Exercise 10
Fundamental solution
In this exercise we investigate the fundamental solution of the transient heat equation.
In particular we try to understand how it can be used to solve the initial value problem.
We consider the initial value problem given by the transient 1D heat equation with constant diffusion constant \(\alpha \equiv 1\):
\[ \begin{aligned} \frac{\partial T(x,t)}{\partial t} - \frac{\partial ^2 T(x,t)}{\partial x^2} &= 0 & \quad , \\ T(x, t=0) &= g(x) & \quad , \end{aligned} \]
where \(g(x)\) is a bounded initial condition.
Our PDE operator to be applied to a function \(\bullet\) is hence given by \[ \mathcal{L}\bullet = \frac{\partial \bullet}{\partial t} - \frac{\partial^2 \bullet}{\partial x^2} \quad . \] and by definition, a solution to the initial value problem has to satisfy \(\mathcal{L} T(x,t) = 0\) at all times \(t > 0\).
- Show that the fundamental solution
\[ G(x,t) = \frac{1}{\sqrt{4 \pi t}} e^{-\frac{x^2}{4t}} \quad , \quad t > 0 \quad . \]
is a solution to the differential operator \(\mathcal{L}\). What do we have to compute to show this?
- We have to show \(\mathcal{L} G(x,t) = 0\) for all \(x\) and \(t > 0\).
- Compute the derivative \(\partial_t\) and \(\partial_{x^2}\) of the fundamental solution and show that they always add up to zero, independent of the position \(x\) and time \(t\).
We can compute the derivatives: \[ \frac{\partial G(x,t)}{\partial t} = \left(-\frac{2\pi}{\left(4\pi\,t\right)^{\frac{3}{2}}} + \frac{1}{\sqrt{4\pi\,t}}\frac{x^2}{4\,t^2}\right)\,e^{-\frac{x^2}{4\,t}} \quad , \]
\[ \frac{\partial G(x,t)}{\partial x} = \left(-\frac{2\pi\,x}{\left(4\pi\,t\right)^{\frac{3}{2}}} \right)\,e^{-\frac{x^2}{4\,t}} \quad , \]
\[ \frac{\partial^2 G(x,t)}{\partial x^2} = \left(-\frac{2\pi}{\left(4\pi\,t\right)^{\frac{3}{2}}} + \frac{1}{4\,t}\frac{2\pi\,x^2}{\left(4\pi\,t\right)^{\frac{3}{2}}}\right)\,e^{-\frac{x^2}{4\,t}} \]
to show that that
\[ \partial_t G(x,t) - \partial_{x^2} G(x,t) = 0 \quad \forall x,\,t\]
- Show that for a given initial condition \(g(x)\), the solution to the initial value problem \(T(x,t)\) is given by the convolution
\[ T(x,t) = (G(\cdot,t)*g(\cdot))(x) = \int_{-\infty}^{\infty} G(x - y, t)\, g(y) \, dy \quad , \quad t > 0 \quad . \]
of the fundamental solution with the initial condition. You can use the following properties of the convolution:
\[ \mathcal{L} (f * g) = (\mathcal{L} f) * g \quad , \] if the differential operator \(\mathcal{L}\) is linear and has constant coefficients and \[ (g * \delta)(x) = g(x) \quad , \] where \(\delta\) is the Dirac delta distribution.
- We have to show \(\mathcal{L} T(x,t) = 0\) for all \(x\) and \(t > 0\).
- and \(\lim_{t \to 0^+} T(x,t) = g(x)\) for all \(x\).
Using property 1:
\[ \begin{aligned} \mathcal{L} T(x,t) & = \mathcal{L} (G(\cdot,t) * g(\cdot))(x) \\ & = (\mathcal{L} G(\cdot,t)) * g(\cdot)(x) \\ & = \left(0 * g(\cdot)\right)(x) = 0 \quad . \end{aligned} \]
Where we used that \(\mathcal{L} G(\cdot,t) = 0\), shown in the first part of the exercise (the fundamental solution is a solution to the differential operator \(\mathcal{L}\)).
Then we still have to show that \(T(x,t)\) converges to \(g(x)\) as \(t \to 0^+\), e.g. we show that the initial condition is satisfied:
\[ \begin{aligned} \lim_{t \to 0^+} T(x,t) & = \lim_{t \to 0^+} (G(\cdot,t)*g(\cdot))(x) \\ & = (\lim_{t \to 0^+}G(\cdot,t)*g(\cdot))(x) \\ & = (g(\cdot) * \delta)(x) \\ & = g(x) \quad . \end{aligned} \]
Here we used that no other term other thant \(G\) depends on time and the convolution operator only acts on the spatial variable, so we can pull the limit into the convolution.
The limit of the fundamental solution \(G(\cdot,t)\) as \(t \to 0^+\) is the Dirac delta distribution \(\delta\) by construction.