Paper 1, Section II, A

Differential Equations | Part IA, 2020

Show that for each t>0t>0 and x∈Rx \in \mathbb{R} the function

K(x,t)=14πtexp⁡(−x24t)K(x, t)=\frac{1}{\sqrt{4 \pi t}} \exp \left(-\frac{x^{2}}{4 t}\right)

satisfies the heat equation

∂u∂t=∂2u∂x2\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}}

For t>0t>0 and x∈Rx \in \mathbb{R} define the function u=u(x,t)u=u(x, t) by the integral

u(x,t)=∫−∞∞K(x−y,t)f(y)dyu(x, t)=\int_{-\infty}^{\infty} K(x-y, t) f(y) d y

Show that uu satisfies the heat equation and lim⁡t→0+u(x,t)=f(x)\lim _{t \rightarrow 0^{+}} u(x, t)=f(x). [Hint: You may find it helpful to consider the substitution Y=(x−y)/4tY=(x-y) / \sqrt{4 t}.]

Burgers' equation is

∂w∂t+w∂w∂x=∂2w∂x2\frac{\partial w}{\partial t}+w \frac{\partial w}{\partial x}=\frac{\partial^{2} w}{\partial x^{2}}

By considering the transformation

w(x,t)=−21u∂u∂xw(x, t)=-2 \frac{1}{u} \frac{\partial u}{\partial x}

solve Burgers' equation with the initial condition lim⁡t→0+w(x,t)=g(x)\lim _{t \rightarrow 0^{+}} w(x, t)=g(x).

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