# Solving PDE $xu_x+(x+t)u_t=1$ with $u(1,t)=t$ [duplicate]

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 $$u(x,0)$$,

$$x(0,s)=s$$

$$y(0,s)=0$$

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

## 1 Answer

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 $$u(x,t) = t/x$$.