### General Info

Remember: for every question you **must** indicate all the people with whom you collaborated. If you didn't collaborate with anyone else on a given problem, write "Collaborators: none."

Please write legibly, or type-up your solutions in LaTeX. The grader will greatly appreciate it.

This problem set will be due *in class, Friday October 13*.

### Problems

*The first five problems on this problem set will concern the following bounday value problem.*

Consider the second order ordinary differential equation
\[ \ddot{y}(t) + b(t) \dot{y}(t) = c(t) y(t) \tag{ODE} \]
where the functions \(b, c\in C^\infty(\mathbb{R};\mathbb{R}) \), with the boundary conditions
\[ y(t_0) = y_0, \qquad y(t_1) = y_1, \qquad t_0 < t_1. \tag{BC}\]
The problems will guide you through the proof of the following existence and uniqueness theorem

TheoremIf for every \(t\in \mathbb{R}\), \(c(t) \geq 0\), then there exists a unique \(y\in C^2([t_0,t_1]; \mathbb{R})\) that solves (ODE) satisfying (BC).

- (Maximum principle) Assume \(c(t) \geq 0\). Prove that a solution \(y\) to (ODE) with boundary conditions (BC) must satisfy
\[ \max_{t\in [t_0, t_1]} y(t) \leq \max( y_0, y_1, 0) \]
and
\[ \min_{t\in [t_0, t_1]} y(t) \geq \min( y_0, y_1, 0).\]
*(Hint: in the case \(c\) is strictly positive, this follows from the second derivative test. To handle the case \(c\) is only non-negative, consider the function \(y^{(\epsilon,\lambda)}(t) := y(t) + \epsilon e^{\lambda t}\). Show that for any sufficiently large \(\lambda\) and any \(\epsilon > 0\) you can apply the second derivative test prove that \(y^{(\epsilon,\lambda)}\) cannot achieve a positive local maximum.)* - (Uniqueness) Prove that, when \(c(t) \geq 0\), if \(y,\tilde{y}\) both solve (ODE) with the same boundary conditions (BC), then \( y = \tilde{y}\).
- (Non-uniqueness) Give an explicit counterexample (by this we mean explicit functions \(b, c, y, \tilde{y}\) and explicit values \(y_0,y_1, t_0,t_1\)) where \(c(t)\) is allowed to be negative, such that \(y\) and \(\tilde{y}\) are
*distinct*solutions to (ODE) with the same (BC). (Existence)

*For this problem, you may make use of the following consequence of Picard's existence theorem for ordinary differential equations:*

**Lemma**Given any \(y_0, z_0\in\mathbb{R}\), there exists a unique solution \(y \in C^2(\mathbb{R};\mathbb{R})\) to the equation (ODE) satisfying the*initial conditions*\( y(t_0) = y_0, \dot{y}(t_0) = z_0\). Furthermore, for any fixed \(t'\in \mathbb{R}\), the mapping \(z_0 \mapsto y(t')\) is continuous.

(*Hint: show using the maximum principle that there exists a solution to (ODE) with*$y(t_0) = y_0$*and*$y(t_1) > y_1$;*similarly show there exists a solution with the same boundary condition at*$t_0$*but with*$y(t_1) < y_1$.)(Non-existence) Give an explicit counterexample (functions \(b,c\) and values \(y_0, y_1, t_0, t_1\)) where \(c(t)\) takes some negative values and the corresponding (ODE) with (BC) admits

*no*solution.Formulate and prove, based on question 1 above, a maximum principle for solutions to equations of the form \[ - \triangle u + b \cdot \nabla u + c u = 0. \] (Here \(b\) is a vector valued function on the domain \(\Omega\), and \(c\) is some given real-valued function.)

Let \(d \geq 3\) and fix \(x\in B(0,1)\). Set \(v^x(y) = \Phi(x-y)\) and define \(w^x = Kv^x\), the Kelvin transform of \(v^x\) (see Homework 2, question 2 for definition).

Prove that:- \(w^x\) extends to a continuous function defined on \(\overline{B(0,1)}\). (What is the value of \(w^x(0)\)?)
- The extended function \(w^x\) is harmonic on \(B(0,1)\). (Why does it suffice to only check that \(w^x\) is twice differentiable at the origin and that its Laplacian vanishes there?) (Do
*not*use the removable singularity theorem here! Do the computations by hand. This is necessary because the proof of the removable singularity theorem uses the solvability of the Laplace equation on the ball in one of the steps.) - \( w^x = v^x\) along \(\partial B(0,1)\).

Now, let \(G(x,y) = v^x(y) - w^x(y)\) (as defined in the previous problem). Prove that:

If \(g\) is a continuous function defined on \(\partial B(0,1)\), then the function defined by the formula \[ u(x) := - \int_{\partial B(0,1)} \partial_{\nu} G(x,y) g(y) ~\mathrm{d}S(y) \] for \(x\in B(0,1)\) is harmonic, and satisfies \[ \lim_{x\to y} u(x) = g(y) \] when the limit is taken over \(x\in B(0,1)\) and \(y\in \partial B(0,1)\). (In other words, it solves \(-\triangle u = 0\) with the boundary condition \( u|_{\partial B(0,1)} = g\). This function \(G\) is in fact the Green's function for the unit ball.)Let \(\Omega\) be an open subset of \(\mathbb{R}^d\). Prove that if \(f\in C^0(\Omega)\) is such that \[ \int_\Omega f \cdot g ~\mathrm{d}x = 0 \] for every \(g \in C^\infty_c(\Omega)\), then \( f \equiv 0 \).

(*Hint: By continuity, if*\(f(x_0) \neq 0\),*there exists some*\(\epsilon > 0\)*such that*\(f\)*is signed on*\(B(x_0, \epsilon)\).*Use this to derive a contradiction using the freedom to choose*\(g\).)Suppose \(u\) is smooth and solves \(\partial_t u = \triangle u\) in \((0,\infty) \times \mathbb{R}^d\). Show that the function \[ v(t,x) := 2t \partial_t u(t,x) + \sum_{i = 1}^d x^i \partial_{x^i} u(t,x) \] is also a solution to the heat equation.

Write down an explicit formula for a solution of the differential equation \[ \partial_t u - \triangle u + \lambda u = f \] on the domain \( (0,\infty)\times \mathbb{R}^d\) with the initial data \[ u(0,x) = g(x).\] \(\lambda\) should be taken to be an arbitrary real-valued constant.

Let \(u\in C^2((0,T)\times \Omega)\cap C^0(\overline{[0,T]\times \Omega})\), where \(\Omega\) is open and bounded, and suppose \(u\) satisfies \(\partial_t u - \triangle u \leq -\epsilon < 0\) for some constant \(\epsilon > 0\).

- Prove that \(u\) cannot attain a local maximum on the set \( (0,T] \times \Omega\).

*(Edit 10.16.2017: see this discussion for a sketch of the solution to this part.)* - Prove, without using the mean value property, that if \(v \in C^2((0,T)\times \Omega)\cap C^0(\overline{[0,T]\times \Omega})\) solves the heat equation, then \(v\) satisfies the (weak) maximum principle.

*(Hint: apply the first part to the function \(u = v - \epsilon t\).)*

- Prove that \(u\) cannot attain a local maximum on the set \( (0,T] \times \Omega\).