2.II.6B

Differential Equations | Part IA, 2006

(i) Consider the equation

∂u∂t+∂u∂x=∂2u∂x2+f(t,x)\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}=\frac{\partial^{2} u}{\partial x^{2}}+f(t, x)

and, using the change of variables (t,x)↦(s,y)=(t,x−t)(t, x) \mapsto(s, y)=(t, x-t), show that it can be transformed into an equation of the form

∂U∂s=∂2U∂y2+F(s,y)\frac{\partial U}{\partial s}=\frac{\partial^{2} U}{\partial y^{2}}+F(s, y)

where U(s,y)=u(s,y+s)U(s, y)=u(s, y+s) and you should determine F(s,y)F(s, y).

(ii) Let H(y)H(y) be the Heaviside function. Find the general continuously differentiable solution of the equation

w′′(y)+H(y)=0w^{\prime \prime}(y)+H(y)=0

(iii) Using (i) and (ii), find a continuously differentiable solution of

∂u∂t+∂u∂x=∂2u∂x2+H(x−t)\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}=\frac{\partial^{2} u}{\partial x^{2}}+H(x-t)

such that u(t,x)→0u(t, x) \rightarrow 0 as x→−∞x \rightarrow-\infty and u(t,x)→−∞u(t, x) \rightarrow-\infty as x→+∞x \rightarrow+\infty

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