [/step] --- ## Proof 2: Distributional space-time convolution [step:Construct the causal fundamental solution $E$ and verify $(\partial_t - \Delta)E = \delta$ in $\mathcal{D}'(\mathbb{R}^{n+1})$] Define the causal extension \begin{align*} E: \mathbb{R}^{n+1} &\to \mathbb{R} \\ (x,t) &\mapsto \begin{cases} \Phi(x,t) & \text{if } t > 0, \\ 0 & \text{if } t \leq 0. \end{cases} \end{align*} [claim:$E$ is locally integrable] $E \in L^1_{\mathrm{loc}}(\mathbb{R}^{n+1})$. [/claim] [proof] For any compact set $K \subset \mathbb{R}^{n+1}$, the integral $\int_K |E(x,t)|\,d\mathcal{L}^{n+1}(x,t)$ is finite because the contribution from $\{t \leq 0\}$ is zero, and for $t > 0$ the unit mass property gives $\int_{\mathbb{R}^n}\Phi(x,t)\,d\mathcal{L}^n(x) = 1$. Therefore $\int_K |E|\,d\mathcal{L}^{n+1} \leq \int_0^{T_K} 1\,d\mathcal{L}^1(t) = T_K < \infty$, where $T_K := \sup\{t : (x,t) \in K \text{ for some } x\}$. [/proof] [claim:$(\partial_t - \Delta)E = \delta$ in $\mathcal{D}'(\mathbb{R}^{n+1})$] For every test function $\psi \in C_c^\infty(\mathbb{R}^{n+1})$, \begin{align*} ((\partial_t - \Delta)E)(\psi) = \psi(0,0). \end{align*} [/claim] [proof] Since $E$ is a [regular distribution](/page/Regular%20Distribution), the distributional derivatives act by integration against the adjoint operators. The distributional time derivative satisfies $(\partial_t E)(\psi) = -E(\partial_t\psi) = -\int_0^\infty\int_{\mathbb{R}^n}\Phi(x,t)\,\partial_t\psi(x,t)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t)$, and the distributional Laplacian satisfies $(\Delta E)(\psi) = E(\Delta\psi)$. For the Laplacian term: since $\Phi(\cdot,t)$ is smooth and decays as a Gaussian for each $t > 0$, [integration by parts](/theorems/210) in $x$ (boundary terms vanish at infinity) gives \begin{align*} \int_{\mathbb{R}^n}\Phi(x,t)\,\Delta\psi(x,t)\,d\mathcal{L}^n(x) = \int_{\mathbb{R}^n}(\Delta\Phi)(x,t)\,\psi(x,t)\,d\mathcal{L}^n(x). \end{align*} For the time derivative term: choose $T > 0$ large enough that $\psi(x,t) = 0$ for $t > T$. Integrate by parts in $t$ on $[\varepsilon, T]$: \begin{align*} -\int_\varepsilon^T\int_{\mathbb{R}^n}\Phi\,\partial_t\psi\,d\mathcal{L}^n\,d\mathcal{L}^1 &= \int_\varepsilon^T\int_{\mathbb{R}^n}(\partial_t\Phi)\,\psi\,d\mathcal{L}^n\,d\mathcal{L}^1 - \left[\int_{\mathbb{R}^n}\Phi(x,t)\,\psi(x,t)\,d\mathcal{L}^n(x)\right]_{t=\varepsilon}^{t=T}. \end{align*} The boundary term at $t = T$ vanishes because $\psi(\cdot, T) = 0$. The boundary term at $t = \varepsilon$ is $-\int_{\mathbb{R}^n}\Phi(x,\varepsilon)\,\psi(x,\varepsilon)\,d\mathcal{L}^n(x)$. Combining the time and Laplacian contributions and using $\partial_t\Phi - \Delta\Phi = 0$ for $t > 0$ (property 2 of the [fundamental solution](/theorems/53)): \begin{align*} ((\partial_t - \Delta)E)(\psi) &= \lim_{\varepsilon \to 0^+}\left(\int_\varepsilon^T\int_{\mathbb{R}^n}\underbrace{(\partial_t\Phi - \Delta\Phi)}_{= 0}\,\psi\,d\mathcal{L}^n\,d\mathcal{L}^1 + \int_{\mathbb{R}^n}\Phi(x,\varepsilon)\,\psi(x,\varepsilon)\,d\mathcal{L}^n(x)\right). \end{align*} The first integral vanishes. For the second: $\psi$ is continuous and bounded, and $\psi(x,\varepsilon) \to \psi(x,0)$ uniformly as $\varepsilon \to 0$. By the delta-limit property (property 4 of the [fundamental solution](/theorems/53)): \begin{align*} \lim_{\varepsilon \to 0^+}\int_{\mathbb{R}^n}\Phi(x,\varepsilon)\,\psi(x,\varepsilon)\,d\mathcal{L}^n(x) = \psi(0,0). \end{align*} Therefore $((\partial_t - \Delta)E)(\psi) = \psi(0,0) = \delta(\psi)$. [/proof] [guided] The causal fundamental solution $E$ extends $\Phi$ to all of $\mathbb{R}^{n+1}$ by setting it to zero for non-positive times: \begin{align*} E(x,t) := \begin{cases} \Phi(x,t) = (4\pi t)^{-n/2}\exp\!\left(-\dfrac{|x|^2}{4t}\right) & \text{if } t > 0, \\ 0 & \text{if } t \leq 0. \end{cases} \end{align*} This extension is locally integrable: $E \in L^1_{\mathrm{loc}}(\mathbb{R}^{n+1})$, since for any compact $K \subset \mathbb{R}^{n+1}$ with $K \subset \mathbb{R}^n \times (-\infty, T_K]$, the unit mass property $\int_{\mathbb{R}^n}\Phi(x,t)\,d\mathcal{L}^n(x) = 1$ gives $\int_K |E|\,d\mathcal{L}^{n+1} \leq T_K < \infty$. Therefore $E$ defines a [regular distribution](/page/Regular%20Distribution) and its distributional derivatives are computed by integration against test functions. The distributional identity $(\partial_t - \Delta)E = \delta$ is the precise sense in which $E$ is a fundamental solution of the heat operator on all of $\mathbb{R}^{n+1}$. To verify this, take $\psi \in C_c^\infty(\mathbb{R}^{n+1})$ with $\psi(x,t) = 0$ for $t > T$. The distributional time derivative acts as $(\partial_t E)(\psi) = -\int_0^\infty\int_{\mathbb{R}^n}\Phi(x,t)\,\partial_t\psi(x,t)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t)$ (the integral starts at $0$ because $E = 0$ for $t \leq 0$). Integration by parts in the spatial variables transfers $\Delta$ from $\psi$ to $\Phi$: \begin{align*} \int_{\mathbb{R}^n}\Phi(x,t)\,\Delta\psi(x,t)\,d\mathcal{L}^n(x) = \int_{\mathbb{R}^n}(\Delta\Phi)(x,t)\,\psi(x,t)\,d\mathcal{L}^n(x), \end{align*} with boundary terms vanishing because $\Phi$ has Gaussian decay. For the time derivative, integrate by parts in $t$ on $[\varepsilon, T]$ (not $[0, T]$, since $\Phi$ is undefined at $t = 0$): \begin{align*} -\int_\varepsilon^T\int_{\mathbb{R}^n}\Phi\,\partial_t\psi\,d\mathcal{L}^n\,d\mathcal{L}^1 = \int_\varepsilon^T\int_{\mathbb{R}^n}(\partial_t\Phi)\,\psi\,d\mathcal{L}^n\,d\mathcal{L}^1 - \left[\int_{\mathbb{R}^n}\Phi(x,t)\,\psi(x,t)\,d\mathcal{L}^n(x)\right]_{t=\varepsilon}^{t=T}. \end{align*} The boundary term at $t = T$ vanishes since $\psi(\cdot, T) = 0$. The boundary term at $t = \varepsilon$ is $-\int_{\mathbb{R}^n}\Phi(x,\varepsilon)\,\psi(x,\varepsilon)\,d\mathcal{L}^n(x)$, which captures the "jump" in $E$ across $\{t = 0\}$. Combining the time and Laplacian contributions and using $\partial_t\Phi - \Delta\Phi = 0$ for $t > 0$ (property 2 of the [fundamental solution](/theorems/53)): \begin{align*} ((\partial_t - \Delta)E)(\psi) &= \lim_{\varepsilon \to 0^+}\left(\int_\varepsilon^T\int_{\mathbb{R}^n}\underbrace{(\partial_t\Phi - \Delta\Phi)}_{= 0}\,\psi\,d\mathcal{L}^n\,d\mathcal{L}^1 + \int_{\mathbb{R}^n}\Phi(x,\varepsilon)\,\psi(x,\varepsilon)\,d\mathcal{L}^n(x)\right). \end{align*} The interior integral vanishes identically. For the boundary term, $\psi(x,\varepsilon) \to \psi(x,0)$ uniformly as $\varepsilon \to 0$ (since $\psi$ is smooth), and the delta-limit property (property 4 of the [fundamental solution](/theorems/53)) gives: \begin{align*} \lim_{\varepsilon \to 0^+}\int_{\mathbb{R}^n}\Phi(x,\varepsilon)\,\psi(x,\varepsilon)\,d\mathcal{L}^n(x) = \psi(0,0). \end{align*} Therefore $((\partial_t - \Delta)E)(\psi) = \psi(0,0) = \delta(\psi)$ for every test function $\psi$, establishing the fundamental solution identity $(\partial_t - \Delta)E = \delta$ in $\mathcal{D}'(\mathbb{R}^{n+1})$. [/guided] [/step] [step:Identify the Duhamel integral as the space-time convolution $u = E * \tilde{f}$] Define the zero-extension \begin{align*} \tilde{f}: \mathbb{R}^{n+1} &\to \mathbb{R} \\ (x,t) &\mapsto \begin{cases} f(x,t) & \text{if } t \geq 0, \\ 0 & \text{if } t < 0. \end{cases} \end{align*} Since $f$ has compact support in $\mathbb{R}^n \times [0,\infty)$, the extension $\tilde{f}$ has compact support in $\mathbb{R}^{n+1}$. The space-time [convolution](/page/Convolution) is \begin{align*} (E * \tilde{f})(x,t) &= \int_{-\infty}^{\infty}\int_{\mathbb{R}^n}E(x - y, t - s)\,\tilde{f}(y,s)\,d\mathcal{L}^n(y)\,d\mathcal{L}^1(s). \end{align*} The integrand is nonzero only when both $E(x-y,t-s) \neq 0$ and $\tilde{f}(y,s) \neq 0$. The first condition requires $t - s > 0$, i.e., $s < t$. The second requires $s \geq 0$. Therefore the $s$-integral reduces to $[0,t)$, and on this domain $E(x-y,t-s) = \Phi(x-y,t-s)$ and $\tilde{f}(y,s) = f(y,s)$. Since the integrand at $s = t$ vanishes ($E(x-y,0) = 0$), the integral over $[0,t)$ equals the integral over $[0,t]$: \begin{align*} (E * \tilde{f})(x,t) = \int_0^t\int_{\mathbb{R}^n}\Phi(x - y, t - s)\,f(y,s)\,d\mathcal{L}^n(y)\,d\mathcal{L}^1(s) = u(x,t). \end{align*} [/step] [step:Apply the heat operator to the convolution and conclude $\partial_tu - \Delta u = f$] Since $\tilde{f}$ has compact support and $E \in L^1_{\mathrm{loc}}(\mathbb{R}^{n+1})$, the convolution $E * \tilde{f}$ is well-defined. The $C^{2,1}$ regularity of $u$ follows from the smoothing property of convolution: the regularity of $E$ for $t > 0$ combined with the compact support and $C^{2,1}$ regularity of $\tilde{f}$ ensures that $u$ inherits $C^{2,1}$ regularity. For distributional [convolutions](/page/Convolution), the derivative identity states: if $T \in \mathcal{D}'(\mathbb{R}^{n+1})$ and $\varphi$ has compact support, then $D^\alpha(T * \varphi) = (D^\alpha T) * \varphi$ for any partial derivative $D^\alpha$. Applying this with $T = E$ and $\varphi = \tilde{f}$, and using linearity: \begin{align*} (\partial_t - \Delta)(E * \tilde{f}) = ((\partial_t - \Delta)E) * \tilde{f} = \delta * \tilde{f}, \end{align*} where the last equality uses $(\partial_t - \Delta)E = \delta$ from the previous claim. Since convolution with $\delta$ is the identity ($\delta * \tilde{f} = \tilde{f}$): \begin{align*} (\partial_t - \Delta)u = \tilde{f}. \end{align*} For $t > 0$, $\tilde{f}(x,t) = f(x,t)$, giving $\partial_t u - \Delta u = f$ in $\mathbb{R}^n \times (0,\infty)$. For the initial condition: when $t \leq 0$, the convolution $(E * \tilde{f})(x,t)$ integrates over $s \in (-\infty, t] \cap [0,\infty)$. For $t \leq 0$, this set is at most $\{0\}$, which has $\mathcal{L}^1$-measure zero, so $u(x,t) = 0$ for $t \leq 0$. In particular, $u(\cdot, 0) = 0$. [guided] Since $\tilde{f}$ has compact support and $E \in L^1_{\mathrm{loc}}(\mathbb{R}^{n+1})$, the [convolution](/page/Convolution) $E * \tilde{f}$ is well-defined. The $C^{2,1}$ regularity of $u$ follows from the smoothing property: $E$ is $C^\infty$ for $t > 0$, and the compact support of $\tilde{f}$ ensures the convolution inherits this regularity. For distributional convolutions, the derivative identity states: if $T \in \mathcal{D}'(\mathbb{R}^{n+1})$ and $\varphi$ has compact support, then \begin{align*} D^\alpha(T * \varphi) = (D^\alpha T) * \varphi \end{align*} for any partial derivative $D^\alpha$. Applying this with $T = E$ and $\varphi = \tilde{f}$, and using linearity of the heat operator $\partial_t - \Delta$: \begin{align*} (\partial_t - \Delta)(E * \tilde{f}) = ((\partial_t - \Delta)E) * \tilde{f} = \delta * \tilde{f}, \end{align*} where the last equality uses $(\partial_t - \Delta)E = \delta$ from the previous step. Since convolution with $\delta$ is the identity ($\delta * \tilde{f} = \tilde{f}$): \begin{align*} (\partial_t - \Delta)u = \tilde{f}. \end{align*} For $t > 0$, $\tilde{f}(x,t) = f(x,t)$, giving $\partial_t u - \Delta u = f$ in $\mathbb{R}^n \times (0,\infty)$. For the initial condition: when $t \leq 0$, the convolution integrates over $s \in (-\infty, t] \cap [0,\infty)$, which for $t \leq 0$ is at most $\{0\}$ (measure zero), so \begin{align*} u(x,t) = 0 \quad \text{for all } t \leq 0. \end{align*} In particular, $u(\cdot, 0) = 0$. The entire argument reduces to the algebraic chain $(\partial_t - \Delta)(E * \tilde{f}) = ((\partial_t - \Delta)E) * \tilde{f} = \delta * \tilde{f} = \tilde{f} = f$. This reveals Duhamel's principle as an instance of inverting a constant-coefficient operator $L$ by convolving with its fundamental solution $E$ satisfying $LE = \delta$. The compact support of $\tilde{f}$ is essential -- without it, the convolution might not be well-defined or the derivative interchange might fail. [/guided] [/step]