Terminology: Question 9-3.6 & Solution

Question 6

For $u_t+u_x=0$ on $\mathbb R^2$ , a classical solution here means a $C^1$ field satisfying the PDE at every point. Compare $u(x,t)=|x-t|,\qquad u_\varepsilon(x,t)=\sqrt{(x-t)^2+\varepsilon^2},\quad\varepsilon>0.$ For this question only, an integral solution is a locally integrable field satisfying $\int_{\mathbb R^2}u(x,t)\bigl(\phi_t+\phi_x\bigr)(x,t)\,dx\,dt=0$ for every smooth compactly supported test function $\phi$ .

Tasks

Check the PDE for $u$ off $x=t$ , and decide whether $u$ is a classical solution on the whole plane.
Verify that each $u_\varepsilon$ is classical and prove a uniform error bound tending to zero as $\varepsilon\downarrow 0$ .
Explain precisely why uniform convergence of these classical solutions does not establish that $u$ is classical. Inspect $u_{\varepsilon,x}$ near the line $x=t$ .
Verify that $u$ is an integral solution using $s=x-t$ , $r=t$ . State which conclusions about $u$ have and have not been proved.

Original worksheet page 1: question and worked solution for 9-3-006

Show solutionHide solution+

Question 6 – Solution

Strategy. A solution label includes a regularity requirement; checking a formula away from a singular line is not enough.

Step 1: Locate the failure of classical regularity. For $x>t$ , $u_x=1,u_t=-1$ ; for $x<t$ , $u_x=-1,u_t=1$ . Thus the PDE holds on both open regions. At $x=t$ , the two one-sided spatial derivatives disagree. The field is not differentiable there, hence is .

Step 2: Verify the smooth approximations. Writing $s=x-t$ , one has $u_{\varepsilon,x}=\frac{s}{\sqrt{s^2+\varepsilon^2}},\qquad u_{\varepsilon,t}=-\frac{s}{\sqrt{s^2+\varepsilon^2}}.$ These are smooth and sum to zero. Also $0\le\sqrt{s^2+\varepsilon^2}-|s| =\frac{\varepsilon^2}{\sqrt{s^2+\varepsilon^2}+|s|}\le\varepsilon.$ Equality occurs at $s=0$ , so the uniform error on the whole plane is exactly $\varepsilon$ .

Step 3: Separate convergence of values from derivatives. For $s\ne 0$ , $u_{\varepsilon,x}\to\operatorname{sgn}(s)$ , while its value at $s=0$ is always zero. This pointwise limit is discontinuous at zero. It cannot be a uniform limit on any closed interval about zero of these continuous derivatives. Uniform convergence of function values does not preserve differentiability, so it does not prove classical solvability.

Step 4: Check the stated integral formulation. Set $\psi(s,r)=\phi(s+r,r)$ . The change of variables has Jacobian one, and $\psi_r=\phi_x+\phi_t$ . Compact support and local integrability allow integration in either order. Therefore $\int_{\mathbb R^2}|x-t|(\phi_t+\phi_x)\,dx\,dt =\int_{\mathbb R}|s|\left(\int_{\mathbb R}\psi_r(s,r)\,dr\right)ds=0.$ The inner integral vanishes by compact support. Thus $u$ is an integral solution under the supplied definition, as well as a classical solution off the line. It remains nonclassical on the whole plane; no uniqueness assertion follows from this verification alone.

Original worksheet page 2: question and worked solution for 9-3-006

Terminology — Question 6

Question 6

Question 6 – Solution