Direct integration of PDEs

**punkstart** · February 16th 2010, 11:55 AM

What are the rules for integrating a PDE of say 2 or more variables? And where can i learn more about these ? For example, i am asked to find the solution by direct integration of

$u_{xyz}(x,y,z) = e^{x} + xy\;$
Now $\int u_{xyz}(x,y,z)\; dx = \int (e^{x} + xy)\; dx = e^{x} + (1/2)x^{2}y + g(y) + h(z) = u_{yz}$ ; And $\int u_{yz} dy = \int [e^{x} + (1/2)x^{2}y + g(y) + h(z)]\; dy$ $= ye^{x} +(1/4)x^{2}y^{2} +\int g(y)\;dy +yh(z) = u_{z}; So\; that$ $u = \int [ye^{x} +(1/4)x^{2}y^{2} +\int g(y)\;dy +yh(z)]\; dz$ $\; = {\bf yze^{x} + (1/4)x^{2}y^{2}z + z\int g(y)\;dy + y\int h(z)\;dz + f(x)}$

Have i done this correctly ? I can't seem to find much information on integrating PDEs online. I am unsure of the arbitrary f(x),g(y) and h(z) functions, and how i have worked with them? Any help is much appreciated!

**Danny** · February 16th 2010, 03:03 PM

Originally Posted by punkstart

What are the rules for integrating a PDE of say 2 or more variables? And where can i learn more about these ? For example, i am asked to find the solution by direct integration of

$u_{xyz}(x,y,z) = e^{x} + xy\;$
Now $\int u_{xyz}(x,y,z)\; dx = \int (e^{x} + xy)\; dx = e^{x} + (1/2)x^{2}y + g(y) + h(z) = u_{yz}$ ; And $\int u_{yz} dy = \int [e^{x} + (1/2)x^{2}y + g(y) + h(z)]\; dy$ $= ye^{x} +(1/4)x^{2}y^{2} +\int g(y)\;dy +yh(z) = u_{z}; So\; that$ $u = \int [ye^{x} +(1/4)x^{2}y^{2} +\int g(y)\;dy +yh(z)]\; dz$ $\; = {\bf yze^{x} + (1/4)x^{2}y^{2}z + z\int g(y)\;dy + y\int h(z)\;dz + f(x)}$

Have i done this correctly ? I can't seem to find much information on integrating PDEs online. I am unsure of the arbitrary f(x),g(y) and h(z) functions, and how i have worked with them? Any help is much appreciated!

I would question your first step. Why are you assuming the function of integration is separable ${\it i.e.}\; g(y) + h(z)$ ?

**punkstart** · February 18th 2010, 07:48 AM

Well...i don't know...My knowledge of integrating a PDE is secondary, i have just picked things in physics, other than that i have somehow managed to skip an education in this field ! Am currently doing a third year applied math course and feel i might have skipped quite a necessary course on this area.Where could i find out more on the integration of partials? What might it look like if the variables were not seperable?

**shawsend** · February 18th 2010, 09:17 AM

This is what I'd do: Assuming the mixed partials are equal and you want to integrate with respect to x, than y, then z, I'd write $u_{xyz}(x,y,z) = e^{x} + xy$ as:

$\int \partial_x\left(\frac{\partial^2 u}{\partial z \partial y}\right)=\int (e^x+xy)\partial x$

$\frac{\partial^2 u}{\partial z\partial y}=e^x+x^2/2 y+f(x,y)$

(in general, you can't assume the arbitrary constant is separable so it's f(x,y))

$\int \partial_y\left(\frac{\partial u}{\partial z}\right)=\int\left(e^x+x^2/2y+f(x,y)\right)\partial y$

$\frac{\partial u}{\partial z}=e^x y+\frac{x^2 y^2}{4}+\int f(y,z)\partial y+h(z)$

Then integrating with respect to z:

$u(x,y,z)=e^x y z+\frac{x^2 y^2 z}{4}+\int\int f(y,z)\partial y \partial z+\int h(z)dz$

Also, I'd recommend "Basic Partial Differential Equations" by Bleecker and Csordas.

I think you can drop that last single integral as it's still just an arbitrary function of z such as $f(z)$ .

**Danny** · February 18th 2010, 10:28 AM

Originally Posted by shawsend

This is what I'd do: Assuming the mixed partials are equal and you want to integrate with respect to x, than y, then z, I'd write $u_{xyz}(x,y,z) = e^{x} + xy$ as:

$\int \partial_x\left(\frac{\partial^2 u}{\partial z \partial y}\right)=\int (e^x+xy)\partial x$

$\frac{\partial^2 u}{\partial z\partial y}=e^x+x^2/2 y+f(x,y)$

(in general, you can't assume the arbitrary constant is separable so it's f(x,y))

$\int \partial_y\left(\frac{\partial u}{\partial z}\right)=\int\left(e^x+x^2/2y+f(x,y)\right)\partial y$

$\frac{\partial u}{\partial z}=e^x y+\frac{x^2 y^2}{4}+\int f(y,z)\partial y+h(z)$

Then integrating with respect to z:

$u(x,y,z)=e^x y z+\frac{x^2 y^2 z}{4}+\int\int f(y,z)\partial y \partial z+\int h(z)dz$

Also, I'd recommend "Basic Partial Differential Equations" by Bleecker and Csordas.

I think you can drop that last single integral as it's still just an arbitrary function of z such as $f(z)$ .

I think what you really want is

$u = e^x y z+\frac{x^2 y^2 z}{4} + f(x,y) + g(x,z) + h(y,z)$

where $f, g\; \text{and}\; h$ are arbitrary functions of their arguments.

**punkstart** · February 19th 2010, 07:12 AM

Originally Posted by Danny

I think what you really want is

$u = e^x y z+\frac{x^2 y^2 z}{4} + f(x,y) + g(x,z) + h(y,z)$

where $f, g\; \text{and}\; h$ are arbitrary functions of their arguments.

Thank you ! That does make more sense. I like this website,my first time, i think i'll stay! Hopefully i can contribute aswell.

**punkstart** · February 19th 2010, 07:16 AM

Originally Posted by shawsend

This is what I'd do: Assuming the mixed partials are equal and you want to integrate with respect to x, than y, then z, I'd write $u_{xyz}(x,y,z) = e^{x} + xy$ as:

$\int \partial_x\left(\frac{\partial^2 u}{\partial z \partial y}\right)=\int (e^x+xy)\partial x$

$\frac{\partial^2 u}{\partial z\partial y}=e^x+x^2/2 y+f(x,y)$

(in general, you can't assume the arbitrary constant is separable so it's f(x,y))

$\int \partial_y\left(\frac{\partial u}{\partial z}\right)=\int\left(e^x+x^2/2y+f(x,y)\right)\partial y$

$\frac{\partial u}{\partial z}=e^x y+\frac{x^2 y^2}{4}+\int f(y,z)\partial y+h(z)$

Then integrating with respect to z:

$u(x,y,z)=e^x y z+\frac{x^2 y^2 z}{4}+\int\int f(y,z)\partial y \partial z+\int h(z)dz$

Also, I'd recommend "Basic Partial Differential Equations" by Bleecker and Csordas.

I think you can drop that last single integral as it's still just an arbitrary function of z such as $f(z)$ .

Thanx,will check the book out! I've studied my degree to this point through the post unaided, very relieved/happy to get help!

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