Solve $xu_x+(x+t)u_t=1$ such that $u(1,t)=t$.

Is the solution defined everywhere?

I had known that this specific problem is related to a Heat Equation problem. I tried solving for its characteristic curves but I had difficulty solving for its parametrized initial condition.

How am I gonna solve for its parametrized initial condition? I know that for an initial condition $u(x,0)$,



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    $\begingroup$ How is it related to the heat equation? $\endgroup$ – Calvin Khor Aug 19 '19 at 8:40
  • $\begingroup$ What are $x$ and $y$ in your question? $\endgroup$ – Viktor Glombik Aug 19 '19 at 13:59

The Method. Set $\mathbf x = (x,t)$. Your PDE is $$ \mathbf a \cdot \nabla u= b$$ where $\mathbf a(x,t) = (x,x+t), b\equiv 1. $ The initial data is prescribed along the curve $\mathbf x_0(z) =(1,z)$, which is $$ u(\mathbf x_0(z)) = u_0(z)$$where $u_0(z) = z$. Note that $\mathbf x_0'(z)^\perp \cdot \mathbf a(\mathbf x_0(z)) =(-1,0)\cdot (1,1+z) = -1 $ so there are no characteristic points. If you first solve the 2 ODEs $$ \partial_s\mathbf X(z,s) = \mathbf a(\mathbf X(z,s)),\quad \mathbf X(z,0) = \mathbf x_0(z), \\ \partial_s Y(z,s) = b,\quad\qquad\ \ \ \ \ Y(z,0) = u_0(z),$$ then you can check that the solution $u=u(\mathbf x)$ to the original problem is given by $$ u(\mathbf X(z,s)) = Y(z,s).$$ Obtaining the form of $u(\mathbf x)$ naturally involves inverting the formula for $\mathbf X$.

Implementation. $$Y(z,s)=z+s.\\\left.\substack{\displaystyle \partial_sX_1 =X_1\qquad \\ \displaystyle\partial_sX_2 =X_1+X_2 }\right\}\implies \mathbf X(z,s) = \binom{e^s}{(z+s)e^s}. $$ Inverting $\mathbf X$: $$ s = \log x,\qquad \\z = t/x-\log x.$$ So $\mathbf X^{-1}(x,t) = (\log x, t/x-\log x).$ This gives the solution $u(x,t) = t/x$.


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