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\centerline{Mathematics 373 -- Applied Mathematics (PDE)}
\centerline{Individual Problem Set 1 -- First Order Linear PDE}
\centerline{January 18, 2001}
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A.  (Review from Multivariable Calculus) Recall that the chain rule 
from 1-variable calculus says that if $g$ is differentiable at $x$ 
and $f$ is differentiable at $g(x)$ then the composition $f(g(x))$ 
is differentiable at $x$ and 
$${d\over dx} f(g(x)) = f'(g(x))g'(x)$$
A first generalization of this to functions of several variables
is the following:  If $f(x,y)$ is a differentiable function
of two variables and $\alpha(t) = (x(t),y(t))$ is a differentiable 
parametric curve with image in the domain of $f$, then
$${d\over dt} f(\alpha(t)) = {\partial f\over\partial x}(x(t),y(t))x'(t) + 
{\partial f\over\partial y}(x(t),y(t))y'(t)
= \nabla f(\alpha(t))\cdot\alpha'(t)\leqno(1)$$
where $\nabla f$ is the gradient vector of $f$.
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\item{1)}  Use (1) to compute ${d\over dt} f(\alpha(t))$ if 
$f(x,y) = x^2 - y^2 + x^3$ and $\alpha(t) = (t^2 - 1,t^3 - t)$.
What does your result say about the curve $\alpha$ and $f$?
\item{2)}  If $\alpha(t) = (x_0,y_0) + t(u,v)$ where $(u,v)$ is
a unit vector, 
$${d\over dt}f(\alpha(t))|_{t=0} = \nabla f(\alpha(0))\cdot\alpha'(0) 
=\nabla f(x_0,y_0)\cdot (u,v)$$
is the {\it directional derivative} of $f$ at $\alpha(0) = (x_0,y_0)$
in the direction of the vector $\alpha'(0) = (u,v)$.  Fixing 
the point $(x_0,y_0)$, consider the directional derivative as a 
function of the direction vector $(u,v)$.
In which direction is
the directional derivative {\it maximized}?  In which direction 
is the directional derivative {\it minimized}?  In which directions
is the directional derivative {\it equal to zero}?  Prove your 
assertions.
\item{3)}  The {\it line integral} of a vector field
${\bf F}(x,y) = (P(x,y),Q(x,y))$ along a parametric curve
$\alpha(t) = (x(t),y(t))$ ($a \le t \le b$) in the plane is
{\it defined to be}
$$\int_\alpha {\bf F} = \int_a^b {\bf F}(\alpha(t))\cdot \alpha'(t)\, dt$$
Show that if ${\bf F} = \nabla f$ for some $f(x,y)$, then
the line integral doesn't actually depend on the path $\alpha$, only
on the starting and final points of the path: $\alpha(a)$ and $\alpha(b)$.
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B.  Solve the PDE $4u_x + 7u_t = 0$ with the Cauchy condition
$u(x,0) = \sin(3x) - \cos(2x)$.  
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C.  Use the change of coordinates method to solve 
$u_x + 2u_y + u = e^{3x-2y}$ with $u(x,0) = 0$.
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D.  Solve $(1+x^2)u_x + u_y = 0$ for $u(x,y)$.  Sketch some of the 
characteristic curves.
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E.  Solve $yu_x + xu_y = 0$ with $u(0,y) = e^{-y^2}$.  In which 
region of the plane is the solution uniquely determined?
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F.  What happens if we try to solve a first order linear PDE
$a(x,y)u_x + b(x,y)u_y = 0$ with the condition $u(x,0) = f(x)$,
but the $x$-axis {\it is a characteristic curve} of the equation?
(Note: There are two cases!)
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{\it Assignment} 
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Individual write-ups due in class on Thursday, January 25.
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