The assumed linear estimate and the first bound give
$$
\lVert A(w)\rVert_{L_t^\infty H_x^1}
\leq\alpha\big(\lVert\psi\rVert_{H^1}
+\beta\tau^{1/2}b^3\big).
$$
Choose $b>2\alpha\lVert\psi\rVert_{H^1}$ and then choose $\tau>0$ so small that $\alpha\beta\tau^{1/2}b^3\leq b/2$. This proves that $A$ maps $X_{b,\tau}$ into itself.

For $w,\widetilde w\in X_{b,\tau}$, linearity of the heat equation and the second cubic estimate give
$$
\lVert A(w)-A(\widetilde w)\rVert_{L_t^\infty H_x^1}
\leq2\alpha\gamma\tau^{1/2}b^2
\lVert w-\widetilde w\rVert_{L_t^\infty H_x^1}.
$$
Shrinking $\tau$ again makes $2\alpha\gamma\tau^{1/2}b^2<1$. Thus $A$ is a contraction of the closed ball $X_{b,\tau}$ in the Banach space $L^\infty((0,\tau);H_0^1(U))$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 105 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-105/4/b/solution