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Let $u_0 \in L^{\infty}(\mathbb{R})$. We say that $u \in L^{\infty}(\mathbb{R} \times [0, +\infty))$ is weak solution of the Burgers equation

$$\left\{\begin{array}{ccl} u_t + uu_x & = & 0;\\ u(x,0) & = & u_0(x), \end{array}\right.$$ if for all $\varphi \in C_c^{\infty}(\mathbb{R} \times [0;1))$ is valid $$\int_{\mathbb{R} \times [0,\infty)} \left[u(x,t)\varphi_t(x,t) + \dfrac{1}{2}u^2(x,t)\varphi_x(x,t)\right]dxdt + \int_{\mathbb{R}} u_0(x)\varphi(x,0)dx = 0.$$

Question: Consider the Burgers equation above with $u_0$ invariant by dilation, that is, there is $\lambda > 0$ such that $$u_0(\lambda x) = u_0(x), \ \ \forall x \in \mathbb{R} \setminus \{0\}.$$

(a) Suppose that $u$ is a weak solution in $\mathbb{R} \times [0,+\infty)$. Show that $v_{\lambda}(x,t) = u(\lambda x,\lambda t)$ is also weak solution of the equation in $\mathbb{R} \times [0,+\infty)$.

(b) Determine all functions $V$ of class $C^1$ such that $V (x/t)$ is a solution of the equation, for all $t>0$ and all $x \in \mathbb{R} $.

Help me with the details of the calculations. I'm having a hard time solving it! :)

EditPiAf
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Santos
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1 Answers1

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(a) The first can be tackled by introducing the change of variable $x'=\lambda x$, $t'=\lambda t$ with $\lambda > 0$ as follows \begin{aligned} \int_{\mathbb{R} \times [0,\infty)} \left[v_\lambda\varphi_t + \tfrac{1}{2}v_\lambda^2\varphi_x\right] dxdt &= \int \lambda \left[u\psi_{t'} + \tfrac{1}{2}u^2\psi_{x'}\right]\frac{dx'dt'}{\lambda^2} \\ &= -\frac1{\lambda} \int u_0 \psi|_{t'=0} dx' \\ &= - \int u_0 \varphi|_{t=0} dx \end{aligned} where we have used the notation $\varphi(x,t) = \psi(\lambda x,\lambda t)$. In other words, $v_\lambda$ is a weak solution too. Here explicit dependence w.r.t. the dependent variables has been omitted for sake of conciseness.

(b) Here we set $u = V(x/t)$ in the PDE $u_t + uu_x = 0$. Since $V$ is continuously differentiable, the chain rule yields $u_t = -\frac{x}{t} V'(x/t)/t$ and $u_x = V'(x/t)/t$, see this post.

EditPiAf
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