lec2 - Math 124A September 30, 2010 Viktor Grigoryan 2...

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Unformatted text preview: Math 124A September 30, 2010 Viktor Grigoryan 2 First-order linear equations Last time we saw how some simple PDEs can be reduced to ODEs, and subsequently solved using ODE methods. For example, the equation u x = 0 (1) has constant in x as its general solution, and hence u depends only on y , thus u ( x,y ) = f ( y ) is the general solution, with f an arbitrary function of a single variable. Today we will see that any linear first order PDE can be reduced to an ordinary differential equation, which will then allow as to tackle it with already familiar methods from ODEs. Let us start with a simple example. Consider the following constant coefficient PDE au x + bu y = 0 . (2) Here a and b are constants, such that a 2 + b 2 6 = 0, i.e. at least one of the coefficients is nonzero (otherwise this would not be a differential equation). Using the inner (scalar or dot) product in R 2 , we can rewrite the left hand side of (2) as ( a,b ) ( u x ,u y ) = 0 , or ( a,b ) u = 0 . Denoting the vector ( a,b ) = v , we see that the left hand side of the above equation is exactly D v u ( x,y ), the directional derivative of u in the direction of the vector v . Thus the solution to (2) must be constant in the direction of the vector v = a i + b j . v = H a , b L- 5 5 x- 5 5 y Figure 1: Characteristic lines bx- ay = c . H a , b L H- b , a L H x , y L = H x , y L H a , b L = H x , y L H- b , a L- 10- 5 5 x- 5 5 y Figure 2: Change of coordinates. The lines parallel to the vector v have the equation bx- ay = c, (3) since the vector ( b,- a ) is orthogonal to v , and as such is a normal vector to the lines parallel to v . In equation (3) c is an arbitrary constant, which uniquely determines the particular line in this family of parallel lines, called characteristic lines for the equation (2). As we saw above, u ( x,y ) is constant in the direction of v , hence also along the lines (3). The line containing the point ( x,y ) is determined by c = bx- ay , thus u will depend only on bx- ay , that is u ( x,y ) = f ( bx- ay ) , (4) 1 where f is an arbitrary function. One can then check that this is the correct solution by plugging it into the equation. Indeed, a x f ( bx- ay ) + b y f ( bx- ay ) = abf ( bx- ay )- baf ( bx- ay ) = 0 . The geometric viewpoint that we used to arrive at the solution is akin to solving equation (1) simply by recognizing that a function with a vanishing derivative must be constant. However one can approach equation (2) from another perspective, by trying to reduce it to an ODE....
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lec2 - Math 124A September 30, 2010 Viktor Grigoryan 2...

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